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mars1129 [50]
2 years ago
9

Lionel makes a graphic organizer to compare the electric field around a positive charge with the electric field around a negativ

e charge.
Which labels belong in regions X, Y, and Z?


X: Points away from the charge

Y: Points in the direction of the force on the positive charge

Z: Points toward the charge

X: Points away from the charge

Y: Points in the direction of the force on the negative charge

Z: Points toward the charge

X: Points toward the charge

Y: Points in the direction of the force on the positive charge

Z: Points away from the charge

X: Points toward the charge

Y: Points in the direction of the force on the negative charge

Z: Points away from the charge
Physics
2 answers:
krek1111 [17]2 years ago
6 0
X- points away from the charge
y- points in the direction of the force on the positive charge
z- points toward the charge
aksik [14]2 years ago
4 0

Answer:

The Answer is A <-------------

Explanation:

Because I took the test on Edge

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A rigid tank internal energy of fluid 800kJ. Fluid loses 500kJ of heat and padle wheel does 100kJ of work. Find final internal e
Nesterboy [21]

Answer:

 U₂ = 400 KJ      

Explanation:

Given that

Initial energy of the tank ,U₁= 800 KJ

Heat loses by fluid ,Q= - 500 KJ

Work done on the fluid ,W= - 100 KJ

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take final internal energy =U₂

We know that

Q= U₂ - U₁ + W

-500 = U₂ - 800 - 100

U₂ = -500 +900 KJ

U₂ = 400 KJ

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2 years ago
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Explanation:

x=VT+at²/2

x=2.7x35+9.8x(35)²/2

x=6097

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3 years ago
Read 2 more answers
Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.
Fofino [41]

Answer:

Part a)

\alpha = 782.6 rad/s^2

Part B)

\omega = 587 rad/s

Part c)

a_t = 24.3 m/s^2

Explanation:

Part a)

As we know that

a = R \alpha

so we will have

a = 1.80 m/s^2

R = 0.230 cm

\alpha = \frac{a}{R}

\alpha = \frac{1.80}{0.230 \times 10^{-2}}

\alpha = 782.6 rad/s^2

Part B)

Angular speed of the yo-yo

\omega = \alpha t

so we have

\omega = 782.6 \times 0.750

\omega = 587 rad/s

Part c)

Tangential acceleration is given as

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Answer:

diffusion is the answer.

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MArishka [77]

Complete Question:

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m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

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