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GrogVix [38]
3 years ago
11

5. Peter had an inflated balloon that he released and flies across the room. The balloon slows down and then stops on top of the

dinning table, As the balloon slows downs the force becomes
A.balanced
B.frictional
C.restricted
D.unbalanced​
Physics
2 answers:
Fiesta28 [93]3 years ago
7 0
Balanced as it reaches terminal velocity
Sergio039 [100]3 years ago
5 0
I think it is c because it was slowing down I don’t know for sure
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The balls in the image above have different masses and speeds. Rank them in terms of momentum, from least to greatest.
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The momentum of an object is equivalent to the product of the object's mass and velocity. Computing the momentum for each ball:
A- 15 * 0.7 = 10.5
B- 5.5 * 1.2 = 6.6
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D- 1.5 * 5.0 = 7.5

Therefore, ball C has the greatest momentum.
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Which of the following statements correctly describes the position of the intake and exhaust valves during most of the power sta
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Jennifer and Katie are leaning on each other. Jennifer weighs 150 and Katie weighs 120. Which one is pushing harder on the other
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A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta
larisa [96]

Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

y = 7.49 m

Part e)

x = 83.23 m

Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

Horizontal displacement in 1.03 s

x = v_x t

x = (15.3)(1.03)

x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

y = 7.94 m

Part c)

Horizontal displacement in 1.71 s

x = v_x t

x = (15.3)(1.71)

x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

Horizontal displacement in 5.44 s

x = v_x t

x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

6 0
2 years ago
An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o
ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0
2 years ago
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