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AleksandrR [38]
3 years ago
12

Please type a three paragraph essay (5-7 sentences per paragraph), and the essay should include the following things:

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
4 0
When I designed my vessel I knew inertia would cause the egg to continue to want to fall. I also knew that it would not accelerate at a great speed so I had to create a vessel that would contract and adjust easily. That's why I used a sponge and cotton to soften the impact of the egg. Then I simply held it together with a rubber band. With this construction I was able to build a vessel that safely carried my egg.

The stored energy in this contraption seemed to lie at the sponge. Were when I picked it up and threw it, I knew that the sponge was more dense then the egg so there was a lot of potential energy waiting to squash the sponge. Of course there was a lot of kinetic energy as I threw it into the air so I didn't throw it to high into the air otherwise the vessels acceleration falling down would have been a higher speed, resulting into a possible catastrophic event. So the mechanical energy was evenly distributed. I used Newton's first law of motion, inertia, to create a vessel that would save the egg from destruction due to inertia.

We use potential and kinetic knowledge to determine whether a car could drive up a hill or how fast a ball would roll down my driveway. Both of these ways represent physics and how we use this to solve equations. All though these questions probably don't get asked by kids. It's important to know that when designing contraptions like cars or even swimming flippers, you need to know how much kinetic or potential energy will be available.
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egoroff_w [7]

Answer:

what do you mean by that.

5 0
2 years ago
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11) A sled is initially given a shove up a frictionless 35º incline. It reaches a maximum height of 2.5
Andrej [43]

Answer:

7 m/s

Explanation:

To solve this problem you must use the conservation of energy.

K1 +U1=K2+U2

That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.

The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:

KE1= \frac{1}{2}mv^{2}}

PE2=mgh

K1=U2

\frac{1}{2} mv^{2} =mgh

Solve for v

v=\sqrt{2gh}

Input known values and you get 7 m/s.

5 0
3 years ago
A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet
densk [106]

Answer:

0.8726  m^3/s

Explanation:

We are to convert 1.85 x 10^3 ft^3/min to m^3/s

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x 10^3 ft3 = 1.85 x 10^3 x 0.0283 m3

     = 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x 10^3 ft^3/min = 52.355/60  m^3/s

        = 0.8726  m^3/s

7 0
3 years ago
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

5 0
3 years ago
A monochromatic light passes through a narrow slit and forms a diffraction pattern on a screen behind the slit. As the wavelengt
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