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Ad libitum [116K]
3 years ago
6

two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignorin

g friction, how fast is the wagon moving 3.50 s later?
Physics
1 answer:
Dovator [93]3 years ago
7 0

Answer:

AFter 3.5 s, the wagon is moving at:   4.62\,\,\frac{m}{s}

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}

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What is the name of the invisible line that runs<br> down the center of the axial region?
wariber [46]

Answer:

An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.

Explanation:

The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.

3 0
3 years ago
A car starts from rest and experiences a constant acceleration of 5 m/s2 until it reaches a speed of 650 meters
Zigmanuir [339]

Answer: Time.

Explanation:

Hey there!!!

The reason is,

According to the question we have,

Initial velocity = 0m/s { as it starts from rest}.

Final velocity = 650 m/s

acceleration = 5m/s^2.

Looking on this formula,

a =  \frac{v - u}{t}

Where "v" represents final velocity, "u" represents initial velocity and "a" represents acceleration. We dont have is time "t".

So, it's obvious that tine is an unknown variable.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

8 0
3 years ago
Olivia is on a swing at the playground.
balu736 [363]

Olivia is on a swing at the playground. The kinetic energy is increasing and her potential energy decreasing at point X.

<h3>What is kinetic energy and potential energy?</h3>

The kinetic energy of an object is the ability to do work by virtue of its motion and potential energy is the ability to do work by virtue of its position.

At point W and Z, Olivia is at the maximum displacement from the mean position, where kinetic energy is zero and potential energy is maximum.

At point Y, it is approaching to increase its potential energy and decreasing kinetic energy. Opposite to this, at point X, kinetic energy is increasing and potential energy is decreasing.

Thus, the kinetic energy is increasing and her potential energy decreasing at point X.

Learn more about  kinetic energy and potential energy.

brainly.com/question/15764612

#SPJ1

5 0
3 years ago
1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the highe
Alex73 [517]

Answer:

\Delta S_u=2.1429\ J.K^{-1}

W_c=416.67\ J

Explanation:

Given:

temperature of source reservoir, T_H=600\ K

temperature of sink reservoir, T_L=350\ K

energy absorbed from the source, Q_{in}=1000\ J

work done, W=250\ J

a.

<u>Now change in entropy of the surrounding:</u>

\Delta S_u=\frac{dQ_L}{T_L}

<em>Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.</em>

\Delta S_u=\frac{Q_H-W}{T_L}

\Delta S_u=\frac{1000-250}{350}

\Delta S_u=2.1429\ J.K^{-1}

b.

<u>We know Carnot efficiency is given as:</u>

\eta_c=1-\frac{T_L}{T_H}

\eta_c=1-\frac{350}{600}

\eta_c=0.4167

<u>Now the Carnot work done:</u>

W_c=Q_H\times \eta_c

W_c=1000\times 0.4167

W_c=416.67\ J .......................(1)

c.

From eq. (1) we have the Carnot work, so the difference:

\Delta W=W_c-W

\Delta W=416.67-250

\Delta W=166.67\ J

Now, we find:

T_L.\Delta S_u=350\times 2.1429

5 0
3 years ago
TRUE OR FALSE : Boyle’s law states that, as the pressure of a gas increases, the volume decreases.
White raven [17]

True, but only if the temperature of the gas doesn't change ...
which is pretty hard to manage when you're compressing it.

I think Boyle's law actually says something like

       (pressure) x (volume) / (temperature)  =  constant.

So you can see that if you want to say anything about two of the
quantities, you always have to stipulate that the statement is true
as long as the third one doesn't change.


6 0
3 years ago
Read 2 more answers
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