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marusya05 [52]
3 years ago
6

A football player kicks a 0.94 kg football with a force of 2.4 N. Calculate the acceleration of the football as the player kicks

the ball. Round your answer to the nearest tenth of a m/s2.
Physics
2 answers:
Anon25 [30]3 years ago
7 0
Hello, I hope this helps :)
So the equation to figure these kinds of questions is F=MA
F refers to force, which in this situation F would be 2.4
M refers to mass, mass would be 0.94
A refers to acceleration, which we are trying to figure out
So if we put in the information we know into the equation, it is now 2.4=0.94*A
So with that equation we can figure that 2.4 divided by 0.94 equals A
So 2.4/0.94= 2.553191489362 :'D
Don't worry, the rounded and correct answer is 2.6
Have a nice day =)
dolphi86 [110]3 years ago
4 0

Answer:

2.6 just answered it on E2020

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A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
cupoosta [38]

Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0
3 years ago
a block (mass 0.6kg) is released from rest at point A at the top of a ramp inclined at 36.9 degress above the horizontal. The bl
bija089 [108]
14.136 J as shown on the photo with two thought processes but overall same calculation

3 0
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A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl
nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

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Why is it better to use the metric system, rather than the English system, in scientific measurement?
fiasKO [112]

A. The English system uses one unit for each category of measurement.

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Determine the amount of potential energy of a 5.0Kg book that is moved to three different shelves on a bookcase. The height of e
rjkz [21]
Formula to find gravitational potential energy:
mgh
m: mass
g: gravitational acceleration
h: height (relative to reference level)

so the P.E. at 1.0.m is (5x9.8x1)= 49J
P.E. at 1.5m is (5x9.8x1.5) =73.5J
P.E. at 2.0m is (5x9.8x2)=98J
8 0
3 years ago
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