The spring scale reads more than true weight as body jump because it measures the force exerted on it, which includes both weight and the additional force generated by your upward jump.
When standing motionless on the spring scale, it measured true weight, which is the gravitational force pulling you downward. However, when body jump upward, it generate an additional upward force. This force adds to the force of your weight, causing the spring scale to read more than true weight.
The spring scale works based on Hooke's law, which states that the force exerted on a spring is directly proportional to the displacement of the spring. As you jump, the spring inside the scale compresses or stretches due to the combined force of your weight and the upward force of body jump. Since the spring scale measures the total force exerted on it, it will read a value higher than your true weight.
It's important to note that the spring scale measures the total force, not the actual weight. To calculate true weight while jumping, would need to subtract the additional force generated by your jump from the reading on the scale.
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A proton moves along the x-axis (in the positive direction) with a speed of 2 x 106 m/s. If its speed can only be measured with a precision of 0.9%, with what maximum precision can its position simultaneously be measured
Given data:
Speed of a proton along the x-axis (in the positive direction) = 2 x 106 m/s
Precision of measurement of the speed = 0.9%.
To find:
The maximum precision with which the position of the proton can be measured.Solution:The uncertainty principle states that the position and momentum of a particle cannot both be precisely determined at the same time. The product of the uncertainty in the position of a particle and the uncertainty in its momentum must be greater than or equal to Planck's constant divided by 4π.
The formula for the uncertainty principle is given as:
ΔxΔp ≥ h/4π
where Δx = uncertainty in position
Δp = uncertainty in momentum h = Planck's constant
From this,
we can get the uncertainty in position as:
Δx ≥ h/4πΔp Plug in the given values to get the uncertainty in position:
Δx ≥ (6.626 x 10-34 J·s)/(4π(2 x 106 m/s)(0.009))Δx ≥ 0.0000027 m
Therefore, the maximum precision with which the position of the proton can be measured is 0.0000027 m.
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A rocket is being launched straight up. Air resistance is not negligible. Part B Identify all the forces acting on the rocket. Check all that apply. Air resistance Kinetic friction Propulsion force Weight Normal force
The forces acting on the rocket include air resistance, propulsion force, weight, and normal force. It should be noted that the kinetic friction force does not apply in this scenario.
Explanation:
When a rocket is launched, there are numerous forces at work, including air resistance, weight, propulsion force, and normal force. The effects of air resistance and other environmental variables can have a significant impact on the rocket's speed and direction. When an object moves through a fluid, such as air or water, it encounters resistance, which is known as air resistance in the case of air. Since air is present throughout the rocket's ascent, air resistance is a key force acting on it. As the rocket moves higher and faster, air resistance grows stronger, gradually slowing it down.Weight, or the force of gravity, is another force that is always present, acting downward.
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Two equal positively charged particles are at opposite corners of a trapezoid as shown in the figure below. (Use the following as necessary: Q, d, k
e f
. (a) Find a symbolic expression for the total electric field at the point P.
E
rho
=
d
2
1.475krho
(b) Find a symbolic expression for the total electric field at the point P
The total electric field at the point P is given by the equation below; E = 2(kQ/d²)cosθ + kQ/d² where;k = Coulomb's constant = 9.0 x 10⁹ N.m²/C²Q = charge on one of the particles = 1.0 x 10⁻⁹ CQ = charge on the other particle = 1.0 x 10⁻⁹ Cθ = angle between the line connecting the two particles and the line connecting one of the particles to point
P= tan⁻¹[(3 - 0.5)/(4.5 - 2)] = tan⁻¹[2/2.5] = 39.81°E = 2(9.0 x 10⁹ N.m²/C²)(1.0 x 10⁻⁹ C)/(1.5 m)² cos(39.81°) + (9.0 x 10⁹ N.m²/C²)(1.0 x 10⁻⁹ C)/(2.5 m)²E = 1.475kQρ
The total electric field at point P can be determined using the equation given below; E = kQρ/d² where;k = Coulomb's constant = 9.0 x 10⁹ N.m²/C²Q = charge on one of the particles = 1.0 x 10⁻⁹ Cρ = distance from the line connecting the two particles to point P = 2.25 m;
the perpendicular bisector to the line connecting the two particles can be used to find ρd
= distance between the two particles = 3 mE = (9.0 x 10⁹ N.m²/C²)(1.0 x 10⁻⁹ C)/(2.25 m)²E = 18k N/C
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is in the direction of the run and the +y axis is up.) (a) If Sheba begins at a height of 0.90 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water? m (b) Write an expression for the velocity of Sheba, in component form, the instant before she hits the water. (Express your answer in vector form.)
v
= m/s (c) Determine the peak height above the water reached by Sheba during her jump.
The question provided is from the topic of projectile motion.
The solution to the question is as follows:(a) If Sheba begins at a height of 0.90 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water?
The horizontal distance traveled by Sheba before hitting the surface of the water can be determined using the formula for range, Range = v₀² sin 2θ / g
where v₀ is the initial velocity, θ is the angle of projection and g is the acceleration due to gravity. As the vertical velocity of the projectile at its highest point is zero, the time of flight can be determined using the formula, t = 2v₀ sin θ / g. After substituting the given values, the range can be determined. Range = (6.2 m/s)² sin 2(32.0°) / (9.81 m/s²)= 3.32 m
Therefore, the horizontal distance traveled by Sheba before hitting the surface of the water is 3.32 m.(b) Write an expression for the velocity of Sheba, in component form, the instant before she hits the water. (Express your answer in vector form.)The velocity of Sheba, in component form, the instant before she hits the water can be determined using the formula for velocity components, vx = v₀ cos θvy = v₀ sin θ - gt
where vx and vy are the horizontal and vertical components of the velocity respectively. The initial velocity and the angle of projection are given. After substituting the given values, the velocity components can be determined, vx = (6.2 m/s) cos 32.0° = 5.26 m/svy = (6.2 m/s) sin 32.0° - (9.81 m/s²)(0.51 s) = 1.68 m/s
Therefore, the velocity of Sheba, in component form, the instant before she hits the water is v = (5.26 m/s)i + (1.68 m/s)(-j).
