What is the wavelength of an earthquake wave if it has a speed of 7 km/s and a frequency of 3 Hz?

What is the difference between mass and weight?

xxTIMURxx [149]

Answer:

C. Mass is independent of the force acting on the mass.

Weight depends on the gravitational force acting on the mass.

Example: A mass of 1 kg on the earth still has a mass of 1 kg on the moon;

whereas the weight of that mass on earth is 9.8 N and only 1/6 of that amount on the moon.

4 0

4 years ago

A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the stri

zvonat [6]

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

Here

n = Number of node

T = Tension

$\mu$ = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 281.2Hz}$

Similarly plug in 2 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 562.4Hz}$

Similarly plug in 3 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)$

$\mathbf{f= 843.7Hz}$

4 0

3 years ago

A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s

AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0

4 years ago

You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w

Vesna [10]

Answer:

$\ \text{m/s}$

Explanation:

$u_1$ = Velocity of one lump = $3x+3y-3z$

$u_2$ = Velocity of the other lump = $-4x+0y-4z$

m = Mass of each lump = $30\ \text{g}$

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

$mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}$

The velocity of the stuck-together lump just after the collision is $\ \text{m/s}$ .

4 0

3 years ago

Each tyre of a car has an area of 100 cm2 in contact with the ground. The car has a mass of 1600 kg. The weight of the car is eq

vlabodo [156]

Answer:

b is the answer

Explanation:

tq friend b is the answer

3 0

3 years ago

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