2 years ago

Work done depends on

Physics

1 answer:

natima [27]2 years ago

3 0

Answer:

C. Both force and displacement

Explanation:

Hope this helps

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How is life determined on math

olga nikolaevna [1]

Answer:

you can use math as a banker, a doctor, a scientist, the president probably uses math, you use math to see how much less juice you gave your sibling, and you can use math to help in collage! (sorry if its wrong tell me if it is)

6 0

3 years ago

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A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

In-s [12.5K]

Vf=vi plus 2 ad
0=7.76 + 2(9.8)d
d=0.395m

4 0

3 years ago

A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that

irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

$h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)$

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

7 0

3 years ago

1. Na jaką wysokość wzniesie się piłka kopnięta pionowo w górę z prędkością 20 m/s ? 2. Piłka o masie 20 dag spada z wysokości 1

Tresset [83]

Answer:

its gonn abe 23

Explanation:

6 0

3 years ago

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One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

3 years ago