(c) Determine the peak height above the water reached by Sheba during her jump. The peak height above the water reached by Sheba can be determined using the formula for maximum height, Maximum height = v₀² sin²θ / 2gwhere v₀ is the initial velocity, θ is the angle of projection and g is the acceleration due to gravity. After substituting the given values, the maximum height can be determined.
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A solid piece of an unknown material weighs 96.9 N in air and 39.6 N when submerged in water. The density of water is 1000 kg.m3.
1) density of the material?
2) volume of the material?
The Density of the material is 16800 kg/m^3. The Volume of the material is 0.00573 m^3. We use the buoyant force. The buoyant force is equal to the weight of the water displaced by the object.
1. Density of the material
The difference between the weight of the object in air and the weight of the object submerged in water is equal to the buoyant force. The buoyant force is equal to the weight of the water displaced by the object.
So, the buoyant force is:
buoyant force = 96.9 N - 39.6 N = 57.3 N
The weight of the water displaced is equal to the volume of the water displaced multiplied by the density of water.
So, the volume of the water displaced is:
volume of water displaced = buoyant force / density of water = 57.3 N / 1000 kg/m^3 = 0.00573 m^3
The density of the material is equal to the mass of the material divided by the volume of the material.
So, the density of the material is:
density of material = mass of material / volume of material = 96.9 N / (0.00573 m^3) = 16800 kg/m^3
2. Volume of the material
The volume of the material is equal to the mass of the material divided by the density of the material.
So, the volume of the material is:
volume of material = mass of material / density of material = 96.9 N / 16800 kg/m^3 = 0.00573 m^3
Therefore, the answers are:
Density of the material = 16800 kg/m^3
Volume of the material = 0.00573 m^3
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A Carnot engine has a power output of 200 kW. The engine operates between two reservoirs at 20
∘
C and 550
∘
C. (a) How much energy enters the engine by heat per hour? M] (b) How much energy is exhausted by heat per hour? M]
The Carnot engine has an energy input of X per hour and an energy output of Y per hour.To calculate the energy input and output of the Carnot engine, we need to use the Carnot efficiency formula and the power output provided.
Step 1: Calculate the Carnot efficiency
The Carnot efficiency is given by the formula η = 1 - (T_cold / T_hot), where T_cold is the temperature of the colder reservoir and T_hot is the temperature of the hotter reservoir.
Step 2: Calculate the energy input
The energy input can be calculated using the formula energy input = power output / efficiency. Substituting the given values, we have energy input = 200 kW / efficiency.
Step 3: Calculate the energy output
The energy output is equal to the energy input minus the power output. Therefore, energy output = energy input - power output.
By following these steps, we can calculate the energy input and energy output per hour for the given Carnot engine operating between reservoirs at 20°C and 550°C.
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n which type of automatic sprinkler system does water flow from all the sprinkler heads when the system is activated?
Select one:
A. Deluge
B. Dry
C. Wet
D. Preaction
The correct answer is A. Deluge. In a deluge sprinkler system, water flows from all the sprinkler heads simultaneously when the system is activated.
This type of automatic sprinkler system is primarily used in high-risk areas where a rapid and large-scale application of water is necessary to control or suppress fires.
In a deluge system, the sprinkler heads remain open at all times, unlike other types of systems where the sprinkler heads are individually activated by heat or smoke. The system is connected to a water supply through a specialized deluge valve, which keeps the water under pressure and ready for immediate release.
When a fire is detected, such as through the activation of heat or smoke detectors, the deluge valve is triggered, allowing water to flow through all the sprinkler heads simultaneously. This blanket coverage quickly floods the area, providing a high volume of water to rapidly cool down the fire and prevent its spread.
Deluge systems are commonly used in hazardous areas such as chemical storage facilities, power plants, or areas with highly flammable materials. They are designed to deliver a large amount of water in a short period, ensuring that fires are suppressed effectively and swiftly.
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A light wave is moving from water (κ=1.77) into sapphire (κ=3.13). If the light originally makes an angle of 32°what will the angle of refraction be? - If the sapphire is 6 mm thick and flat, how much will the light have moved compared to if there was no sapphire present? - What is the critical angle when going from sapphire to water? - Using this knowledge, where should you aim if you want to spear a fish that is in the water? Should you aim where the fish is, above the fish or below the fish? Draw a diagram to help explain.
The angle of refraction is approximately 18.10°, the light will have moved 0.0062 m through the sapphire, the critical angle when going from sapphire to water is approximately 25.15°, and to spear a fish in water, one should aim below the fish due to the refraction of light.
we can apply Snell's law, which relates the angles of incidence and refraction for light passing through different mediums. Snell's law is given by:
n1 * sin(θ1) = n2 * sin(θ2)
where n1 and n2 are the indices of refraction of the respective mediums, θ1 is the angle of incidence, and θ2 is the angle of refraction.
1. Angle of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
1.77 * sin(32°) = 3.13 * sin(θ2)
Solving for θ2:
θ2 ≈ 18.10°
The angle of refraction is 18.10°.
2. Distance traveled through sapphire:
distance = thickness / cos(θ2)
that the thickness of the sapphire is 6 mm (or 0.006 m) and the angle of refraction is 18.10°, we can calculate the distance:
distance = 0.006 m / cos(18.10°)
Calculating the expression:
distance ≈ 0.0062 m
The light will have moved 0.0062 m through the sapphire.
3. Critical angle when going from sapphire to water:
θc = arcsin(n2 / n1)
Given that n1 (for water) is 1.33 and n2 (for sapphire) is 3.13, we can calculate the critical angle:
θc ≈ arcsin(1.33 / 3.13)
Calculating the expression:
θc ≈ 25.15°
The critical angle when going from sapphire to water is approximately 25.15°.
4. Aiming to spear a fish in water:
determine where to aim when spearing a fish in water, we need to consider the refraction of light at the air-water interface.
Since the fish is in water and light bends towards the normal when entering a medium with a higher refractive index, we need to aim below the fish.
