It will take the Hubble Telescope 94,247 seconds or 26.18 hours to complete one full rotation around the Earth.
Rotation time of the Space Hub TelescopeThe Hubble Telescope is in a circular orbit around the Earth at a constant speed of 20 m/s. The distance between the telescope and the center of the Earth is:
300 km + 100 m = 300,100 m.
Orbital period T of the Hubble Telescope = T = 2πr / v
where
r is the radius of the orbit v is the orbital speed.In this case, r = 300,100 m and v = 20 m/s, so:
T = 2π(300,100 m) / (20 m/s) = 94247 seconds
Thus, it will take the Hubble Telescope approximately 94,247 seconds, or about 26.18 hours, to complete one full rotation around the Earth.
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A horizontal rope is attached from a truck to a 1500-kg car. As the truck tows the car on a horizontal straight road, the rope will break if the tension is greater than 2696 N. Ignoring friction, what is the maximum possible acceleration of the truck if the rope does not break?
A horizontal rope is attached from a truck to a 1500-kg car. As the truck tows the car on a horizontal straight road, the rope will break if the tension is greater than 2696 N.
The maximum tension that the rope can handle is 2696 N. To avoid breaking the rope, the tension in the rope when the truck accelerates should not exceed this value. We can use Newton's second law to relate the tension in the rope to the acceleration of the car
Tension = (mass of car) × (acceleration)
Where Tension is the tension in the rope, and the mass of the car is 1500 kg.
Therefore, the maximum acceleration of the truck can be found by dividing the maximum tension by the mass of the car.
Acceleration = Tension / (mass of car)
Acceleration = 2696 N / 1500 kg
Acceleration = 1.8 m/s²
Hence, the maximum possible acceleration of the truck is 1.8 m/s².
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Need help solving this problem
a) The tension in the rope is 123.9 N.
b) The moment of inertia of the wheel is 0.09 kg m².
c) The angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.
(a) To determine the tension in the rope, we need to analyze the forces acting on the object. There are two forces: the force of gravity pulling the object down the incline and the tension force pulling the object up the incline.
The force of gravity can be broken down into two components: one parallel to the incline and one perpendicular to the incline.
The parallel component causes the object to accelerate down the incline, while the perpendicular component is balanced by the normal force of the incline.
The tension force is responsible for the object's acceleration down the incline, so we can set up the following equation:
T - mg sin(theta) = ma
where T is the tension force, m is the mass of the object, g is the acceleration due to gravity, theta is the angle of the incline, and a is the acceleration of the object down the incline.
Putting in the given values, we get:
T - (12.5 kg)(9.81 m/s²)(sin(37°)) = (12.5 kg)(2.00 m/s²)
Solving for T, we get:
T = 123.9 N
Therefore, the tension in the rope is 123.9 N.
(b) To determine the moment of inertia of the wheel, we can use the following equation:
I = (1/2)MR²
where I is the moment of inertia, M is the mass of the wheel, and R is the radius of the wheel.
Putting in the given values, we get:
I = (1/2)(12.5 kg)(0.12 m)²
= 0.09 kg m²
Therefore, the moment of inertia of the wheel is 0.09 kg m².
(c) To determine the angular speed of the wheel after 3.50 s, we can use the following equation:
ω = ω₀ + αt
where ω is the final angular speed, ω₀ is the initial angular speed (which is zero in this case), α is the angular acceleration, and t is the time.
We can find the angular acceleration using the following equation:
α = a/R
where a is the acceleration of the object down the incline (which we already know) and R is the radius of the wheel.
Putting in the given values, we get:
α = 2.00 m/s² / 0.12 m
= 16.7 rad/s²
Putting in the values for α and t, we get:
ω = 0 + (16.7 rad/s²)(3.50 s)
= 58.5 rad/s
Therefore, the angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.
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16 of 20:
Select the best answer for the question.
16. What happens to a substance at critical temperatures?
O A. The substance changes its state if it continues gaining or losing thermal energy.
O B. The substance can't lose any more thermal energy.
O C. The substance can't change its state, only its temperature.
