A 2500 kg mustang car accelerates from rest to 30 m/s in 6 sec.What is acceleration of the mustang car?

One gallon of paint (volume = 3.79 X 10-???? m3) covers an area of 25.0 m2. What i the thickne s of the fresh paint on the wall?

Drupady [299]

Answer:

Explanation:

Given

Volume of paint is $V=3.79\times 10^{-3}\ m^3$

Area of cover $A=25\ m^2$

Suppose paint to be a rectangular box with thickness t and volume V

therefore we can write as

$V=A\times t$

$t=\frac{V}{A}$

$t=\frac{3.79\times 10^{-3}}{25}$

$t=1.516\times 10^{-3}\ m$

$t=1.516\ mm$

6 0

3 years ago

The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r

Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity. Therefore:

marble > manhole cover > basketball > wedding ring

7 0

3 years ago

A closed system:

adoni [48]

The answer is b i just did the test

3 0

2 years ago

f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t

inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

$h(t)=35t-0.83t^2$

We know that the rate of change of displacement is equal to the velocity of an object.

$v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t$

Velocity of the arrow after 3 seconds will be :

$v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s$

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0

3 years ago

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

lisabon 2012 [21]

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

6 0

2 years ago