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3 years ago

Physics laws of motion

Physics

1 answer:

expeople1 [14]3 years ago

8 0

The acceleration is $0.095 m/s^2$

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. Therefore we can write:

$\sum F = ma$

where:

$\sum F$ is the resultant force acting on the object

m is its mass

a is its acceleration

In this problem, we have the following forces acting on the system:

$F_1 = 2500 N$ (forward)

$F_2 = -2400 N$ (backward)

So, Newton's second law can be rewritten as:

$F_1 -F_2 = ma$

where:

m = 1050 kg is the mass of all the students

Solving the formula for a, we find the acceleration of the system:

$a=\frac{F_1-F_2}{m}=\frac{2500+(-2400)}{1050}=0.095 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

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A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to

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Velocity of the mass $m_1$ just before the collision

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$v_1 = \sqrt{2(9.8)(3)}$

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Therefore the momentum just before collision would be

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Since the momentum is conserved we have that

$153.36= 60u_1$

$u_1 = \frac{153.36}{60}$

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The velocity of mass $m_2$ after the collision is given by

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4 years ago

The Earth’s surface, on average, carries a net negative charge (while the clouds and lower atmosphere carry a net positive charg

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Answer:

(a) Q = -6.765 * 10⁵ C ; σ = 1.33 * 10⁻⁹ C/m²

(b) 0.23 N/C

Explanation:

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E = kQ/r²

Where

k = Coulombs constsnt

Q = charge

r = radius of sphere

To find charge Q, we make Q subject of the formula:

Q = (E * r²)/k

Hence, charge, Q, at the surface of the earth, having radius, r = 6.371 * 10⁵ and electric field, E = 150 N/C is:

Q = [150 * (6.371 * 10⁶)²] / (9 * 10⁹)

Q = 6.765 * 10⁵ C

Since we're told that the charge at the earth's surface is negative,

Q = -6.765 * 10⁵ C

Surface charge density, σ, given as:

σ = |Q|/A

Where

|Q| = magnitude of charge

A = surface area.

Surface area, A, of the earth is given as:

A = 4πr²

A = 4π * (6.371 * 10⁶)²

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σ = 6.765 * 10⁵/510064471909788

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E = 149.77 N/C

The difference between the electric field at the surface of the earth and at a height of 5km is:

159 - 149.77 N/C = 0.23 N/C

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