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Verizon [17]
3 years ago
13

Is this right? I need help, thank you

Physics
1 answer:
erastovalidia [21]3 years ago
7 0

Velocity is correct, the symbol for acceleration is a. That is all I know.

(As in f=m*a)

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Help please 20 points ? ASAP
KengaRu [80]

Answer:

21: Mass

22: Unbalanced Force

23: Friction

24: Balanced

Explanation:

3 0
3 years ago
An engineering firm is designing a ski lift. The wire rope needs to travel with a linear velocity of 2.0 meters per second, and
Mazyrski [523]

Answer:

The diameter of the bull-wheel is 3.82

Explanation:

Given that,

Velocity = 2.0 m/s

Angular velocity = 10 rev/m

\omega=10\times\dfrac{2\pi}{60}

\omega=1.0472\ rad/s

We need to calculate the diameter of bull-wheel

Using formula of angular velocity

v= r\omega

r=\dfrac{v}{\omega}

Put the value into the formula

r=\dfrac{2.0}{1.0472}

r=1.91\ m

The diameter of the bull-wheel

D=2r

D=2\times1.91

D=3.82\ m

Hence, The diameter of the bull-wheel is 3.82 m.

6 0
3 years ago
What radiation do remote controls use?
aleksandr82 [10.1K]

Answer:infrared radiation

Explanation:

Most remote control uses infrared radiation

5 0
3 years ago
How do you change matter into other phases of matter?
Vera_Pavlovna [14]
Hey there!

There's many ways to do it - like melting and evaporating.

For example, we'll use water. Plain old water in a water bottle. Right now, it's in its liquid state of matter, but say you put it in the freezer for an hour. That would change its state of matter to solid, since it would be solid ice. Now, if you were to put it out in the sun on a blazing hot day for a couple of hours, it would evaporate and become water vapor, a gas. Lastly, if you can cool that water vapor it becomes a liquid again.

Hope this helps!
3 0
3 years ago
When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di
AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

F=kx

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

k= \frac{F}{x}

k= \frac{mg}{x}

k= \frac{(0.35)(9.8)}{12*10^{-2}}

k = 28.58N/m

Perioricity in an elastic body is defined by

T = 2\pi \sqrt{\frac{m}{k}}

Where,

m = Mass

k = Spring constant

T = 2\pi \sqrt{\frac{0.35}{28.58}}

T = 0.685s

Therefore the period of the oscillations is 0.685s

4 0
3 years ago
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