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3 years ago

Shows an object suspended from two ropes. The weight of the object is 360 N. The magnitude of the tension

Physics

1 answer:

konstantin123 [22]3 years ago

6 0

Answer:

Tension T in each rope will be 254.56 N.

Explanation:

From the picture attached,

Weight suspended by the two ropes will be supported by the vertical components of the tension in the ropes.

Vertical components of both the ropes = Tsin(45)° + Tsin(45)°

= 2Tsin(45)°

= $2T(\frac{1}{\sqrt{2}})$

= $T\sqrt{2}$

Since, $T\sqrt{2}=360$

T = $\frac{360}{\sqrt{2} }$

T = 254.56 N

Therefore, tension T in each rope will be 254.56 N.

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Explanation:

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2 years ago

How much Power does a truck use to complete 4,100 J of Work in 30<br> seconds? *

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3 years ago

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Neporo4naja [7]

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3 years ago

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3 years ago

A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie

Scilla [17]

Answer:

Induced current, I = 0.5 A

Explanation:

It is given that,

number of turns, N = 20

Area of wire, $A=50\ cm^2=0.005\ m^2$

Initial magnetic field, $B_i=2\ T$

Final magnetic field, $B_f=6\ T$

Time taken, t = 2 s

Resistance of the coil, R = 0.4 ohms

We know that due to change in magnetic field and emf will be induced in the coil. Its formula is given by :

$\epsilon=\dfrac{-d\phi}{dt}$

Where

$\phi=BA$

$\epsilon=\dfrac{-d(NBA)}{dt}$

$\epsilon=NA\dfrac{B_f-B_i}{t}$

$\epsilon=20\times 0.005\times \dfrac{6-2}{2}$

$\epsilon=0.2\ V$

Let I is the induced current in the wire. It can be calculated using Ohm's law as :

$\epsilon=I\times R$

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.2}{0.4}$

I = 0.5 A

So, the magnitude of the induced current in the coil is 0.5 A. Hence, this is the required solution.

5 0

3 years ago