The angular distance from the North Celestial Pole to the point on the sky known as the summer solstice is approximately 23.5 degrees.
The summer solstice marks the point in the Earth's orbit around the Sun when the Northern Hemisphere experiences its longest day and shortest night. During this time, the North Pole is tilted towards the Sun at its maximum angle of 23.5 degrees. As a result, the Sun appears to reach its highest point in the sky for observers in the Northern Hemisphere. The North Celestial Pole, also known as the North Star or Polaris, is the point in the sky directly above the Earth's North Pole. It serves as a fixed reference point for celestial navigation. The summer solstice occurs when the Sun's declination reaches its maximum positive value of +23.5 degrees. This angle represents the tilt of the Earth's axis in relation to its orbit around the Sun. Therefore, the angular distance from the North Celestial Pole to the summer solstice point is approximately 23.5 degrees.
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A 52 kg skateboarder is standing on the edge of a 35 m tall half-pipe.
How much energy will the skateboarder have when he drops in the pipe?
What will his kinetic energy be when he reaches the bottom?
Calculate his speed using the energy from the question above.
Answer:
The speed I got is 26.2m/s (3sf)
Explanation:
When the skateboarder is standing at the edge of the half-pipe, they would have max gravitational potential energy of about 17800J (3sf), calculated by using this formula;
Gravitational potential energy (Eₚ) = mgh
Eₚ = 52 × 9.8 × 35
Eₚ = 17836J
Eₚ = 17800J (3sf)
As the skateboarder drops, this energy will also decrease because they are losing height and gets converted into kinetic energy. Hence, the kinetic energy increases. When they reach the bottom (assuming they haven't landed and stop moving), the skateboarder will reach max kinetic energy.
To calculate the speed from this energy, we can use this formula;
Eₖ = 1/2 × m × v²
Substitute the values;
17800 = 1/2 × 52 × v²
17800 = 26 × v²
v² = 17800/26
v = √684.6
v = 26.2m/s (3sf)
1.
Occupations that produce "work"
Babysitter
Banker
Orange picker
Cake decorator
Delivery driver
Lumberjack
Weatherman
Moving man
Truck driver
Receptionist
Fisherman
Rabbi
Carpet installer
Bricklayer
The occupations which are listed and involved some form of work are Orange picker, Delivery driver, Lumberjack, Moving man and Truck driver.
In physics, the concept of "work" has a specific definition that relates to the transfer of energy. Occupations that involve physical tasks and the transfer of energy can be considered as producing work in the context of physics. Here are some occupations from the list that can be associated with work in physics:
1. Orange picker: This occupation involves physical labor to pick oranges, which requires exerting force and doing mechanical work against gravity.
2. Delivery driver: Delivery drivers perform work when they lift and carry packages, loading and unloading them from vehicles, which involves applying force over a distance.
3. Lumberjack: Lumberjacks engage in physically demanding work, such as cutting down trees and splitting wood, which requires the application of force and energy.
4. Moving man: Moving professionals lift and transport heavy furniture and boxes, which involves doing work against gravity and overcoming the resistance of objects.
5. Truck driver: While driving itself may not involve work in the physics sense, truck drivers may perform physical tasks like loading and unloading cargo, which can involve exerting force and doing work.
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1.0 mol of an ideal gas starts at 1.0 atm and 77F and does 1.0
kJ of work during an adiabatic expansion. Calculate the final
volume of the gas. Express your answer in litres. In your
calculation, f
The volume of 1.0 mol of an ideal gas starting at 1.0 atm and 77°F is 24.4 L.
The volume of 1.0 mol of an ideal gas starting at 1.0 atm and 77°F can be calculated using the ideal gas law equation: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Using the given values of P = 1.0 atm, T = 77°F = 298.15 K, and n = 1.0 mol, we can rearrange the equation to solve for V:V = nRT/P = (1.0 mol)(0.08206 L·atm/(mol·K))(298.15 K)/(1.0 atm) = 24.4 L
So the volume of 1.0 mol of an ideal gas starting at 1.0 atm and 77°F is 24.4 L.
