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3 years ago

6

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?
a. The tension in the rope is everywhere the same.
b. The magnitudes of the forces exerted on the two objects by the rope are the same.
c. The forces exerted on the two objects by the rope must be in opposite directions.
d. The forces exerted on the two objects by the rope must be in the direction of the rope.

1 answer:

julia-pushkina [17]3 years ago

5 0

Answer: They are all true

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

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

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

d. The forces exerted on the two objects by the rope must be in the direction of the rope.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)

- Cutiepatutie ☺❀❤

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konstantin123 [22]

Explanation:

The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.

The net resistance is:

R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)

R = 20Ω

Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:

V = IR

120 V = I (20Ω)

I = 6 A

So the voltage drops are:

V = (4Ω) (6A) = 24 V

V = (10Ω) (6A) = 60 V

That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:

V = 120 V − 24 V − 60 V

V = 36 V

So the currents are:

I = 36 V / 11 Ω = 3.27 A

I = 36 V / 22 Ω = 1.64 A

I = 36 V / 33 Ω = 1.09 A

If we wanted to, we could also show this using Kirchhoff's laws.

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3 years ago

b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?

umka2103 [35]

Answer:

Twice.

Explanation:

The momentum of an object is given by :

p = mv

Where

m is mass and v is the velocity

If the mass of the ball were doubled, m'=2m and v'=v=3 m/s

New momentum,

p'=m'v'

p'=2m × v

p'=2mv

or

p'=2p

So, the new momentum becomes twice the initial momentum.

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2 years ago

A go-kart and rider have a mass of 14 kg. If the cart accelerates at 6 m/s² during a 40 m sprint in 100 seconds, how much power

Nutka1998 [239]

Answer:

<em>P=33.6 \ W</em>

Explanation:

<u>Mechanical Work and Power</u>

Work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being $\vec F$ the force vector and $\vec s$ the displacement vector, the work is calculated as:

$W=\vec F\cdot \vec s$

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

$W=F.s$

Mechanical Power is the amount of energy transferred or converted per unit of time. The SI unit of power is the watt, equal to one joule per second.

The power can be calculated as:

$\displaystyle P=\frac {W}{t}$

Where W is the work and t is the time.

If an object of mass m has an acceleration a, the net force is:

F=m.a

The go-kart and rider have a mass of m=14 kg and accelerate at $6 m/s^2$ , thus the net force applied is:

$F=14\cdot 6 = 84\ N$

The work done by the cart when traveling d= 40 m is:

$W=84\cdot 40$

$W=3,360\ J$

Finally, the power for t= 100 seconds is:

$\displaystyle P=\frac {3,360}{100}$

P=33.6 \ W

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3 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

<u>Now the force due to second charge:</u>

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

<u>Now the force due to third charge:</u>

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

<u>Now the its horizontal component:</u>

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

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3 years ago

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andrezito [222]

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