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Ostrovityanka [42]

2 years ago

8

Which of the following choices best represents a potential stressor?

1 answer:

Julli [10]2 years ago

6 0

an unexpected visitor as it begins making you feel uncomfortable but it's get calm

You might be interested in

When a car is coasting downhill, how are its potential and kinetic energies changing?

olasank [31]

When a car is coasting downhill, the kinetic and potential energies are increasing and decreasing respectively.

<h3>What are kinetic and potential energy?</h3>

Kinetic energy is the energy possessed by an object because of its motion, equal to one half the mass of the body times the square of its speed.

Potential energy, on the other hand, is the energy possessed by an object because of its position (in a gravitational or electric field), or its condition (as a stretched or compressed spring, as a chemical reactant, or by having rest mass).

According to this question, a car going downhill will begin to speed because there is lesser friction. This suggests that the kinetic energy increases while the potential energy decreases.

Learn more about potential energy at: brainly.com/question/24284560

#SPJ1

3 0

2 years ago

How long will it take this wave to travel 3000 m in the x-direction?

REY [17]

2.4 secs I think

?????????????????????????

5 0

3 years ago

The pic says everything, thank you !

Alenkasestr [34]

Answer:

1/5 or .2

Explanation:

you put miles over seconds 3/15 then simplify to 1/5 to get miles per second

4 0

3 years ago

0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes t

Musya8 [376]

Answer:

The magnitud of the velocity is

$8.46m/s$

and the direccion:

$-28.3$ degrees from the horizontal.

Explanation:

Fist we define our variables:

$m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?$

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

$m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}$

Clearing for the velocity of the stone after the crash:

$v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}$

Substituting known values:

$v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)$

The magnitud of the velocity is :

$|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s$

and the direction:

$tan^{-1}(-8.75/16.25)=-28.3$

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.

3 0

3 years ago

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.600 rev

natulia [17]

Answer:

The final velocity $\omega_f = 0.4235 \frac{rev}{s}$

Explanation:

Given data

Mass of merry go round $M_m$ = 120 kg

Radius = 1.8 m

Initial angular velocity $\omega_i$ = 0.6 $\frac{rev}{sec}$

Mass of boy $M_{boy}$ = 25 kg

We know that the final velocity is given by

$\omega_f = \frac{\frac{1}{2}M_m \omega_i }{M_{boy} + \frac{1}{2} M_m }$

Put all the values in above formula we get

$\omega_f = \frac{\frac{1}{2}(120) 0.6}{25 + \frac{1}{2} (120) }$

$\omega_f = 0.4235 \frac{rev}{s}$

This is the final velocity.

3 0

3 years ago