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murzikaleks [220]
3 years ago
14

To move a 51kg cabinet across a floor requires a force of 200N to start it moving, but then only 100N to keep it moving a steady

speed. (a) Find the maximum coefficient of static friction between the cabinet and the floor. (b) Find the coefficient of kinetic friction between the cabinet and the floor.
Physics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

a) \mu_s =0.40

b) \mu_k =0.20

Explanation:

Given:

Mass of the cabinet, m = 51 kg

a) Applied force = 200 N

Now the force required to move the from the state of rest (F) = coefficient of static friction (\mu_s)× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 200N  = \mu_s × 499.8 N

⇒\mu_s = \frac{200}{499.8}=0.40

b) Applied force = 100 N

since the cabinet is moving, thus the coefficient of kinetic(\mu_k) friction will come into action

Now the force required to move the from the state of rest (F) = coefficient of kinetic friction (\mu_k)× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 100N  = \mu_k × 499.8 N

⇒\mu_k = \frac{100}{499.8}=0.20

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Gwar [14]

Answer:

0.95 seconds

Explanation:

t = Time taken

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Time taken by the ball to reach the maximum height

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s

Maximum height

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m

Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m

So, the distance that will be covered on the way down is 4.46 m

Now

u = 0

s = 4.46

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s

Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.

Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds

7 0
3 years ago
A 45.0 kg ice skater stands at rest on the ice. A friend tosses the skater a 5.0 kg ball. The skater and the ball then move back
Allushta [10]

Answer:

u = 5 m/s

Explanation:

given,

Mass of ice skater, M = 45 Kg

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velocity of skater and the ball,V = 0.5 m/s

speed of the ball = ?

using conservation of momentum

m u + M u' =( M + m ) V

initial speed of ice skater is zero

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u = 5 m/s

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Hatshy [7]

Answer:

The charge on point A is q_a = 2.4 × 10^{-5} C

The charge on point B is q_b = 1.2 × 10^{-5} C

Explanation:

Given data

Distance (r) = 0.19 m

Magnitude of the charge on A is twice that of the charge on B  i.e.

q_A = 2 q_B

F = 46 N

We know that force between the charges is given by

F = \frac{k Q_A Q_B}{r^{2} }

65 = \frac{(9) (10^{9}) (2q_b^{2} ) )}{0.2^{2} }

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q_a = 2 × 1.2 × 10^{-5}

q_a = 2.4 × 10^{-5} C

Therefore the charge on point A is q_a = 2.4 × 10^{-5} C

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Otrada [13]

Answer:

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rusak2 [61]

Answer:

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Explanation:

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