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Sedaia [141]
3 years ago
6

Later in the game, the quarterback throws a pass to the wide receiver with a defender in hot pursuit. If the pass does not arriv

e to the wide receiver in two seconds, the pass will be intercepted. If the receiver is 30 yards away and the pass is thrown at a 10 degree angle from the ground, how fast must the ball be thrown to reach the receiver
Physics
1 answer:
nikdorinn [45]3 years ago
4 0

Answer:

Explanation:

In projectile motion , formula for range is as follows

R = u² sin 2 α / g , where u is initial velocity of throw , α is angle of throw

Given R , range = 30 yards , α = 10°

30 = u² sin 20 / 9.8

u² x .342 = 294

u² = 859.65

u = 29.32 m / s

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Your car burns gasoline as you drive up a large mountain. What energy transformation is the car performing?
Sonja [21]

Your car is performing a transformation of energy of:

Chemical energy to Mechanical energy

The chemical is the gasoline which is then converted to fire as the car runs thus creating the movement of the car which is mechanical energy.

7 0
3 years ago
Read 2 more answers
Find the equivalent resistance of this
MAXImum [283]

Explanation:

1/R = 1/R1 + 1/ R2

1/R = 1/960 + 1/640

1/R = 5 / 1920

<h3> R = 384 ohm </h3>

So , Req = 384 + R3

Req = 384 + 180

<h3> Req = 564 ohm</h3>

\huge\red{A}\pink{N}\orange{S}{W}\blue{E}\green{R}

<h2> Req = 564 ohm</h2>

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2 years ago
The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi
ahrayia [7]

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

R = \rho \frac{L}{A}

Where,

R= Resistance of the conductor

\rho =Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2

A_m = 9.274*10^{-3}m^2

Using the equation from Resistance we have that at the muscle the value is:

R_m = \rho \frac{L}{A}

R_m = \frac{13*(0.4)}{9.274*10^{-3}}

R_m = 560.70\Omega

At the same time we can make the same process to calculate the resistance of the fat, then

A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2

A_m = 2.0357*10^{-3}m^2

The resistance of the fat would be,

R_f = \rho \frac{L}{A}

R_f = \frac{25*(0.4)}{2.0357*10^{-3}}

R_f = 4912.3\Omega

Then the total resistance in a set as the previously writen, i.e, in parallel is:

R=\frac{R_mR_f}{R_m+R_f}

R = \frac{(560.70)(4912.3)}{4912.3+560.70}

R = 502.62\Omega

We can here apply Ohm's law, then

I= \frac{V}{R}

I = \frac{1.5}{502.62}

I = 2.984*10^{-3}A

I = 2.984mA

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