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saveliy_v [14]
3 years ago
11

The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this

impressive position is traveling with an initial velocity of 8.0 m/s before accelerating. Then the rider travels 40.0 m at a constant acceleration of 2.00 m/s2. What is the rider's velocity after the acceleration?
Physics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

Hence the rider's velocity after the acceleration is 14.97m/s

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Answer:

where is the example...?

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2 years ago
A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 1.9 m/s^2. The acceleration of gravity is 9.8
Keith_Richards [23]

Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

F = m\cdot(g-a)

We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

900 N = m\cdot g \implies m = \frac{900N}{9.8 \frac{m}{s^2}}

So

F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN

3 0
3 years ago
PLS ANSWER FAST WILL GIVE BRAINLEST!!!
Kipish [7]

Answer:

2 Newtons

Explanation:

F = ma

Therefore, your mass would be 1kg and your acceleration would be 2m/s/s

Plug the numbers into the equation:

(1kg)(2m/s/s)

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2 Newtons

8 0
3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t
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7 0
3 years ago
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