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steposvetlana [31]

3 years ago

12

A quarterback throws a football at 40.0 km/hr to a receiver 50.0 yd away. How much time does it take the football to reach the r

eceiver?

1 answer:

Fiesta28 [93]3 years ago

8 0

Convert the yards to km and devide your answer by 40 to get the amount of time it took to reach the receiver.
Good luck ! Hope it helped.

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The voltage in a circuit is given by the equation V = IR . In this equation , Vis the voltage, the current, and Ris the resistan

Mumz [18]

Answer:

The choice D.

$R = \frac{v}{I}$

I hope I helped you^_^

8 0

3 years ago

A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1

alexgriva [62]

Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

6.4% = vertical height / 12.42

vertical height = 6.4% * 12.42

vertical height = 0.8 miles

The distance travelled by the car (s) is:

s² = 0.8² + 12.42²

s² = 154.9

s = 12.45 miles

Acceleration (a) = 2.93 ft/s^2 = 0.00055 mile/s²

initial velocity (u) = 0, final velocity = 203 mph

Using:

s = ut + 0.5at²

12.45 = 0.5(0.00055)t²

t =213 s

5 0

3 years ago

A point charge of 9.00 × 10−9 C is located at the origin of a coordinate system. A positive charge of 3.00 × 10−9 C is brought i

dlinn [17]

A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x 10⁷ m.

<h3>What is electric potential energy?</h3>

The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.

The electric potential energy is given by the relation,

V = kQ/r

where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.

Substitute the values into the expression to get the distance between the charges.

8.09 × 10⁻⁷ = 9 x 10⁹ x 3 x 10⁻⁹ / r

r =3.34 x 10⁷ m

Thus, the distance between the charges will be 3.34 x 10⁷ m.

Learn more about electric potential energy.

brainly.com/question/12645463

#SPJ1

8 0

2 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

A boy is using a rope to pull a sled to the left on a horizontal surface. The rope is parallel to the surface. What are the dire

trasher [3.6K]

Answer:

Explanation:

Check attachment

3 0

4 years ago