5. A wire, resistance 23.6 carries a current. 5.9 A. What is the

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

2 years ago

During metamorphism, what is the major effect of chemically active fluids?

xz_007 [3.2K]

Answer:

Option b

Explanation:

Metamorphism is the process where the variation of the geological texture resulting from the different arrangement of the minerals or the variation of minerals in protoliths, i.e., pre- existing rocks take place such that there occurs no change in state of the protolith, i.e., it does not melt into magma.

The change takes place as a result of the presence of chemically active fluids, heat and pressure.

There is a reaction between the chemically active fluid and the rock through which it passes and promotes the movement of the dissolved ions of silicate and promotes the growth of the mineral grains.

7 0

2 years ago

A bike rider pedals with constant acceleration to reach a velocity of 7.8 m/s over a time of 4.2 s. during the period of acceler

Artyom0805 [142]

To calculate the initial velocity of the bike, we use the following equation

$d=\frac{1}{2} (u+v)t$ .

or

$u=\frac{2d}{t} -v$

Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.

Given, $u =7.8 m/s$ , $d= 19 m$ and $t=4.2 s$ .

Substituting these values in above equation, we get

$u = \frac{2 \times 19}{4.2 \ s} -7.8 m/s = 9.05 \ m/s - 7.8 \ m/s \\\\ u= 1.2 m/s$ .

Thus, the initial velocity of the bike is 1.2 m/s.

3 0

3 years ago

All isotopes of a particular element have the same atomic number. How then do the isotopes of a particular element differ?

xz_007 [3.2K]

Isotopes of an element will contain the same number of protons and electrons but will differ in the number of neutrons they contain. In other words, isotopes have the same atomic number because they are the same element but have a different atomic mass because they contain a different number of neutrons

7 0

3 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid