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3 years ago

Anaya Goodloe decides to spike the football after a big touchdown (on Earth) and it

Physics

1 answer:

Llana [10]3 years ago

6 0

Answer:

The football leaves with the velocity, u = 15.68 m/s

Explanation:

Given data,

The football bounces back up off the ground and is airborne for, t = 3.2 s

Let the football bounces back up off the ground in the vertical direction

The formula for time of flight is given by,

t = 2u /g

∴ u = gt / 2

Substituting the values,

u = 9.8 x 3.2 / 2

u = 15.68 m/s

Hence, the football leaves with the velocity, u = 15.68 m/s

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When the car was stopped by the tree, its change in velocity during the collision was-6 meters/second. This change in

HACTEHA [7]

Answer: -3 meters/second^2.

Explanation:

3 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

3 years ago

A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no

ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, $\vec{F_{net}} = \vec{F'} - \vec{F}$

$\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N$

Thus the force will be along the direction of force whose magnitude is higher

Therefore,

$\vec{F_{net}} = 7 N$ towards North

4 0

2 years ago

Iii A load of 2.5 N is hung on the spring.

Anna11 [10]

N stands for Newton
2- upload the graph..

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2 years ago

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of t

Lelu [443]

Answer:

c) At a distance greater than r

Explanation:

If G= Gravitational constant

M= Mass of earth

r= distance from earth center

then orbital speed is ;

v = $\sqrt{\frac{GM}{r} }$

==> v²=GM/r

If speed of first satellite = V₁

==> V₁² = GM/r

==> r = GM/V₁²

If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.

So our answer is c

5 0

2 years ago