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3 years ago

PLEASE HELP!!!

Physics

1 answer:

d1i1m1o1n [39]3 years ago

6 0

Answer: 95.7 N

Explanation: The tension on the chandelier is equal to the its weight.

Weight is the force directed downwards and tension is the force exerted of the opposite side.

T = W of the chain + W of the chandelier

= m1( g) + m2 (g)

= 0.37 kg ( 9.8 m/s2) + 9.4 kg (9.8 m/s2)

= - 95.7 N it means the force is directed downwards

since tension is equal weight

T = 95.7 N

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What is the mass of 3 m3 of a substance having density 1200 kg/m3

adell [148]

Answer:

3600 kg

Explanation:

From the question,

Density = Mass/Volume

D = M/V.............................. Equation 1

Where D = Density of the substance, M = mass of the substance, V = Volume of the subtance.

Make M the subject of the equation

M = D×V ............................ Equation 2

Given: D = 1200 kg/m³, V = 3 m³.

Substitute these values into equation 2

M = 1200×3

M = 3600 kg.

Hence the mass of the substance is 3600 kg

4 0

3 years ago

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows: