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Serjik [45]
2 years ago
12

a block of mass 5 kg is being pulled along a board horizontally with a constant velocity; the coefficient of friction between th

e two surface is 0.25, then what is the frictional forces that opposes the motion​
Physics
1 answer:
antoniya [11.8K]2 years ago
6 0

The block is in equilibrium, so the net forces parallel and perpendicular to the surface are

∑ F[para] = F[normal] - mg = 0

∑ F[perp] = F[pull] - F[friction] = 0

where mg is the weight of the block. It follows that

F[normal] = mg = (5 kg) (9.8 m/s²) = 49 N

and so the frictional force has magnitude

F[friction] = 0.25 F[normal] = 12.25 N ≈ 12 N

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Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulatin
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Explanation:

(a)   Since, it is given that the blocks are identical so distribution of charge will be uniform on both the blocks.

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Therefore, final charge on the block A is 4.35 nC.

(b)  As it is given that the positive charge is coming on block A . This means that movement of electrons will be from A to B.

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3 years ago
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A water heater that has the shape of a right cylindrical tank with a radius of 1 foot and a height of 4 feet is being drained. h
Natali5045456 [20]

<span> </span>For any prism-shaped geometry, the volume (V) is assumed by the product of cross-sectional area (A) and height (h). 

<span> V = Ah </span>

<span>
Distinguishing with respect to time gives the relationship between the rates. 
dV/dt = A*dh/dt</span>

<span> in the meantime the area is not altering </span>

<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>

<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min 

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5 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.
sattari [20]

Answer:

d=691.71km

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

t=\frac{d}{v_t}-\frac{d}{v_l}

Here d is the distance at which the earthquake take place and v_t, v_l is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km

8 0
3 years ago
Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to
Degger [83]

The acceleration of the  box up the ramp is 9.65 m/s².

<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

F(net) = ma

where;

  • m is the mass of the box
  • a is the acceleration of the box

The net force on the box is calculated as follows;

F(net) = F - Ff

F(net) = F - μmgcosθ

where;

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  • μ is coefficient of friction

F(net) = 170 -  (0.3 x 15 x 9.8 x cos55)

F(net) = 144.7

The acceleration of the box is calculated as;

a = F(net) / m

a = (144.7) / (15)

a = 9.65 m/s²

Thus, the acceleration of the  box up the ramp is 9.65 m/s².

Learn more about acceleration here: brainly.com/question/14344386

#SPJ4

8 0
1 year ago
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