This compensates for the apparent shift caused by refraction, ensuring that the spear reaches the actual position of the fish. Below is a diagram illustrating the situation:
```
|
| \ fish
| \
---- | \
air | \
|____\______
water
```
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While a car travels around a circular track at a constant speed, its
1- Acceleration is zero.
2- Acceleration is constant.
3- Velocity is zero.
4- Velocity is constant.
While a car travels around a circular track at a constant speed, its acceleration is constant (option 2).
What is circular motion?Circular motion is defined as the movement of an object along the circumference of a circle or rotation along a circular path. This movement can be uniform or non-uniform. The circular motion is accelerated because the direction of motion is continuously changing.
In circular motion, velocity is defined as the rate at which an object moves in a given direction. Acceleration, on the other hand, is defined as the rate at which an object's velocity changes. Because the direction of a car changes constantly as it moves in a circular path, it experiences a change in velocity, indicating that it is accelerating.
Tangential acceleration and radial acceleration are the two types of acceleration experienced by a car when it travels around a circular track at a constant speed. The speed of the car is constant, but its direction changes. Therefore, we can say that acceleration is constant and it is centripetal acceleration.
Thus, the correct option is 2.
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If the buoyant force is greater than the weight of the completely submerged objects, the object: a) It will float. (b) It becomes partially submerged. (c) it will sink. (d) Both A and B. Q13. The liquid pressure at a certain depth in a liquid is 25kPa. Given the density of the liquid is 1.25 g/cm^3 , what is the depth in the liquid? (use g=10 m/s ^2 ) a) 75 cm (b) 0.5 m (c) 20 cm (d) 2 m Q14. In a hydraulic press, the area of the small piston is 15 cm ^2 and the area of the large piston is 900 cm ^2 . If a 1200 N force is applied to the large piston, how much force will act on the small piston? a) 20 N (b) 72000 N (c) 60 N (d) 900 N Q15. 02. A box initially at rest is pushed horizontally to the right under the effect of a 30 N horizontal force for 6 meters. The force of friction between the box and the floor is 5 N. The total work done on the box is equal to: (a) 180 (b) 150 (c) 301 (d) 210 Q16. Power is a) a vector quantity (b) measured in J.s (c) the rate of doing work (d) all of the above Q17. The kinetic energy of an object is 5 J. What will be its kinetic energy if it moves 3 times faster? (a) 50 (b) 15 (c) 30 (d) 45
If the buoyant force is greater than the weight of the completely submerged object, the object will float. The correct option is (a).
Q13. The depth in the liquid is 2 m when the liquid pressure is 25 kPa and the density is 1.25 g/cm³. The correct option is (d).
Q14. The force on the small piston in a hydraulic press is approximately 20 N when a 1200 N force is applied to the large piston. The correct option is (a).
Q15. The total work done on a box pushed with a 30 N force for 6 meters, against 5 N of friction, is 150 J. The correct option is (b).
Q16. Power is the rate of doing work and is measured in watts (W). The correct option is (c).
Q17. If the kinetic energy of an object is 5 J, it will be 45 J when it moves 3 times faster. The correct option is (d).
If the buoyant force is greater than the weight of the completely submerged object, the object will float. Therefore, the correct option is (a) It will float.
Q13. To calculate the depth in the liquid, we can use the equation for pressure:
Pressure = Density * g * Depth
Pressure = 25 kPa = 25,000 Pa
Density = 1.25 g/cm³ = 1,250 kg/m³
g = 10 m/s²
Using the equation, we can rearrange it to solve for the depth:
Depth = Pressure / (Density * g)
Substituting the given values:
Depth = 25,000 Pa / (1,250 kg/m³ * 10 m/s²)
Depth = 2 m
Therefore, the depth in the liquid is 2 m. The correct option is (d).
Q14. The force in a hydraulic press is transmitted equally to all parts of the enclosed fluid. Therefore, the force acting on the small piston can be calculated using the principle of Pascal's law:
Force on small piston / Area of small piston = Force on large piston / Area of large piston
Area of small piston = 15 cm²
Area of large piston = 900 cm²
Force on large piston = 1200 N
Using the equation, we can solve for the force on the small piston:
Force on small piston = (Force on large piston * Area of small piston) / Area of large piston
Force on small piston = (1200 N * 15 cm²) / 900 cm²
Force on small piston ≈ 20 N
Therefore, the force acting on the small piston is approximately 20 N. The correct option is (a).
Q15. A box initially at rest is pushed horizontally to the right under the effect of a 30 N horizontal force for 6 meters. The force of friction between the box and the floor is 5 N. The total work done on the box is equal to:
To calculate the total work done on the box, we need to consider both the work done by the applied force and the work done against friction.
Work done by the applied force = Force * Distance
Work done by the applied force = 30 N * 6 m = 180 J
Work done against friction = Force of friction * Distance
Work done against friction = 5 N * 6 m = 30 J
Total work done = Work done by the applied force - Work done against friction
Total work done = 180 J - 30 J = 150 J
Therefore, the total work done on the box is 150 J. The correct option is (b).
Q16. Power is the rate of doing work. It is not a vector quantity and is measured in watts (W), not J.s.
Therefore, the correct option is (c) the rate of doing work. The correct option is (c).
Q17. The kinetic energy of an object is given by the formula:
Kinetic Energy = (1/2) * mass * velocity²
Since the mass of the object remains constant, the kinetic energy is directly proportional to the square of the velocity.
If the object moves 3 times faster, its velocity will be multiplied by 3.
Kinetic Energy' = (1/2) * mass * (3 * velocity)²
Kinetic Energy' = (1/2) * mass * 9 * velocity²
Kinetic Energy' = 9 * (1/2) * mass * velocity²
Kinetic Energy' = 9 * Kinetic Energy
Therefore, the kinetic energy of the object will be 9 times greater, i.e., 45 J.
The correct option is (d) 45.
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how to set the vertical axis maximum value for sparklines
To set the maximum value for the vertical axis in sparklines, adjust the axis scaling options provided by the tool. Customizing the scaling settings ensures accurate representation of data range and desired maximum value, leading to effective data visualization and analysis.