O D. The substance changes its state only if it gains thermal energy.
At the critical temperatures of the substance, the substance can't lose any more thermal energy. Hence, option B is correct.
At the critical temperature, the properties of the liquid and gas phases become difficult to differentiate, and the substance exhibits particular type of behavior such as infinite compressibility and a lack of surface tension.
Additionally, at the critical temperature, the substance reaches its maximum vapor pressure, and any further increase in temperature and pressure will not cause it to change its state but only its density and hence, it cannot lose any more thermal energy from itself.
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The mass of 12 protons is approximately equal to ?
1 atomic mass unit
12 atomic mass units
The mass of 1 electron
The mass of 12 electrons
Answer:
Option 1, 1 atomic mass unit
At 3:00 P.M., a bank robber is spotted driving north on I-15 at milepost 129. His speed is 135.0 mi/h. At 3:37 P.M., he is spotted at milepost 182 doing 103.0 mi/h. (Assume a straight highway). Assume north to be positive. What is the bank robber’s displacement during this time interval? Enter a positive value if the displacement is toward north and enter a negative value if the displacement is toward south.
To find the displacement of the bank robber during this time interval, we need to calculate the distance he traveled in each direction and then find the vector difference between his initial and final positions.
From 3:00 P.M. to 3:37 P.M., the bank robber traveled for 37 minutes = 0.617 hours at a speed of 135.0 mi/h. Therefore, he traveled a distance of:
d1 = v1 x t1 = 135.0 mi/h x 0.617 h = 83.3 miles north
From 3:00 P.M. to 3:37 P.M., the bank robber traveled for 37 minutes = 0.617 hours at an average speed of 103.0 mi/h. Therefore, he traveled a distance of:
d2 = v2 x t2 = 103.0 mi/h x 0.617 h = 63.5 miles north
To find the displacement, we need to subtract the initial position from the final position. The initial position is at milepost 129 and the final position is at milepost 182. Therefore, the displacement is:
displacement = final position - initial position
displacement = 182 - 129 = 53 miles north
Since the displacement is in the positive direction, we enter a positive value. Therefore, the bank robber's displacement during this time interval is 53 miles north.
What evidence did Wegener NOT use to support his idea of continental drift?
A. Mountain ranges on different continents lined up when coastlines were matched up.
B. Fossils of plants and animals in climates where their survival would have been impossible. C. The thickness of layers of ice in the Antarctic.
D. Rock strata on different continents lined up when coastlines were matched up.
The evidence that Wegener did NOT use to support his idea of continental drift is "the thickness of layers of ice in the Antarctic.
option C
What is Wegener's primary evidence for continental drift?Wegener's primary evidence for continental drift included the fit of the coastlines of different continents, the distribution of fossils across different continents, and the alignment of rock strata on different continents.
So the thickness of layers of ice in the Antarctic, was not used by Wegener to support his idea of continental drift. While this evidence is important for supporting the theory of glaciation, it is not relevant to the theory of continental drift.
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'Assume all angles to be exact. A beam of light is incident on a plane mirror at an angle of 33 ∘ relative to the normal. What is the angle between the reflected ray and the surface of the mirror?
Answer:
When a beam of light is incident on a plane mirror, the angle of incidence (i) is equal to the angle of reflection (r) relative to the normal line to the mirror's surface. Therefore, if the incident angle is 33 degrees, the reflected angle will also be 33 degrees.
The angle between the reflected ray and the surface of the mirror can be calculated by subtracting the angle of reflection from 90 degrees (which is the angle between the normal line and the mirror's surface). So, the angle between the reflected ray and the surface of the mirror is:
90 degrees - 33 degrees = 57 degrees
Therefore, the angle between the reflected ray and the surface of the mirror is 57 degrees.
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Electric Forces need help will mark brainliest.
The net electric force on charge q1 is 15.47 N towards the left.
What is the net electric force on q1?The net electric force on charge q1 is calculated by applying Coulomb's law of electrostatic force.
