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mestny [16]
2 years ago
11

The three notches in the graph occur where the driver changed gears. Describe the changes in velocity and acceleration of the ca

r while in first gear. Is the acceleration just before a gear change larger or smaller than the acceleration just after the change?
Explain your answer
Physics
1 answer:
kaheart [24]2 years ago
6 0
<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>

<h3>B ) The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>C ) the direction of motion can be changed without changing the direction of the applied force .</h3>

D ) the system would continue to move without any further, after and initial force has set in motion.

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If you drop an object from rest, the distance it falls is given by (1/2)at2, where a is the acceleration of the object and t is
klio [65]

If,

d \:  =  \:   \frac{1}{2} a {t}^{2}

then, with 3x time t, (suppose, new distance is h)

h \:  =  \:   \frac{1}{2} a {(3t)}^{2}

=  \frac{1}{2} a9 {t}^{2}

= 9 \:  \frac{1}{2} a{t}^{2}

= 9d

Therefore, new distance h will be 9 times bigger than distance d.

answer: c

7 0
3 years ago
A 2.1 kg block is dropped from rest from a height of 5.5 m above the top of the spring. When the block is momentarily at rest, t
ella [17]

Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE =  ¹/₂mv²

70.59 =  ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

8 0
3 years ago
Read 2 more answers
A small 24 kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy?
USPshnik [31]
J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј
3 0
3 years ago
Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.
lbvjy [14]

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

8 0
3 years ago
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500
kenny6666 [7]

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e \frac{1}{2} kx^2  = \frac{1}{2} mv^2

kx^2 = mv^2

4500* 0.04^2 = 200*10^{-3} *v^2

7.2 =200*10^{-3}*v^{2}

v^{2}   =\frac{7.2}{200*10^{-3}}

v   =\sqrt{\frac{7.2}{200*10^{-3}}}

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

5 0
3 years ago
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