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2 years ago

In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

Physics

1 answer:

gulaghasi [49]2 years ago

8 0

Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

$\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}$

Squaring both side

$\frac{L_2}{L_1}=1.108$

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A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

2 years ago

The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change

mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

$\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})$

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m$

(b)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m$

(c)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m$

(d) The three lines belong to the ultraviolet region.

8 0

2 years ago

A hoop is rolling (without slipping) on a horizontal surface so it has two types of kinetic energy: translational kinetic energy

LenaWriter [7]

Answer:

$\dfrac{T}{K}=1$

Explanation:

The translational kinetic energy of the hoop is given by :

$K=\dfrac{1}{2}Mv^2$ ..................(1)

M is the mass of the hoop

v is the velocity of the hoop

The rotational kinetic energy of the hoop is given by :

$T=\dfrac{1}{2}I\omega^2$

Since, $I=MR^2$

$\omega=\dfrac{v}{R}$

$T=\dfrac{1}{2}\times MR^2\times (\dfrac{v}{R})^2$ ..............(2)

From equation (1) and (2) :

$\dfrac{T}{K}=1$

Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy is 1.

8 0

2 years ago

A moving sidewalk has a velocity of 1.7m/s north. if a man walks 1.1m/s, how long does it take him to travel 15m north in relati

alexira [117]

I think the answer will be a

3 0

2 years ago

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a hei

DiKsa [7]

Answer;

- 15 J

Explanation;

-Potential energy is defined as mechanical energy, stored energy, or energy caused by its position.

-For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m /s² at the surface of the earth) and h is the height in meters.

Potential energy of the rabbit at the peak of its height is

PE = (3)(10)(0.5) = 15 J

(around 14.7 but because energy is lost, it is less than that)

3 0

2 years ago

Read 2 more answers