A ball with topspin rotates in the opposite direction as a ball with underspin. What would be the direction of the Magnus effect
on a ball with topspin? How do you think this would affect the ball’s motion?

Answer: The mass on Moon will be 98. 1 kg.

Explanation:

As per the known fact, the mass is independent of acceleration due to gravity of planet.

Answer:A ball is attached by a tether to a stake in the ground. The ball travels in a circle at the full length of the tether. The pull of the tether acts in a direction perpendicular to the ball's velocity, and changes its direction.

Explanation: i big brained thats y

The bottom of the hill, the cart's kinetic energy would be 147150 joules.

In physics, an object's kinetic energy is the energy it has as a result of motion. It is explained as the amount of effort required to move a mass-based body from rest to the indicated velocity. The body keeps its kinetic energy, which it acquired during acceleration, unless its speed changes.

Let KE denote Kinetic Energy

As an equation, the relationship would be:

[tex]$m g h=\frac{1}{2} m v^2$$[/tex]

Kinetic Energy = m g h

Assuming the acceleration due to gravity near the earth's surface exists [tex]$9.81 \frac{\mathrm{m}}{\mathrm{s}}$[/tex], our equation becomes:

[tex]$$\begin{aligned}& K E=500 \mathrm{~kg} \cdot 9.81 \frac{\mathrm{m}}{\mathrm{s}} \cdot 30 \mathrm{~m} \\& K E=147150 \text { Joules }\end{aligned}$$[/tex]

Once a result, as the cart nears the base of the hill, where its gravitational potential energy is transformed into kinetic energy, its kinetic energy would be equal to its initial potential energy.

At the bottom of the hill, the cart's kinetic energy would be 147150 joules.

The complete question is:

A roller coaster car with a mass of 500 kg at the top of a hill that is 30 m high. Without friction, what would its kinetic energy be as it reached the bottom of the hill?

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Answer:

Approximately [tex]3.59\times 10^{22}\; {\rm N}[/tex] (assuming that the orbit is circular.)

Explanation:

When an object is in a circular orbit of radius [tex]r[/tex] with a (linear) speed of [tex]v[/tex], the centripetal acceleration of this object would be [tex]a = (v^{2} / r)[/tex]. If the mass of the object is [tex]m[/tex], magnitude of the net force on this object would be [tex]F = m\, a = (m\, v^{2} / r)[/tex].

Look up the mass of the Earth: [tex]m \approx 5.9722\times 10^{24}\; {\rm kg}[/tex].

Look up the radius of the orbit (mean distance between the Earth and the Sun): [tex]r \approx 1.4960\times 10^{11}\; {\rm m}[/tex] (one Astronomical Unit.)

It is given that the linear orbital speed is [tex]v = 3.00 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex].

Therefore, magnitude of the centripetal force (net force) on the Earth would be:

[tex]\begin{aligned}F &= m\, a = \frac{m\, v^{2}}{r} \\ &\approx \frac{(5.9722 \times 10^{24}\; {\rm kg})\, (3.00 \times 10^{4}\; {\rm m\cdot s^{-1}})^{2}}{1.4960 \times 10^{11}\; {\rm m}} \\ &\approx 3.59 \times 10^{22}\; {\rm N} \end{aligned}[/tex].

The scale used by Michael is precise but not accurate when it measures the mass.

While measuring the quantities, we use different devices. These devices has to be precise as well as accurate.

Being precise means that it has to give the same reading, which is the case for us because we are having the scale that is measuring the value of the scale consistently that is actually 2 kg less than the actual mass of the object.

Being accurate means that the reading should be either very very close to the actual reading or equal to the actual reading, which is not the case to us because the scale is not showing the correct value of the mass.

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The change in the car's gravitational potential energy from A to B is calculated as 4.90 J.

Gravitational potential energy may be characterized as a type of energy that is typically possessed or acquired by an object due to an alteration in its position when it is present in a gravitational field. In more simple terms, it can be revealed that gravitational potential energy is an energy that is associated with gravitational force or gravity.

