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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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faust18 [17]

Answer:

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Explanation:

8 0
3 years ago
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What would happen to the moon if there was no sun and earth ?
iragen [17]

Answer:

High tides would be much smaller than they are now, and low tides would be even lower. This is because the sun would be influencing the tides, not the moon; however, the sun has a weaker pull, which would decrease the tides. ... Winds could become much faster and much stronger without the moon.

Explanation:

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4 0
3 years ago
A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat
KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

L_s = Latent heat of steam

s_w = Specific heat of water

L_i = Latent heat of ice

v = Velocity of ice

\Delta T = Change in temperature

Amount of heat required for steam

Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)

Heat released from water at 100 °C

Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6

Heat released from water at 0 °C

Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)

Total heat released is

Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m

The kinetic energy of the bullet will balance the heat

K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s

The velocity of the ice would be 2452.79432 m/s

6 0
4 years ago
The sun will probably end up as a white dwarf. How will it be different then from what it is today
just olya [345]

Answer:

The differences that will be observed are;

1) The Sun will become faint and will no longer be yellow but rather appear white and will no longer be visible (become invisible) by unassisted vision as we can see the Sun today

2) The size of the Sun will shrink to a size comparable to the size of the Earth

3) The Sun will cool down and will no longer radiate as much heat

4) The nuclear reactions that generate energy on the Sun's will seize and the and the heat from the Sun will be from residual thermal energy

5) The core, which is the hottest part of the Sun will no longer be hydrogen but carbon and oxygen

Explanation:

6 0
3 years ago
A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the
Nezavi [6.7K]

We use the equation of motion for vertical component,

s_{y} = u_{y} t+\frac{1}{2} gt^2.

Here, s_{y}   is displacement of bullet, u_{y}  is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

s_{y} =\frac{1}{2}gt^2

Given, s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m

Substituting the values, we get time of flight

2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s

4 0
3 years ago
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