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Dmitry [639]
2 years ago
12

A potato gun is fired horizontally from a height of 1.5 meters with the potato launched at 25 m/s What is the time of flight of

the potato?​
Physics
1 answer:
CaHeK987 [17]2 years ago
3 0

Answer:

0.55 s

Explanation:

From the question given above, the following data were obtained:

Height (h) = 1.5 m

Horizontal velocity (v) = 25 m/s

Time of flight (t) =?

The time of flight of the potato talks about the total time spent by the potato in the air i.e the time taken for the potato to get to the ground..

Thus, we can obtain the time of flight of the potato as illustrated below:

Height (h) = 1.5 m

Acceleration due to gravity (g) = 9.8 m/s²

Time of flight (t) =?

H = ½gt²

1.5 = ½ × 9.8 × t²

1.5 = 4.9 × t²

Divide both side by 4.9

t² = 1.5 / 4.9

Take the square root of both side

t = √(1.5 / 4.9)

t = 0.55 s

Thus, the time of flight is 0.55 s

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Mercury is in the 80th position in the periodic table. How many protons does it have?
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4 0
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A student carried out an experiment adding different weights to a spring and recording the results. Look at the table of results
MAXImum [283]

Answer:

0.25 m.

Explanation:

We'll begin by calculating the spring constant of the spring.

From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:

Force (F) = 0.1 N

Extention (e) = 0.125 m

Spring constant (K) =?

F = Ke

0.1 = K x 0.125

Divide both side by 0.125

K = 0.1/0.125

K = 0.8 N/m

Therefore, the force constant, K of spring is 0.8 N/m

Now, we can obtain the number in gap 1 in the diagram above as follow:

Force (F) = 0.2 N

Spring constant (K) = 0.8 N/m

Extention (e) =..?

F = Ke

0.2 = 0.8 x e

Divide both side by 0.8

e = 0.2/0.8

e = 0.25 m

Therefore, the number that will complete gap 1is 0.25 m.

5 0
3 years ago
Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of
Aneli [31]

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0
3 years ago
According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s
Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

\tau_{net} = 0

now we have

\frac{dL}{dt}= 0

now we also know that

Area = \frac{1}{2}r^2d\theta

so rate of change in area is given as

\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}

so we will have

\frac{dA}{dt} = \frac{1}{2}r^2\omega

\frac{dA}{dt} = \frac{L}{2m}

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0
3 years ago
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