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Aloiza [94]
3 years ago
8

A soccer player applies a force of 48.4 N to a soccer ball while kicking it. If the ball has

Physics
2 answers:
Katarina [22]3 years ago
8 0
Yea C is the answer the first answer is right
AlekseyPX3 years ago
6 0

Answer:

C. 110 m/s2

Explanation:

Force = Mass x Acceleration

Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:

Force/Mass = (Mass x Acceleration)/Mass

Acceleration = Force/Mass

Now we just have to plug in our values and calculate!

Acceleration = 48.4/0.44

Acceleration = 110m/s/s

It is option C. 110 m/s2

Hope this helped!

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What force is required to move 7 M if the work done is 9 J
Murrr4er [49]

Answer:

1.29 N

Explanation:

The equation for force (with work and distance) is:

Force=\frac{Work}{distance}

We can plug in the given values into the equation:

Force=\frac{9J}{7m}$\approx1.29 N

5 0
2 years ago
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A point from which the position of other objects can be described is called what?
Alisiya [41]

Answer:

Reference Point

Explanation:

3 0
3 years ago
The conservation of momentum is most closely related to
BartSMP [9]

Answer:

<h2>B) Newton's 2nd law</h2>

Explanation:

<h2>From; force= mass × acceleration </h2><h2> f= m×a </h2><h2>where a(acceleration)= velocity/time</h2><h3> force = mv/t</h3><h3>But momentum(p) = Mass × velocity </h3><h2>hence force =p/t </h2><h3>that is Momentum = force × time ( Newton's 2nd law)</h3>
6 0
1 year ago
A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete
Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is 6.25\times 10^{- 3} ft/s^{2}

Solution:

As per the question:

Constant velocity of the horse in the horizontal, v_{x} = 1 ft/s

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

20^{2} + 10^{2} = l^{2}

l^{2}= 500

Now,

x^{2} + y^{2} = 500                            (1)

Differentiating equation (1) w.r.t 't':

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} = - y\frac{dy}{dt}

where

\frac{dx}{dt} = Rate of change of displacement along the horizontal

\frac{dy}{dt} = Rate of change of displacement along the vertical

v_{x} = velocity along the x-axis.

v_{y} = velocity along the y-axis

xv_{x} = -yv_{y}

v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s

|v_{y}| = 0.5\ ft/s

Acceleration of the hay bale is given by the kinematic equation:

v_{y}^{2} = u_{y} + 2ay

(-0.5)^{2} =0 + 2ay

0.25 = 2ay

\frac{0.25}{2y} = a

a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}

7 0
3 years ago
What is the fundamental frequency (in Hz) of a 0.632 m long tube, open at both ends, on a day when the speed of sound is 344 m/s
Greeley [361]

Answer:

f=272.15Hz

Explanation:

Given data

Length of tube L=0.632 m

Speed of sound v=344 m/s

To find

Fundamental frequency f

Solution

The fundamental frequency of the tube can be given as:

f=\frac{v}{2L}\\ f=\frac{344m/s}{2(0.632m)}\\ f=272.15Hz

4 0
3 years ago
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