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3 years ago

Question 3 of 10

Physics

1 answer:

Gemiola [76]3 years ago

7 0

Answer:

c. earth is a satellite of the sun

Explanation:

a satellite is an object that moves around a larger object therefore the earth is a satellite of the sun

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A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?

Irina-Kira [14]

F = 400 Pa x 2.5 m2
F = 1 kN

4 0

3 years ago

Which conclusion, based on measurements of the ball in the lab in this lesson, is correct? a. The density of the ball was 1.48 g

ser-zykov [4K]

Answer:

u dummy

Explanation:

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3 years ago

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s

Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375 m / s

3 0

4 years ago

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33Ã—Ã—10âˆ

AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, $F=3.33\times 10^{-21}\ N$

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

$F=k\dfrac{q^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$

$q=1.21\times 10^{-16}\ C$

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

$n=\dfrac{q}{e}$

$n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}$

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0

4 years ago

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon’

Yuki888 [10]

Answer:

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

$s = ut + \frac{1}{2}at^2$

where u = initial velocity = 0 m/s

a = acceleration = $250 m/s^2$

t = time = 0.02 s

Therefore:

$s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m$

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

$distance = speed * time$

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

$v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s$

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

0.05 + 0.15 = 0.2 m

8 0

3 years ago