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Svetach [21]
3 years ago
6

A football player running at two meters per second dives towards a football flying towards him with a velocity of five meters pe

r second. what is the relative velocity of the football player and the football?
a. 0 m/s
b. 2 m/s
c. 5 m/s
d. 7 m/s
Physics
2 answers:
Lady_Fox [76]3 years ago
5 0

Answer: option d.

Explanation:

In this situation, we can suppose that the football player has a velocity of 2m/s in the positive x-axis, and the ball has a velocity of -5m/s in the x-axis.

If we step in the point of view of the football player, all the previous x-axis now moves with the same velocity that the player had, but in the other direction, so all the axis now moves with a velocity of -2m/s. And the ball that also is in the axis also gets this "boost" in velocity.

So now the velocity of the ball is -5m/s -2m/s = -7m/s where the minus sign is because of the situation that I crafted.

So the correct option is the option D

Alexus [3.1K]3 years ago
3 0
I'm not positive but maybe d
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Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

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3 years ago
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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

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3 0
3 years ago
Carol drops a stone in a mine shaft 122.5 metres deep.How
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Answer:

 T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

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time taken  = ?

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using equation of motion

s = u t + \dfrac{1}{2}gt^2

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t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 122.5}{9.8}}

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time taken by the sound to travel

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    t = 0.36 s

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T = 5 + 0.36

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