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pentagon [3]
2 years ago
13

a small table has a mass of 4kg, stands on four legs, each leg having an area of 0.001 m2. what is the pressure exerted by the t

able on the floor?
Physics
1 answer:
Gemiola [76]2 years ago
3 0

Answer:

P = 10 kPa

Explanation:

Given that,

The mass of a small table, m = 4 kg

The area of each leg = 0.001 m²

We need to find the pressure exerted by the table on the floor. Pressure is equal to the force per unit area. So

P=\dfrac{mg}{4\times A}\\\\P=\dfrac{4\times 10}{4\times 0.001}\\P=10000\ Pa\\\\or\\\\P=10\ kPa

So, the required pressure is 10 kPa.

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Answer:

m_1 / m_2 = sqrt (1 / 2)

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- Initial velocity of both skaters V_i = 0

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- Velocity of skater  after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

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What is the ratio m1/m2 of their masses

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- Apply conservation of momentum for two skaters just before and after the push as follows:

                                              P_i = P_f

                                  0 = m_1*V_1 - m_2*V_2

- Evaluate:                 m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

                               0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify:                      0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

                               0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify:                      0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

                                    (V_1 / V_2)^2 = 2

                                    (V_2 / V_1)^2 = 1 / 2

- simplify:                     (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

                                  m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

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