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REY [17]
3 years ago
11

PLEASE HELP..... Which of these statements is false

Physics
1 answer:
olga_2 [115]3 years ago
3 0
The 3rd one is false. The Earth is FARTHEST from the sun in early July.
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How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
Explain how machines can be useful if the output is always less than the input work
Anni [7]

Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.

6 0
3 years ago
What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass
Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

8 0
2 years ago
A ticker timer makes 50 dots per second. When a body is pulled by a tap through the timer, the distance between the third and fo
fenix001 [56]

Answer:

The acceleration is 1 cm/s^2.

Explanation:

The acceleration is defined as the rate of change of velocity.

Here, initial velocity, u = 3/1 = 3 cm/s

final velocity, v = 4/1 = 4 cm/s

time, t = 1 s

Let the acceleration is a.  

Use first equation of motion

v =  u + at

4 = 3 + 1 x a

a = 1 cm/s^2

8 0
3 years ago
This is a two part question:
adelina 88 [10]

Answer:

1. 310 N

2. 310 N

Explanation:

1. Determination of the net force.

Force applied (Fₐ) = 430 N

Force experience (Fₑ) = 120 N

Net force (Fₙ) =?

Fₙ = Fₐ – Fₑ

Fₙ = 430 – 120

Fₙ = 310 N

Hence, the net force acting on the crate is 310 N

2. Determination of the net force.

Force applied (Fₐ) = 375 N

Force of friction (Fբ) = 65 N

Net force (Fₙ) =?

Fₙ = Fₐ – Fբ

Fₙ = 375 – 65

Fₙ = 310 N

Hence, the net force acting on the crate is 310 N.

8 0
2 years ago
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