All categories

3 years ago

A peacock is flying around and its velocity

Physics

1 answer:

rodikova [14]3 years ago

4 0

Answer:

2 m.

Explanation:

From the question given, the following data were obtained:

Initial velocity (u) = 2 m/s

Initial time = 2 s

Final velocity (v) = 4 m/s

Final time (t) = 3 s

Next, we shall determine the change in velocity. This can be obtained as follow:

Initial velocity (u) = 2 m/s

Final velocity (v) = 4 m/s

Change in velocity (Δv) =?

Change in velocity (Δv) = v – u

Change in velocity (Δv) = 4 – 2

Change in velocity (Δv) = 2 m/s

Next, we shall determine the change in time. This can be obtained as follow:

Initial time (t1) = 2 s

Final time (t2) = 3 s

Change in time (Δt) =..?

Change in time (Δt) = t2 – t1

Change in time (Δt) = 3 – 2

Change in time (Δt) = 1 s

Finally, we shall determine the displacement (Δx) of the peacock as follow:

Velocity (Δv) = 2 m/s

Time ((Δt) = 1 s

Displacement (Δx) =?

Velocity = Displacement /Time

ΔV = Δx/Δt

2 = Δx /1

2 = Δx

Δx = 2 m

Therefore, the displacement of the peacock is 2 m

You might be interested in

Where might water be found on the moon

Cerrena [4.2K]

Water Could be Found in the frozen Ice on the moon. it would most likely be underground however

8 0

3 years ago

Read 2 more answers

Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en

Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0

3 years ago

A 1000 kg racecar, which is capable of a top speed of 125 m/s, is sitting in a

k0ka [10]

Sitting = no movement
KE=0

5 0

3 years ago

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart

Ber [7]

Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
$\sum F = ma$
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force $F_f$ acting in the opposite direction. So Newton's second law can be rewritten as
$F-F_a = ma$
or
$F=ma + F_f$

since the frictional force is 15 N and we want to achieve an acceleration of $a=1.50 m/s^2$ , we can substitute these values to find what is the force the man needs:
$F=(30 kg)(1.5 m/s^2)+15 N=60 N$

8 0

3 years ago

Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we

Scilla [17]

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>

Mass 1 = m₁
Mass 2 = m₂
Gravitational constant = G
Distance apart = r
Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>

Mass 1 = 2m₁
Mass 2 = 3m₂
Gravitational constant = G
Distance apart (r) = r
New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

6 0

2 years ago