According to the question,

The formula for gravitational potential energy is as follows:

where m is the mass (in kg) of the body. G is the magnitude of the gravitational field that the body is exposed to. H is the height at which the body is located.

First, you have to calculate the gravitational potential energy of the car on point A, i.e. U₁= (2.0 kg)(9.8 m/s²)(0 m)= 0 J.

Second, the Gravitational potential energy of the car on point B is found as, U₂= (2.0 kg)(9.8 m/s²)(0.25 m)= 4.90 J.

Now, you must have to calculate the change in the car’s gravitational potential energy: ΔU= U₂-U₁= 4.90 J - 0 J= 4.90 J.

Therefore, the change in the car's gravitational potential energy from A to B is calculated as 4.90 J.

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After 2.00 seconds, the astronaut is moving at the speed of 0.67 m/s. The result is obtained by using the concept of Newton's second law.

The Newton's second law states that "The acceleration is directly proportional to the net force acting on an object and inversely proportional to the object's mass." It can be expressed as

a = ∑F/m

or

∑F = ma

Where

An astronaut moves after receiving a force. We have

Find the speed of the astronaut moving!

The force makes the astronaut moves from rest to a certain speed.

The speed can be calculated by the Newton's second law.

∑F = ma

∑F = m (v₁ - v₀)/t

30.0 = 90.0 (v₁ - 0)/2.00

60.0 = 90.0v₁

60.0/90.0 = v₁

v₁ = 2/3

v₁ = 0.67 m/s

Hence, the astronaut is moving at the speed of 0.67 m/s.

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As the radius of curvature and focal length of the mirror is bottom half of the mirror is same as earlier, so the image formation by this mirror will not change. As a result, the bottom half of the spherical mirror will reflect an image that is an exact replica of the image it projected prior to being chopped.

Only the aperture area will be cut in half when a mirror is cut in half horizontally; the radius of curvature remains unchanged.

The spherical mirror has a radius of curvature that is twice its focal length. Because the radius of curvature remains unchanged when the mirror is cut horizontally, the focal length of the bottom portion will be the same as in the initial situation.

The location of the image created by the lower portion of the mirror is therefore the same as in the preceding situation according to the mirror equation.

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Assuming that range aliasing can be tolerated and there are no other targets between the moon and Earth, a pulse repetition frequency of around 10-20 MHz would likely be sufficient for mapping the moon surface.

The maximum pulse repetition frequency (PRF) that can be used for mapping the surface of the moon is determined by the need to avoid range ambiguities and to achieve the desired resolution. The PRF is the number of pulses emitted per second. A high PRF allows for high resolution mapping but also increases the chance of range ambiguities.

If range ambiguities can be tolerated, a high PRF can be used to achieve a high resolution map. The altitude of the spacecraft, the size of the antenna, and the desired resolution of the map are also factors that determine the maximum PRF. A general estimate is that a PRF of around 10-20 MHz would be sufficient for mapping the surface of the moon with a high resolution.

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If we neglect the effect of air friction, then the motion of the ball a short time later will be a combination of horizontal and vertical motion.

If you stand on tall cliff and throw ball so that it is initially moving purely horizontally and neglecting the effect of air friction, motion of the ball a short time later will be combination of horizontal and vertical motion. This is because the only force acting on ball is gravity, which will pull the ball downwards. Therefore, ball will experience a vertical acceleration due to gravity, while continuing to move horizontally at constant velocity.

Motion of the ball is a parabolic path, which is the trajectory of projectile under the influence of gravity. Horizontal component of the motion will remain constant, while vertical component will be affected by gravity and will decrease until ball reaches maximum height and then it will increase again until the ball hit the ground.

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The dog run 28 km back and forth between its owner and the road, at the speed of 10.0 km/hr.

