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2 years ago

A 140 N block rests on a table. The suspended mass has a weight of 77 N.

Physics

2 answers:

finlep [7]2 years ago

6 0

Answer:

a) Force required to keep both the masses in rest will be - 77 N

b) The coefficient of friction will be 0.55

Explanation:

a) As the weight of 77 N is suspended vertically so, there must act 77 N force on the 140 N weight to keep both masses in rest. Hence, the answer is -77 N.

b) We know that:

140µ = 77

µ = 77/140

µ = 0.55

Hence, the minimum co-efficient of friction required to keep both bodies in rest will be 0.55.

Basile [38]2 years ago

5 0

I'm assuming that this is what your diagram looks like because these types of problems always look like this (lol).

Comment if you need clarification!

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3 years ago

What clues are useful in reconstructing pangaea

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3 years ago

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One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from

Luba_88 [7]

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

$s_1 = v_1t = 8t$

Similarly the distance traveled by the 2nd skater after t seconds is

$s_2 = v_2t = 10t$

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$s_2 = s_1 + 80$

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$10t = 8t + 80$

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2 years ago

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2 years ago

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A metal block suspended from a spring balance is submerged in water. You observe that the block displaces 55 cm3 of water and th

DiKsa [7]

Answer:

8977.7 kg/m^3

Explanation:

Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3

Reading of balance when block is immersed in water = 4.3 N

According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.

Buoyant force = weight of the water displaced by the block

Buoyant force = Volume of water displaced x density of water x g

= 55 x 10^-6 x 1000 x .8 = 0.539 N

True weight of the body = Weight of body in water + buoyant force

m g = 4.3 + 0.539 = 4.839

m = 0.4937 kg

Density of block = mass of block / volume of block

= $\frac{0.4937}{55\times10^{-6}}$

Density of block = 8977.7 kg/m^3

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