F(net) = F(12) + F(13)
The force on q1 due to charge 2 is calculated as;
F(12) = (9 x 10⁹ x 8.99 x 10⁻⁶ x 5.16 x 10⁻⁶ )/(0.22²)
F(12) = 8.63 N
The force on q1 due to charge 3 is calculated as;
F(13) = -(9 x 10⁹ x 8.99 x 10⁻⁶ x 89.9 x 10⁻⁶ )/(0.55²)
F(13) = -24.1 N
The net force on q1 is calculated as;
F(net) = -24.1 N + 8.63 N = -15.47 N
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A 2000 kg car moves at a speed of 30 m/s. To reach this speed, it was necessary to burn 0.1 l of gas. Burning gas provides 30 MJ/l of source energy. Determine the energy efficiency (in %) of this car.
The energy efficiency of the car is approximately 16.7%.
The energy efficiency of a car is the ratio of the useful work output (in this case, the kinetic energy of the car) to the total energy input (in this case, the energy released by burning the gasoline). The equation for energy efficiency is:
Efficiency = Useful work output / Total energy inputThe useful work output can be calculated as the kinetic energy of the car using the equation:
KE = 0.5mv²where m is the mass of the car and v is its velocity.
Substituting the given values:
KE = 0.5 x 2000 kg x (30 m/s)² = 900,000 JThe total energy input is the energy released by burning 0.1 L of gasoline, which is:
Total energy input = 0.1 L x 30 MJ/L = 3 MJ = 3,000,000 JSubstituting these values into the equation for efficiency:
Efficiency = (900,000 J / 3,000,000 J) x 100% = 0.3 x 100% = 16.7%Therefore, the energy efficiency of the car is approximately 16.7%.
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Children and parents spend less time together in middle childhood than in the preschool years.
a. True
Ob. False
E
C
A
Apr 30
Children and parents spend less time together in middle childhood than in the preschool years. Yes this true. Hence option A is correct.
Children are people who are between the ages of birth and puberty, or between the early years of development and puberty. It may also be a reference to a human embryo. The term "child" as used in law often refers to a minor, also known as a person under the age of majority. In general, children are subject to fewer rights and obligations than adults. They are considered incapable of making important judgements.
In their own species, a parent is someone who cares for their children. In humans, a parent is a child's primary carer. A individual whose gamete produced a child—a male through the sperm and a female through the ovum—is said to be the biological parent. First-degree relatives and sharing 50% genetic material are the biological parents.
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A torque of T= 4Nm is required to rotate the intermediate cylinder at W= 30 rad/min .
Calculate the viscosity of the
oil. All cylinders are 450 mm long. Neglect the end effects.
R=0.15m, t=0.003 m.
Result: 1.258 N.s/m
The viscosity of the oil with a torque 4Nm and a rotational speed is 30 rpm is 0.2003 N.s/m.
From the given,
torque = 4Nm
rotational speed = 30 rpm = (30 × 3.14)/60 = 3.14 rad/sec
radius = 0.15 m
thickness (h) = 0.003 m
To find linear velocity,
V = R×ω
= 0.15×3.14
V =0.471 m/s
The dragging force,
F = 2μA(V/h) (V is linear velocity and h is the thickness )
Area = area of cylinder = 2πRH, R is radius and H is the height of the cylinder.
F = 2μ(2πRH) (V/h)
= 2μ(2×3.14×0.15×0.45) (0.471/0.003)
= 133.10 μ (μ is the viscosity of the oil)
F = 133.1μ
Torque (τ) = Force × radius
4 Nm = 133.1μ × 0.15
μ = 4 / (133.1×0.15)
= 0.2003 N.s/m
The viscosity of the oil is 0.2003 N.s/m.
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The hydroelectric dam on the lake has been reintroducing water from its processes before the water has cooled to normal lake temperature, causing a rise in water temperature around the dam. An assessment of the microorganism populations affected by the high water temperature is needed, along with recommendations for the correct ranges of water temperature.
what field of science is this?
This problem is related to environmental science, specifically to the subfield of aquatic ecology.
The process of reintroduction of water from the hydroelectric dam has an influence on the temperature of the water in the nearby lake, which has an impact on the populations of microorganisms. The study of the interactions between the natural world and human society falls under the purview of environmental science, which is relevant to this topic.
