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11 months ago

Newton's second law shows that there is a(n) ________

Physics

1 answer:

USPshnik [31]11 months ago

4 0

Answer:

Objects accelerate when the net force acting on them divides by their mass, according to Newton's second law of motion. The second law states that force and acceleration are directly related, while mass and acceleration are inversely related.

Explanation:

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A 5.0 Newton force is applied to the right of a 6.0 kg object to move across a smooth surface. The coefficient of friction is ze

AlekseyPX

Answer:

$0.833\ \text{m/s}^2$

Explanation:

m = Mass of block = 6 kg

F = Force on block = 5 N

$\mu$ = Coefficient of friction = 0

a = Acceleration of the block

From Newton's second law we have

$F=ma\\\Rightarrow a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{5}{6}\\\Rightarrow a=0.833\ \text{m/s}^2$

The acceleration of the block is $0.833\ \text{m/s}^2$

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1 year ago

Problem 28.3 Suppose that you are planning a trip in which a spacecraZ is to travel at a constant velocity for exactly six month

iragen [17]

Answer:

v = 0.999981c m/s

Explanation:

Using the time dilation equation

$T = \frac{T_{0} }{\sqrt{1 - \frac{v^{2} }{c^{2} } } }$

T = stationary time = 100 years

T₀ = 11/12 years = 0.917 years

v = speed of travel in the space = ?

c = speed of light = 3 * 10⁸ m/s

$100 = \frac{0.917 }{\sqrt{1 - \frac{v^{2} }{(3*10^{8} )^{2} } } }\\$

$(0.917/100) ^{2} = 1 - \frac{v^{2} }{ 9 * 10^{16} }$

v = 299987395.57 m/s

v = 2.99 * 10⁸ m/s

v = 0.999981c m/s

3 0

1 year ago

What is a proton?

eimsori [14]

The answer is A. A positive particle inside the nucleus

7 0

1 year ago

Read 2 more answers

HELP PLEASE. WILL GIVE BRAINLIEST!!!

nikitadnepr [17]

Answer:

its A

Explanation:

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1 year ago

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A 40 kg skier starts at the top of 12 meter high slope. At the bottom she is traveling 10 m/s how much energy does she lose to f

Mashcka [7]

To solve this problem we start from the premise that energy is conserved. This means that the energy that the skier has at the beginning of the slope is equal to the energy he has at the end of the slope.

When the skier is at the top of the slope (point 1) his kinetic energy is minimal and his potential energy is maximum. Let's call $E_1$ to the energy in point 1.

$E_1 = mgh$

Where:

mgh = potential energy

When the skier is at the lowest point of the slope (point 2) his kinetic energy is maximum and his potential energy is minimal. Let's call $E_2$ the energy in point 2.