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3 years ago

12

Two point charges are separated by a certain distance. How does the strength of the electric field produced by the first charge,

at the position of the second charge, change if the second charge is doubled?

1 answer:

olga55 [171]3 years ago

8 0

Answer:

The field will remain the same

Explanation:

This is because electric field given as

E1= kq1/r²

And that of second charge

E² = kq2/r²

Is not affected by the size of the second charge q2

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If a 60 kg student is standing on the edge of a cliff. Find the students gravitational potential of the cliff is 30 m high

kipiarov [429]

Answer:

17640J

Explanation:

Give mass m= 60kg

Height h =30m

Since the student is standing on the edge of a cliff , acceleration due to gravity g is 9.8m/s^2

The student’s gravitational potential energy can be found by the formula

mgh

That’s 60 x 9.8 x 30

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3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

<h2>In the y-axis: </h2>

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

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3 years ago

You obtain a spectrum of an object in space. The spectrum consists of a number of sharp, bright emission lines. Is this object a

Andreyy89

Answer:

In hot gases , the atoms keeps colliding with each other and sometimes the energy liberated during collision takes the electron to a higher level,thus, .The object is a cloud of hot gas and finally the electron returns back emitting photon

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3 years ago

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A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling

KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

$\frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$ ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

$\frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ ...... (2)

Now, equating both equations (1) and (2) as follows.

$Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ = $Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$

$\gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}$

= $\frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})$

= $\frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})$

= 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

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