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leonid [27]
3 years ago
9

A bus carrying 10 people has over turned on a remote hillside during an intense thunderstorm. What three factors could contribut

e to creating a delay in advanced care
Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:

1. <u>Lack of communication:</u> Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.

2. <u>Steep Nature of the Hill:</u> This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.

3. <u>Thunderstorm:</u> This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.

Explanation:

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3 years ago
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When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th
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Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

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3 years ago
The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t
ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

r=\frac{mv}{qB}         (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

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