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3 years ago

To get an accurate data reading, scientists must be sure to

Physics

1 answer:

joja [24]3 years ago

8 0

D.
Always use the right tool to get accurate measurements

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Which would not be a reason for using wind power?

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C it takes a lot of capital to get the windmills and the fields of windmills set up

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A vehicle accelerates from 0 to 30 m/s in 10 seconds on a straight road

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2 years ago

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A cannon fires a cannonball of mass 16.0 kg by applying a force of 2750 N along the 1.25 m length of the barrel. (a) How much wo

Zarrin [17]

To develop this problem it is necessary to apply the concepts related to Work and energy conservation.

By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,

$W = F*d$

Where,

F= Force

d = Distance

On the other hand we know that the potential energy of a body is given based on height and weight, that is

$PE = -mgh$

The total work done would be given by the conservation and sum of these energies, that is to say

$W_{net} = W+PE$

PART A) Applying the work formula,

$W = F*d\\W = 2750*1.25\\W = 3437.5J$

PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then

$PE = -mgh*sin25$

$PE = -16*9.81*1.15sin25$

$PE = -82.9J$

The net work would then be given by

$W_{net} = W+PE$

$W_{net} = 3437.5J-82.9J$

$W_{net} = 3354.6J$

Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J

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3 years ago

A 2.7-kg cart is rolling along a frictionless, horizontal track towards a 1.1-kg cart that is held initially at rest. The carts

olga2289 [7]

Answer:

Part a)

P = 8.23 kg m/s

Part b)

v = 3.05 m/s

Explanation:

Part a)

momentum of cart 1 is given as

$P_1 = m_1v_1$

$P_1 = (2.7)(3.7) = 9.99 kg m/s$

Momentum of cart 2 is given as

$P_2 = m_2v_2$

$P_2 = (1.1)(-1.6) = -1.76 kg m/s$

Now total momentum of both carts is given as

$P = P_1 + P_2$

$P = 8.23 kg m/s$

Part b)

Since two carts are moving towards each other due to mutual attraction force and there is no external force on two carts so here momentum is always conserved

so here we will have

$P_i = P_f$

$(2.7 kg)v = 8.23$

$v = 3.05 m/s$

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3 years ago

If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is

Leokris [45]

73 Newton is the correct answer

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