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slega [8]
4 years ago
13

How can you tell high frequency on a wave

Physics
1 answer:
Svetradugi [14.3K]4 years ago
5 0
A wave travelling with high frequency has short wavelength and high pitched !
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A skateboarder is accelerating forward. How would his acceleration change if he had more mass ?
kupik [55]

Answer:

His acceleration would decrease

Explanation:

because there is more friction.

8 0
4 years ago
MA seconds class lever always more than 1​
Hoochie [10]
It is highly helpful to know that the mechanical advantage (M.A.) of Class two levers is usually greater than one. It is because the overall length of the Effort Arm is higher than the overall length of Load Arm. It is easily known by MA is effort arm/load arm.

!! ⬇️

The mechanical advantage of a lever of the second order is always greater than one because its effort arm is always longer than the load arm i.e. Effort arm > Load arm.

Second class lever has mechanical advantage always more than one as load is in between fulcrum and effort making the effort arm longer than the load arm.

First Class Lever -- the effort and the load on either side of the fulcrum. Some examples would be a crowbar or a seesaw. The effort is only less than the load if the load is closer to the fulcrum. The lever then acts as a force magnifier and the mechanical advantage is greater than one.
8 0
3 years ago
Earth has an __ of about 30 percent, reflecting much of the sun's radiation<br> back into space.
kotegsom [21]

Answer: albedo

Explanation: ?

5 0
3 years ago
What are the name of variables that do not change through out and experiment?
kifflom [539]

Answer:

The answer is C

Explanation:

3 0
3 years ago
Read 2 more answers
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is
padilas [110]

Answer:

P_{C} = 3.2\, atm

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T

Bulb B (3 L, 4 atm) - Before opening:

P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T

Bulbs A & B (5 L) - After opening:

P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T

After some algebraic manipulation, a formula for final pressure is derived:

P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}

And final pressure is obtained:

P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}

P_{C} = 3.2\, atm

5 0
3 years ago
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