All categories

3 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Physics

1 answer:

Nostrana [21]3 years ago

3 0

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
\ J$

$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

You might be interested in

Write forty million in scientific notation.

Ivanshal [37]

Answer: 4 x 106

That’s how you write forty million in scientific notation.

7 0

3 years ago

Read 2 more answers

The relationship that exists between gravity and distance and mass respectively.

ruslelena [56]

Answer:

D...............................

7 0

3 years ago

1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of

VLD [36.1K]

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

6 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

4 years ago

Identify all graphs that represent motion at

Vesna [10]

Answer:

Good Luck!

Follow Me

Explanation:

3) For a displacement time graph, straight line denotes constant speed. For a velocity time graph, the graph parallel to time axis denotes constant speed. Hence, the correct option is a) and d).

3 0

3 years ago