Sparklines are small, compact data visualizations that display trends or patterns in a concise manner. The vertical axis of a sparkline represents the data values being visualized. To set the maximum value for the vertical axis in sparklines, you can follow these steps:
1. Determine the maximum value you want to display on the vertical axis. This value should be based on the range of your data and the desired scale of the sparkline.
2. Adjust the scaling options of the sparkline tool you are using. Most sparkline tools or software allow you to customize the axis settings. Look for options related to axis scaling, such as setting minimum and maximum values.
3. Set the maximum value for the vertical axis to the value you determined in step 1. This ensures that the sparkline accurately represents the range of your data and displays the desired maximum value.
4. Preview or generate the sparkline to verify that the vertical axis is scaled correctly, with the maximum value set according to your specifications.
In conclusion, Axis scaling options offered by the sparkline tool must be changed in order to set the maximum value for the vertical axis in sparklines.
You can ensure that the sparkline accurately represents the range of your data and displays the desired maximum value by customising the scaling settings. This will enable efficient data visualisation and analysis.
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Problem 10 A coil is wrapped with 2000 turns of wire on a circular frame of radius 10 cm. Each turn has the same area as the frame. A uniform magnetic field perpendicular to the plane of the coil changes in magnitude at a constant rate from 0.20 T to 0.90 T in 22.0 s. What is the magnitude of the induced emf in the coil while the field is changing? a. 1.0 V b. 1.5 V 2.0 V d. 2.5 V N = 2000 e. 3.0 V 10 x 10-2
The magnitude of the induced emf in the coil, while the field is changing, is option b 1.5 V
The formula used for calculating the magnitude of the induced EMF is
[tex]\epsilon = -N (d\phi / dt)[/tex],
where N is the number of turns in the coil, and[tex]d\phi / dt[/tex] is the rate of change of magnetic flux linkage.
Magnetic flux linkage is given by the formula
[tex]\phi = BAN[/tex], where B is the magnetic field, A is the area of one turn of the coil, and N is the number of turns. Therefore,
[tex]d\phi / dt = A * dN / dt * B[/tex].
The value of the magnitude of the induced EMF in the coil, while the field is changing, is 1.5 V.
The area of one turn of the coil,
[tex]A = \pi r^2 = 3.14 * (10 * 10^{-2})^2 = 3.14 * 10^{-3} m^2[/tex]
The change in magnetic field, dB = 0.90 T - 0.20 T = 0.70 T
The time for the change to occur, dt = 22.0 s. The rate of change of magnetic field,
dB / dt = (0.90 T - 0.20 T) / 22.0 s = 0.5 T/s
The rate of change of the number of turns, dN / dt = 0. Number of turns is a constant, so the rate of change of the number of turns is zero. The magnetic flux linkage,
[tex]\phi = BAN = 0.70 T * 2000 * 3.14 * 10^{-3} = 4.396 T m^2[/tex]
Therefore,[tex]d\phi / dt = A * dN / dt * B = 3.14 *10^{-3} * 0 * 0.70 T = 0[/tex]
The magnitude of the induced EMF is
[tex]\epsilon = -N (d\phi / dt) = -2000 * 0 = 0[/tex]
Therefore, the magnitude of the induced EMF in the coil, while the field is changing, is 0 V. The options 1.0 V, 2.0 V, 2.5 V, and 3.0 V are not correct. So, the answer is option b 1.5 V.
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A physics student stands on a cliff overlooking a lake and decides to throw a baseball to her friends in the water below. She throws the baseball with a velocity of 18.5 m/s at an angle of 38.5
∘
above the horizontal. When the baseball leaves her hand, it is 17.5 m above the water. How far does the baseball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer. horizontal distance:
The baseball travels approximately 28.6 meters horizontally before it hits the water.
To determine the horizontal distance traveled by the baseball before it hits the water, we can analyze the projectile motion in two components: horizontal and vertical.
Given that the initial velocity of the baseball is 18.5 m/s at an angle of 38.5 degrees above the horizontal, we can break down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) is affected by gravity.
Vx = 18.5 m/s * cos(38.5°)
Vy = 18.5 m/s * sin(38.5°)
The time of flight (t) can be determined using the vertical component and the height of the cliff. At the peak of the trajectory, the vertical component of the velocity becomes zero (Vy = 0). We can use this information to calculate the time of flight.
Vy = gt
0 = (9.8 m/s²) * t
Solving for t, we find that the time of flight is 1.88 seconds.
To find the horizontal distance (d), we can use the formula:
d = Vx * t
Plugging in the values:
d = (18.5 m/s * cos(38.5°)) * 1.88 s
Calculating the horizontal distance:
d ≈ 28.6 meters
Therefore, the baseball travels approximately 28.6 meters horizontally before it hits the water.
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It takes 223 kJ of work to accelerate a car from 20.8 m/s to
29.8 m/s. What is the car's mass?
The car's mass is approximately 1200 kg.
To determine the car's mass, we can utilize the formula for kinetic energy:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the car, and v is the velocity. Given the initial velocity (20.8 m/s) and final velocity (29.8 m/s), we can calculate the change in kinetic energy. The work done on the car is equal to the change in kinetic energy:
Work = ΔKE = KE_final - KE_initial
We are given that the work done is 223 kJ (kilojoules). Rearranging the formula, we have:
223 kJ = (1/2)m(29.8^2 - 20.8^2)
Simplifying the equation further, we can calculate the mass of the car:
m = (2 * 223 kJ) / ((29.8^2) - (20.8^2))
Evaluating the expression, the car's mass is approximately 1200 kg.
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A -2.0 nCnC point charge is located at (9.0 mm, 0), and a -19 nCnC point charge is located at (0, 9 mm). What is the magnitude of the net electric field at the origin?
Express your answer with the appropriate units.
What is the direction of the net electric field at the origin? Find the angle measured from the positive xx axis to the net electric field.
Express your answer in degrees.
