All categories

3 years ago

A 2000 kg car experiences a constant braking force

1 answer:

4 0

In that case, there are three possible scenarios:

-- If the braking force is less than the force delivered by the engine,
then the car will continue to accelerate, and the brakes will eventually
overheat and erupt in flame.

-- If the braking force is exactly equal to the force delivered by the engine,
then the car will continue moving at a constant speed, and the brakes will
eventually overheat and erupt in flame.

-- If the braking force is greater than the force delivered by the engine,
then the car will slow down and eventually stop. If it stops soon enough,
then the absorption of kinetic energy by the brakes will end before the
brakes overheat and erupt in flame. Even if the engine is still delivering
force, the brakes can be kept locked in order to keep the car stopped ...
They do not absorb and dissipate any energy when the car is motionless.

You might be interested in

A 0.155 kg arrow is shot upward

Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0

3 years ago

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

What is the energy of a rock with a mass of 10.2 kg on a cliff that is 300 m height?

Anuta_ua [19.1K]

The potential energy of the rock is 30,000 J

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):

$E=PE+KE$

where

PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field

KE is the kinetic energy, which is the energy possessed by the object due to its motion

In this problem, the rock is at rest, so its kinetic energy is zero:

KE = 0

Therefore, the energy of the rock is just equal to its potential energy, which is:

$E=PE=mgh$

where

m = 10.2 kg is the mass of the rock

$g=9.8 m/s^2$ is the acceleration of gravity

h = 300 m is the height of the rock above the ground

Substituting and solving, we find

$PE=(10.2)(9.8)(300)=30,000 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago

What is one way that early scientific practice differed from modern scientific practice?

mote1985 [20]

Early hypotheses were not based on observations.
Early hypotheses were not tested by experimentation.
Early hypotheses were formed from scientific questions.
Early hypotheses were influenced by creative thinking

6 0

3 years ago

Read 2 more answers

What is addictive about vaping?

AnnZ [28]

Answer:
The last one, nicotine from tobacco

4 0

3 years ago