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zysi [14]
3 years ago
11

Write an original definition for periodic

Physics
1 answer:
shepuryov [24]3 years ago
6 0
I didn't want to type all of this, so instead I showed proof and gave you the answer. 

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An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul
PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight  due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ =  400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

= 100 N .

weight at height = 100 N

5 0
3 years ago
A teacher asks her students to jump off of the ground. Once the students complete the task, she says, "All of you just made Eart
Crazy boy [7]

Answer:

Explained below

Explanation:

A) Newton's first law of motion states that an object will remain at rest or continue in its current state of motion except it is acted upon by another force.

Now using this law, when you jump off the ground, the earth will move a tiny bit and accelerate due to the force applied by the jumping.

B) Newton's 2nd law states that the acceleration of a system is directly proportional to the net external force acting on that system, is in the same direction with it and also inversely proportional to the mass.

In this case, when one jumps, an external force is exerted on the earth and we are told it is directly proportional to the acceleration of the system which in this case it's the earth, then it means that there is some motion by the earth even though you didn't see it move.

C) Newton's third law of motion states that to every action, there is an equal and opposite reaction.

In this case the motion of the jumper will lead to an equal and opposite reaction of the earth.

8 0
2 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
What are the differently types of motion in a bicycle?
Ksivusya [100]
The two types of motion exerted in bicycle are:
1. rotary motion
2. linear motion
7 0
3 years ago
Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
6 0
3 years ago
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