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Ganezh [65]
3 years ago
12

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

nitude of the force is different in the two cases, while the directional angle θ is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is 0.340. The weight of the block is 51.0 N, and the directional angle for the force is θ = 44.0°. Determine the magnitude of when the block slides (a) up the wall and (b) down the wall.
Physics
1 answer:
sweet-ann [11.9K]3 years ago
3 0

Answer:

Explanation:

Given

coefficient of kinetic friction(\mu _k)=0.34

inclination \theta =44

weight of block=51 N

(a) When block is moving upward friction force acts downward

thus

Fsin\theta -W-f_r=0

as block is moving with constant velocity thus F_{net} is zero

f_r=\mu _kN=0.34\times Fcos\theta

F\left ( \sin \theta -\mu \cos \theta \right )=W

F=\frac{51}{0.45}=113.31 N

(b)When Block slides down the wall friction changes its direction to oppose the block

Fsin\theta -W+f_r=0

F\left ( \sin \theta +\mu \cos \theta \right )=W

F=\frac{W}{\left ( \sin \theta +\mu \cos \theta \right )}

F=\frac{51}{0.939}=54.299 N

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Which is present when iodine changes from brown to blue or purple?
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5 0
3 years ago
It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f
Elena-2011 [213]

Answer:

B) the change in momentum.

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6 0
3 years ago
A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,
Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a)  Kinetic energy of block = potential energy in spring  

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg.  Spring constant k is unknown, but you can find it from given data:  

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.  

From the energy equation above, solve for v,

v = v √(k/m)  

= 0.15 √(300/1)

= 2.598 m/s.

b)  Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.  

This is the same momentum carried by bullet as it strikes the block.  Therefore, if u is bullet speed,  

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

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6 0
3 years ago
The density of gold is 19 300kg/m cube. what is the mass of gold cube with the length 0.2015m?
Sergeeva-Olga [200]

Answer:

157.9 kg

Explanation:

Density: This can be defined as the ratio of the mass of a body and it's volume.

The S.I unit of density is kg/m³.

From the question,

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D = m/v............................ Equation 1

Where D = Density of gold, m = mass of gold, v = volume of gold.

make m the subject of the equation

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Since the gold is a cube,

v = l³................... Equation 3

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Substitute equation 3 into equation 2

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Substitute into equation 4

m = 19300(0.2015)³

m = 157.9 kg.

4 0
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