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2 years ago

Times will the kinetic one

Physics

1 answer:

mixer [17]2 years ago

3 0

Answer:

The new kinetic energy is 9 times of the initial kinetic energy.

Explanation:

The kinetic energy of an object is given by the formula as follows :

$E=\dfrac{1}{2}mv^2$

Where

m is mass and v is speed of the body

If the speed of the body is tripled. Let v' is the new speed, v' = 3v. The new kinetic energy is E' and it is given by :

$E'=\dfrac{1}{2}mv'^2\\\\\text{Put v' = 3v}\\\\E'=\dfrac{1}{2}m\times (3v)^2\\\\=\dfrac{1}{2}m\times 9v^2\\\\=9\times (\dfrac{1}{2}mv^2)\\\\E'=9\times E$

Hence, the new kinetic energy is 9 times of the initial kinetic energy.

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Ket [755]

Answer:

1) D

2) A

Explanation:

1) Each group has the same number of valence electrons, which are the outer electrons.

2) Ionic bonds are between a metal and non - metal, the metal being sodium and the non - metal being chlorine.

3 0

3 years ago

The angle of incidence and the angle of reflection are measured from the_____, the perpendicular to the surface.

kogti [31]

I believe it is angles

6 0

2 years ago

In an engine governor, the two spheres (total mass of 1.0kg) are at 0.05m and rotating at 37rad/s If the engine increases the an

kirill115 [55]

Here we can say that there is no external torque on this system

So here we can say that angular momentum is conserved

so here we will have

$I_1\omega_1 = I_2\omega_2$

now we have

$I_1 = mr^2$

$I_1 = (1kg)(0.05^2)$

$I_1 = 25\times 10^{-4} kg m^2$

similarly let the final distance is "r"

so now we have

$I_2 = mr^2$

$I_2 = 1r^2$

now from above equation we have

$(25\times 10^{-4})37 = (r^2)(58)$

$r = 0.04 m$

so final distance is 0.04 m between them

8 0

2 years ago

A rescue pilot drops a survival kit while her plane is flying at an altitude of 1.5 km with a forward velocity of 100.0 m/s. If

tresset_1 [31]

1750 meters.
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:
d = 1/2 A T^2
where
d = distance
A = acceleration
T = time
Solving for T, gives
d = 1/2 A T^2
2d = A T^2
2d/A = T^2
sqrt(2d/A) = T
Substitute the known values and calculate.
sqrt(2d/A) = T
sqrt(2* 1500m / 9.8 m/s^2) = T
sqrt(3000m / 9.8 m/s^2) = T
sqrt(306.122449 s^2) = T
17.49635531 s = T
Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity.
17.50 s * 100 m/s = 1750 m

3 0

3 years ago

Read 2 more answers

There is a hydraulic system that by means of a 5 cm diameter plunger to which a 5 N force is applied and that force is transmitt

KengaRu [80]

Explanation:

Pressure is the same for both plungers.

P = P

F / A = F / A

F / (¼ π d²) = F / (¼ π d²)

F / d² = F / d²

5 N / (0.05 m)² = F / (1 m)²

F = 2000 N

None of the options are correct.

4 0

3 years ago