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2 years ago

Which will increase the kinetic energy of an object more, increasing its mass or increasing its speed? Why?

1 answer:

4 0

Increasing its velocity will add to the kinetic energy more as the formula for kinetic energy is 0.5*m*v^2. (The speed will be squared making it greater)

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Facts about light years

Anton [14]

A light-year is how astronomers measure distance in space.
-Light can travel 300,000 kilometres in one seconds!
Light year is the distance that light travel can travel in one year.
-Light can also travel 9,00,000,000,000 kilometers.
-One light is about 10 trillion km.
-Almost 300 years ago ,a Danish scientist named olaus roemer,was the fist man to estimate the distance light travels in a year -and using that he figure out the approximate speed of light!

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2 years ago

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Which is a example of speed? A)Cindy biked westward at 30km/h B)A Rock is the the left of a flowerpot C) A dog runs an average o

finlep [7]

I think the answer is A) Cindy biked westward at 30km/h. I'm sorry if I'm wrong. If that's not correct then it's definitely C) A dog runs an average of 8mp/h. I hope this helps. :)

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3 years ago

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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

2 years ago

What is work done in physics and how can it be calculated

shusha [124]

Work is done when a force is applied to an object moves that object. the work is calculated by multiplying the force by the amount of movement of an object

8 0

2 years ago

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A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along

Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

$F(x)=(3\ N/m^2)x^2+(6\ N/m)x$

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

$W=\int\limits {F{\cdot} dx}$

$W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J$

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

3 0

2 years ago

Read 2 more answers