Aquatic ecology is a branch of environmental science that focuses on understanding freshwater and marine habitats, as well as the organizms that live there. A thorough analysis of the microorganism populations impacted by the high water temperature is required to solve the issue at hand, as well as suggestions for the suitable ranges of water temperature that will support a healthy ecosystem's functioning.
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How much current is in a circuit with a 1.5 V battery and three
2-ohm resistances (bulbs) in series?
Answer: 0.25A
Explanation: To calculate the current in the circuit, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R):
I = V / R
In this case, the total resistance of the circuit is the sum of the individual resistances of the three bulbs in series, which gives us:
R = 2 + 2 + 2 = 6 ohms
The voltage of the battery is given as 1.5V.
So, using Ohm's Law, we can calculate the current in the circuit as:
I = V / R = 1.5 / 6 = 0.25 amps
Therefore, the current in the circuit is 0.25 amps.
The current in the circuit is 0.25 A.
Voltage across the circuit, v = 1.5 V
Resistance in each resistors, R = 2Ω
Since, the resistors are connected in series combination, their effective resistance,
R' = 3R
R' = 3 x 2
R' = 6Ω
According to Ohm's law, if the temperature and all other physical factors remain constant, the voltage across a conductor is directly proportional to the current that is flowing through it.
So, according to Ohm's law,
V = IR
Therefore, current flowing through the given circuit,
I = V/R
I = 1.5/6
I = 0.25 A
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In 7.5 s, 4200 waves pass a given point. What is the frequency of the waves?
In 7.5 s, 4200 waves will pass a given point. Then, the frequency of the waves is 560 Hz.
Frequency is the number of cycles or oscillations of a wave that occur in one second, measured in Hertz (Hz). It is the rate at which a wave completes a full cycle, which is usually measured as the number of wavelengths that pass a fixed point in a unit of time.
The formula for frequency is;
frequency = waves/time
Where "waves" is the number of waves that pass a given point and "time" is the time it takes for those waves to pass.
In this case, we know that 4200 waves pass a given point in 7.5 seconds, so we can plug those values into the formula;
frequency = 4200/7.5
Simplifying;
frequency = 560
Therefore, the frequency of the waves is 560 Hz.
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The platform has a tension rating of 2000N. The platform itself is 4 meters long and has a mass of 200 kg. The rope on the left side is attached directly to the left edge, while on the right side it is 1 meter inside the edge. If a 70 kg person stands on the far right edge, will the platform be safe?
Answer:
Since the total tension on the ropes (4630.5 N) is greater than the maximum tension that the platform can withstand (2000 N), the platform is not safe. Therefore, the person should not stand on the far right edge of the platform.
Explanation:
To determine whether the platform is safe, we need to calculate the tension on the ropes and compare it to the maximum tension that the platform can withstand (2000N).
First, let's calculate the weight of the platform:
Weight of the platform = mass x gravity
Weight of the platform = 200 kg x 9.8 m/s^2
Weight of the platform = 1960 N
Next, let's calculate the weight of the person standing on the platform:
Weight of the person = mass x gravity
Weight of the person = 70 kg x 9.8 m/s^2
Weight of the person = 686 N
The total weight on the platform is:
Total weight = weight of platform + weight of person
Total weight = 1960 N + 686 N
Total weight = 2646 N
Now, let's calculate the tension on the ropes. Since the rope on the left side is attached directly to the left edge, it will bear the entire weight of the platform and the person. The rope on the right side will bear a portion of the weight, equal to the distance of the person from the edge divided by the length of the platform:
Tension on left rope = Total weight = 2646 N
Tension on right rope = (Distance from person to right edge / Length of platform) x Total weight
Tension on right rope = (3 m / 4 m) x 2646 N
Tension on right rope = 1984.5 N
The total tension on the ropes is:
Total tension = Tension on left rope + Tension on right rope
Total tension = 2646 N + 1984.5 N
Total tension = 4630.5 N
3. Ana is skating down the ice at the Anoka vs Blaine hockey game. She is skating at 10
meters/second while driving towards the net. She accelerates (shoots) the puck to a final speed
of 40 meters/second over a time of .5 seconds (how long it is on the stick). What was the puck's
acceleration?
