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podryga [215]
3 years ago
5

7. Starting at rest, a car accelerates at 5.5m/s/s for 12s. What is its

Physics
1 answer:
Musya8 [376]3 years ago
6 0

Answer:

66 m/s

Explanation:

v=u+at

= 0 + 5.5 * 12

= 66 m/s

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Why is visual constancy important?
weqwewe [10]

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C.

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4 0
2 years ago
A scared elephant has a mass of 7000 kg. The mouse that frightened the elephant is 0.02 kg. The distance between the elephant an
arsen [322]

a. 9.34\cdot 10^{-9} N

The gravitational force between two objects is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M,m are the masses of the two objects

r is the separation between the objects

In this problem, we have:

M = 7000 kg is the mass of the elephant

m = 0.02 kg is the mass of the mouse

r = 1 m is the separation

Substituting into the equation, we find the gravitational force exerted by the elephant on the mouse:

F=(6.67\cdot 10^{-11}) \frac{(7000)(0.02)}{1^2}=9.34\cdot 10^{-9} N

b. -9.34\cdot 10^{-9}N

We can solve this part by keeping in mind Newton's third law of motion, which states that:

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

In our problem, we can identify the elephant as object A and the mouse as object B: this means that the gravitational force exerted by the elephant on the mouse is the action, while the gravitational force exerted by the mouse on the elephant is the reaction. The two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the mouse on the elephant is equal to

F=-9.34\cdot 10^{-9}N

where the negative sign simply means that the direction is opposite.

6 0
3 years ago
You are asked to design a spring that will give a 1020 kg satellite a speed of 2.25 m/s relative to an orbiting space shuttle. Y
julsineya [31]

Answer:

(a) 2.45×10⁵ N/m

(b) 0.204 m

Explanation:

Here we have that to have a velocity of 2.25 m/s then the relationship between the elastic potential energy of the spring and the kinetic energy of the rocket must be

Elastic potential energy of the spring =  Kinetic energy of the rocket

\frac{1}{2} kx^2 = \frac{1}{2} mv^2

Where:

k = Force constant of the spring

x = Extension of the spring

m = Mass of the rocket

v =  Velocity of the rocket

Therefore,

\frac{1}{2} kx^2 = \frac{1}{2} \times   1020 \times 2.25^2

or

kx^2 =  1020 \times 2.25^2 = 10,226.25\\So \ that \ the \ force \ on \ the \ satellite\ kx = \frac{10226.25}{x}

(b) Since the maximum acceleration is given as 5.00×g we have

Maximum acceleration = 5.00 × 9.81 = 49.05 m/s²

Hence the force on the rocket is then;

Force = m×a = 1020 × 49.05 = ‭50,031 N

kx = \frac{10226.25}{x} = 50031 \ N

Therefore,

x = \frac{10226.25}{ 50031} = 0.204 \ m

(a) From which

k = \frac{10226.25}{x^2} = \frac{50031}{x} = \frac{50031}{0.204} = 244,772.13 \ N/m or

Force constant of the spring, k = 2.45×10⁵ N/m.

6 0
3 years ago
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