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3 years ago

What's the velocity for each trial? And what's the average velocity?

Physics

1 answer:

adelina 88 [10]3 years ago

5 0

Since each time trial is the same the average will be the direct answer, and the formula for velocity is distance divided by time, therefore it will come out to, 1.92307692. Whatever your teacher what the rounding process to be will vary but the straight up answer is there.

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec

borishaifa [10]

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

$\lambda = \frac{c}{f}$

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

$f = 2Hz$

$c = 3*10^8km/s$

Replacing we have,

$\lambda = \frac{c}{f}$

$\lambda = \frac{3*10^8km/s}{2Hz}$

$\lambda = 1.5*10^8m$

<em>Therefore the wavelength of this wave is $1.5*10^{8}m$ </em>

8 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

Which of these is an example of a mechanical wave?

Oduvanchick [21]

i think the answer is B police siren.

8 0

3 years ago

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If your car runs out of gas and you must push it 250 m to the nearest gas station. How much work is done on the car if it is pus

tankabanditka [31]

Answer:

1750 Joules.

Explanation:

Work done = force * distance

= 7 * 250

= 1750 Joules.

6 0

3 years ago

Which scientist began organizing elements into the periodic table?

Gre4nikov [31]

Answer:

B. Dmitri Mendeleev

Explanation:

he invented it

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3 years ago

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