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skelet666 [1.2K]
2 years ago
10

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

ht line. For this situation, which of the following statements are true? Check all that apply. 1. The tension in the rope is everywhere the same. 2. The magnitudes of the forces exerted on the two objects by the rope are the same.3. The forces exerted on the two objects by the rope must be in opposite directions.4. The forces exerted on the two objects by the rope must be in the direction of the rope.
Physics
1 answer:
Virty [35]2 years ago
6 0

Answer:

1. The tension in the rope is everywhere the same.

2. The magnitudes of the forces exerted on the two objects by the rope are the same.

3. The forces exerted on the two objects by the rope must be in opposite directions.

Explanation:

"Massless ropes" do not have a<em> "net force"</em> which means that it is able to transmit the force from one end of the rope to the other end, perfectly. It is known for its property of having a total force of zero. In order to attain this property, the magnitude of the forces exerted on the two stationary objects by the rope are the same and in opposite direction. <u>So this explains number 2 & 3 answers.</u>

Since the objects that are held by the rope are stationary, then this means that the tension in the rope is also stationary. This means that the tension in the rope everywhere is the same (provided that the rope is still or in a straight line, as stated in the situation above, and is being held by two points). <u>So, this explains number 1.</u>

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Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

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2 years ago
If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is th
cluponka [151]

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

7 0
2 years ago
5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by
Anna007 [38]

Answer:

a) W=2.425kJ

b) \Delta E=2.425kJ

c) T_f=20.06^{o}C

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

W=Fd

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

W=mgh

so now e can substitute:

W=(25kg)(9.7 m/s^{2})(10m)

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

\Delta E=2.425kJ

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

\Delta Q=mC_{p}(T_{f}-T_{0})

Where:

Q= heat transferred

m=mass

C_{p}=specific heat

T_{f}= Final temperature.

T_{0}= initial temperature.

So we can solve the forula for the final temperature so we get:

T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}

So now we can substitute the data we know:

T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C

Which yields:

T_{f}=20.06^{o}C

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

\Delta Q=-2.425kJ

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2 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
1 year ago
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