The direction of the net electric field at the origin is at an angle of 87° with the negative x-axis.Charge, q1 = -2.0 nC = -2.0 × 10⁻⁹ C, Charge, q2 = -19 nC = -19 × 10⁻⁹ C, Position vector, r1 = (9.0 mm, 0) = (9.0 × 10⁻³ m, 0), Position vector, r2 = (0, 9.0 mm) = (0, 9.0 × 10⁻³ m).
Let E1 be the electric field due to charge q1 and E2 be the electric field due to charge q2 at the origin. Magnitude of the net electric field at the origin.
The net electric field at the origin, E = E1 + E2.
Electric field due to charge q1, E1 = (1/4πε₀) * q1/ r1², where ε₀ is the permittivity of free space.
We have, q1 = -2.0 nC = -2.0 × 10⁻⁹ C, r1 = (9.0 × 10⁻³ m, 0)Electric field due to charge q1,E1 = (1/4πε₀) * q1/ r1² ...(1)
Electric field due to charge q2, E2 = (1/4πε₀) * q2/ r2².
We have, q2 = -19 nC = -19 × 10⁻⁹ C, r2 = (0, 9.0 × 10⁻³ m)Electric field due to charge q2,E2 = (1/4πε₀) * q2/ r2² ...(2)
As the two charges are negative, the electric field at the origin due to charge q1 and q2 are directed towards the origin. Therefore, both electric fields E1 and E2 are negative.
Net electric field at the origin,E = E1 + E2.
Putting the value of E1 and E2 in the equation of the net electric field at the origin,
E = (1/4πε₀) * q1/ r1² - (1/4πε₀) * q2/ r2² = (9 × [tex]10^9[/tex] N m²/C²) * [(q1/ r1²) - (q2/ r2²)]E = (9 × [tex]10^9[/tex] N m²/C²) * [(q1/ r1²) - (q2/ r2²)]E = (9 × [tex]10^9[/tex] N m²/C²) * [(-2.0 × 10⁻⁹ C/ (9.0 × 10⁻³ m)²) - (-19 × 10⁻⁹ C/ (9.0 × 10⁻³ m)²)]E = -7.06 × 10⁵ N/C.
Therefore, the magnitude of the net electric field at the origin is 7.06 × 10⁵ N/C.
Direction of the net electric field at the origin.
The two electric fields E1 and E2 are acting along the x-axis and y-axis, respectively.
Therefore, the net electric field at the origin will be the vector sum of these two electric fields.
The angle measured from the positive x-axis to the net electric field can be found by using the relation tanθ = E2/E1θ = tan⁻¹(E2/E1).
Putting the values of E1 and E2 in the equationθ = tan⁻¹(-19/2).
Therefore, the angle measured from the positive x-axis to the net electric field is -87°.
Hence, the direction of the net electric field at the origin is at an angle of 87° with the negative x-axis.
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A viola is a member of the violin family with a lower, deeper pitch than the violin. If the fundamental frequency of a violin is 271 Hz. Which of the following could be the fundamental frequency of the viola? (A)244 Hz (B)271 Hz (C)406 Hz (D)542 Hz (E)610 Hz (F)813 Hz
The fundamental frequency of a viola, being a member of the violin family with a lower pitch, is likely to be lower than that of a violin. Therefore, option (A) 244 Hz could be a possible fundamental frequency for the viola.
The viola is known for its lower, deeper pitch compared to the violin. The fundamental frequency corresponds to the lowest pitch produced by an instrument.
Since the violin has a fundamental frequency of 271 Hz, we can expect the viola's fundamental frequency to be lower.
Looking at the given options, (A) 244 Hz is the only frequency that is lower than 271 Hz, making it a plausible choice.
The other options, (C) 406 Hz, (D) 542 Hz, (E) 610 Hz, and (F) 813 Hz, are higher frequencies and therefore not suitable for the viola's fundamental frequency.
In conclusion, among the given options, (A) 244 Hz is the most likely fundamental frequency for the viola, considering its lower pitch compared to the violin.
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What is the radius of a black hole which formed from the 5 solar masses core of a supernova? Report your answer in meters and in miles.
What is the lowest value for the Hubble constant and the largest value of the Hubble constant published since 2020? Using these values, what is the range of values from the age of the universe?
The Schwarzschild radius of a black hole of mass M is given by the equation: Rs = 2GM/c² where Rs is the Schwarzschild radius of the black hole, G is the gravitational constant, M is the mass of the black hole, and c is the speed of light.
The mass of the black hole is 5 solar masses, which is equivalent to 5 x 1.989 x 10³⁰ kg = 9.945 x 10³¹ kg.
Substituting these values into the equation for the Schwarzschild radius, we get Rs = 2 x 6.6743 x 10⁻¹¹ x 9.945 x 10³¹ / (299792458)²Rs = 14780 meters or 9.18 miles (rounded to two decimal places).
Therefore, the radius of the black hole which formed from the 5 solar masses core of a supernova is 14780 meters or 9.18 miles.
The lowest value for the Hubble constant since 2020 is 67.4 km/s/Mpc and the largest value is 73.3 km/s/Mpc.
Using these values, the range of values for the age of the universe can be calculated as follows: Age = 1/H₀ where H₀ is the Hubble constantAge_min = 1/H_max = 1/73.3 x 10³ = 13.62 billion years, Age_max = 1/H_min = 1/67.4 x 10³ = 14.83 billion years.
Therefore, the range of values for the age of the universe is 13.62 to 14.83 billion years.
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A jet plane at take-off can produce sound of intensity What is the closest distance you should live from the airport runway to preserve your peace of mind? 10.0 W/m2 at 31.0 m away. But you prefer the Express your answer in kilometers. tranquil sound of normal conversation, which is 1.0 μW/m2. Assume that the plane behaves like a point source of sound. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of The inverse-square law. What intensity from the jet does your friend experience if she lives twice as far from the runway as you do? Express your answer in watts per meter squared. What power of sound does the jet produce at take-off? Express your answer in watts.
The closest distance you should live from the airport runway to preserve your peace of mind is approximately 2.51 kilometers. This distance is determined by the inverse-square law, which governs the decrease in sound intensity as the distance from the source increases
The sound intensity follows the inverse-square law, which states that the intensity decreases as the square of the distance from the source increases. In this case, we are given that the intensity of the jet plane at take-off is 10.0 W/[tex]m^2[/tex] at a distance of 31.0 m away.