Speed of the man throughout the whole distance, v₁ = 5 km/hr

After 1 km, the speed of the man is same but the speed of dog, v₂ = 10 km/h

The distance between the road and the dog's owner after he unleash the dog, d = (15 - 1) = 14 km

Time taken by the owner to reach the road, t = 14/5 = 2.8 hrs

Distance travelled by the dog in 2.8 hrs, = 2.8 × 10 = 28 km

So the dog run a total of 28 km distance.

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The SI unit of K so that the equation Velocity = K x density is correct is given as

K=m^4/s-kgK= m4/s−kg

Velocity is described as the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

The SI unit of velocity is m/s and the Si unit of density is kg/m^3kg/m3 .

Adding the SI units of both quantities, we have that  :

m/s=K\times kg/m^3m/s=K×kg/m3

In terms of dimensions,

[LT^{-1}]=K\times [ML^{-3}][LT−1]=K×[ML−3]

Dimension of K is :

[K]=[M^{-1}L^4T^{-1}][K]=[M−1L4T−1]

SI unit of K is :

K=m^4/s-kgK=m4/s−kg

Hence,  the required solution is  K=m^4/s-kgK=m4/s−kg

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A hamiltonian function representing an electron in the presence of a stationary positive point charge is

H = (p^2)/(2m) - (e^2)/(4πε_0r)

Generally, The Hamiltonian function representing an electron in the presence of a stationary positive point charge is given by:

H = kinetic energy + potential energy = (p^2)/(2m) - (e^2)/(4πε_0r)

where

Note that the first term represents the kinetic energy of the electron, and the second term represents the potential energy of the electron due to the attractive electrostatic force from the positive point charge.

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although the electric field at the origin owing to the quarter-circle arc is zero, the electric field at the origin due to a point charge q is non-zero.

An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. It can also refer to a system of charged particles' physical field. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically. Electric charge or magnetic fields with variable amplitudes can produce an electric field. The area of space surrounding an electrically charged particle or object in which the charge body perceives force is known as the electric field.

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The electric field strength between the conducting plates would be 3.6 x 10⁵ V/m.

What is electric field strength?

Electric field strength (E) is a measure of the force exerted on a charged particle by an electric field. It is measured in units of volts per meter (V/m) or newtons per coulomb (N/C). An electric field can be created by a static charge or by a changing magnetic field. The strength of the electric field is affected by the distance from the source of the field, the charge of the source, and the permittivity of the medium. Electric fields are used in many applications, including in electric motors, generators, and particle accelerators. It is also used in the study of charges and the behavior of conductors and insulators.

The electric field strength (E) can be determined by using the equation: E = W / qd

Where,

W = energy of the ion (36.0 keV)

q = charge of the ion (doubly charged ion, so 2e)

d = distance between the plates (1.60 cm)

To convert keV to J and cm to m:

1 keV = 1.602176634 x 10⁻¹⁶ J

1cm = 0.01m

Therefore,

E = (36.0 x 1.602176634 x 10⁻¹⁶ J) / (2e x 0.01m)

E = (57.634 x 10⁻¹⁶J/C) / (2 x 1.602176634 x 10⁻¹⁹ C x 0.01m)

E = 3.6 x 10⁵ V/m

So, the electric field strength between the plates is 3.6 x 10⁵ V/m.

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The velocity with which the ball left the ground when it is dropped from a tall building is calculated to be 25.82 m/s.

Height to which a ball bounces back s = 34 m

From the equations of kinematics, we know, the relation connecting height, velocities and acceleration as,

v² - u² = 2 a s

where, v is the final velocity

u is the initial velocity

a is the acceleration

s is the height travelled

Here, u = 0 and a is the acceleration due to gravity

v² = 2 × 9.8 × 34

v² = 666.4

v = 25.82 m/s

Thus, the velocity with which the ball left the ground is calculated to be 25.82 m/s.

