Ask question

Ask question

All categories

2 years ago

7

Two rods are made of brass and have the same length. The cross section of one of the rods is circular, with a diameter of 2a. Th

e other rod has a square cross section, where each side of the square is a length 2a. One end of the rods is attached to an immovable fixture which allows the rods to hang vertically. To the free end of each rod, a block of mass m is attached. Which rod, if either, will stretch more after the block is attached?
A. The one with the circular cross section will stretch more.

B. The one with the square cross section will stretch more.

C. Both will stretch by the same amount.

D. One cannot say which will stretch more without knowing the numerical values of a and m.

2 answers:

Ivanshal [37]2 years ago

8 0

Answer:

A. The one with the circular cross section will stretch more.

Explanation:

According to the given data:

Two rods are made of brass and have the same length

Both rods having circular and square cross-section

Diameter of circular cross-section given is 2 a

therefore, Cross-section = $A_c=\frac{\pi (2a)^2}{4}=\pi a^2$

If the length of square=2 a

then, Cross-section = $A_{s}$ = (2a)²=>4a²

Change in Length of rod = PL / AE

δL $\alpha \frac{1}{A}$

Now, we are considering other factors same

the area of cross-section of square rod is more than Area of cross-section of circular rod

thus, the one with the circular cross section will stretch more

svetlana [45]2 years ago

8 0

Answer:

The correct option is;

A. The one with the circular cross section will stretch more.

Explanation:

Here we have the cross section as being

1. Circular, with diameter, D = 2·a

2. Square cross section with each side length = 2·a

The area of the circular rod is then

Area of circle = π·D²/4 which is equal to

π×(2·a)²/4 = π·4·a²/4 = π·a²

The area of the rod with square cross section is

Area of square = Side² which gives

Area of cross section = (2·a)² = 4·a²

Therefore, since π = 3.142, the cross sectional area of the circular rod is less than that of the one with a square cross section

That is, π·a² = 3.142·a² < 4·a²

We note that the elongation or extension is directly proportional to the force applied as shown as follows

$\frac{P}{A} = E\frac{\delta}{L}$

Where:

P/A = Force and

δ = Extension

The force is inversely proportional to the area, therefore a rod with less cross sectional area experiences more force and more elongation.

You might be interested in

These nerve cells are needed in muscle and tendons to detect degree of stress and stretching as a sensory activity. A. Ganglia.

SOVA2 [1]

Answer:

D.Proprioceptors

Explanation:

"Proprioceptors is Sensory receptors found in muscle an tendons that detect their degree of stretch"

4 0

2 years ago

Generally, what will happen to most liquids if the temperature is increased?

GalinKa [24]

Temperature doesn't really affect solubility on liquids so it can only be D besides it's already a liquid....

5 0

3 years ago

Read 2 more answers

What is the intensity of 60dB sound?

polet [3.4K]

Answer:

The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

Explanation:

Given;

intensity of the sound level, dB = 60 dB

The intensity of the sound in W/m² is calculated as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\$

where;

I₀ is threshold of hearing = 1 x 10⁻¹² W/m²

I is intensity of the sound in W/m²

Substitute the given values and for I;

$dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2$

Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

5 0

2 years ago

What can we say about Gamma waves on<br> the electromagnetic spectrum?

sladkih [1.3K]

Answer: Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions. Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.

Explanation: :)

6 0

1 year ago

Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the

777dan777 [17]

Answer:

Part a)

$L = 2683.2 kg m^2/s$

Part b)

$v' = 8.60 m/s$

Part c)

$W = 4326.7 J$

Explanation:

Part a)

As we know that there is no external torque on the system of two twins

so here we will use

$L = mv r + mvr$

$L = 2(78 \times 4.30 \times 4)$

$L = 2683.2 kg m^2/s$

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

$2(m v r) = 2( m v' \frac{r}{2})$

$v' = 2v$

$v' = 8.60 m/s$

Part c)

Work done by both of them = change in kinetic energy

so we have

$W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)$

$W = m(v'^2 - v^2)$

$W = 78(8.60^2 - 4.3^2)$

$W = 4326.7 J$

5 0

2 years ago