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iogann1982 [59]
2 years ago
5

In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire

ction of the frictional force?
A. 20 newtons in the direction of the applied force
B. 20 newtons opposite to the direction of the applied force
C. 20 newtons perpendicular to the direction of the applied force
D. 20 newtons in two directions, perpendicular and in the direction of the applied force
E. No friction is acting on the block.
Physics
2 answers:
ololo11 [35]2 years ago
8 0
The answer is B because it is true my teacher literally told me this last week! Good thing I remember
sladkih [1.3K]2 years ago
5 0

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

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For 1983 through 1989, the per capita consumption
Tresset [83]

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y = 43.55 + 2.15t

Explanation:

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Thus, expressing the information as ordered pairs, we have; (3,37. 1) and (9,50).

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8 0
3 years ago
Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm
Ghella [55]

Answer:

Total electric potential, V=1.32\times 10^6\ volts

Explanation:

It is given that,

First charge, q_1=3\ \mu C=3\times 10^{-6}\ C

Second charge, q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C

Distance of first charge from origin, r_1=1.15\ cm=0.0115\ m

Distance of second charge from origin, r_2=2\ cm=0.02\ m

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}

V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})

V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})

V = 1321826.08 V

or

V=1.32\times 10^6\ volts

So, the total electric potential at the origin is 1.32\times 10^6\ volts. Hence, this is the required solution.

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