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To design a race car with a balloon engine for the best possible performance. must take into consideration Newton's laws of motion. the formula Velocity (speed) = Distance / Time (v = d / t).

The Complete Question.

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The final position of the rocket after 1 hour is (2.64 × 10⁴, 3.5 × 10⁴, 0).

Initial velocity, = (0, 2× 10⁴, 0)

Initial location, = (1.2× 10⁴, 1.5 × 10⁴, 0)

Net force acting, F₁ = (6× 10⁴, 0, 0) N

Mass of the rocket, m₁ = 1.5 × 10⁴

Initial momentum, p₁ = mv₁ = (1.5 × 10⁴) × (0, 2× 10⁴, 0)

p₁ = (0, 3 × 10⁸, 0)

Final momentum after 1 hour(3600 sec), P₂ = p₁ + F₁×Δt

P₂ = (0, 3 × 10⁸, 0) + (6× 10⁴, 0, 0)×3600

P₂ = (2.16 × 10⁸, 3 × 10⁸, 0)

Final position of this time,

r₂ = r₁ + (P₂)/m₁

r₂ = (1.2 × 10⁴, 1.5 × 10⁴, 0) + (2.16 × 10⁸, 3 × 10⁸, 0)/(1.5 × 10⁴)

r₂ = (2.64 × 10⁴, 3.5 × 10⁴, 0)

--The given question is incomplete, the complete question is:

"Suppose that you are navigating a spacecraft far from other objects. The mass of the spacecraft is 1.5 × 10⁴ kg. The rocket engines are shut off, and you're coasting along with a constant velocity of (0, 2× 10⁴, 0) m/s. As you pass the location (1.2× 10⁴, 1.5 × 10⁴, 0) you briefly fire side-thruster rockets, so that your spacecraft experiences a net force of (6× 10⁴, 0, 0) N for 1 hour. After turning off the thrusters, you then continue coasting with the rocket engines turned off. The ejected gases have a mass that is small compared to the mass of the spacecraft. you then continue coasting with the rocket engines turned off. what will be the components of your position vector an hour later? start by finding the x component, rf,x?"

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The magnitude of the net force on the student is zero.

The force generated by the student is 500N.

The magnitude of the normal force is 500 N

The net force on the student is determined as follows:

The formula of the force of gravity is given by :

W = mass × acceleration due to gravity

The mass of the student is 50 kg

The magnitude of the net force in the student is:

Force = 50 * 10

W = 500 N

The force generated by the student is equal to the force of gravity acting o the student since the student is moving at a constant speed.

The force generated by the student = 500N

The net force = weight - force generated by the student

Net force = 500 N - 500 N

Net force = 0 N

The magnitude of the normal force is equal to the weight of the student.

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Roughly 1,700 km/h. The Earth rotates faster than sound near the equator.

How quickly rotates the Earth close to the equator?

The Earth's circumference is around 40,075 kilometres at the equator; multiplying this by the length of the day yields a spin speed of approximately 1670 kilometres per hour.

How many times does the equator revolve in a day?

Each 24-hour day, the Earth rotates once around its axis. The speed of Earth's spin at its equator is around 1,000 miles per hour (1,600 km per hour). Every day of your existence, this day-night spin has taken you outside and under the sun and stars.

Can people feel the Earth's spin?

The speed of Earth is rapid. It orbits the Sun at a speed of roughly 67,000 miles (107,000 kilometres per hour), spinning (rotating) at about 1,000 miles (1600 kilometres) per hour. Due to the consistent speeds, we are not aware of any motion.

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Around 1,700 km/h.Correct!. Its circumference is approximately 40,075 kilometers at the equator; multiplying this by the length of the day yields a value of 1670 kilometers per hour for the equator-bound speed of rotation.

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The first time pirates used airplanes in peacetime was to transport letters. Considered to be the founding flight of what would become American Airlines, this flight was the first regularly scheduled flight.

