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3 years ago

A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.

Physics

1 answer:

Bond [772]3 years ago

4 0

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

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What object requires less force to overcome inertia B.80 kg piece of furniture or A.1000 kg vehicle

Tju [1.3M]

Less force will be necessary to overcome inertia for the 80 kg piece of furniture.

Force is a factor that has the power to alter an object's motion. A massed object's velocity can be changed or accelerated by a force. A push or a pull is a straightforward method to explain forces.

The term "moment of inertia" refers to the quantity that describes how a body resists angular acceleration and is calculated by multiplying each particle's mass by its square of the distance from the rotational axis.

I = mr², where m is the mass of the object and r is the distance to the rotation axis.

Therefore, The inertia is directly proportional to the mass of the object.

So, as the mass increases the inertia increases.

Therefore, 80 kg piece of furniture will require less force to overcome inertia.

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2 years ago

A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a

marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$ .

where y(t) represent the height from the ground. For our problem, the initial height will be:

$y_0 \ = \ 1000 m$ .

The initial velocity:

$v_0 = - 20 \frac{m}{s}$ ,

take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:

$a=-10\frac{m}{s}$

So, the equation for our problem its:

$y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2$ .

Taking t=6 s:

$y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - 180 m$ .

$y(6 \ s) = \ 1000 m \ - 300 m$ .

$y(6 \ s) = \ 700 m$ .

So this its the height of the ball 6 seconds after being thrown.

6 0

3 years ago

What is the equation for the potential energy stored in a spring when it is stretched or compressed?

gregori [183]

Answer:

if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).

8 0

3 years ago

Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the

AlexFokin [52]

The wavelength of the wave is 0.055 m

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency

$\lambda$ is the wavelength

For the sound wave in this problem we have

v = 340 m/s is the speed

f = 6,191 Hz is the frequency

Solving for $\lambda$ , we find the wavelength:

$\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m$

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4 years ago

Salmon often jump waterfalls to reach their

PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

height of the waterfall, h = 0.432 m
distance of the Salmon from the waterfall, s = 3.17 m
angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$

The minimum velocity of the Salmon jumping at the given angle is calculated as;

$X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s$

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

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2 years ago