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2 years ago

I need help with the problem below

Physics

1 answer:

sdas [7]2 years ago

4 0

Answer:

Explanation:

a) F = ma

a = F/m

a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s

b) t = v/a

t = 200 / 7.2 x 10⁻⁶

t = 2.8 x 10⁷ s about 10½ months

c) v² = u² + 2as

s = (v² - u²) / 2a

s = (200² - 0²) / (2( 7.2 x 10⁻⁶))

s = 2.8 x 10⁹ m nearly 7 times around the earth

And all this assumes NO FRICTION.

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On a frictionless surface how much force is necessary to accelerate a 0.49 kg object to the left at 4.8 m/s2?

Alenkasestr [34]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

force = 0.49 × 4.8 = 2.352

We have the final answer as

Hope this helps you

8 0

3 years ago

.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c

ohaa [14]

Answer:

a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

F t = 2 (m v)

therefore the average force is

F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

Em_f = K = ½ m v²

energy is conserved in this system

Em₀ = Em_f

m g h = ½ m v²

v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

F = 2m √(2gh) /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

F = m a

a = F / m

a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

a = 2 √ (2 9.8 1.5) / 10⁻³

a = 1.1 10⁵ m / s²

6 0

3 years ago

If you are in an elevator that speed up then the apparent weight is

Sholpan [36]

Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

$W_{net}$ = mg + ma

where: mg is the weight of the object, and ma is the apparent weight.

Apparent weight (ma) = $W_{net}$ - mg

3 0

2 years ago

Which of the following statements describes a battery?

LiRa [457]

Answer:

The negative electrode of a battery has an excess of positive charge

Explanation:

This is because in every battery, there is a negative electrode ( cathode ) and only positive charge is deposited on it.

For other statements:

Batteries donot store electric charge but they store chemical energy

Some batteries donot use metals for the flow of electrons, but some use hydrogen gas at a pressure of 1 atmosphere.

8 0

3 years ago

How much does the Earth weigh?<br><br><br><br> Have an AWESOME day<br> -NaomiTheGenuis

babymother [125]

Answer:

The Earth weighs about:
13,170,000,000,000,000,000,000,000 pounds

or

5.97 billion trillion metric tons

Explanation:

Have a great rest of your day
#TheWizzer

3 0

2 years ago

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