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2 years ago

8

When you jump, you exert a pushing force against the ground. Gravity pulls you back down. Why can a person jump higher on the mo

on than on Earth?

1 answer:

Dennis_Churaev [7]2 years ago

7 0

Answer:

This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher

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A bicycle travels 22 km in 0.50 hr. What is its average speed?

algol [13]

Answer:

22/0.50=44

Explanation:

average speed= distance / time

8 0

2 years ago

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by θ = 0.220e3t, where

OLga [1]

Answer: $a_{t}=3.96$

$a_{c}=0.8712$

Explanation:

Given

$\theta =0.220e^{3t}$

r=2cm

Now angular velocity is given by $\omega =\frac{\mathrm{d}\theta}{\mathrm{d}t}$

$\omega =0.66e^{3t}$

Now linear velocity(v) is given = $\omega r$

$v=1.32e^{3t}$

Now tangential component of acceleration is given by

$a_{t}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=3.96e^{3t}$

at t=0

$a_{t}=3.96cm/s^2$

radial component of acceleration is given by

$a_{c}=\omega ^{2}r$

$a_{c}=0.4356e^{6t}\times 2$

$a_{c}=0.8712e^{6t} cm/s^{2}$

at t=0

$a_c=0.8712 cm/s^{2}$

5 0

3 years ago

True or False: a physical change may produce a new substance.

expeople1 [14]

Answer:

false

Explanation:

only chemical changes can produce new substances

8 0

2 years ago

Read 2 more answers

olasank [31]

This question involves the concept of Colomb's Law and electrostatic force.

The electrostatic force will be "0.2 N".

<h3>COLOMB'S LAW:</h3>

According to Colomb's Law, every charge exerts an electrostatic force on the other charge, which is directly proportional to the product of the magnitudes of both the charges and inversely proportional to the square of the distance between them.

$F=\frac{kq_1q_2}{r^2}$

where,

F = electrostatic force
k = Colomb's constant = 9 x 10⁹ N.m²/C²
q₁ = magnitude of first charge
q₂ = magnitude of second charge
r = distance between charges

In case of +2 C charges:

q₁ = q₂ = 2 C
r = ?
F = 0.1 N

Therefore,

$0.1\ N = \frac{(9\ x\ 10^9\ N.m^2/C^2)(2\ C)(2\ C)}{r^2}\\\\r^2=\frac{(9\ x\ 10^9\ N.m^2/C^2)(2\ C)(2\ C)}{0.1\ N}\\\\r^2=3.6\ x\ 10^{11}\ m^2$

Now, for the second case:

F = ?
q₁ = 2 C
q₂ = 4 C
r² = 3.6 x 10¹¹ m²

Therefore,

$F=\frac{(9\ x\ 10^9\ N.m^2/C^2)(2\ C)(4\ C)}{3.6\ x\ 10^{11}\ m^2}$

<u>F = 0.2 N</u>

Learn more about Colomb's Law here:

brainly.com/question/9774180

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1 year ago

Equation of Motion Problems

barxatty [35]

Answer:

Below in bold.

Explanation:

1. Speed = distance / time

= 100 / 9.84

= 10.16 m/s.

Speed = 200/19.32 = 10.35 m/s.

Speed = 400/43.49 = 9.20 m/s.

7 0

2 years ago