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4 years ago

15

How much work must be done to stop a 1075-kg car traveling at 115 km/h ? Express your answer to two significant figures and incl

ude the appropriate units

1 answer:

jok3333 [9.3K]4 years ago

6 0

Answer:

Work done on the car will be 548440.9463 J

Explanation:

We have given mass of the car m = 1075 kg

As the car finally stops so final velocity of the car $v_f=0m/sec$

Initial velocity of car u = 115 km/hr $=115\times \frac{5}{18}=31.943m/sec$

Work done is equal to change in kinetic energy

So work done $=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$

= $=\frac{1}{2}m(v_f^2-v_i^2)=\frac{1}{2}\times 1075\times (0^2-31.943^2)=-548440.946J$

As the work done is negative so work is done on the car to stop it .

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3 years ago

Scientists use what scale to compare the hardness of minerals?

frozen [14]

Mohs hardness scale. hope this helps

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3 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

3 years ago

Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going

4vir4ik [10]

Answer:

B. The force would be the same in both cases.

Explanation:

According to Newton's 3rd law where impulse was equated to the momentum formula.

F × t = M × V

where,

Force = F

V = velocity

Since, Impulse is force multiplied by time, whereas the time of contact is the same for both, therefore the impulse is the same in magnitude for the two trucks.

Case 1: Hitting the other car

Case 2: Hitting the brick wall

In Case 1, both the cars are identical and have same velocity whereas in the Case 2, the wall is stationary.

The case of hitting the brick wall have the same impact force as hitting the other car, because they have the same change in momentum.

Therefore, The force would be the same in both cases.

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4 years ago

A 30 kg box sits on the floor it requires 275N of force to get it moving once it is moving it only takes 225N of force.what are

gavmur [86]

1) The coefficient of static friction is 0.935

2) The coefficient of kinetic friction is 0.765

Explanation:

1)

When a force is applied on a box sitting on the floor, the force that must be applied in order to make the box moving is equal to the maximum force of static friction between the floor and the box, which is:

$F = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 275 N

Therefore, the coefficient of static friction is

$\mu_s = \frac{F}{mg}=\frac{275}{(30)(9.8)}=0.935$

2)

Once the box is in motion, the force that must be applied in order to make the box moving at constant velocity is equal to the force of kinetic friction between the floor and the box, which is:

$F = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 225 N

Therefore, the coefficient of kinetic friction is

$\mu_k = \frac{F}{mg}=\frac{225}{(30)(9.8)}=0.765$

Learn more about friction:

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8 0

3 years ago