Please need help ASAP. Would really appreciate it.

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

Reil [10]

Answer:

$\Delta l=0.015m$

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area $A=0.5mm^2=0.5\times 10^{-6}m^2$

Young's modulus $\gamma=2\times 10^{11}Pa$

Force F = 1500 N

So stress $=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa$

We know that young's modulus $=\frac{stress}{strain}$

So $2\times 10^{11}=\frac{3\times 10^{9}}{strain}$

$strain=1.5\times 10^{-2}=0.015m$

Now strain $=\frac{\Delta l}{l}$

$0.015=\frac{\Delta l}{1}$

$\Delta l=0.015m$

6 0

3 years ago

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The magnitude of the angular momentum of the two-satellite system is best represented by

zloy xaker [14]

The magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

<h3>What is angular momentum.?</h3>

The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.

It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.

The magnitude of the angular momentum of the two-satellite system is best represented as;

L=∑mvr

L=m₁v₁r₁-m₂v₂r₂

Hence, the magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

To learn more about the angular momentum, refer to the link;

brainly.com/question/15104254

#SPJ4

3 0

2 years ago

A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north

sergiy2304 [10]

Answer:

Acceleration of the car will be $a=0.1375m/sec^2$

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car $F_R=7340N$

So net force on the car $F_{net}=F-F_R=7560-7340=220N$

According to second law of motion we know that F=ma

So $220=1600\times a$

$a=0.1375m/sec^2$

7 0

3 years ago

A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?

Rama09 [41]

The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J

4 0

3 years ago

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A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

By substituting the values, we get

$\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}$

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0

3 years ago