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bogdanovich [222]
3 years ago
8

The diagram shows a charge moving into an electric field. The charge will most likely leave the electric field near which letter

? OW OX OY OZ ​
Physics
1 answer:
Simora [160]3 years ago
5 0

you haven't attached the diagram, but i assume that this diagram is what you were talking about

Answer:

near Y

Explanation:

the electric field lines goes from a positive charge to a negative charge. This means that a positive charge would move in the same direction of the field lines, while a negative charge would move in the opposite direction of the field lines. the field lines are created from +vely charged plate to -vely charged plate so the negative charged particles moves towards the lower plate which is positively charged, and opposite to the direction of field lines.

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The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential diffe
bazaltina [42]

To solve this problem we will apply the concepts related to the calculation of power, from the two electrical forms:

P = VI

P = \frac{V^2}{R}

Here,

V = Voltage

I = Current

R= Resistance

P = Power

PART A)

R = \frac{V^2}{P}

Replacing,

R = \frac{(120V)^2}{150W}

R = 96\Omega

Resistance of bulb is 96\Omega

PART B)

I = \frac{P}{V}

I = \frac{150W}{120V}

I = 1.25A

The bulb will draw 1.25A current

7 0
3 years ago
Which of the following is a contact force?
shusha [124]

hope this helps.........

7 0
3 years ago
Suppose 8.41 moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 395 to 279 K. Determine (a)
sergeinik [125]

Answer:

a) W=12166.20876 J

b) U= -12166.20876 J

Explanation:

No. of moles, n = 8.41

Change of temperature, ΔT = T1 - T2

                                         = 395 - 279

                                         = 116 K

For monatomic gas, γ = 5/3

γ -1 = 2 /3

Solution:

(a)

Work done,W= \frac{nR}{\gamma-1}(T_1-T_2)

plugging values we get

W= \frac{8.314\times8.41}{2/3}(116)

Ans:   12166.20876 J

Work done, W = + 12166.20876 J

(b)

From first law of thermodynamics, dQ = U + W

but, dQ = 0 ( adiabatic process)

Hence, U = - W

                 = - 12166.20876 J

Ans:

Change in internal energy, U = - 12166.20876 J

8 0
3 years ago
A drum rotates around its central axis at an angular velocity of 19.4 rad/s. If the drum then slows at a constant rate of 8.57 r
lys-0071 [83]

Answer:

Explanation:

Using equations of motion:

(a)

v=u+at

∴0=19.4−8.57t

∴t=19.4/8.57

=2.3s

B. Using s= ut + 1/2 at²

19.4(2.3)-1/28.57(2.3)²

= 21.92rad

4 0
3 years ago
The surface area of the earth's crust is 5.10*10^8km^2. The average thickness of the earth's crust is 35km. The mean density of
Mars2501 [29]

Answer:

The answer is V = 1.785 \times 10^{10}\ Km^3.

Explanation:

You are asking only for the total volume, so the important data here is the surface area of the earth's crust,

A = 5.10 \times 10^8\ Km^2

and the average thickness of the earth's crust,

h = 35\ Km.

Imagine that you can stretch the surface area as if it were a blanket, now if you want to calculate its volume, you just need to multiply its area by its thickness,

V = A*h = 1.785 \times 10^{10}\ Km^3.

4 0
3 years ago
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