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Luden [163]
3 years ago
10

A baseball has a mass of 0.15kg .What is the net force on the ball if it's acceleration is 40.0m/sw

Physics
1 answer:
kvasek [131]3 years ago
7 0

By Newton's second law, we have

F=ma

So, in order to give a 0.15kg body an acceleration of 40m/s^2, you need a force of

F=0.15\cdot 40 = 6N

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Presently, the speed of light in a vacuum is defined to be exactly 299,792,458 m/s (approximately 186,282 miles per second). . An early experiment to measure the speed of light was conducted by Ole Romer, a Danish physicist, in 1676. Using a telescope, Ole observed the motions of Jupiter and one of its moons, Io
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Why do metals make good electrical conductors?
alexdok [17]

Answer:

Explained

Explanation:

Metals are good conductors of electricity because they contain free electrons in their atoms. The outer shell of atom's of metal have free electrons. These free electrons are responsible of electrical conductivity of metals. These electron are not bounded by the attraction forces of the nucleous. They are free to wonder in lattice of positive ion and thus allow electrical conductivity.

3 0
3 years ago
13. A transformer has a primary coil with 600 turns and a secondary coil with 300 turns. If the output voltage is 320 volts, wha
vivado [14]
His is a step down transformer since n(primary) is greater than n(seconcary). You relate the input voltage with the ouput voltage with the following equation: 

<span>Vout = n2/n1*Vin (n2/n1 is essentially your 'transfer function' that dictates what a specified input would produce) </span>

<span>Solving the equation: </span>

<span>Vin = Vout*n1/n2 = (320V)*(600/300) = 640 V </span>

<span>This is checked by seeing if Vin is greater than Vout, which it is for a step down transformer.</span>
5 0
3 years ago
If a force of 3000 N is applied to a large rock, but the rock does not move, how much work is done on the rock?
kkurt [141]

Answer:

No work was done.

W = 0

Explanation:

Work is said to be done whenever a force of one newton moves a body of one kilogram through a distance of one meter. Meaning the applied force has to move the body from a point of rest through certain distance.

Work = force × distance

So, in the case of this question, we only have the force been applied, but no distance was covered. Hence, no work was done.

W = 3000× 0 meter

W = 0

8 0
3 years ago
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25
Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: d_{1x} = 5.2 cos 25^{\circ}=4.7 mi

North: d_{1y} = 5.2 sin 25^{\circ}=2.2 mi

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: d_{2x}=0

North: d_{2y} = 5.0 mi

So the components of the total displacement are

East: d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi

North: d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi

5 0
3 years ago
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