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3 years ago

7

A force of 500 N accelerates a 100 kg crate at a 2 m/s along a horizontal surface. What is the friction force acting on the obje

ct?

1 answer:

WITCHER [35]3 years ago

8 0

$F_{net} = F_x - f_k$ . Your net force will be the crate's weight multiplied 2. and force in the x-direction will be 500N. You can then solve the equation for the friction force.

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A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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Which situation is the best analogy for the doppler effect?

Rudiy27

The best scenario to describe the doppler effect would be listening to the siren of a passing ambulance or fire truck

then it is coming towards you, the pitch is higher, it gets higher as it approaches and peaks as it gets right in front of you. then it drop at once when it passes you and continues to drop till it fades away. this is a classic descrption of the doppler effect

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Brainliest to first to answer. What is the maximum stress which a material can withstand when it is pulled apart?

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Answer:

stress tension tensile strength

Explanation:

The maximum stress which a material can withstand when it is pulled apart is its: stress tension tensile strength.

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Hey! I found this question quite interesting. Check it out - https://www.meritnation.com/ask-answer/question/to-raise-a-200kg-st

Yanka [14]

Answer:

yes yes so do I

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