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3 years ago

What 2 things need to be known about an object in order to determine its kinetic energy?

Physics

1 answer:

NikAS [45]3 years ago

7 0

I believe the answer is the mass of the object and the speed at which it is moving.

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The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th

Paul [167]

According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.

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3 years ago

Read 2 more answers

At 5.0 minutes, the temperature of the water reaches 100 °C. The volume of the water in the urn

serious [3.7K]

Answer:

fdvevddvevkejokef0jeovdlvkjeuiyv

Explanation:

ddf4edscd

4 0

2 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago

I got this information for a lab but I don't know how to do the hypothesis and the conclusion please can you guys help me with i

pochemuha

Answer:

A hypothesis is what you think will happen.

A conclusion is the results of an experiment summarized.

Hope this helps.

8 0

3 years ago

1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H

JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

$dV = \overrightarrow{E} . d\overrightarrow{r}\\\\$

For equipotential surface, dV = 0 so

$0 = \overrightarrow{E} . d\overrightarrow{r}\\\\$

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

5 0

2 years ago