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Scorpion4ik [409]
3 years ago
13

I need help finding the answer

Physics
1 answer:
miss Akunina [59]3 years ago
5 0

Speed = (distance) / (time)

Speed = (2.3 m) / (3 sec)

Speed = (2.3/3) (m/s)

<em>Speed = 0.766... m/s</em>

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Three horses are side-by-side on a merry-go-round: one at the edge, one near the axis, and one in between. Each horse has the sa
sineoko [7]

Answer:

one at the edge

Explanation:

The relation between the linear velocity and the angular velocity is given by

v = r x ω

Where, v be the linear velocity, ω be the angular velocity and r be the radius of the circular path.

As the angular velocity is constant, thus, the linear velocity depends on the radius of circular path.

So, the horse which is near to the edge has maximum radius of circular path in which it is rotating. So, the horse which is at the edge of the merry go round has maximum linear speed.

4 0
3 years ago
Describe the importance of the neutron in a atomic nuclei ​
RUDIKE [14]

Neutrons are required for the stability of nuclei, with the exception of the single-proton hydrogen nucleus. Neutrons are produced copiously in nuclear fission and fusion. They are a primary contributor to the nucleosynthesis of chemical elements within stars through fission, fusion, and neutron capture processes.

Hope it helps!

5 0
2 years ago
Read 2 more answers
2 decaliters + 800 deciliters = __________ liters?
Stolb23 [73]
Answer is 100 liters
8 0
3 years ago
Read 2 more answers
A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

0-0.9^2=2a(30*0.3)

Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0
3 years ago
You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2​
kumpel [21]

Answer:

6

Explanation:

120/30=6

8 0
2 years ago
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