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Art [367]
3 years ago
6

This diagram represents a top-down view of an experiment on a table. The 250g and 100g masses are falling and are pulling the bl

ock.
What is the location and minimum number of masses that would have to be added to the picture above to cause the block to move only toward the bottom (straight down)?

A. 1000g mass on bottom
B. 500g mass on bottom
C. 250g mass on the left
D. 250g mass on left and 200g mass on bottom

Physics
2 answers:
nata0808 [166]3 years ago
7 0
The answer should be B: 500g of mass on bottom.
g100num [7]3 years ago
3 0

Answer:

C. 250g mass on the left

Explanation:

As the given figure there are two forces on the box placed on the table

1) due to 250 gram there will be a force which will pull it towards right

2) due to 100 g mass which will pull it downwards

now if we wish to move the block only in downward direction then we have to add a force on the block that will counterbalance the force towards right direction

so here it must have a force towards left which will balance the force of 250 g mass towards right

so it would be

C. 250g mass on the left

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A car and a lorry are about to collide. When they collide the two vehicles become tightly locked together. The lorry is going at
BartSMP [9]

Answer:

The speed of the vehicles immediately after the collision is 5.84 m/s.

Explanation:

The speed of the vehicles after the collision can be found by conservation of linear momentum:

p_{i} = p_{f}

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Where:

m₁: is the mass of the car = 0.5 ton = 500 kg

m₂: is the mass of the lorry = 9.5 ton = 9500 kg

v_{1_{i}}: is the initial speed of the car = 40 km/h = 11.11 m/s

v_{2_{i}}: is the initial speed of the lorry = 20 km/h = 5.56 m/s

v_{1_{f}}: is the final speed of the car =?

v_{2_{f}}: is the final speed of the lorry =?    

Since the two vehicles become tightly locked together after the collision v_{1_{f}} = v_{2_{f}}:

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = v(m_{1} + m_{2})

v = \frac{m_{1}v_{1_{i}} + m_{2}v_{2_{i}}}{m_{1} + m_{2}} = \frac{500 kg*11.11 m/s + 9500 kg*5.56 m/s}{500 kg + 9500 kg} = 5.84 m/s

Therefore, the speed of the vehicles immediately after the collision is 5.84 m/s.

I hope it helps you!  

8 0
2 years ago
Water flows through a multisectional pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s
Stolb23 [73]

Answer:

P_1 - P_2 = 2295 pascal

Explanation:

As we know by Bernoulli's principle

P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2

P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2} \rho v_1^2

here we know that

\rho = 1000 kg/m^3

also we know that

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now we have

P_1 - P_2 = \frac{1}{2}(1000)(3.0^2 - 2.1^2)

P_1 - P_2 = 2295 pascal

5 0
3 years ago
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