The acceleration of the puck was 60 m/s².
What was the puck's acceleration?The acceleration of the puck is calculated by applying the formula for acceleration.
a = (v - u) / t
Where;
a is accelerationv is final velocityu is initial velocityt is time.The acceleration of the puck is calculated as follows;
a = (40 m/s - 10 m/s) / 0.5 s
a = 60 m/s²
Thus, the acceleration of the puck is determined from the change in velocity of the puck to change in time of motion of the pcuk.
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A system underwent an adiabatic process during which the internal energy decreased 613 J, the temperature decreased 461 K, the pressure decreased to a quarter of its original value, & the number of moles remained constant. Determine the total heat flow of the system. And complete table
The total heat flow of the system is -2413 J.
How to calculate the heat flowV1 = nRT1/P1
V2 = nRT2/P2 = 4nRT1/P1
Since the number of moles is constant, the work done by the system is:
W = -PΔV = -P(V2 - V1) = -P(3V1) = -(3nRT1)
Substituting the given values, we get:
W = -1800 J
Finally, we can use the first law of thermodynamics to calculate the heat flow:
Q = ΔU + W = -613 J - 1800 J = -2413 J
Therefore, the total heat flow of the system is -2413 J. Note that this is a negative value, indicating that the system lost heat during the process.
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Sweating is one of the main mechanisms with which the body dissipates heat. Sweat evaporates with a latent heat of 2430 kJ/kg at body temperature, and the body can produce as much as 1.4 kg of sweat per hour. If sweating were the only heat dissipation mechanism, what would be the maximum sustainable metabolic rate, in watts, if 80% of the energy used by the body goes into waste heat?
The maximum sustainable metabolic rate of the body is 1140.8 W.
Latent heat of the sweat, L = 2340 x 10³ kJ/kg
Amount of sweat produced, m' = 1.4 kg/hr = 3.9 x 10⁻⁴ kg/s
Given that 80% of the energy used by the body goes into waste heat.
Power is equal to the heat dissipated by the body per unit time.
P = m'L
0.8P = 3.9 x 10⁻⁴x 2340 x 10³
P = 3.9 x 10⁻⁴x 2340 x 10³/0.8
P = 912.6/0.8
P = 1140.8 W
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3. Ana is skating down the ice at the Anoka vs Blaine hockey game. She is skating at 10 meters/second while driving towards the net. She accelerates (shoots) the puck to a final speed of 40 meters/second over a time of .5 seconds (how long it is on the stick). What was the puck's
The acceleration of the puck was 60 m/s².
What was the puck's acceleration?The acceleration of the puck is calculated by applying the formula for acceleration.
a = (v - u) / t
Where;
a is accelerationv is final velocityu is initial velocityt is time.The acceleration of the puck is calculated as follows;
a = (40 m/s - 10 m/s) / 0.5 s
a = 60 m/s²
Thus, the acceleration of the puck is determined from the change in velocity of the puck to change in time of motion of the pcuk.
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The missing question;
What was the puck's acceleration?
You are designing a machine for a space exploration vehicle. It contains an enclosed column of oil that is 1.50 m tall, and you need the pressure difference between the top and the bottom of this column to be 0.125 atm. (a) What must be the density of the oil? (b) If the vehicle is taken to Mars, where the acceleration due to gravity is 0.379g, what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?
Answer:
Explanation:
(a) To find the density of the oil, we can use the formula for pressure difference in a fluid column:
ΔP = ρgh
where ΔP is the pressure difference, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.
Plugging in the given values, we have:
0.125 atm = ρgh = ρ(9.81 m/s^2)(1.50 m)
Solving for ρ, we get:
ρ = 0.125 atm / (9.81 m/s^2 x 1.50 m) ≈ 0.00847 g/cm^3
Therefore, the density of the oil must be approximately 0.00847 g/cm^3.
(b) On Mars, the acceleration due to gravity is 0.379 times that of Earth, or g_Mars = 0.379g_Earth. The pressure difference between the top and bottom of the oil column will be:
ΔP_Mars = ρgh_Mars = ρg_Earth(0.379)(1.50 m)
Using the density we found in part (a), we have:
ΔP_Mars = (0.00847 g/cm^3)(9.81 m/s^2)(0.379)(1.50 m) / (1 atm/101325 Pa)
ΔP_Mars ≈ 0.019 atm
So, the pressure difference between the top and bottom of the oil column on Mars will be approximately 0.019 atm, or about 0.15 times the pressure difference on Earth.
Answer:
The pressure difference (in Earth's atmosphere) between the top and bottom of the oil column on Mars is 0.045 atm.
Explanation:
(a) To find the density of the oil, we can use the formula for pressure difference in a fluid column: ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.
We know that the pressure difference is 0.125 atm, the height of the column is 1.50 m, and the acceleration due to gravity on Earth is 9.81 m/s². Plugging in these values, we get:
0.125 atm = ρ(9.81 m/s²)(1.50 m)
Solving for ρ, we get:
ρ = 0.00803 g/cm³
Therefore, the density of the oil must be 0.00803 g/cm³.
(b) If the vehicle is taken to Mars, where the acceleration due to gravity is 0.379g, we can use the same formula to find the pressure difference:
ΔP = ρgh
We know that the height of the column is still 1.50 m, but the acceleration due to gravity is now 0.379g. Plugging in these values, we get:
ΔP = (0.00803 g/cm³)(9.81 m/s²)(0.379)(150 cm)
Solving for ΔP, we get:
ΔP = 0.045 atm
Therefore, the pressure difference (in Earth's atmosphere) between the top and bottom of the oil column on Mars is 0.045 atm.
Which particles are considered nucleons? O A. Electrons and neutrons O B. Neutrons and protons OC. Protons and electrons D. Only electrons
The particles that are considered nucleons are neutrons and protons. Option B.
What are nucleons?Nucleons are particles found in the nucleus of an atom, specifically protons and neutrons.
Protons have a positive charge, while neutrons have no charge, and together they make up the majority of the mass of an atom.
Electrons, on the other hand, are much smaller than protons and neutrons, and they orbit around the nucleus in shells. Since they are not found in the nucleus, electrons are not considered nucleons.
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A bicycle travels 10.40 km due east in 0.280 h, then 8.80 km at 15.0° east of north in 0.310 h, and finally another 10.40 km due east in 0.280 h to reach its destination. The time lost in turning is negligible. Assume that east is in the +x-direction and north is in the +y-direction.What is the direction of the average velocity for the entire trip? Enter the answer as an angle in degrees north of east.
° north of east
Answer:
19
Explanation:
A 104 kg man is skiing across level ground at a speed of 8.7 m/s when he comes to the small slope 1.2 m higher than ground level shown in the following figure. 1.2 m (a) If the skier coasts up the hill, what is his speed (in m/s) when he reaches the top plateau? Assume friction between the snow and skis is negligible. m/s (b) What is his speed (in m/s) when he reaches the upper level if a 75 N frictional force acts on the skis? m/s †
The speed of skier's when he reaches the top plateau is 5.26 m/s, and the speed of skier's when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.
We can use the conservation of mechanical energy to solve this problem. Initially, the skier has kinetic energy, and at the top of the slope, he will have both potential and kinetic energy.
Since friction is negligible, the only force acting on the skier is the force of gravity. The work done by this force will be equal to the change in the skier's potential energy as he climbs up the slope. Therefore;
mgh = (1/2)mv²
where m = 104 kg is mass of the skier, g = 9.8 m/s² is the acceleration due to gravity, h = 1.2 m is height of the slope, and v is the skier's velocity when he reaches the top.
Solving for v, we get;
v = √(2gh) = √(29.81.2) = 5.26 m/s
Therefore, the skier's speed when he reaches the top plateau is 5.26 m/s.
