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2 years ago

3. A 1500 kg car moving at 30 m/s strikes a 6000 kg van initially at rest. If the car

Physics

1 answer:

MissTica2 years ago

3 0

Answer:

7.5m/s

Explanation:

Force= mass × velocity

Energy is conserved, the car and van should have the same overall force.

1500kg × 30m/s= 6000kg × final velocity

Final velocity = 7.5m/s

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If the mass of a 1.8 g paperclip was able to be completely converted to energy, how much energy would you obtain?

Anton [14]

Answer:

$E=1.62\times 10^{14}\ J$

Explanation:

Given that,

The mass of the paperclip, m = 1.8 g = 0.0018 kg

We need to find the energy obtained. The relation between mass and energy is given by :

$E=mc^2$

Where

c is the speed of light

So,

$E=0.0018\times (3\times 10^8)^2\\\\E=1.62\times 10^{14}\ J$

So, the energy obtained is $1.62\times 10^{14}\ J$ .

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2 years ago

Calculate the total resistance for a 650ohm , a 350 ohm , and a 1000 ohm resistor connected in series

Mekhanik [1.2K]

Answer:

2000 ohms

Explanation:

Resisters in series just add.

Rt = R1 + R2 + R3

R1 = 650 ohm

R2 = 350 ohm

R3 = 1000 ohm

Rt = 650 + 350 + 1000

Rt = 2000 ohms.

5 0

3 years ago

Read 2 more answers

What are the different ways that the simulation shows you that the equation is balanced, visually? For each balanced reaction, i

Nastasia [14]

Answer:

Explanation:

3 0

3 years ago

A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is

Allushta [10]

Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

$Re_{m} =\frac{p_{m}V_{m}L_{m} }{u_{m} }$

$Re_{p} =\frac{p_{p}V_{p}L_{p} }{u_{p} }$

Equaling both Reynold's number:

$\frac{p_{p}V_{p}L_{p} }{u_{p} }=\frac{p_{m}V_{m}L_{m} }{u_{m} }$

Clearing Vm:

$V_{m} =\frac{p_{p}V_{p}L_{p} u_{m} }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s$

b) The drag force is:

$\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2} } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2} } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2} } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2} } =10.32N$

6 0

3 years ago

Read 2 more answers

The force generated by a single muscle fiber can be increased by increasing is called

mars1129 [50]

Answer:

The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials

Explanation:

The force generated by a muscle fiber is the result of the shortening of the skeletal muscle, and this force is also know as muscle tension. The larger motor units shorten along with the smaller units to produce the muscle force. The time lapsed between the beginning of the action potential in the muscle and the beginning of the contraction is the latent period. Action potential is the result of the difference electrical potential as a result of passage of an impulse along the membrane of a muscle or nerve cell.

4 0

3 years ago