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Answer:

1) [tex]t=4.16s[/tex]

2) [tex]s=44.10m[/tex]

3) [tex]v_{final}=40.84\frac{m}{s}[/tex]

Explanation:

a) How long till it hits the ground?

Since the person threw the ball perfectly horizontal, the ball has 0 vertical velocity at the start. In fact, it takes the same time for the ball to reach the ground if they dropped the ball instead of throwing. The acceleration in a vertical drop is gravity ([tex]a=9.81\frac{m}{s^2}[/tex])

Basically, we'll solve first part as if he just dropped the ball.

Use this linear motion formula:

[tex]s=v_{inital}(t_{final} -t_{initial})+1/2a(t_{final}-t_{inital})^2[/tex]

Since [tex]v_{initial} =0[/tex] we can remove that part from equation:

[tex]s=1/2a(t)^2[/tex]

[tex]85=1/2*9.81(t)^2[/tex] Plug and solve

[tex]170=9.81(t)^2[/tex]

[tex]17.33=t^2[/tex]

[tex]t=4.16s[/tex]

b) How far away does it land?

We will use the same formula above for horizontal motion.

[tex]s=v_{inital}(t_{final} -t_{initial})+1/2a(t_{final}-t_{inital})^2[/tex]

[tex]s=v(t)+0[/tex]  (Remove the last part since [tex]a=0[/tex])

[tex]s=10.6(4.16)[/tex]

[tex]s=44.096m[/tex]

c) How fast was it traveling vertically at the end?

Use this other linear motion formula for this step. s=85m (height of building).

[tex]v_{final} ^2-v_{initial} ^2=2as[/tex]

[tex]v_{final} ^2-0^2=2(9.81)*(85)[/tex] Started with 0 vertical velocity.

[tex]v_{final} ^2=1667.7[/tex]

[tex]v_{final}=40.837m/s[/tex]

According to the question of speed, a) 16.05 seconds does the ball hit the ground, b) 169.33 m far from the building will it land, c) 37.68 m/s is the velocity of the ball just before it hits the ground.

What is speed?
Speed is the rate at which something moves or an object's rate of motion. It can be measured in metres per second, kilometres per hour, and miles per hour, and is often used to describe the motion of objects, such as vehicles, animals, or natural phenomena.

a. The time it takes for the ball to hit the ground is given by the equation t = (2×h)/v, where h is the height of the building (85 m) and v is the initial velocity of the ball (10.6 m/s). Plugging in the given values, we get t = (2×85)/10.6 = 16.05 seconds.

b. The distance from the building at which the ball will land is found using the equation x = vt, where v is the initial velocity (10.6 m/s) and t is the time it takes for the ball to hit the ground (16.05 seconds). Plugging in the given values, we get x = 10.6×16.05 = 169.33 m.

c. The velocity of the ball just before it hits the ground is determined by the equation v = √(2gh), where g is the acceleration due to gravity (9.8 m/s2) and h is the height of the building (85 m). Plugging in the given values, we get v = √(2×9.8×85) = 37.68 m/s. This is the velocity of the ball just before it impacts the ground.

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A ball is thrown upwards at 20.0 m/s from the top of a roof that is 10.0 meters above the ground and t = 4.84 seconds

The equation used to solve this problem is

s = vt - (1/2)gt^2

where s = the distance travelled (in this case, 10.0 meters), v = the initial velocity (in this case, 20.0 m/s), g = the acceleration due to gravity (in this case, 9.81 m/s^2), and t = the time (in this case, the time it takes for the ball to hit the ground).

Rearrange the equation to solve for t:

t = (2s/g) + (v^2/g^2)

Substitute the given values into the equation:

t = (2(10.0)/(9.81)) + (20.0^2/(9.81^2))

Simplify the equation:

t = 0.811 + 4.03

t = 4.84 seconds

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The total displacement, in meters, of the ball going through its entire motion: traveling from the ground to the top and then falling back to the ground is 0.

 An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a lecturer moves to the right in relation to a whiteboard. Displacement describes this shift in location.

