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Ierofanga [76]
3 years ago
12

HELLPPP ASAPPPP!!!!!

Physics
1 answer:
OLga [1]3 years ago
4 0

Answer:

initial velocity=19 ms^{-1}

greatest height=48.05 m

Explanation:

Applying motion equations,

upward motion (g= -10m/s2)

where

u = initial velocity

v = final velocity

g = gravitational acceleration

s = distance

v^{2}=u^{2} +2gs\\31^{2} = u^{2}+2*(-10)*(-30)\\u^{2}=361\\u=19 ms^{-1}

2)when the rock is at the highest point the velocity = 0

v^{2}=u^{2} +2gs\\0=361-2*10*S\\S=18.05 m

total high to the highest point = 18.05 +30

                                                  = 48.05 m

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PLEASE ANSWER, I NEED HELP
Scorpion4ik [409]

1) The gravitational force between Ellen and the moon is 1.56\cdot 10^{-3} N

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

1)

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

In this problem, we have:

m_1 = 47 kg is the mass of Ellen

m_2 = 7.35\cdot 10^{22} kg is the mass of the moon

r=3.84\cdot 10^8 m is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N

2)

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

a=\frac{F}{m}

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

And about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0
3 years ago
Two forces that are not equal in size are
Bumek [7]
They are unbalanced forces ..... Hope this helps :3
4 0
3 years ago
A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the
lina2011 [118]

Answer:

The the analysis for the free fall part should be done under the constant acceleration.

Explanation:

In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.

Therefore, the the analysis for the free fall part should be done under the constant acceleration.

8 0
3 years ago
Almost everything in the universe orbits around a central object. The planets orbit the sun and moons orbit planets. The force M
DanielleElmas [232]
Gravity is why orbits are happening.
3 0
3 years ago
Read 2 more answers
How much force would be needed to cause a 4.6kg object to accelerate at 9.2m/s/s? *
poizon [28]

Answer:

<h2>42.32 N</h2>

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 we have

force = 4.6 × 9.2

We have the final answer as

<h3>42.32 N</h3>

Hope this helps you

5 0
3 years ago
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