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help ASAP 2i, 2ii, 2iii
2. A surfer is drifting west at 2 m/s. He catches a wave, and it accelerates him north at 6 m/s² for 3 seconds. i. What is the final velocity of the surfer after this acceleration?(8) ii. How far nor
A surfer is drifting west at 2 m/s. He catches a wave, and it accelerates him north at 6 m/s² for 3 seconds.(i)The final velocity of the surfer after the acceleration is 16 m/s northward.(II) The surfer will have traveled 21 meters northward when he stops accelerating.
i. To find the final velocity of the surfer after the acceleration, we can use the formula:
v = u + at
where:
v is the final velocity,u is the initial velocity,a is the acceleration, andt is the time.Given:
Initial velocity (u) = 2 m/s (westward)
Acceleration (a) = 6 m/s² (northward)
Time (t) = 3 seconds
The initial velocity is in the westward direction, so we can consider it as negative. Let's calculate the final velocity (v):
v = u + at
v = -2 m/s + (6 m/s² * 3 s)
v = -2 m/s + 18 m/s
v = 16 m/s (northward)
Therefore, the final velocity of the surfer after the acceleration is 16 m/s northward.
ii. To find the distance traveled northward during the acceleration, we can use the equation:
s = ut + (1/2)at²
where:
s is the distance traveled,
u is the initial velocity,
a is the acceleration, and
t is the time.
Given:
Initial velocity (u) = 2 m/s (westward)
Acceleration (a) = 6 m/s² (northward)
Time (t) = 3 seconds
Since the initial velocity is westward, we can consider it as negative. Let's calculate the distance traveled (s):
s = ut + (1/2)at²
s = -2 m/s * 3 s + (1/2) * 6 m/s² * (3 s)²
s = -6 m + (1/2) * 6 m/s² * 9 s²
s = -6 m + 27 m
s = 21 m (northward)
Therefore, the surfer will have traveled 21 meters northward when he stops accelerating.
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Question 8 A 2.7 m long string vibrates as a three loop standing wave. The amplitude is 0.98 cm and wave speed is 90 m/s. Part A Find the frequency of the vibration. μA ? f= Value Submit Provide Feed
The frequency of the vibration is approximately 50 Hz, calculated using the formula v = fλ, where v is the wave speed and λ is the wavelength determined by the length of the string and the number of loops.
In a standing wave, the length of the string can be related to the wavelength of the wave by the equation:
λ = 2L/n,
where λ is the wavelength, L is the length of the string, and n is the number of loops.
In this case, the length of the string L is given as 2.7 m, and the number of loops n is 3. Plugging in these values, we can solve for the wavelength:
λ = 2(2.7 m)/3
λ = 1.8 m.
The wave speed v is given as 90 m/s. The frequency f can be calculated using the formula:
v = fλ.
Rearranging the equation, we have:
f = v/λ = 90 m/s / 1.8 m
f = 50 Hz.
Therefore, the frequency of the vibration is approximately 50 Hz.
The frequency of the vibration is approximately 50 Hz, calculated using the formula v = fλ, where v is the wave speed and λ is the wavelength determined by the length of the string and the number of loops.
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using the method of -substitution, where (enter a function of ) (enter a function of ) (enter a number) (enter a number) (enter a function of ).
Explanation of how to solve a problem using the method of substitution. Please provide the problem that you need help with so that I can provide a detailed explanation.
Here is the general process for solving a problem using the method of substitution.
Step 1: Identify the function that can be expressed in terms of the other function.
Step 2: Express one of the variables in terms of the other by rearranging the equation.
Step 3: Substitute the expression for the variable in terms of the other variable into the other equation.
Step 4: Simplify the equation by combining like terms and solve for the remaining variable.
Step 5: Use the value of one of the variables to find the value of the other variable.
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the north pole of one magnet experiences an attractive force to the north pole of a second magnet. select one: true false
Answer: false
Explanation: Similar poles always repel and only different poles attract (south pole and north pole attract) but (north pole and north pole/ south pole and south pole repel).
if the cg of an aircraft is moved from the aft limit to beyond the forward limit, how will it affect the cruising and stalling speed?
When the center of gravity (CG) of an aircraft is moved from the aft limit to beyond the forward limit, it will affect the cruising and stalling speed as follow
When the center of gravity (CG) is moved from the aft limit to beyond the forward limit, the aircraft will become more unstable. The weight of the aircraft is mostly at the forward end and the tail portion becomes lighter. As a result, it will cause the aircraft to pitch down and result in difficulty in controlling the airplane.
This shift in CG to the forward limit will decrease the cruising speed of the aircraft as it requires more power to maintain the required altitude and speed. Also, the stalling speed of the aircraft will be lower, which means that the aircraft will stall at a lower speed.
This is because the forward limit shifts the neutral point behind the center of gravity, making the aircraft more unstable and reducing the speed at which the airplane stalls. As the CG shifts forward, the margin between the stall speed and the maximum operating speed reduces, making it more difficult to recover from a stall.
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√√₁ + + Ct². Using dimensional analysis obtain the dimensions and SI units for A, t 3) (5 points) If v = Av² t + B B, and C?
