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ivolga24 [154]
4 years ago
9

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

Physics
1 answer:
Llana [10]4 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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Raindrops are falling straight down at 11 m/s when suddenly the wind starts blowing horizontally at a brisk 5.0 m/s. From your p
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Explanation:

Vertical component of raindrop speed = 11m/s

Horizontal component of wind = 5m/s

In this case, all we have to do is to use trigonometric ratios of angles to sides as in a triangle

Doing this, we see that

tan (theta) = 5/11

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3 years ago
Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict
GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = m

Mass of the heavier cart = 2m

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ KE =  \frac{mv^2}{2}

⇒ KE =  \frac{mv^2}{2}\times \frac{m}{m}

⇒ KE =  \frac{m^2v^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(mv)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ K=\frac{(F\times t)^2}{2m}           ...equation (i)

KE (K1) of mass 2m.

⇒ K_1=\frac{(F\times t)^2}{2\times 2m}

⇒ K_1=\frac{(F\times t)^2}{4m}         ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ \frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}

⇒ \frac{K_1}{K} =\frac{2m}{4m}

⇒ \frac{K_1}{K} =\frac{2}{4}

⇒ \frac{K_1}{K} =\frac{1}{2}

⇒ K_1=\frac{K}{2}

⇒ K_1=\frac{1}{2}K

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0
3 years ago
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