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Inga [223]
3 years ago
10

Which is an example of potential energy? Question 9 options: an electric current lighting a light bulb releasing the strings on

a guitar a roller coaster speeding down a hill the pulled back string on an archer's bow
Physics
1 answer:
Feliz [49]3 years ago
4 0
The pulled back string on an archer’s bow.
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A student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The s
guapka [62]

Answer:

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.

b. A meterstick to measure the distance of the track that the car travels on.

Explanation:

Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.

In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.

Therefore, to test the validity of the claim, the student should use the following measuring tools;

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.

b. A meterstick to measure the distance of the track that the car travels on.

Hence, with these two devices the student can effectively measure or determine the validity of the claim.

5 0
3 years ago
1. Negative acceleration is also called _____.
FinnZ [79.3K]

Answer:

B

B

B

Explanation:

3 0
4 years ago
When a cold air mass replaces a warm air mass, it's called a(n) _____ front?
Anvisha [2.4K]
When a cold air <span>mass replaces a warm air mass, this is called a cold front. Some characteristics of a called front before passing are winds coming from south or southwest area, warm temperature, falling pressure, and drizzles. When it passes, the winds are shifting, there is a sudden drop of temperature, minimum pressure followed by a sharp rise. After passing, the winds head to the west or northwest area, temperature is steadily dropping and the pressure is rising steadily.</span>
3 0
3 years ago
Read 2 more answers
Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was
tigry1 [53]

Answer:

\Delta V=316167V

Explanation:

The difference of electric potential between two points is given by the formula \Delta V=Ed, where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is E=\frac{F}{q}, where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have \Delta V=\frac{Fd}{q}, so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force N=W-F. On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) d=\frac{at^2}{2}, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

a_1=\frac{2d}{t_1^2}

a_2=\frac{2d}{t_2^2}

These relate with the forces by Newton's 2nd Law:

W=ma_1

N=ma_2

Putting all together:

N=W-F=ma_1-F=ma_2

Which means:

F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})

And finally we substitute:

\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})

Which for our values means:

\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V

7 0
3 years ago
An electric current of 200 A is passed through a stainless steel wire having a radius R of 0.001268 m. The wire is L = 0.91 m lo
denis23 [38]

Answer:

The value is   T_m  =  435.2 \  K

Explanation:

From the question we are told that  

   The  current is  I  =  200 \ A

   The radius is  R =  0.001268 \  m

   The  length of the wire is  L  =  0.91 \  m\

    The  resistance is  R  =  0.126 \  \Omega

    The  outer surface temperature is  T _o  =  422.1 \  K

    The average thermal conductivity is  \sigma  =  22.5 W/mK

   

Generally the heat generated in the stainless steel wire is mathematically represented as  

    Q =  \frac{Power}{ \pi r^2L}

     Q =  \frac{I^2 R}{ \pi r^2L}

=>   Q =  \frac{200^2 * 0.126}{3.142 *  (0.001268)^2 * 0.91}

=>   Q =  1.096*10^{9}\  W/m^3

Generally the middle temperature is mathematically represented as

      T_m  =  T_o  + \frac{Q * r^2 }{ 6  * \sigma }

       T_m  =  422.1  +  \frac{1.096*10^{-9} * 0.001268^2}{6 * 22.5}

       T_m  =  435.2 \  K

4 0
3 years ago
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