The dimensions and SI units for A, t, and C in the equation v = Av²t + B are:
- A has dimensions of [1]/[L] and its SI unit is 1/meter (1/m).
- t has dimensions of [T] and its SI unit is seconds (s).
To obtain the dimensions and SI units for the variables A, t, and C in the equation v = Av²t + B, we can analyze the equation using dimensional analysis.
The equation is:
v = Av²t + B
Let's assign dimensions and units to each term:
- v has dimensions of velocity, [L]/[T] (length per time), and its SI unit is meter per second (m/s).
- Av²t has dimensions of [A][L]²[T], where [A] represents an unknown dimension and [L] and [T] are length and time dimensions, respectively. The SI unit will depend on the dimensions of A and t.
- B has dimensions of velocity, [L]/[T] (length per time), and its SI unit is also meter per second (m/s).
Equating the dimensions on both sides of the equation, we have:
[L]/[T] = [A][L]²[T] + [L]/[T]
To balance the dimensions, the dimensions of [A] must be [1]/[L] and the dimensions of [t] must be [T].
Therefore, the dimensions and SI units for A, t, and C in the equation v = Av²t + B are:
- A has dimensions of [1]/[L] and its SI unit is 1/meter (1/m).
- t has dimensions of [T] and its SI unit is seconds (s).
- C does not appear in the given equation, so we cannot determine its dimensions or SI units based on the given information.
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when 6.50 ×105j of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from 1.9 m3 to 4.1 m3 .
Calculate the work done by the gas.
Calculate the change in internal energy of the gas.
Graph this process on a PV diagram.
When 6.50 × 10^5 J of heat is added to a gas in a cylinder with a frictionless piston at atmospheric pressure, gas volume increases from 1.9 m^3 to 4.1 m^3. We need to calculate the work done and change in internal energy.
To calculate the work done by the gas, we can use the equation:
Work = Pressure × Change in Volume.
Given that the pressure is maintained at atmospheric pressure, we can substitute the values:
Work = Atmospheric Pressure × (Final Volume - Initial Volume).
Work = 1 atm × (4.1 m^3 - 1.9 m^3).
Next, we calculate the change in internal energy of the gas using the first law of thermodynamics:
Change in Internal Energy = Heat Added - Work Done.
Given that heat added is 6.50 × 10^5 J and we have already calculated the work done, we can substitute the values:
Change in Internal Energy = 6.50 × 10^5 J - Work Done.
To graph this process on a PV diagram, we plot pressure (P) on the y-axis and volume (V) on the x-axis. We mark the initial point at 1.9 m^3 and atmospheric pressure. Then, we mark the final point at 4.1 m^3 and atmospheric pressure. The process is represented by a straight line connecting these two points.
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In which one of the following lists are different types of electromagnetic waves ranked in order of increasing wavelength? a. X-rays, gamma rays, visible light, radio waves b. Visible light, radio waves, gamma rays, x-rays c. Gamma rays, x-rays, visible light, radio waves d. Gamma rays, x-rays, radio waves, visible light
Electromagnetic rays ranked in order of increasing wavelength are:
c. Gamma rays < X-rays < Visible light < Radio Waves
Electromagnetic waves are generated when an electric field comes in contact with magnetic field. They represent a family of waves showing similar properties.
Gamma rays have the shortest wavelength, ranging between 10⁻¹¹ to 10⁻¹³m while Radio rays have the longest wavelength, ranging from 10³ to 10⁻¹m
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An airplane has an air speed of 300km/h and is heading due west. If it encounters a wind blowing south at 50km/h, what is the resultant ground velocity of the plane?
The ground velocity of the airplane is (-300, -50) km/h.
The airplane's airspeed is 300 km/h and is directed due west. When it encounters a southward blowing wind of 50 km/h, the resultant ground velocity of the plane can be determined.
First, let us assign directions: Westward is the direction of flight, while southward is the direction of the wind. As a result, the velocity of the wind is negative. Here are the steps to compute the ground velocity of the airplane:
Step 1: Determine the vector components. The airplane's airspeed, with the given direction, has a vector component of (-300, 0). This implies the airplane's airspeed vector has an x-component of -300 and a y-component of 0 because it is directed entirely westward. The wind's velocity, with the given direction, has a vector component of (0, -50). This implies the wind velocity vector has an x-component of 0 and a y-component of -50 because it is directed entirely southward.
Step 2: Add the vector components to obtain the ground velocity. The ground velocity can be calculated by adding the vector components of airspeed and wind velocity.