In this case, there is also a frictional force acting on the skier, which does negative work on the skier as he moves up the slope. The work done by the frictional force is equal to the force of friction multiplied by the distance traveled;
W = Fd = μmgd
where μ = F/N is the coefficient of kinetic friction, N is the normal force acting on the skier (equal to the skier's weight), and d is the distance traveled up the slope (equal to the height of the slope, 1.2 m).
The net work done on the skier will be equal to the change in his mechanical energy;
Wnet = ΔK + ΔU = (1/2)m[tex]V_{f}[/tex]² - (1/2)m[tex]V_{i}[/tex]² + mgh
where vi = 8.7 m/s is the skier's initial velocity, [tex]V_{f}[/tex] is his final velocity at the top of the slope, and ΔK and ΔU are the changes in kinetic and potential energy, respectively.
Since the net work done on the skier is equal to the work done by the gravitational force minus the work done by the frictional force, we have;
Wnet = mgh - μmgd
Substituting the expressions for Wnet and mgh, we get:
(1/2)m[tex]V_{f}[/tex]² - (1/2)m[tex]V_{i}[/tex]² = μmgd
Solving for [tex]V_{f}[/tex], we get:
[tex]V_{f}[/tex] = √([tex]V_{i}[/tex]² + 2μgd) = √(8.7² + 20.729.8×1.2) = 4.23 m/s
Therefore, the skier's speed when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.
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An electric heater uses 0,5 MJ when connected to a 220 V supply for 30 minutes. Calculate the power rating of the heater and the current drawn from the supply.
The power rating of the heater is 277.8 W and the current drawn from the supply is 0.07015mA.
The power is obtained from the ratio of work and time and the power is measured in watts(W). The current is defined as the flow of electrons in the circuit and it is measured in ampere(A).
From the given,
The energy of the electric heater = 0.5MJ = 0.5×10⁶ J
Voltage (V) = 220 V
time (t) = 30 minutes = 30×60 s = 1800 s.
To find power (P):
Power = Energy / Time
= 0.5×10⁶ / 1800
= 277.77 W
Power of the electric heater = 277.8 W
To find current from the given:
Power (P) = V×I×t
Current (I) = P/(V×t)
= 277.8 / (220×1800)
= 0.7015 mA
The current drawn from the supply is 0.7015 mA.
Hence, the power rating of the heater is 277.8 W and the current drawn from the supply is 0.7015 mA.
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the magnitude of kilogram is same all over the world give reason
The SI unit is a unit of measurement that is a globally recognized system that is used for technical and scientific activities.
The kilogram, which corresponds to the mass of the international prototype of the kilogram, is the SI base unit of mass.
The weight of a specific international prototype constructed of platinum-iridium and stored at the International Bureau of Weights and Measures is referred to as one kilogram.
Initially, it was specified as being equal to the mass of one litre (10-3 cubic metres) of pure water. A mass of 1 kg weighs around 2.20 pounds (lb) at the surface of the Earth.
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I NEEDDD HELP LIKE RIGHT NOW ITS SIMPLE
What is the best description of originality?
nonobvious
special or interesting
convergent
having a low probability, unique
Answer: a
Explanation:
Calculate forces and accelerations for a horizontal spring system. A 0.350 kg object attached to a spring of force constant 1.30 x 10 N/m is free to move on the frictionless horizontal surface. If the object is released from rest at x = 0.100 m, find the force on it and its acceleration at x = 0.100 m, x-0.0500m, x = 0. x= -0.0500m and x = -0.100 m.
The force and acceleration alter as the object moves away from its equilibrium position.
How to calculate force and acceleration?The force on the object at any point is given by Hooke's Law:
F = -kx
where F = force, k = force constant of the spring, and x = displacement of the object from its equilibrium position.
The acceleration of the object at any point is given by Newton's Second Law:
a = F/m
where a = acceleration, F = force, and m = mass of the object.
