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Strike441 [17]
1 year ago
5

A projectile is launched from ground level with an initial speed of 41.5 m/s at an angle of 31.3° above the horizontal. It strik

es a target in the air 1.63 s later.
What is the vertical distance from where the projectile was launched to where it hits the target?
Physics
1 answer:
Ahat [919]1 year ago
8 0

Answer:

6.26

Explanation:

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Which of the following is a chemical equation that accurately represents what happens when a sulfur and oxygen are produced from
Ksenya-84 [330]
It's c................
5 0
3 years ago
(1 pt) A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14
Elenna [48]

Answer:

w = 5832.372 Joules

Explanation:

Mass of water, m = 20 kg

The water was pulled up to a height of 35 meters, i.e. h = 35 m

It takes 14 minutes to pull up the water through the height, 35 m

speed = distance/ time = 35/14 = 2.5 m/min

The bucket's height, y = speed * time = 2.5t meters

6 kg of water drips out of the bucket throughout the 14 minutes

The rate at which the water drips drips out = (6/14) = 0.4286 kg/min

Mass of water that drips out in time, t = 0.4286t kg

The mass of water remaining = (20 - 0.4286t) kg

Change in Workdone, Δw = mgΔy

Δy = 2.5 Δt

Δw = mg *  2.5 Δt

dw =  (20 - 0.4286t)g2.5 dt

integrating both sides

dw = (50g - 1.07gt)dt

w = \int\limits^a_b {(50g-1.07gt)} \, dx where b = 0, a = 14

w = 50gt - 1.07g(t²)/2      g = 9.8 m/s²

w = 490t - 5.243t²

w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)

w = 6860 - 1027.628

w = 5832.372 Joules

3 0
3 years ago
Which of the following are required for scientific questions
icang [17]
The answer is D-Testable

Hope this helps
6 0
3 years ago
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
Ne4ueva [31]

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

                                 t = v / a

                                 t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

                                 t = v / g

                                 t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0
3 years ago
On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou
Vilka [71]
The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g.............> (T /2π)^2 = L/g

g = L/(T /2π)^2...........> g = 22.657 m/s^2
3 0
3 years ago
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