The total displacement of an object is the distance and direction between its initial and final positions. If a ball is thrown upward, travels to the top and then falls back to the ground, its total displacement is zero.

This is because the ball travels upward and then downward, covering the same distance in opposite directions. The upward motion and the downward motion cancel each other out, resulting in a net displacement of zero.

So, the ball will be back to the ground where it was originally thrown.

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Answer:

the answer it's flas

Explanation:

I wish I help you

The minimum size of the snowshoes needed to enable her to walk without sinking will be 1.37 m².

Assumption:

(i) Her weight is uniformly distributed on the imprint area

(ii) the weight of the shoes is negligible

Now, given that mass of women, m = 70 kg

For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be:

A = W/P

here, W = mg = 70 kg . 9.81 m/s² = 70x9.81 N

P = 0.5 kPa = 0.5 x 1000 N/m²

so, A = [tex]\frac{70x9.81 N}{0.5(1000)N/m^2}[/tex]

or A = 1.37 m²

Therefore, the minimum size of the snowshoes needed to enable her to walk without sinking will be 1.37 m².

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The x-component of A vector is - 2.39 unit.

The y-component of A vector is  - 4.02  unit.

Vector is a colloquial term  that refers to some quantities that cannot be expressed by a single number (a scalar) or to elements of some vector spaces.

The magnitude of A vector is 4.68 unit.

The  direction of A vector is 30.7° to the left of the negative y-axis.

So, A vector is in 3rd co-ordinate.

Hence, the x-component of A vector is = - Asin30.7°

= - 4.68 × sin30.7°

= - 2.39 unit.

Hence, the y-component of A vector is = - Acos30.7°

= - 4.68 × cos30.7°

= - 4.02  unit.

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When the red barge collides with the blue barge, which has a mass of 1000000 kg, its final velocity with in collision elastic becomes 0.311 m/s.

The given data:

Using the law of conservation of momentum.

Following the formula, the red barge's final velocity (v3) is determined.

m1×v1+ m2×v2= (m1+m2)v3

Put the values;

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500 = 1150000×v3

v3 = 0.311 m/s

Thus,

After one head-on elastic hit where the projectile is substantially more substantial than the target, the target particle's velocity will be nearly twice as fast as the projectile, but really the projectile's velocity will essentially be unaffected.

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It will take the model rocket 0.2 seconds to accelerate from rest to a velocity of 208 ft/sec.

In order to answer the question of how long it will take a model rocket to accelerate from rest to a velocity of 208 ft/sec, we need to understand the acceleration of the rocket.

According to the given parameters, the acceleration of the model rocket is proportional to the difference between 260 ft/sec and the rocket's velocity.

Therefore, when the rocket is initially at rest and its initial acceleration is 260 ft/sec, the time it takes to accelerate to 208 ft/s can be calculated as follows:

Acceleration = (260 ft/sec - 208 ft/sec) / time

Rearranging the equation, we get:

Time = (260 ft/sec - 208 ft/sec) / acceleration

Plugging in the values given, we get:

Time = (260 ft/sec - 208 ft/sec) / 260 ft/sec

Time = 0.2 seconds

Therefore, it will take the model rocket 0.2 seconds to accelerate from rest to a velocity of 208 ft/sec.

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878 Watts of power is needed to move a person (mass 72 kg) up an escalator (angle 34 degrees with the horizontal) with a speed of 1.5 m/s (neglect friction).

Moving sidewalks move people horizontally or at a little slope, elevators or lifts take people and cargo up and down, escalators are moving staircases from one story of a building to another. A really compact commercial elevator called a du.mb. waiter elevator is designed to convey things rather than humans.

The given data:

m = 72 kg

w = fd

w mgd

w = 72 kg × 9.8 × 1.5 m

w = 878w

The pressure preventing movement against one another of hard substrates, liquids layers, and material components is known as friction. There are many sorts of friction.

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