Which element would have properties most like helium (He)?

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position

antiseptic1488 [7]

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_{m}}$ ) is

$v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_{m}$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm$

6 0

3 years ago

The magnetic field in the region between the poles of an electromagnet is uniform at any time, (1 point) but is increasing at th

olga nikolaevna [1]

Answer:

B)

The magnitude of induced emf in the conducting loop is 0.99 mV.

Explanation:

Rate of increase in magnetic field per unit time = 0.090 T/s

Area of the conducting loop = 110 cm^2 = 0.0110 m^2

Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.

Induced e.m.f is given as:

EMF = (-N*change in magnetic field/time)*Area

EMF = rate of change of magnetic field per unit time * Area

EMF = 0.090 * 0.0110

EMF = 0.00099 V

EMF = 0.99 mV

5 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

Light incident on a lake surface is partly reflected and partly refracted.What is the differences between the reflected ray and

iren2701 [21]

Answer: As per the question, a ray of light is incident on a surface and it is partly reflected and refracted. The incident light is an unpolarised light. The reflected light is partially polarised.

If the angle of incidence becomes equal to the Brester angle (polarising angle), then the reflected light is completely plane polarised.

5 0

3 years ago

Read 2 more answers

Can someone give me an objective and subjective statement example please

shusha [124]

Answer:

Objective: It is raining. Subjective: I love the rain!

Explanation:

Anything objective sticks to the facts, but anything subjective has feelings. Objective and subjective are opposites.

(Hope this helps can I pls have brainlist (crown)☺️)

7 0

2 years ago