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pashok25 [27]
3 years ago
11

You are directed to set up an experiment in which you drop, from shoulder height, objects with similar surface areas but differe

nt masses, timing how long it takes for each object to hit the floor. Of the following explanations, which best describes your findings?
a. the most dense object hits the ground first
b. the less dense object hits the ground first
c. they will hit the ground at the same time
Physics
1 answer:
tangare [24]3 years ago
3 0

Answer: c. they will hit the ground at the same time

Explanation:

The volume of both objects is almost the same, so the force of friction will be the same in each one, so we can discard it.

Now, when yo drop an object, the acceleration of the object is always g = 9.8m/s^2 downwards, independent of the mass of the object.

So if you drop two objects with the same volume but different mass, because the acceleration is the same for both of them, they will hit the ground at the same time, this means that the density of the object has no impact in how much time the object needs to reach the floor.

So the correct option is c

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Does a light bulb at a temperature of 2500K produce as white a light as the sun at 6000K
sattari [20]

Answer:

No. ... UV light causes sunburn, whereas visible light does not.

Explanation:

7 0
3 years ago
An energy plant produces an output potential of 1500 kV and serves a city 143 km away. A high-voltage transmission line carries
jekas [21]

Answer:

2123.55 $/hr

Explanation:

Given parameters are:

V_{plant} = 1500 KV

L = 143 km

I = 500 A

\rho = 2.4 \Omega / km

So, we will find the voltage potential provided for the city as:

V_{wire} =IR = I\rho L = 1500*2.4*143 = 514.8 kV

V_{city} = V_{plant}- V_{wire} = 1500-514.8 = 985.2 kV

Then, we will find dissipated power because of the resistive loss on the transmission line as:

P = I^2R = I^2\rho L=500^2*2.4*143 = 8.58*10^7 W

Since the charge of plant is not given for electric energy, let's assume it randomly as x =  \frac{\dollar 0.081}{kW.hr}

Then, we will find the price of energy transmitted to the city as:

Cost = P * x = 8.58*10^7 * 0.081 * 0.001 = 6949.8 $/hr

To calculate money per hour saved by increasing the electric potential of the power plant:

Finally,

I_{new} = P/V_{new} = I/1.2\\P_{new} = I_{new}^2R_{wire}\\Cost = P_{new}/1.44=6949.8/1.44 = 4826.25 $/hr

The amount of money saved per hour = 6949.8 - 6949.8/1.44 = 2123.55 $/hr

Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.

3 0
3 years ago
NEED THIS NOW PLZ!!!
SVETLANKA909090 [29]

Answer: velocity

Explanation: Hope this helps :)

8 0
3 years ago
What happens to the energy put into a machine that does not get used to do useful work?
Nonamiya [84]
"Most gets converted to heat energy" is the one among the following choices given in the question that describes what <span>happens to the energy put into a machine that does not get used to do useful work. The correct option among all the options that are given in the question is the second option or option "B". I hope it helps you.</span>
3 0
3 years ago
Read 2 more answers
A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is 96.0 kg, and the hill is
mr_godi [17]

Answer:

 The force applied 275 N in a direction parallel to the hill

Explanation:

Newton's second law is adequate to work this problem, in the annex we can see a free body diagram, where the weight (W) is vertical, the friction force (fr) is parallel to the surface and the normal (N ) is perpendicular to it. In general for these problems a reference system is taken that is parallel to the surface and the Y axis is perpendicular to it.

Let us decompose the weight into its two components, the angle T is taken from the axis and

            Wx = W sin θ

            Wy = W cos T

We write Newton's second law

              ∑ F = m a

X axis

          The cyclist falls at a constant speed, which implies that the acceleration is zero

              fr - W sin θ = 0

              fr = mg sin θ

              fr = 96 9.8 without 17

              fr = 275 N

When the cyclist returns to climb the hill, he must apply the same force he has to overcome the friction force that always opposes the movement .  The force applied 275 N in a direction parallel to the hill

5 0
3 years ago
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