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Alchen [17]
2 years ago
14

A family leaves home at 1:00pm and arrives at their vacation spot 200 miles away at 5:00pm on the same day. What is their averag

e speed?
Physics
1 answer:
Shkiper50 [21]2 years ago
4 0
The answer is 40 because you just have to do 200 divided by 5
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Why do water and table salt have very different properties?
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D)

Explanation:

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Studies have shown that driving after drug use was more common among drivers who were
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The force of kinetic friction between a box sliding on a surface depends on the ____ .
vazorg [7]

Answer: speed of the box

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A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the
Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

<u>Given</u>:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

           due to the movement in opposite direction)

           Change in time period = Δt = 1.50 s

<u>Required</u>:

           Change in angular velocity = Δω = ?

           Average angular acceleration = αav = ?

<u>Solution</u>:

          <u>Angular velocity (Δω):</u>

               Δω = ωf - ωi

               Δω = -2.70 - 2.70

               Δω = -5.4 rad/s.

          <u> Average angular acceleration (αav):</u>

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0
2 years ago
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3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number
Rina8888 [55]

ANSWER

\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

F=\frac{GMm}{r^2}

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\  \\ F=4.67*10^{-9}\text{ }N \end{gathered}

That is the answer.

6 0
1 year ago
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