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3 years ago

What would be the path of the moon if somehow all gravitational forces on it vanished to zero?

1 answer:

6 0

The Moon would take off in a straight line, at a constant speed, with the
same speed and direction it had at the instant that gravity got switched off.

(Of course, if ALL gravity got switched off, then the Moon would fall apart.
The radioactive and volcanic heat activity in its core would continue, and
suddenly there wouldn't be anything to hold it together. Same for the Earth.)

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The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of

kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

$P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}$

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

$=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m$

Δh_air=714m

7 0

3 years ago

Accelerations are produced by equal forces or unequal forces?

sukhopar [10]

Answer: unequal forces

Explanation: in order for something to accelerate it must speed up or slow down . When unequal forces react it casuse a change in motion which is also know as acceleration .

5 0

3 years ago

2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or

krek1111 [17]

a) The work done by the tree is $-1.44\cdot 10^5 J$

b) The amount of force applied is 2880 N

Explanation:

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

$W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

3 years ago

I can't seem to get the right angular acceleration and also not sure how to do part b. Help will be much appreciated.

Tcecarenko [31]

Answer:It’s 5 I believe

Explanation: it says to round to the nearest thousandths, so it’ll be 5.

8 0

2 years ago

A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the

ladessa [460]

The units are not consistent - 1 m/s is not the same as 1 km/h.

First thing to do would be to convert from one unit of speed to the other, say km/h to m/s. There are 1000 meters (m) for every kilometer (km) and 3600 seconds (s) for every hour (h), so

$1\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}\approx0.278\,\dfrac{\mathrm m}{\mathrm s}$

So in fact 1 km/h is about 4 times slower than 1 m/s.

4 0

3 years ago