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2 years ago

A 7.30 kg sign hangs from two wires. The

Physics

1 answer:

Alex2 years ago

5 0

Answer:

$28.0\; {\rm N}$ to the right.

Explanation:

Since the sign is not moving, the net force on this sign should be $0\; {\rm N}$ . For that, the horizontal component ( $x$ -component) of external forces on this sign should be $0\; {\rm N}$ .

Sources of external forces on this sign include tension from the wires, as well as gravitational pull (weight) from the earth. The gravitational pull from the earth is entirely vertical ( $y$ -component,) with a magnitude of $0\; {\rm N}$ in the horizontal direction. Thus, the only external forces on this sign in the $x$ -component would be from the two wires.

The question states that the $x$ -component of the force from the first wire is $28.0\; {\rm N}$ to the left. Thus, for the net force in the $x$ -direction to be $0\; {\rm N}$ , the force from the other wire in the $x\!$ -component needs to be $28.0\; {\rm N}\!$ to the right (same magnitude but opposite direction.)

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A sprinter runs 50m. You are given a stopwatch. Describe how you would measure: 1. the average speed of the sprinter over the en

Oksanka [162]

Answer:

Explanation:

Distance = 50 m

1. To find the average speed, first start the stop watch as the sprinter starts running and then stop it when he reaches the finish line.

Now note the time taken by the sprinter to run for 50 m.

The average speed of the sprinter is defined as the ratio of total distance covered to the total time taken.

Average speed = total distance / total time

2. To find the instantaneous speed, check the seed of the sprinter as he is at the finish line.

6 0

3 years ago

When bridges are built, special joints must be used because the material of the bridge shrinks, and without these joints, the ma

egoroff_w [7]

Answer:

Option D.

Explanation:

The correct answer is Option D.

The shrinkage of the bridge material is because of thermal contraction.

Thermal contraction and thermal expansion are the phenomena of the bridge material which takes place due to the change in temperature of the atmosphere.

When the temperature of the surrounding increases expansion of the bridge material takes place and when temperature decreases the contraction of the material takes place.

This phenomenon sometimes damages the structure because due to continuous expansion and contraction of materials strength of the bridge decreases.

8 0

4 years ago

3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the

andreev551 [17]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0

3 years ago

A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the obje

iogann1982 [59]

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now we have

$I = mk^2$

here k = radius of gyration of object

also for pure rolling we have

$v = R\omega$

so now we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})$

$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$

$v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}$

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

$mk^2 = mR^2$

k = R

For spherical shell

$mk^2 = \frac{2}{3}mR^2$

$k = \sqrt{\frac{2}{3}} R$

For solid sphere

$mk^2 = \frac{2}{5}mR^2$

$k = \sqrt{\frac{2}{5}} R$

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

7 0

3 years ago

Hey, can someone help me with question 19.

Anit [1.1K]

HI!

Transformers are devices used in electrical circuits to change the voltage of electricity flowing in the circuit.

Meter are typically calibrated in billing units, the most common one being the kilowatt hour (kWh)

Distribution Panel (also known as panel board, breaker panel, or electric panel) is a component of an electricity supply system that divides an electrical power feed into subsidiary circuits, while providing a protective fuse or circuit breaker for each circuit in a common enclosure.

I hope this helped !

3 0

3 years ago