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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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a train leaves a station heading east at 50 mph from that same station a car drives north at a speed of 30mph after 3 hours how

Alecsey [184]

Answer:

d = 175 miles

Explanation:

Train is moving towards East with constant speed of 50 mph

While car is moving at speed of 30 mph

so after t = 3 hours

the distance moved by the train is given as

$d_1 = vt$

$d_1 = (50)(3) = 150 miles$

at the same time the distance moved by the car is given as

$d_2 = 30(3) = 90 miles$

now we know that both car and train is moving perpendicular to each other

So the distance between them after t = 3 hours is given as

$d= \sqrt{d_1^2 + d_2^2}$

$d = \sqrt{150^2 + 90^2}$

$d = 175 miles$

8 0

3 years ago

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

3 0

3 years ago

When you drop a paper clip, why doesn't it fall toward you instead of toward<br> Earth?

svetoff [14.1K]

Because gravity has been known to define as a force of attraction between things that have mass.

5 0

2 years ago

Describe how newtons third law is related to fluid pressure.

Hatshy [7]

Fish swimming forward in the water, the water gets pushed backward because the fish moving forward is forcing the water to move backward, the motion forward and backward are the same, they are opposite and equal.

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3 years ago

What do the law of superposition and the law of inclusion have in common? (1 point) 1. Both laws are about matching fossils in d

r-ruslan [8.4K]

Answer:

its b

i took the test

Explanation:

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3 years ago