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Liula [17]
2 years ago
14

A footballer kicks a football of mass 430 g and it flies off at a speed of 8 m/s. What is the KE of the football?

Physics
1 answer:
klasskru [66]2 years ago
3 0

Answer:

13.76 Joules

Explanation:

The SI units for joules are defined to be kilogram*meters per second. Therefore, we must turn the grams to kilograms by dividing the answer by 1,000. This is due to 1,000 grams being in a single kilogram. Kinetic energy is defined to be:

KE=\frac{1}{2} mv^2

Therefore:

KE=\frac{1}{2} (0.43kg)(8m/s)^2=13.76J

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A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
3 years ago
What would happen to the moon if earth stopped exerting the force of gravity on it?
Mila [183]
There are two equal forces of gravity between the Earth and the Moon.
One force pulls the Moon toward the Earth.
The other force pulls the Earth toward the Moon.

If only this gravity suddenly switched off, then the moon would
continue to orbit the Sun, very much as it does now.

If ALL gravity suddenly switched off, then . . .

-- the Moon would stop orbiting the Earth and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- the Earth would stop orbiting the Sun and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- all the gases surrounding the Earth ... which we call "air" ... would
start drifting away, and expanding into a giant cloud of gas, and stop
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-- the Sun would completely fall apart, expand into a giant cloud of gas,
and stop being a star.
6 0
3 years ago
Read 2 more answers
A spring is originally 0.4m. When a 23N force is applied it stretches to a total of 0.7m. How far is the spring been extended? W
sergejj [24]

Recall the force applied to a spring and its extension/compression:

F = kΔx

Δx is the change in the spring's length (measured from its original length).

x = 0.7 m

x₀ = 0.4 m

Δx = x-x₀ = 0.7 m - 0.4 m = 0.3 m

F = 23 N


Substitute and solve for k:

23 = k×0.3

<h3>k = 7.67 N/m</h3>
8 0
3 years ago
Find the velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south.
mars1129 [50]
Velocity is 6.9 x 1.5 = 10.35 m/s due South
7 0
3 years ago
At a given moment, a plane passes directly above a radar station at an altitude of 6 km. Let θ be the angle that the line throug
Rom4ik [11]

- 187.237 km/hr fast is θ changing 12 min after the plane passes over the radar station

<u>Explanation:</u>

Let the distance x and angle θ be defined as in the figure below. Then

                  \tan \theta=\frac{6}{x}

Now, differentiate with respect to t, we get

                 \sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}

Now, calculate the travel distance from radar station to plane after 12min

Distance,  x=800 \times \frac{12}{60}=160

Substituting ‘x’ value, we get

                \tan \theta=\frac{6}{160}=\frac{3}{80}

Find the rate of change of theta after 12 min,

                \frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}

We know, the formula for,

                \sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}

So, then, \frac{d x}{d t}=800 \mathrm{km} / \mathrm{hr}(\text { let assume })

               \frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800

               =-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800

               =-\frac{1}{1.0014} \times \frac{6}{25600} \times 800

             =-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}

When express the value in km/he, we get, the change in theta as

              =-187.237 \mathrm{km} / \mathrm{hr}

5 0
3 years ago
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