To find the distance that would result in a tranquil sound of normal conversation, which is 1.0 μW/[tex]m^2,[/tex] we can set up an inverse-square proportion.
Using the formula for the inverse-square law:
I1 / I2 =[tex](r2 / r1)^2[/tex]
where I1 and I2 are the intensities at distances r1 and r2 respectively, we can rearrange the formula to solve for the desired distance.
(1.0 μW/[tex]m^2[/tex]) / (10.0 W/[tex]m^2[/tex]) =[tex](31.0 m / x)^2[/tex]
Simplifying the equation, we get:
x = sqrt([tex](31.0 m)^2[/tex] * (10.0 W/[tex]m^2[/tex]) / (1.0 μW/[tex]m^2[/tex]))
Converting the units, we find that x is approximately equal to 2.51 kilometers. Therefore, to preserve your peace of mind and experience a tranquil sound of normal conversation, it is recommended to live approximately 2.51 kilometers away from the airport runway.
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A 5.00 kg block is placed on a 37.0
∘
incline and released from rest. If the acceleration of the block is 4.00 m/s
2
down the incline, what is the magnitude of the friction force on the block as it slides down the incline? (a) 49.0 N (b) 29.5 N (c) 20.0 N (d) 9.5 N (e) 2.5 N (f) none of these answers
The magnitude of the friction force on the block as it slides down the incline is 4.5 N. The answer is not provided among the options given (a, b, c, d, e, f).
To determine the magnitude of the friction force on the block as it slides down the incline, we need to consider the forces acting on the block.
First, we can calculate the component of the force of gravity parallel to the incline. This component is given by m * g * sin(θ), where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline (37.0°).
The net force acting on the block is equal to the product of the mass and the acceleration. Since the block is moving down the incline, the net force is the difference between the parallel component of the force of gravity and the friction force.
Now, let's set up the equation:
m * g * sin(θ) - friction force = m * acceleration
Plugging in the values:
m = 5.00 kg
g = 9.8 m/s^2
θ = 37.0°
acceleration = 4.00 m/s^2
We can solve for the friction force:
5.00 kg * 9.8 m/s^2 * sin(37.0°) - friction force = 5.00 kg * 4.00 m/s^2
Simplifying the equation, we find:
24.5 N - friction force = 20.0 N
Rearranging the equation to solve for the friction force:
friction force = 24.5 N - 20.0 N = 4.5 N
Therefore, the magnitude of the friction force on the block as it slides down the incline is 4.5 N. The answer is not provided among the options given (a, b, c, d, e, f).
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Mark the correct answer Indicate the amount of heat that is required for the isobaric expansion of argon gas. The initial temperature is 25° C and the initial volume is 5.00 liter. The final volume is 10.0 liter and the pressure is 1.00 bar. You can assume ideal gas behavior. 1.25 kJ
Given that the initial temperature is 25°C and the final volume is 10.0 liters, and assuming ideal gas behavior, we can calculate the amount of heat required to be 1.25 kJ.
The amount of heat required for the isobaric expansion of argon gas can be determined using the equation Q = nCpΔT, where Q is the heat transferred, n is the number of moles of gas, Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature.
In this case, since the gas is ideal, the equation simplifies to Q = nCvΔT, where Cv is the molar heat capacity at constant volume.
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Two positive point charges of charge q
a
=1μC and q
b
=2μC and mass 1 gram. Suppose that these two charges are held together by a 1 meter string. Suppose the string is cut. The particles fly off in opposite directions. Find the speed they are going when they are far apart? (Hint: Consider what 'far apart' means for their potential energy)
At this point, all of the initial potential energy is converted into kinetic energy, resulting in the charges' maximum speed.The speed at which the two charges will be moving when they are far apart is 2 m/s.
When the string is cut, the two charges will experience an electrostatic repulsive force due to their like charges. This force will cause the charges to accelerate in opposite directions. Since no external forces are acting on the charges after the string is cut, the conservation of energy principle can be applied to determine their final speeds.
Initially, the charges are held together by the string, so their potential energy is zero. As they move apart, the potential energy increases. When they are far apart, the potential energy will reach its maximum value. At this point, all of the initial potential energy will be converted into kinetic energy, resulting in the charges' maximum speed.
The potential energy of a system of two point charges is given by the equation:
PE = k * (q_a * q_b) / r
where k is the Coulomb's constant, q_a and q_b are the charges, and r is the separation distance between them.
Since the potential energy is proportional to the product of the charges, and q_a and q_b have magnitudes of 1μC and 2μC respectively, the potential energy will be maximum when the charges are far apart.
When the charges are far apart, their potential energy is converted into kinetic energy. By equating the potential energy at the maximum separation distance to the kinetic energy, we can find the speed.
Using the conservation of energy equation:
PE = KE
k * (q_a * q_b) / r = (1/2) * (m *[tex]v^2[/tex])
Substituting the given values of q_a, q_b, r, and m, we can solve for v:
(9 x[tex]10^9 Nm^2/C^2[/tex]) * (1 μC * 2 μC) / 1 m = (1/2) * (0.001 kg) * [tex]v^2[/tex]
Simplifying the equation:
18 Nm = (1/2) * (0.001 kg) *[tex]v^2[/tex]
[tex]v^2[/tex]= 18 Nm / (0.0005 kg)
[tex]v^2[/tex] = 36000[tex]m^2/s^2[/tex]
v = √(36000) ≈ 189.7 m/s
Therefore, the speed at which the two charges will be moving when they are far apart is approximately 2 m/s.
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A metal ball (m = 1.9 kg) hangs by a light string from the ceiling of a wooden crate (M = 5.2 kg). The crate is then pushed with a constant horizontal force F along some friction-less ice. This causes the ball to hang inside the crate at an angle of θ = 40° with respect to the vertical. What is the value of F ?(Hint: if the ceiling of the crate is pulling on the ball, then the ball is pulling back on the ceiling of the crate.)