Using these equations, calculate the force and acceleration at each of the specified points:
At x = 0.100 m:
F = -kx = -(1.30 x 10 N/m)(0.100 m) = -0.130 N
a = F/m = (-0.130 N)/(0.350 kg) = -0.371 m/s² (the negative sign indicates that the acceleration is in the opposite direction to the displacement)
At x = 0:
F = -kx = -(1.30 x 10 N/m)(0) = 0 N (the spring is at its equilibrium position)
a = F/m = 0 N/0.350 kg = 0 m/s²
At x = -0.0500 m:
F = -kx = -(1.30 x 10 N/m)(-0.0500 m) = 0.065 N (note that the force is positive because the displacement is negative)
a = F/m = (0.065 N)/(0.350 kg) = 0.186 m/s²
At x = -0.100 m:
F = -kx = -(1.30 x 10 N/m)(-0.100 m) = 0.130 N
a = F/m = (0.130 N)/(0.350 kg) = 0.371 m/s²
So, the force and acceleration change with the displacement of the object from its equilibrium position.
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a bicycle travels 4.00 km due east in 0.650 h, then 12.10 km at 15.0° east of north in 0.500 h, and finally another 4.00 km due east in 0.650 h to reach its destination. the time lost in turning is negligible. assume that east is in the +x-direction and north is the +y-direction. what is the direction of the average velocity for the entire trip?
please i'm begging someone help me
The direction of the average velocity is 10⁰.
What is the direction of the average velocity?The direction of the average velocity of the bicycle is calculated as follows;
The sum of the horizontal component of the velocity is calculated as;
Vx = 4 km x cos(0) / 0.65h + 12 km x cos (15) / 0.5h + 4 km x cos(0) / 0.65 h
Vx = 6.154 km/h + 23.18 km/h + 6.154 km/h
Vx = 35.488 km/h
The sum of the vertical component of the velocity is calculated as;
Vy = 4 km x sin(0) / 0.65h + 12 km x sin (15) / 0.5h + 4 km x sin(0) / 0.65 h
Vy = 0 km/h + 6.21 km/h + 0 km/h
Vy = 6.21 km/h
The direction of the average velocity is calculated as;
θ = tan⁻¹ (Vy/Vx)
θ = tan⁻¹ (6.21/35.488)
θ = 9.9⁰ ≈ 10⁰
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A gas expands from I to F in the figure. The energy added to the gas by heat is 465 J when the gas goes from I to F along the diagonal path. What is the change in internal energy of the gas?
The gas's internal energy that expands from I to F changes by 1015 J.
How to determine internal energy?Since the gas is expanding and energy is added to it by heat, the change in internal energy of the gas can be calculated using the first law of thermodynamics, which states that:
ΔU = Q - W
where ΔU = change in internal energy,
Q = heat added to the system, 465 J and
W = work done by the system.
Assuming that the process is quasi-static use approximation known as the "staircase approximation."
By adding up the work done in each step, find that the total work done by the gas is approximately -550 J. The negative sign indicates that work is done on the gas.
Therefore, using the first law of thermodynamics, we can calculate the change in internal energy of the gas as:
ΔU = Q - W
ΔU = 465 J - (-550 J)
ΔU = 1015 J
Therefore, the change in internal energy of the gas is 1015 J.
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28g of copper pellets are removed from a 300∘C oven and immediately dropped into 90 mL of water at 25∘C in an insulated cup.
The final temperature of the water and copper pellets is 43.8°C.
We can use the equation:
[tex]mc\Delta T = mwCw\Delta T[/tex]
Substituting the values given in the problem:
m c = 28 g
m w = 90 mL = 90 g
c = 0.385 J/g°C
Cw = 4.18 J/g°C
ΔT = T f - 25°C
where T f is the final temperature of the water and copper pellets.
Simplifying the equation:
28 g x 0.385 J/g°C x (T f - 300°C) = 90 g x 4.18 J/g°C x (T f - 25°C)
Solving for T f:
T f = (90 g x 4.18 J/g°C x 25°C + 28 g x 0.385 J/g°C x 300°C) / (90 g x 4.18 J/g°C + 28 g x 0.385 J/g°C)
T f = 43.8°C
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--The complete Question is, What are the final temperature of the water and the copper pellets after thermal equilibrium is reached, assuming no heat is lost to the surroundings? --
