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Anuta_ua [19.1K]
3 years ago
15

A football player kicks a ball with a mass of 0.5 kg. The average acceleration of the football was 15 m/s/s. How much force did

the kicker apply to the football?
Physics
1 answer:
Nikolay [14]3 years ago
8 0

Answer:

i think its 8.2

Explanation:

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One positive effect of recycling aluminum cans
GarryVolchara [31]

Your answer should be “1.”

MARK ME BRAINLIEST PLEASE!!!!!!!

3 0
4 years ago
Cars cross a certain point on the highway in accordance with a Poisson process with rate = 3 per minute. If Al runs across the h
Svetradugi [14.3K]

Answer:

The probability is 0.2212

Solution:

As per the question:

Poisson rate, \lambda = 3/min = \frac{3}{60}\ s = 0.05\ s

Now,

Let

X: time of waiting for the next vehicle on the highway

Now,

To find the probability of the next vehicle to arrive after 10 s

The probability distribution function is given by:

f(x) = \lambda e^{- \lambda x}

Now,

P(X < x) = 1 - e^{- \lambda x}

P(X \leq x) = 1 - e^{- 0.05 x}

For X> 0,

P(X > x) = e^{- \lambda x}

P(X < 5) = 1 - e^{- 5\lambda} = 1 - e^{- 0.25} = 1 - 0.7788 = 0.2212

4 0
3 years ago
At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is
Katyanochek1 [597]

Answer:

4.47\ \text{km/h}

Explanation:

\dfrac{da}{dt} = Rate at which the distance between A and starting point of B is changing = -20 km/h

\dfrac{db}{dt} = Rate at which the distance of B is changing = 15 km/h

\dfrac{dc}{dt} = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = 20\times 4=80\ \text{km}

a = Distance remaining to the start point of B = 110-80=30\ \text{km}

b = Distance covered by B in 4 hours = 15\times 4=60\ \text{km}

Distance between A and B after 4 hours

c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}

c^2=a^2+b^2

Differentiating with respect to time we get

c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}

The rate at which the distance between the ships is changing at 4 PM is 4.47\ \text{km/h}.

7 0
3 years ago
Why are different constellations<br> of stars seen during different<br> seasons?
slamgirl [31]
Actually, they're not.  There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around.  And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night. 

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ?  Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
5 0
4 years ago
Read 2 more answers
Your nerf dart gun is able to shoot a suction cup dart at a speed of 5 m/s. Your room is 3 meters across. You want to shoot your
postnew [5]
1,766 m according to my app. see foto.

4 0
4 years ago
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