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3 years ago

15

an athlete swims the distance from one end of a 50m pool to the other end in a time of 25 s what is The Athlete's average speed

1 answer:

vredina [299]3 years ago

3 0

Answer:

2.5 N

because Average speed is equal to distance divided by time

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A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to

Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV

V = W/q ....(1)

Using equation (1), the potential difference between points A and B is:

$V_{1}=\frac{W_{1} }{q}$

Substitute the suitable values in the above equation.

$V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }$

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

$V_{2}=\frac{W_{2} }{q}$

Substitute the suitable values in the above equation.

$V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }$

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0

3 years ago

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0

3 years ago

PLEASE HELP ME ASAP!!!!!<br> WHAT AM I SUPPOSED TO DO??

igomit [66]

Write what you can improve in your fitness test

3 0

3 years ago

Read 2 more answers

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

3 years ago

The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0o to

Allushta [10]

Answer:

0.572 Hz

Explanation:

given,

length of simple pendulum, l = 0.76 m

mass of the bob, m = 365 g = 0.365 Kg

angle made from the vertical, = 12°

frequency, f = ?

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{0.76}}$

$f = \dfrac{1}{2\pi}\times 3.59$

f = 0.572 Hz

The frequency at which pendulum vibrates is equal to 0.572 Hz

3 0

4 years ago