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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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a person with a mass of 75kg jumps on the trampoline the trampoline creates a fprce of 375n on them what is the acceleration of
Anarel [89]

Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:

            -4.8 m / s²

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

            F = m a

Where the bold letters indicate vectors, F is the force, m the masses and the acceleration

The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system

 

               F_t -W = m a

Whera F_t is the trampoline force

Body weight is

                W = mg

We substitute

              F_t - mg = ma

              a =\frac{F_t - m g}{m}

Let's calculate

              a = \frac{375 - 75 \ 9.8 }{75}

              a = -4.8 m / s²

The negative sign indicates that the acceleration is directed downward.

In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is

            -4.8 m / s²

Learn more here:  brainly.com/question/19860811

7 0
3 years ago
When the cart moves on an incline at constant speed, it is in equilibrium; i.e., the net force on it is zero. Does it require mo
Amanda [17]

Answer:

It requires more tension to pull up the track

Explanation:

Net force must be zero to maintain constant velocity.

Weight force will always be pointed down the slope. Call it W

Friction force (Call it Ff) will be down slope when movement is up slope.

Friction force will be up slope when movement is down slope.

W and Ff are always positive numbers

Call the pulling force T

If Up slope is considered the positive direction

Moving up slope

Tu - Ff - W = 0

Tu = Ff + W

Moving down slope

Td + W - Ff = 0

Td = Ff - W

Ff + W > Ff - W therefore Tu > Td

5 0
3 years ago
two small charged objects repel each other with a force 2.53 when separated by a distance 0.11. if the charge on each object is
Mekhanik [1.2K]

The force between the charges will be 10.9 N.

<h3>What is the electrostatic force?</h3>

The electrostatic attraction particles that make up the electrostatic force behave as both an attracting and a repulsive force. The force is inversely proportional to the square of the distance between them.

F ∝ 1 / r²

F = k / r²

Fr² = k

When two tiny charged objects are 0.11 of a distance apart, they repel one another with a force of 2.53. If the distance between the two particles is cut to 0.053 and the charge on each item is decreased to 67.9% of its initial amount. Then the force will be

F₁r₁² = F₂r₂²

2.53 × (0.11)² = F × (0.053)²

F = 10.9 N

More about the electrostatic force link is given below.

brainly.com/question/9774180

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3 0
1 year ago
Suppose that a simple pendulum consists of a small 81 g bob at the end of a cord of negligible mass. If the angle θ between the
MariettaO [177]

Answer:

(a) 0.115 m

(b) 2.08 x 10^-5 J

Explanation:

mass of bob, m = 81 g = 0.081 kg

The equation of oscillation is given by

θ = 0.068 Cos {9.2 t + Ф}

Now by comparison

The angular velocity

ω = 9.2 rad/s

(a) \omega^{2} =\frac{g}{L}

where, L be the length of the pendulum

L =\frac{g}{\omega ^{2}}

L =\frac{9.8}{9.2 \times 9.2}

L = 0.115 m

(b) A = L Sinθ

A = 0.115 x Sin 0.068

A = 7.8 x 10^-3 m

Maximum kinetic energy

K = 0.5 x mω²A²

K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6

K = 2.08 x 10^-5 J

8 0
3 years ago
10) Allen was curious about dog interaction at the dog park. He wondered if big dogs liked to play with big dogs and if little d
GalinKa [24]

Answer: B) hypothesis

Explanation:

8 0
3 years ago
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