A quarterback, Patrick, throws a football down the field in a long arching trajectory to wide receiver, Tyreek. The football and Tyreek are traveling in the same direction, started at the same spot, and the football was thrown at the same instant that Tyreek began running. Furthermore, both Tyreek and the football have horizontal components of speed of 22.6 mph. Under these circumstances, no matter what angle the football is thrown at, it will land on Tyreek (whether he catches it or not). True or false

Answer:

So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.

Explanation:

A option.

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm

The gravitational force exerted by the moon on the satellite is such that

F = G M m / R ² = m a   →   a = G M / R ²

where a is the satellite's centripetal acceleration, given by

a = v ² / R

The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that

v = 2πR / T   →   a = 4π ² R / T ²

Substitute this into the first equation and solve for T :

4π ² R / T ² = G M / R ²

4π ² R ³ = G M T ²

T ² = 4π ² R ³ / (G M )

T = √(4π ² R ³ / (G M ))

T = 2πR √(R / (G M ))

The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:

[tex] F = \frac{GmM}{R^{2}} [/tex]    (1)

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

R: is the distance between the satellite and the center of the moon

m: is the satellite's mass

M: is the moon's mass

The gravitational force is also equal to the centripetal force:

[tex] F = ma_{c} [/tex]   (2)

The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):

[tex] a_{c} = \frac{v^{2}}{R} [/tex]   (3)

And from the tangential velocity we can find the period:

[tex] v = \omega R = \frac{2\pi R}{T} [/tex]   (4)

Where:

ω: is the angular speed = 2π/T

By entering equations (4) and (3) into (2), we have:

[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex]   (5)

By equating (5) and (1), we get:

[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]

[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]  

[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]

[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]

Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

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