V_g = V_air + V_windV_g = (-300, 0) + (0, -50) = (-300, -50)
Therefore, the ground velocity of the airplane is (-300, -50) km/h. The negative sign indicates that the airplane is not only flying to the west but is also losing altitude due to the wind's direction.
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Find the potential difference. Magnetic Field = 250 T radius =
36mm (circular path)
The potential difference can be found using the formula ΔV = Bvr where B is the magnetic field, v is the velocity of the charged particle and r is the radius of the circular path.
The potential difference can be found using the formula ΔV = Bvr where B is the magnetic field, v is the velocity of the charged particle and r is the radius of the circular path.
Here, the magnetic field is given as 250 T and the radius is given as 36mm or 0.036m.
However, the velocity of the charged particle is not given. Therefore, the potential difference cannot be determined.
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Red light with a wavelength of 725 nm enters glass with index of refraction 1.52.Located within the glass block is a single slit.and a screen is placed 1.25 m away.If the width of the slit is 18.5m.find the width of the central maximum (a) in degrees and (b) in centimeters. 5. A soap film in air has a thickness of 175 nm.If the index of refraction of the soap film is 1.35,what colour of light will be transmitted through?(i.e.,What colour does the bubble appear to be from the other side?
4. Red light with a wavelength of 725 nm enters glass with index of refraction 1.52.Located within the glass block is a single slit.and a screen is placed 1.25 m away.If the width of the slit is 18.5m (a)The width of the central maximum is approximately 0.275°.(b)The width of the central maximum is approximately 6.052 × 10^(-6) cm.(5)In this case, the wavelength falls within the range of green light, so the soap bubble will appear green from the other side
(4) To calculate the width of the central maximum, we can use the formula:
Width of central maximum = (wavelength * distance to screen) / (slit width * index of refraction)
Given:
Wavelength of red light (λ) = 725 nm = 725 × 10^(-9) m
Distance to screen (D) = 1.25 m
Slit width (w) = 18.5 μm = 18.5 × 10^(-6) m
Index of refraction (n) = 1.52
(a) Width of central maximum in degrees:
To convert the width to degrees, we can use the small angle approximation:
Width in degrees ≈ (Width of central maximum / Distance to screen) * (180° / π)
Substituting the values into the formula:
Width of central maximum = (725 × 10^(-9) m * 1.25 m) / (18.5 × 10^(-6) m * 1.52) ≈ 6.052 × 10^(-4) m
Width in degrees ≈ (6.052 × 10^(-4) m / 1.25 m) * (180° / π) ≈ 0.275°
So, the width of the central maximum is approximately 0.275°.
(b) Width of central maximum in centimeters:
To convert the width to centimeters, we can simply divide by 100:
Width in centimeters = (Width of central maximum) / 100 ≈ 6.052 × 10^(-4) m / 100 ≈ 6.052 × 10^(-6) cm
So, the width of the central maximum is approximately 6.052 × 10^(-6) cm.
(5) To determine the color of light transmitted through the soap film, we need to consider the interference of light waves. When light reflects off the top and bottom surfaces of the film, interference occurs. Depending on the thickness of the film and the wavelength of light, certain colors will be enhanced or canceled out.
Given:
Thickness of soap film (d) = 175 nm = 175 × 10^(-9) m
Index of refraction of soap film (n) = 1.35
To find the color of light transmitted, we can use the equation:
2 * n * d = m * λ
where:
m is the order of the interference (m = 1 for the first-order maximum)
λ is the wavelength of light
Rearranging the equation to solve for λ:
λ = (2 * n * d) / m
Substituting the values:
λ = (2 * 1.35 * 175 × 10^(-9) m) / 1
λ ≈ 4.73 × 10^(-7) m
The wavelength of light transmitted is approximately 4.73 × 10^(-7) m.
By comparing the wavelength to the visible light spectrum, we can determine the corresponding color. In this case, the wavelength falls within the range of green light, so the soap bubble will appear green from the other side.
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geoff counts the number of oscillations of a simple pendulum at a location where the acceleration due to gravity is 9.80 m/s2 and finds that it takes 7.05 s for 16.0 complete cycles. Calculate the length of the pendulum.
The length of the pendulum is 0.44338 m.
Acceleration due to gravity, g = 9.80 m/s2, Number of oscillations, n = 16, Time taken, t = 7.05 s.
Let l be the length of the pendulum. Now, one complete oscillation means when the pendulum starts from its extreme position (i.e., extreme position A), moves to the other extreme position (i.e., extreme position B), and returns back to position A.
