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3 years ago

A wagon is accelerating down a hill. Which of the following statements are true?

Physics

2 answers:

Andrews [41]3 years ago

4 0

Answer:

As the wagon approaches the bottom of the hill, the potential energy decreases and the kinetic energy increases.

Explanation:

There are two types of energy possessed in the wagon i.e. potential energy and the kinetic energy. As a wagon is accelerating down a hill, its potential energy decreases and the kinetic energy increases.

We know that the energy possessed due to its height above ground level is called its potential energy while the energy possessed due to its motion is called its kinetic energy.

So, the correct option is (A) " As the wagon approaches the bottom of the hill, the potential energy decreases and the kinetic energy increases".

Ket [755]3 years ago

3 0

<span>A. As the wagon approaches the bottom of the hill, the potential energy decreases and the kinetic energy increases. Potential energy and kinetic energy are opposites. At the top of the hill the wagon has max potential energy and no kinetic energy. At the bottom of the hill the wagon has max kinetic energy and no potential energy. So as the wagon approaches the bottom of the hill potential energy decreases and kinetic energy increases.</span>

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When you vomit your stomach forces a fluid flow from your mouth. Treating your stomach, esophagus, and mouth as a continuous ver

dalvyx [7]

Answer:

The gauge pressure as calculated is 5862.36 Pa

Solution:

As per the question:

Radius, r = 1 cm = 0.01 m

Length, l = 60 cm = 0.6 m

Velocity of discharge of fluid from the mouth, v = 1.5 m/s

Now,

By using the continuity equation:

Av = A'v' (1)

where

A = Area of the mouth

A' = Area of the tube

v' = Velocity of the fluid inside the stomach

Since, the question assumes the stomach, mouth and esophagus as continuous vertical tube, then:

A = A'

Thus

From eqn (1):

v = v' = 1.5 m/s

Now,

With the help of Bernoulli's equation:

$P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'$

$P + \frac{1}{2}\rho v^{2} + \rho g(h - h') = P' + \frac{1}{2}\rho v^{2} +$

$\rho g(h - h') = P' - P$

where

h - h' = l = 60 am = 0.6 m

P = Pressure at mouth = 1 atm = $1\times 10^{5}\ Pa$

g = acceleration due to gravity

Density of water, $\rho = 997\ kg/m^{3}$

P' = Pressure inside the stomach

P' - P = Gauge Pressure

$P' - P = 997\times 9.8\times 0.6 = 5862.36\ Pa$

5 0

4 years ago

What is the equation for calculating the electrical force, Fe, between two charges?

juin [17]

Fe=k|q1q2| / r^2

(k is the electrostatic constant which has a value of 8.99x10^9 Nm^2/c^2, r is the distance between the charges and q is the quantity of the charges.)

4 0

4 years ago

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A bike travels at a constant speed of 3.1 m/s for 6 s. how far does it go?

IgorC [24]

Ah for this problem you are thinking quite a bit hard on. The problem is actually simpler than it looks. The problem states that a bike travels at a constant speed of 3.1 m/s for 6 s and asks how far will it go?. To figure this out you simply need to take 3.1 times 6 s because every second the bike travels 3.1 m. So the answer to this problem would be 18.6 m

8 0

4 years ago

If the radius of an object in circular motion is doubled, what change will occur in the centripetal force?

cestrela7 [59]

Answer:

The centripetal force will be 1/2 as big as it was. (option c)

Explanation:

Recall that centripetal force ( $F_c$ ) is defined as: $F_c=m\,* \frac{v^2}{r}$ where "v" is the tangential velocity of the object in circular motion, "r" is the radius of rotation and "m" is the object's mass.

So if we start with such formula with a given mass, radius, and tangential velocity, and then we move to a situation where everything stays the same except for the radius which doubles, then the new centripetal force ( $F'_c$ ) will be given by: $F'_c=m\,* \frac{v^2}{2r}$

and this is half (1/2) of the original force:

$F'_c=m\,* \frac{v^2}{2r}\\F'_c=m\,* \frac{v^2}{r}*\frac{1}{2} \\F'_c=F_c\,*\,\frac{1}{2}$

which is expressed by option "c" of the provided list.

8 0

3 years ago

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goblinko [34]

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3 years ago