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1 year ago

Which statement accurately explains why the kinetic energy of the first rider is greater?

Physics

2 answers:

iogann1982 [59]1 year ago

6 0

Answer:

the 2nd option is the answer

Explanation:

Dvinal [7]1 year ago

5 0

Hey

The formula of kinetic energy is 1/2mv^2

So it depends on mass and velocity

As mass increases , kinetic energy increase .

So option b , the first rider had more mass is correct z

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Describe the factors that determine power in your own words (not from a website)

Bond [772]

As it was categorized in the physics category it must be the mechanical power.

Power is basically the amount of work done per unit time or in simpler words the amount of energy transferred per unit time.

The SI unit for power is Watt. Which is basically J/s or Nm/s as wor has the unit of Nm.

In electrical components power can also be categorized as the amount voltage supplied multiplied by the current passed through the component.

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2 years ago

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Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri

Scilla [17]

Answer:

(a) 7 m/s

(b) 931 rad/s

Explanation:

Gravity would be exerting on the 2 masses

$G_1 = Mg = 15.3*9.81 = 150.093N$

$G_2 = mg = 8.29*9.81= 81.32N$

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

$F = G_1 - G_2 = 68.77N$

This force would generate torque on the solid pulley

$T = FR = 68.77 * 0.0075 = 0.5158 Nm$

We can also calculate the pulley moments of inertia, with it being solid

$I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2$

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

$\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2$

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

$\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad$

(c)The time it takes to get to this distance is

$\theta = \frac{\alpha t^2}{2}$

$t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513$

$t = \sqrt{0.513} = 0.716s$

(b)The final angular speed of the disk is

$\omega = \alpha t = 1300.58*0.716 = 931 rad/s$

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

$v = \omega R = 931*0.0075 \approx 7m/s$

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2 years ago

A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the

WARRIOR [948]

Answer:

The head on Collison because you have both cars going (for the sake of it 30mph) and they both collide the energy from that is 60mph because the speed is combined with the two cars.

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2 years ago

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A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

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2 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0

3 years ago