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krek1111 [17]
3 years ago
14

Which statement is true for objects in dynamic equilibrium?

Physics
2 answers:
allsm [11]3 years ago
7 0

Answer:d

Explanation:

I think I've done this

balu736 [363]3 years ago
6 0

Answer:

A objects have zero acceleration

Explanation:

I did this on Plato and it was right

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Mrs. Smith can walk 1.4 m/s. If it takes her 8.5 seconds to get to the teacher lounge, how far is the teacher lounge from her ro
AlexFokin [52]

Answer:

6 meters away

Explanation:

6*1.4= 8.4 which is pretty close

5 0
3 years ago
Two similar fans are operating in a room. Fan 1 makes a squeaking noise while running. Fan 2 is silent.
REY [17]

Answer:

a

Explanation:

because it has more energy

8 0
3 years ago
Read 2 more answers
Bart ran 5000 meters from the cops and had an average speed of 6 meters/second before he got caught. How long did he run?
atroni [7]

Answer:

thats is a long time

Explanation:

maybe 1 hour

4 0
3 years ago
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How can an object overcome static friction?
larisa86 [58]

Answer:

Applying enough force in one direction to move the object, making kinetic energy.

Explanation:

Simpleness

4 0
3 years ago
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
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