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Dennis_Churaev [7]
4 years ago
10

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

her pulls in the same direction with a force of 360 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Physics
1 answer:
m_a_m_a [10]4 years ago
6 0

Answer:0.27

Explanation:

Given

One worker Pushes with force F_1=430 N

other Pulls it with a rope of rope F_2=360 N

mass of crate m=290 kg

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

f_r=F_1+F_2

where f_r is the friction force

f_r=430+360

f_r=790 N

f_r=\mu N

where \muis the coefficient of static friction

N=mg

790=\mu 29\cdot 9.8

\mu =0.27

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2 years ago
1. Calculate the height of tree, 250 m away that produces
gtnhenbr [62]

Answer:

Explanation:

1.5/30 = x/250

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3 years ago
A particular 12 V car battery can send a total charge of 110 A·h (ampere-hours) through a circuit, from one terminal to the othe
DiKsa [7]
<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

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E = 4.752 x 10⁶ J

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8 0
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Answer:

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Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = \frac{I_l}{I_R} = \frac{x}{y} \frac{4}{10}, therefore the angle is the arctangent of 4/10:

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