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3 years ago

Wich type of solar radiation is the least powerful

Physics

1 answer:

ElenaW [278]3 years ago

6 0

X rays with the air in the atmosphere, solar radiation is the least powerful

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A student finds a resistor in an old transistor radio. When the resistor is attached to a 1.5 V cell, the flow of current is mea

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Answer:

yes

Explanation:

because no matter what voltage you connect it up to, the current always increases proportionally to the voltage.

this is because it follows ohm's law V = I X R.

(voltage = current X resistance)

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the value of the resistor is always 60 ohm's, no matter which combination of voltage and current you used and measured, so it follows ohm's law, therefore being an ohmic resistor

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3 years ago

Which part of earths core do convection currents occur

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I think is near the ozone layer, i am not sure

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3 years ago

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Gwar [14]

The third one.

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2 years ago

Jim runs off a diving board and lands in the water 3 meters from the end of the board. When he runs at the same speed on a highe

satela [25.4K]

Answer:

Explanation:

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4 years ago

A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass

saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

$W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)$ (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

$d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m$

Furthermore, the angle α is:

$\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°$

Then, you solve the equation (1) for W and replace the values of all parameters:

$W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J$

The work done by the student is 609.97J

6 0

4 years ago