Let’s calculate the time taken by the pendulum to complete one oscillation, i.e., the time period of the pendulum.t1 = time taken for 1 oscillation.
t1 = t/n = 7.05/16.0=0.44125 s
Now, the time period is given by, T=2π √(l/g) Where T is the time period of the pendulum. π = 3.1415 (approx)
Putting the given values of g and T in the above equation,
T = 2π √(l/g) = 2 x 3.1415 √(l/9.80)
Now, substituting T1 in the above equation, we get:
0.44125 = 2 x 3.1415 √(l/9.80)√(l/9.80)
= 0.44125/(2π)
= 0.07010049l
= (√(0.07010049 × 9.80))2l = 0.44338 m
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Two resistors, 20 ohms and 30 ohms, are connected in parallel. This combination is connected in a series to an 8 ohms resistor and a battery of e.m.f. 12 volts. What is the current along 20 ohms?
Ohm's law is a fundamental principle in physics and electrical engineering that relates the current flowing through a conductor to the voltage applied across it. It states that the current (I) passing through a conductor is directly proportional to the voltage (V) across the conductor, and inversely proportional to the resistance (R) of the conductor.
The current along 20 ohms in a circuit with two resistors, 20 ohms and 30 ohms, which are connected in parallel, and this combination is connected in a series to an 8 ohms resistor and a battery of e.m.f. 12 volts can be calculated as follows: The equivalent resistance for the two resistors connected in parallel by using the formula 1/Rt = 1/R1 + 1/R2, where R1 and R2 are the resistors in parallel.1/Rt = 1/20 + 1/30 = 3/60 + 2/60 = 5/60Rt = 60/5 = 12 ohms.
The equivalent resistance of the two parallel resistors is 12 ohms.
Now, calculate the total resistance of the circuit by adding the equivalent resistance of the parallel resistors and the 8 ohms resistor.
Rtotal = 12 + 8 = 20 ohmsThe total resistance of the circuit is 20 ohms.
Finally, calculate the current along the 20 ohms resistor using Ohm's law, which states that I = V/R, where V is the voltage of the battery and R is the resistance. I = V/R = 12/20 = 0.6 A.
The current along the 20 ohms resistor is 0.6 A.
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Image Diagram A car is following another car along a
straight road. The first car has a rear window tilted at
45° to the horizontal. Draw a ray diagram showing the
position of the Sun that would cause sunlight to
reflect into the eyes of the driver of the second car.
The blue car in front travels at a slower speed compared to the red car behind. Eventually, the red car would have to overtake the blue car because it is much faster. First, let's compute the time it takes before the red car catches up to the blue car. The solution is as follows:
30 m = (60 km/h - 50 km/h)*(1000 m/1 km)*(1 h/3,600 s)*(t)
t = 10.8 seconds
After 10.8 seconds, the red car catches up to the blue car. With this amount of time, the blue car would still cover additional distance. That would be equal to:
Distance = Speed*time
Distance = (50 km/h)*(1 h/3600 s)*(10.8 s)
Distance = 0.15 km
Perception distance is the distance traveled by the vehicle when the driver perceive the hazard situation.
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Two light rays are incident from air into an unknown liquid at
the same point. If θ=25o and
β=37o, find the angle of refraction a? show solution
Two light rays are incident from air into an unknown liquid at the same point. If θ=25degree and β=37degree. The angle of refraction is 143°.
To find the angle of refraction (a) given the incident angle (β) and the refractive index of the unknown liquid, we can use Snell's Law:
n₁ * sin(β) = n₂ * sin(a)
where:
n₁ is the refractive index of air (approximately 1.00),
n₂ is the refractive index of the unknown liquid, and
β and a are the incident and refracted angles, respectively.
Given that β = 37° and θ = 25°, we can substitute these values into Snell's Law:
n₁ * sin(37°) = n₂ * sin(a)
Since the incident angle θ and the refracted angle a are related by the equation θ + a = 180° (or π radians), we can also write:
sin(θ) = sin(180° - a)
Now we can solve these equations simultaneously to find the angle of refraction a.
sin(37°) = sin(180° - a)
Taking the inverse sine of both sides:
37° = 180° - a
a = 180° - 37°
a = 143°
Therefore, the angle of refraction is 143°.
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How much is the energy of a single photon of the blue
light with a frequency of 7.5 x 1014 Hz?
Group of answer choices
4.97 x 1015 J
8.84 x 10-49 J
4.97 x 10-19 J
1.13 x 1048
The energy of a single photon of the blue light with a frequency of 7.5 x 10¹⁴ Hz is 4.97 x 10⁻¹⁹ J.
