A football player runs from his own goal line to the opposing team's goal line, returning to his forty-yard line, all in 22.4 s.
1 answer:
I assume L=120 yards as the length of the football field.
1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
And the time taken is t=22.4 s, so the average speed of the player is
2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
and so, the average velocity is
1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
And the time taken is t=22.4 s, so the average speed of the player is
2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
and so, the average velocity is
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Answer:
The resistance is 0.124 ohm.
Explanation:
It is common for domestic electrical installations to use copper wire with a diameter of 2.05 mm. Determine the resistance of such a wire with a length of 24.0 m.
diameter, d = 2.05 mm
radius, r = 1.025 mm
Length, L = 24 m
resistivity of copper = 1.7 x 10^-8 ohm m
Let the resistance is R.
Answer:
<h2>206.67N</h2>Explanation:
The sum of force along both components x and y is expressed as;
The magnitude of the net force which is also known as the resultant will be expressed as
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
Similarly,
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N