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natka813 [3]
3 years ago
5

A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 26.6 s. C

alculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.) HINT
Physics
1 answer:
Advocard [28]3 years ago
8 0

Answer:

a) Average Speed = 190yards/26.6s = 7.14 yards/s

b) Average Velocity = 10/26.6 = 0.38 yard/s

Complete Question;

A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 26.6 s. Calculate his average speed and the magnitude of his average velocity. (Assume a 100 yard football field; note that the player's X-yard line is X yards from his own goal line. Enter your answers in yards/s.)

Explanation:

a) Speed = distance/time

time t = 26.6s

distance;

d1 = 100 yards (own goal line to the opposing team's goal line)

d2 = 90 yards (returning to his ten-yard line)

d = d1 + d2

d = 100 + 90 = 190 yards

Speed = 190yards/26.6s = 7.14 yards/s

b) velocity = displacement/time

time = 26.6s

displacement = 100-90 yards = 10yards

Velocity = 10/26.6 = 0.38 yard/s

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4 0
3 years ago
For how long should a force of 50,0 N be applied to change the momentum of an object
patriot [66]

Answer:

2,4 second

Explanation:

Change the momentum = F.t

12 = 50 .t

12/50 = t

2,4

6 0
2 years ago
A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8
Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planetm =6.75\times10^{24}\ kg

Mass of star m'=2.75\times10^{29}\ kg

Radius of the orbitr =8.05\times10^{7}\ km

Using centripetal and gravitational force

The centripetal force is given by

F = \dfrac{mv^2}{r}

F=m\omega^2r

We know that,

\omega=\dfrac{2\pi}{T}

F=m(\dfrac{2\pi}{T})^2r....(I)

The gravitational force is given by

F = \dfrac{mm'G}{r^2}....(II)

From equation (I) and (II)

m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

T^2=\dfrac{4\pi^2R^3}{m'G}

T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}

T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}

T =3.34\times10^{7}\ s

T= 387.62\days

Hence, The orbital period of the planet is 387.62 days.

4 0
3 years ago
Read 2 more answers
1 A boy kicks a ball with 40m/s at an angle of 30° with the ground. find the range the ball travels after 6 s.
Nat2105 [25]

Answer:

208m

Explanation:

since the angle is with the ground which gives a horizontal component to the velocity = v(cos)30°

so the velocity equals 40(cos)30°= 34.64m/s

range(distance) = speed × time

34.64m/s × 6s = 207.8m = 208m

5 0
2 years ago
A ball is dropped from a roof of a building and strikes the ground in 3 seconds. If a second ball is thrown horizontally from th
Scrat [10]

it will hit the ground maybe in like 1 second I guess

4 0
3 years ago
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