The energy (E) of a photon can be calculated using the equation:
E = h * f
where:
E = energy of the photon
h = Planck's constant (approximately 6.626 x 10⁻³⁴ J s)
f = frequency of the light wave
E = (6.626 x 10⁻³⁴ J s) * (7.5 x 10¹⁴ Hz)
E = 4.97 x 10⁻¹⁹ J
Therefore, the energy of a single photon of blue light with a frequency of 7.5 x 10¹⁴ Hz is approximately 4.97 x 10⁻¹⁹ J.
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A thin rod, 0.77 m long, is pivoted such that it hangs vertically from one end. You want to hit the free end of the rod just hard enough to get the rod to swing all the way up and over the pivot. Part A How fast do you have to make the end go? Express your answer with the appropriate units. Ć ☐☐ μA Value Units V =
The speed needed to make the free end go in order to swing all the way up and over the pivot is 1.98 m/s.
Part A of the given problem asks to calculate the speed needed to make the free end go to swing all the way up and over the pivot. Let the pivot point be P, the center of mass of the rod be C and the free end of the rod be A. The rod will swing over the pivot if the height of the center of mass C becomes zero. Using the law of conservation of energy, the initial potential energy of the rod is converted to the final kinetic energy of the rod. At the highest point, the kinetic energy will become zero and all the potential energy will become zero. Hence, the potential energy at the initial point will be equal to the potential energy at the highest point: mg(0.77) = (1/2)(0.20)v²Solving this equation, we get: v = 1.98 m/s Therefore, the speed needed to make the free end go in order to swing all the way up and over the pivot is 1.98 m/s.
The term "speed" means. The rate at which an object moves in any direction. The ratio of distance to time traveled is what is used to measure speed. Because it only has a direction and no magnitude, speed is a scalar quantity.
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what focal length should this camera's lens have for the image of the bridge to cover the entire detector?
To ensure that the image of the bridge covers the entire detector, the camera's lens should have a focal length equal to the diagonal of the detector.
This is known as the focal length of the diagonal.
To ensure that the image of the bridge covers the entire detector, the camera's lens should have a focal length equal to the diagonal of the detector.
This is known as the focal length of the diagonal.
To ensure that the image of the bridge covers the entire detector, the camera's lens should have a focal length equal to the diagonal of the detector.
This is known as the focal length of the diagonal.
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1 Which has the LEAST momentum?
Group of answer choices
a 2 kg ball moving at 8 m/s
a 750 g ball moving at 15 m/s
a 80 kg ball moving at 25m/s
a 12 kg ball moving at 1
A 750 g ball moving at 15 m/s will have the least momentum.
The momentum can be defined as the product of mass and velocity. It is a vector quantity that shows how difficult it is
to stop a moving object. The momentum of an object depends on two factors, mass and velocity. The least momentum
will have the object with the least mass and velocity. Therefore, a 750 g ball moving at 15 m/s will have the least
momentum. The momentum can be calculated as follows: p = m × v where, p is momentum, m is mass and v is
velocity. We have given that; a 2 kg ball moving at 8 m/s, p = 2 kg × 8 m/s = 16 kgm/s .A 750 g ball moving at 15 m/s, p =
0.75 kg × 15 m/s = 11.25 kgm/s. A 80 kg ball moving at 25m/s, p = 80 kg × 25 m/s = 2000 kgm/s .A 12 kg ball moving at
1 m/s, p = 12 kg × 1 m/s = 12 kgm/s. The momentum of the 750 g ball moving at 15 m/s is 11.25 kgm/s. Therefore, the
least momentum will be for a 750 g ball moving at 15 m/s.
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10. If an electrical appliance becomes live, the appliance
A) stops working and is safe to touch
B) continues working and is safe to touch
(C) continues working and is dangerous to touch
D) Is safe to touch because of the fuse.
Answer:
C
Explanation:
If an electrical appliance becomes alive, the appliance continues working and is dangerous to touch
A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is considered to be stable and does not change over time. The student claims, "The only experimentally measurable external force exerted on the planet is the force due to gravity from the star. " Is the student’s claim supported by the evidence? What reasoning either supports or contradicts the student’s claim? Yes. Other external forces are exerted on the planet, but they are of negligible magnitude
The student's claim that "The only experimentally measurable external force exerted on the planet is the force due to gravity from the star" is partly supported by the evidence. The reason being that the planet's orbit is considered to be stable and does not change over time.
The planet's motion and orbit are affected by gravity. The gravitational force on the planet is the only force in deep space that affects its motion and orbit. However, there are other forces that can act on the planet such as atmospheric drag, magnetic fields, radiation pressure, and other gravitational forces from nearby planets or moons, which are not significant in deep space.
