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2 years ago

3.

1 answer:

8 0

The force of gravity produces acceleration in all C. freely falling objects and this is known as acceleration due to gravity

Explanation:

A body is said to be in free fall when there is only one force acting on the body: the force of gravity.

Gravity is a force that acts downward, i.e. towards the Earth's centre.

If we are near the Earth's surface, the magnitude of the force of gravity on a body is given by

$F=mg$

where:

m is the mass of the body

g is known as the acceleration of gravity , whose value near the Earth's surface is $9.8 m/s^2$ ).

We can apply Newton's second law on an object in free-fall, to find its acceleration. In fact, we have:

$F=ma$

where F is the force acting on the body and a is its acceleration.

Solving for the acceleration,

$a=\frac{F}{m}$

And substituting F,

$a=\frac{mg}{m}=g=9.8 m/s^2$

Therefore, every object in free-fall accelerates at $9.8 m/s^2$ towards the ground.

Learn more about free fall here:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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The rate at which an object moves is the object's __________.

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Answer:

speed

Explanation:

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2 years ago

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Air resistance is a nonconservative force. It always opposes the motion of an object. An airplane flies from New York to Atlanta

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Answer:

is always negative

Explanation:

Air resistance is a non-conservative force it opposes the motion of an object.It is a force that is always directed opposite to the displacement.

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3 years ago

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The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

belka [17]

Given that,

Energy $H=2.7\times10^{31}\ W$

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

$H=Ae\sigma T^4$

$A=\dfrac{H}{e\sigma T^4}$

$4\pi R^2=\dfrac{H}{e\sigma T^4}$

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}$

$R=5.0\times10^{10}\ m$

(b). Given that,

Radiates energy $H=2.1\times10^{23}\ W$

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}$

$R=5.42\times10^{6}\ m$

Hence, (a). The radius of the star is $5.0\times10^{10}\ m$

(b). The radius of the star is $5.42\times10^{6}\ m$

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2 years ago

When you encounter a yellow light as you approach an intersection what is the safest approach to take?

Harlamova29_29 [7]

This isn't a physics question really, but you should slow down unless you don't believe you can stop in time or don't believe it is safe to stop quickly (say you see someone driving very close behind you and you don't think they're paying attention). In general, if you think that by staying at your current speed your back tires will cross the ending lines of the intersection by the time the light turns red, it is safe to go through the yellow light. However, this is a thing you'll develop a feel for as you're driving, when in doubt, just slow down, just watch slamming on your breaks if there is someone behind you, sometimes people will see the yellow light when they're behind you and they'll speed up behind you so they'll "make it" before the light turns yellow. While this isn't illegal, since they're not technically running the red light, you should never speed up going up to a yellow light, if you need to speed up to make it before it turns red, you shouldn't make the light, just stop, this is especially bad if someone is directly in front of you and is likely to stop at the yellow light while the person behind them speeds up. This causes a lot of accidents.

5 0

2 years ago

A 2,537-kg truck moving at 14 m/s strikes a car waiting at a traffic light, hooking bumpers. The two continue to move together a

slavikrds [6]

Answer:

1902.75 kg

Explanation:

From Law of conservation of momentum,

m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1

make m₂ the subject of the equation,

m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2

Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity

Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)

Substituting into equation 2.

m₂ =[2537(8) - 2537(14)]/(0-8)

m₂ = (20296-35518)/-8

m₂ = -15222/-8

m₂ = 1902.75 kg.

Thus the mass of the car = 1902.75 kg

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3 years ago