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3 years ago

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A spring that is stretched by hanging a 5kg mass. It's equilibrium length was 0.5 meters. Now the length of the spring is 1.6 me

ters. What is the spring constant?

1 answer:

gtnhenbr [62]3 years ago

5 0

45 N/m as attached, please comment

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A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

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Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

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3 years ago

The sum of two or more vectors is called a<br> vector.<br> Answer kere

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Answer:

Resultant vector is the sum of two or more vectors

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3 years ago

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

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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!

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