These forces can cause a change in the planet's motion and orbit. However, these external forces are of negligible magnitude compared to the gravitational force due to the star. Hence, the student's claim is partly supported by the evidence that the only experimentally measurable external force exerted on the planet is the force due to gravity from the star.
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A cannon fires a cannonball with an initial velocity of 300 m/s at 54.5° above the horizontal. The cannonball impacts a raised plateau 42.0 s after firing. Let the +x-direction be directly ahead of the cannon and the +y-direction be upward. Find the x- and y-coordinates of the cannonball's impact point, relative to its firing point (in m). X = m y = m Need Help? Read It
Given data Initial velocity, u = 300 m/s Angle of projection, θ = 54.5°Time of flight, t = 42 s For projectile motion, we have the following kinematic equation s:u_x = u cos θu_y = u sin θx = u_x t + (1/2) a_x t^2y = u_y t + (1/2) a_y t^2u_x is the horizontal component of the initial velocity of the projectile, and u_y is the vertical component of the initial velocity of the projectile. a_x is the horizontal component of the acceleration of the projectile, and a_y is the vertical component of the acceleration of the projectile.
We can find u_x and u_y using trigonometric ratios. u_x = u cos θ = (300 m/s) cos 54.5° = 172.7 m/su_y = u sin θ = (300 m/s) sin 54.5° = 247.7 m/sa_x = 0 (assuming no air resistance) a_y = - g = - 9.81 m/s^2 (taking downward direction as positive) From the equation of motion in y direction, we have y = u_y t + (1/2) a_y t^2 ⇒ y = 247.7 m/s × 42 s + (1/2) (- 9.81 m/s^2) (42 s)^2⇒ y = 10470.6 m
The maximum height reached by the projectile = H = u_y^2 / (2 a_y) = (247.7 m/s)^2 / (2 × 9.81 m/s^2) = 3139.4 mWe can also find the horizontal distance traveled by the projectile using the equation of motion in x direction:x = u_x t = 172.7 m/s × 42 s = 7253.4 m Therefore, the x-coordinate of the impact point is 7253.4 m, and the y-coordinate of the impact point is 10470.6 m. Thus, the coordinates of the impact point are:X = 7253.4 mY = 10470.6 m.
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To find the x- and y-coordinates of the cannonball's impact point, we need to break down the initial velocity into its x and y components. The x-coordinate is found using the equation Vx = V * cos(θ), and the y-coordinate is found using the equation y = Vy * t + (1/2) * a * t^2. Plugging in the given values, the x-coordinate is 7587.9 m and the y-coordinate is 10049.4 m, relative to the firing point.
Explanation:To find the x- and y-coordinates of the cannonball's impact point, we need to break down the initial velocity into its x and y components. The component in the x-direction is given by the equation Vx = V * cos(θ), where V is the initial velocity and θ is the angle of firing. Plugging in the given values, we get Vx = 300 * cos(54.5°) = 300 * 0.60182 = 180.546 m/s.
Next, we need to find the time it takes for the cannonball to reach the impact point. Since the cannonball was fired horizontally in the x-direction, the time of flight can be found using the equation t = d / Vx, where d is the horizontal displacement. Rearranging this equation to solve for d, we have d = t * Vx. Plugging in the given values, we get d = 42.0 s * 180.546 m/s = 7587.9 m.
The y-component of the velocity is given by the equation Vy = V * sin(θ), so Vy = 300 * sin(54.5°) = 300 * 0.79862 = 239.586 m/s. The vertical displacement can be found using the equation y = Vy * t + (1/2) * a * t^2, where a is the acceleration due to gravity (-9.8 m/s^2) and t is the time of flight. Plugging in the given values, we get y = 239.586 m/s * 42.0 s + (1/2) * (-9.8 m/s^2) * (42.0 s)^2 = 10049.4 m.
Therefore, the x-coordinate of the impact point is 7587.9 m and the y-coordinate is 10049.4 m, relative to the firing point.
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The electric field in units of N/C at a distance of 8.1 cm from an isolated point particle with a charge of 2 X10–9 C is: 2.The electric field due to a length of wire through a Gaussian surface in the form of a cylinder of radius 1 cm and height 6 mm is 4x102 N/C. What is the charge (in units of pC) of the wire inside the Gaussian surface?
The charge of the wire can be calculated as;q = ΦEϵ0 = (4x10² N/C)(8.85x10^-12 C²/N m²) = 3.54x10^-9 C = 3.54 pCIn units of pC, the charge will be 12.57 pC (picoCoulombs)
The electric field in units of N/C at a distance of 8.1 cm from an isolated point particle with a charge of 2 X10–9 C is 2.47x104 N/C.
