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2 years ago

Pleaaaase help meeee

Physics

1 answer:

Lana71 [14]2 years ago

8 0

Answer:

Explanation:

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Hey I need help do animals take in oxygen and give off carbon dioxide or Plants take in oxygen and give off carbon dioxide or pl

Sophie [7]

Answer:

Animals take in oxygen and give off carbon dioxide.

Explanation:

This is because plants take in our carbon dioxide and give off the oxygen that we as people and animals need to breathe.

3 0

2 years ago

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A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer

777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

$K = \frac{1}{2}mv^{2}$

By substituting the values

$1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}$

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, $v = \left (\frac{4.9}{1000} \right )\times \left ( \frac{3600}{1} \right )$

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

4 0

2 years ago

The term necrosis means

aksik [14]

Necrosis - the death of most or all of the cells in an organ or tissue due to disease, injury, or failure of the blood supply.

Hope this helps, please mark me brainliest

6 0

2 years ago

The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it

soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

$U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})$

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, $v_a$ as 0, 9.81 for g then

$58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s$

7 0

2 years ago

You pull a block of mass m across across a frictionless table with a constant force. you also pull with an equal constant force

KiRa [710]

For any mass m:

a = F/m
v = √2*F/m*s = √2F/sm = k/√m
Momentum = mv = k√m
Energy = 1/ mv² = 1/2 m.k²/m = 1/2k²

SO
Both will have same energy
The larger mass will have greater momentum

4 0

3 years ago