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Mademuasel [1]
3 years ago
15

A girl runs 40m due south in 40 seconds an d then 20m due north in 10 secondsCalculate.

Physics
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

Average speed = 1.2 m/s

Average velocity = 0.4 m/s

Explanation:

Average speed = total distance/total time

Average speed = (40 + 20)/(40 + 10)

Average speed = 60/50

Average speed = 1.2 m/s

Average velocity = displacement/time

Now, she ran 40 m south and ran 20 m back north which is in the direction of where she began the journey.

Thus;

Displacement = 40 - 20 = 20 m

Average velocity = 20/50 = 0.4 m/s

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during a dodge ball game a student throw a ball at another player what forces act on the ball as it flie through the air
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Gravity acts to accelerate the ball downward, and air resistance acts in a way to slow the ball along it's instantaneous velocity (no matter which way it's moving air applies a force in the opposite direction)
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3 years ago
A sample of 5.6 L of a gas is at a pressure of 1.5 atm. If the volume of the gas is compressed to 4.8 L, what will the new press
Effectus [21]

Answer:

1.75atm

Explanation:

According to Boyle's law, the pressure P of a fixed mass of gas is inversely proportional to it's volume V provided that the temperature remains constant.

P\alpha \frac{1}{V}\\hence\\PV=constant

This implies the following;

P_1V_1=P_2V_2=...=P_nV_n............(1) Provided temperature is kept constant.

Given;

P_1=1.5atm\\V_1=5.6L\\P_2=?\\V_2=4.8L

From equation (1), we can write;

P_1V_1=P_2V_2\\hence\\1.5*5.6=P_2*4.8\\\\P_2=\frac{1.5*5.6}{4.8}\\\\P_2=1.75atm

Since all the units are consistent, there is no need for conversion.

3 0
3 years ago
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"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a
My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0
3 years ago
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