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2 years ago

50 points please help ASAP! !

Physics

1 answer:

Contact [7]2 years ago

4 0

Answer:

The sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.

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What are the animal life like in the tundra

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The tundra is one of the coldest biomes so I would say that the animals have to be adapted. Maybe fur coats‍♀️they have to be warm and they also need food..I dunno I would think about how they live within the biome—(I added a picture of a tundra Biome)

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2 years ago

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A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m

kenny6666 [7]

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, $N - mg = F_c$

$N-mg=\frac{mv^2}{r}$

Now substitute the given values, we get

$N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}$

$N=15.68+46.08$

N = 61.76 N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

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2 years ago

Which equation below is correctly balanced?

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Answer:

D H2 + O2 ---> H2O

Explanation:

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Determine the magnetic flux through a square loop of side a if one side is parallel to, and a distance b from a straight wire th

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2 years ago

A 62.0 kgkg skier is moving at 6.90 m/sm/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch

Margarita [4]

Answer:

the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

the internal energy generated in crossing the rough patch is 820.26 J

Explanation:

Given that,

Mass, m = 62 kg,

Initial speed, = 6.90 m/s

Length of rough patch, L = 4.50 m,

coefficient of friction, = 0.3

Height of inclined plane, h = 2.50 m

$g=9.8 m/s^2$ is the acceleration due to gravity

According to energy conservation equation,

Part (b)

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

$F_f=\mu mg$

Therefore, the work done by friction is:

$W=-F_f d =-\mu mg d$

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

$W=-(0.300)(62.0)(9.8)(4.50)=-820.26 J$

So, the internal energy generated in crossing the rough patch is 820.26 J

Part (a)

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

$E=K_i +U_i = \frac{1}{2}mu^2 + mgh$

After crossing the rough patch, the new mechanical energy is

$E'=E+W$

where

W = -820.26 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

$E' = K_f = \frac{1}{2}mv^2$

where v is the final speed.

According to the law of conservation of energy, we can write:

$E+W=E'$

So we find v:

$\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\\\v=\sqrt{u^2+2gh+\frac{2W}{m}}\\\\=\sqrt{6.9^2+2(9.8)(2.50)+\frac{2(-820.26)}{62.0}}\\\\=8.38 m/s$

Thus, the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

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