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Nimfa-mama [501]
3 years ago
11

The small ball of mass m and its supporting wire become a simple pendulum when the horizontal cord is severed. Determine the rat

io k of the tension T in the supporting wire immediately after the cord is cut to that in the wire before the cord is cut.

Physics
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

See the attached image and the explanation below

Explanation:

We must draw a schematic of the described problem, after the sketch it is necessary to make a free body diagram, at the time before and after cutting the cord.

These free body diagrams can be seen in the attached image.

First we perform a sum of forces on the x & y axes before cutting the cord, to be able to find the T tension of the wire. (This analysis can be seen in the attached image).

In this way we get the T-wire tension equation, before cutting.

Now we make another free body diagram, for the moment when the wire is cut (see in the attached diagram).

It is important to clarify that when the cord is cut, the system will no longer be in statically, therefore newton's second law will be used for summation of forces which will be equal to the product of mass by acceleration.

Finally with equations 1 and 2 we can find the K ratio.

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Dillon wants to go to the park to play basketball. To get to the park, he must travel 9 km East, and then 6 km South, then 9 Km
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Answer:

24 km

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3 years ago
In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
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Answer:

η = 40 %  

Explanation:

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Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

\eta=\dfrac{W(net)}{Q(heat\ addition)}

Now by putting the values in the above equation ,then we get

\eta=\dfrac{400}{1000}

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
If you can run at a speed of 8 miles per hour and you want to run to the store 16 miles away, how
MArishka [77]

Answer:

16÷8=2

Explanation:

if you run 8 mi an hour than in 16 mi you would have run 2 hours

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