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Drupady [299]
3 years ago
7

A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car

and the gravitational force of the Earth on the car compare at this point?A. N > mg B. N < mg C. N = mg
Physics
1 answer:
dybincka [34]3 years ago
4 0

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

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Tere would be too many so there would be more fighting. Thye could overpopulate the world and destroy our stuff for example if it is a pig they can easily destory crops. We would all have to go hunting. There would also be less food so that is why they would start going out and destroying stuff.

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3 years ago
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3 A rocket of mass 1200 kg is travelling at 2000 m/s. It fires its engine for 1 minute. The forward thrust provided by the rocke
natita [175]

The impulse shared by the object equals the difference in momentum of the object. In equation form,

F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.

<h3>How to calculate  thrust provided by the rocket engines is 10 kN (10 000 N).?</h3>

a)There is this impulse-momentum change equation.

where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.

We consider everything exits happen along a straight line, and gravitation does not participate.

So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.

$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$

New velocity after  engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.

To learn more about Impulse-momentum, refer

brainly.com/question/20586658

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5 0
2 years ago
Virginia beach is 15 kilometers wide and 50 kilometers long. if 2 cm of rain falls on virginia beach, how many cubic meters of r
icang [17]
Assume that the shape of Virginia beach is rectangular.

Note that
1 km = 10³ m
1 cm = 10⁻² m

The area is
A = (15 km)*(50 km)
   = (15 x 10³ m)*(50 x 10³ m)
   = 7.5 x 10⁸ m²

Because 2 cm of rain fell, the volume is
V = (7.5 x 10⁸ m)*(2 x 10⁻² m) = 1.5 x 10⁶ m³

Answer:  1.5 x 10⁶ m³
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3 years ago
Most photographs will be made using a shutter speed of 1/60 or faster.
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True

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A trapeze artist swings in simple harmonic motion with a period of 3.8 s.
pogonyaev
As we know that time period of simple pendulum is given as

T = 2π √L/g

here we know that

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now from above equation we know that

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so the length of the cable must be 3.6 m
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