The electric field can be calculated using Coulomb's Law.
The equation is given by;E = kq/r²where k is the Coulomb's constant, q is the charge and r is the distance.
Here, the electric field can be calculated as;E = (9x10^9 N m²/C²)(2x10^-9 C)/(0.081 m)²E = 2.47x10^4 N/C2.
The charge (in units of pC) of the wire inside the Gaussian surface is 12.57 pC.
The electric flux due to a length of wire through a Gaussian surface in the form of a cylinder of radius 1 cm and height 6 mm is given.
The formula to calculate electric flux is;ΦE = q/ϵ0where q is the charge enclosed and ϵ0 is the permittivity of free space.
Therefore, the charge of the wire can be calculated as;q = ΦEϵ0 = (4x10² N/C)(8.85x10^-12 C²/N m²) = 3.54x10^-9 C = 3.54 pCIn units of pC, the charge will be 12.57 pC (picoCoulombs)
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a substance has a volume of 10.0 cm3 and a mass of 89 grams. what is its density? 8.9 g/cm 3 99 g/cm 3 0.11 g/cm 3 79 g/cm 3
The density of the given substance is 8.9 g/cm³. The given substance has a volume of 10.0 cm³ and a mass of 89 grams.
We are to find the density of the given substance. The formula for density is:density = mass/volumeWe can now substitute the given values into the formula and get: density = mass/volume=>density = 89 g/10.0 cm³We simplify this expression as shown below: density = 8.9 g/cm³
Therefore, the density of the substance is 8.9 g/cm³.The response to this question has used only 100 words. To give a more detailed response, we can provide the following explanation:Explanation:When we want to find the density of a substance, we need to know its mass and volume. Density is the ratio of the mass of a substance to its volume, and it is expressed in grams per cubic centimeter (g/cm³).
The formula for density is: density = mass/volume Where density is in g/cm³, mass is in grams (g), and volume is in cubic centimeters (cm³).To find the density of the given substance, we have been given its mass and volume. We simply substitute these values into the formula and simplify to get the density. Therefore, the density of the given substance is 8.9 g/cm³.
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What is the equivalent energy Eb of the mass defect of an atom of 18O?
a) Option A
b) Option B
c) Option C
d) Option D
The equivalent energy Eb of the mass defect of an atom of 18O is 2.58 x 10-11 J.
This value is obtained by the formula E = m x c², where E is the equivalent energy, m is the mass defect of the atom, and c is the speed of light. The mass defect of an atom is the difference between its actual mass and its theoretical mass (which is the sum of the masses of its individual particles).
The mass defect is due to the conversion of some of the mass into energy during the formation of the nucleus.
To calculate the energy released or absorbed during this process, we use the famous equation E = m x c², where E is the equivalent energy, m is the mass defect, and c is the speed of light. The equivalent energy Eb of the mass defect of an atom of 18O is given by
Eb = Δm × c² where Δm is the mass defect of the atom and c is the speed of light.
Eb = 0.0308 × (2.998 × 108)²= 2.58 x 10-11 J
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Newton's law of universal gravitation states that the gravitational force exened by an object on any other object anywhere in the universe by Gmm F= where G is the universal gravitational constant (6.67 x 10-11 N.m 2kg 2), ms is mass 1, m2 is mass 2, and r is the distance between the two masses (from conter to contor). If the distance between the two masses doubles, the gravitational force between the two masse O remains the same O is reduced to 1/4. O is reduced to 1/9, O doubles O quadruples.
Answer:
Gravitational force between the two will reduce to [tex](1/4)[/tex] the original value.
Explanation:
The distance between the two objects was originally [tex]r[/tex]. The gravitational force between the two objects would be:
[tex]\displaystyle F = \frac{G\, m_{1}\, m_{2}}{r^{2}}[/tex].
If the distance between the two is doubled, the new distance will become [tex]2\, r[/tex]. The new gravitational force between the two will become:
[tex]\begin{aligned}\frac{G\, m_{1}\, m_{2}}{(2\, r)^{2}} &= \frac{G\, m_{1}\, m_{2}}{4\, r^{2}} = \frac{1}{4}\, \left(\frac{G\, m_{1}\, m_{2}}{r^{2}}\right)\end{aligned}[/tex].
In other words, the force between the two objects will become one-quarter of the initial value.
a quiet sound is produced by a loud speaker. loudness of the sound is increased. which property of the sound is increased?
Answer:amplitude
Explanation:
