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Naily [24]
3 years ago
15

Kiley, Jermaine, and Gunther are playing tug-of-war. Kiley and Jermaine are on the same side of the rope, and Gunther is on the

other side. Kiley is exerting a force of 20 N, and Jermaine is exerting a force of 25 N.
Which force would result in a balanced force for the game?
Physics
2 answers:
Ivanshal [37]3 years ago
8 0

-45N because you add the side together and make it equal 0

Stolb23 [73]3 years ago
7 0

Kiley and Germaine pull together, resulting in a force of 45N which pulls the rope that way ===> .

In order to balance the forces on the rope, Gunther must pull the rope with a force of <em>45N this way  <===</em>, all by himself.

The net force on the rope will then be zero, and the rope will not accelerate.  (Although the tension inside the rope will be 90N.)

By the way . . . I'd love to see how the kids manage to all get on the "sides" of the rope.  I would have expected them to just do it the easy way and stand at the "ends" of the rope.

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
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Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

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v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

          = 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

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