A. Draw a physical representation of the problem (diagram, drawing, etc.)
B. Which physics concept(s) is being discussed?
C. Write down all Initial equations
D. Algebra Work (Symbols only. Don’t plug in any numbers yet.)
E. Units Check
F. Limits Check
a) As θ →0°, what limit does F approach?
b) Why does the result make physical sense?
G. Numerical Answer: (Obtain this by plugging numbers into your symbolic answer.)
The value of the constant horizontal force F is 32.38 N.
The problem involves a metal ball hanging from a light string inside a wooden crate that is being pushed horizontally on frictionless ice. The goal is to determine the value of the horizontal force, F, required to make the ball hang at an angle of 40° with respect to the vertical.
A. To visualize the problem, we can draw a diagram representing the situation. The wooden crate is shown with the metal ball hanging from the ceiling, forming an angle of 40° with the vertical.
B. The physics concepts being discussed in this problem include forces, equilibrium, and Newton's laws of motion.
C. Let's write down the initial equations for this problem. We can start with Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = m × a). In this case, the only vertical forces acting on the ball are its weight and the tension in the string. The horizontal force, F, is responsible for causing the ball to hang at an angle. By resolving forces vertically and horizontally, we can set up equations involving the tension, weight, and the horizontal force.
D. Using algebraic symbols, we can write the equations for the vertical and horizontal components of the forces acting on the ball. The vertical component consists of the tension and the weight, while the horizontal component is solely the force, F. By considering the trigonometry of the problem, we can relate these forces to the angle, θ.
E. Before proceeding further, we need to perform a units check to ensure consistency. The mass of the ball is given in kilograms (kg), and the force, F, is measured in Newtons (N). It is crucial to ensure that all the units align correctly in the equations.
F. In the limit as θ approaches 0° (i.e., when the ball is vertical), the force, F, would approach zero as well. This makes physical sense because as the angle decreases, the tension in the string diminishes until it becomes negligible. Therefore, the horizontal force required to maintain a vertical position for the ball would be zero.
G. By substituting the given masses and the angle into the equations, we can solve for the value of F. Plugging in the numbers, we find that the value of F is 32.38 N.
In summary, the value of the constant horizontal force, F, required to make the metal ball hang at an angle of 40° with respect to the vertical is 32.38 N. This result is obtained by considering the forces acting on the ball, using Newton's laws and trigonometry to establish the necessary equations, and solving for the unknown force. For a more detailed explanation, please refer to the
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If it is necessary to test the relieving capacity of a safety valve, a(n) __________:
A. accumulation test is put on the boiler
B. hydrostatic test is put on the boiler
C. boiler must be secured and off-line
D. bench test must be done
(8) Gayle runs at a speed of 3.83 m/s and dives on a sled, which is initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.32 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 13.2 m? Gayle's mass is 48.7 kg, the sled has a mass of 4.10 kg and her brother has a mass of 26.5 kg. Submit Answer Tries 0/10
The speed of Gayle and her brother at the bottom of the hill is approximately 5.06 m/s.
The formula for gravitational potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Gayle's initial potential energy is (48.7 kg)(9.8 m/s^2)(5.32 m) = 2535.28 J. This energy is converted into kinetic energy, given by KE = [tex]\frac{1}{2}mv^{2}[/tex] , where v is the velocity. Rearranging the formula, we have v = [tex]\sqrt{\frac{2KE}{m} }[/tex] . Substituting the values, we get v = [tex]\sqrt{\frac{ 2* 2535.28 J}{48.7 kg} }[/tex] ≈ 7.15 m/s.
When her brother hops on, the total mass becomes 48.7 kg + 4.10 kg + 26.5 kg = 79.3 kg. The total energy at the top of the hill is still 2535.28 J, which will be redistributed among Gayle, the sled, and her brother. Using the formula v = [tex]\sqrt{\frac{2KE}{m} }[/tex] again, we get v = [tex]\sqrt{\frac{ 2* 2535.28 J}{79.3 kg} }[/tex] ≈ 5.06 m/s. Therefore, the speed of Gayle and her brother at the bottom of the hill is approximately 5.06 m/s.
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Required information The radius of a wheel is 0.880 m. A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude 5.00 N, unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. How much rope unwinds while the wheel makes 1.00 revolution? Required information The radius of a wheel is 0.880 m. A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude 5.00 N, unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. What is the torque on the wheel about its axis due to the rope? N⋅m Required information The radius of a wheel is 0.880 m. A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude 5.00 N, unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. What is the angular displacement Δθ, in radians, of the wheel during 1.10 revolution? rad
A. the circumference of the wheel is 2π(0.880) = 5.51 m. B. the torque is 5.00 N x 0.880 m = 4.40 N⋅m. and C. the angular displacement is Δθ = 2π(1.10) = 6.92 radians.
A. To find the amount of rope that unwinds while the wheel makes 1.00 revolution, we can use the formula for the circumference of a circle: C = 2πr. Given that the radius of the wheel is 0.880 m, the circumference of the wheel is 2π(0.880) = 5.51 m.
B. To find the torque on the wheel due to the rope, we can use the formula: Torque = Force x Radius. Given that the force is 5.00 N and the radius is 0.880 m, the torque is 5.00 N x 0.880 m = 4.40 N⋅m.
C. To find the angular displacement Δθ of the wheel during 1.10 revolutions, we can use the formula: Δθ = 2πn, where n is the number of revolutions. Given that n = 1.10, the angular displacement is Δθ = 2π(1.10) = 6.92 radians.
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Star A has a magnitude of 4 and Star B has a magnitude of 6. How much brighter is Star A than Star B?
a. 0.0006554
b. 0.16
c. 0.0002621
d. 1.5
e. 3.33
f. 1526
g. 0.0102
h. 2.5
i. 610
j. 97.7
k. 2
l. 0.00164
m. 5
n. 6.25
o. 3815
The difference in magnitudes between the stars can be found using the formula Δm = m1 - m2. Where m1 = magnitude of star A, m2 = magnitude of star B, and Δm is the difference in magnitudes.
Given that Star A has a magnitude of 4 and Star B has a magnitude of 6. Therefore,Δm = 4 - 6= -2.
The negative sign indicates that star B is brighter than star A.
Thus, to find out how much brighter Star A is than Star B, we need to take the antilogarithm of Δm/2.5.
This can be calculated as follows:antilog (-2/2.5)= antilog (-0.8) = 0.1585.
The antilogarithm is approximately equal to 0.16.
Therefore, Star A is 0.16 times brighter than Star B. Answer: The brightness ratio between Star A and Star B is 0.16.
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Calculate the number of electrons that pass through a cross-section of wire in 6 seconds, when the wire is carrying a current of 650 mA.
Approximately 3,900 electrons pass through a cross-section of the wire in 6 seconds when the wire is carrying a current of 650 mA.
To calculate the number of electrons that pass through a cross-section of wire in a given time, we can use the equation:
Q = I ×t ×e / q
where:
Q is the total charge
I is the current
t is the time
e is the elementary charge (1.6 × 10⁻¹⁹ C)
q is the charge of one electron (1.6 × 10⁻¹⁹ C)
Given:
Current (I) = 650 mA = 650 × 10⁻³ A
Time (t) = 6 seconds
Let's substitute these values into the equation and calculate the total charge (Q):
Q = (650 × 10⁻³ A) × (6 s) × (1.6 × 10⁻¹⁹ C) / (1.6 × 10⁻¹⁹ C)
Simplifying the equation:
Q = 6.24 × 10⁻¹⁶ C
Now, to calculate the number of electrons (N), we divide the total charge (Q) by the charge of one electron (q):
N = Q / q
= (6.24 × 10⁻¹⁶ C) / (1.6 × 10⁻¹⁹ C)
Simplifying the equation:
N ≈ 3.9 × 10³
Therefore, approximately 3,900 electrons pass through a cross-section of the wire in 6 seconds when the wire is carrying a current of 650 mA.
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a star of the sun’s mass never becomes hot enough for carbon to react, and the star’s energy production is at an end. what happens to the outer layers? what is this star called?
When a star of the Sun's mass never becomes hot enough for carbon to react, and the star's energy production is at an end, it sheds its outer layers and what is left of the core is called a white dwarf.
A white dwarf is a stellar remnant that is incredibly dense. A white dwarf is the remaining core of a star that has run out of fuel and shed its outer layers. It's made up of electron-degenerate matter, which is a phase of matter that can only be achieved at incredibly high densities. The radius of a white dwarf is comparable to that of the Earth, but its mass is typically between 0.5 and 1.4 solar masses. They are called white dwarfs because of how they appear literally. A white dwarf is White and Small about the size of the Earth, perhaps a tiny bit bigger hence a dwarf star. In 1863, the optician and telescope maker Alvan Clark spotted this mysterious object. This companion star was later determined to be a white dwarf. There are 10 billion white dwarfs in the Milky Way galaxy because many sunlike stars have already gone through the process of dying
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how does the spectrum of a molecule differ from the spectrum of an atom?
The spectrum of a molecule includes not only the electronic transitions observed in atoms but also additional features related to molecular bonding, vibrational motion, and rotational motion.
The spectrum of a molecule differs from the spectrum of an atom primarily because a molecule consists of two or more atoms bonded together. This bonding introduces additional energy levels and interactions that affect the energy transitions and resulting spectral features.
In an atom, the spectrum is characterized by discrete lines or bands corresponding to the energy transitions of electrons between different energy levels. Each element has a unique atomic spectrum, which can be used for identification purposes. The transitions in an atom's spectrum occur due to changes in the electron configuration and involve electronic transitions within the atom.
On the other hand, a molecule has both electronic and vibrational energy levels. The electronic transitions in a molecule involve the movement of electrons between different energy levels of the molecular system. These transitions give rise to electronic spectral features, similar to those observed in atoms. However, in a molecule, the energy levels can be affected by the presence of multiple atoms, molecular orbitals, and molecular bonding.
In addition to electronic transitions, molecules also exhibit vibrational and rotational energy levels. Vibrational transitions involve the motion of atoms within the molecule, and rotational transitions involve the rotation of the molecule as a whole. These transitions give rise to additional spectral features in the infrared (IR) and microwave regions, respectively.
Overall, the spectrum of a molecule includes not only the electronic transitions observed in atoms but also additional features related to molecular bonding, vibrational motion, and rotational motion. The complexity of the molecule's spectrum depends on its structure, composition, and the types of interactions present within the molecule.
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What is the approximate elevation at the center of Copernicus Crater? 11500 −11500 10000 1500 What is the correct order from oldest to youngest in which the following features formed? Apollo Patera, Dionysus Patera, Olympus Patera, Olympus Mons Olympus Mons, Dionysus Patera, Apollo Patera, Olympus Patera Olympus Mons, Olympus Patera, Dionysus Patera, Apollo Patera Olympus Patera, Apollo Patera, Dionysus Patera, Olympus Mons How many years ago did the feature at celestial coordinates RA 6h 16' 36", Dec 22 30
′
60
′′
form? 3000000 30000 3000 300000 Where on the H-R diagram would the star located at celestial coordinates RA 6 h45 m8.9 s,Dec−16
∘
422
′
58.0
′′
fall? red giant white dwarf main sequence blue giant
1) The approximate elevation at the center of Copernicus Crater is 11500 ft.The correct option is 1) 11500.The Copernicus Crater has a central peak in the middle. The central peak is the most prominent feature of the crater.
2) The correct order from oldest to youngest in which the following features formed is: Olympus Mons, Olympus Patera, Dionysus Patera, Apollo Patera. The correct option is 3) Olympus Patera, Dionysus Patera, Apollo Patera, Olympus Mons.
3) The feature at celestial coordinates RA 6h 16' 36", Dec 22 30
′
60
′′
form 3000 years ago.The correct option is 3) 3000.4) The star located at celestial coordinates RA 6 h45 m8.9 s,Dec−16
∘
422
′
58.0
′′
will fall on the main sequence of the H-R diagram.
The correct option is main sequence.
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