PHYSICS COLLEGE

A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?

Answers

Answer 1

Answer:

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

$S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n t=\sqrt{1.245*10^(-3) }\n t=0.035s$

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

$v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s$

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

$D=|vt|\nD=|0.346m/s*4.465s|\nD=1.54489m$

Answer 2

Answer:

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

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You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

$\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)$

where,

λ = wavelength = ?

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An airplane is in level flight over Antarctica, where the magnetic field of the earth is mostly directed upward away from the ground. As viewed by a passenger facing toward the front of the plane, is the left or the right wingtip at higher potential? Does your answer depend on the direction the plane is flying?

Answers

Answer:

It does not depend on direction of plane and the left windtips more potential

Explanation:

Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction

The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.How many capillaries are there?

Answers

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

$A_1v_1=nA_2v_2$ (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

$A=\pi r^2\n\nA_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\n\nA_2=\pi r_2^2=\pi (5*10^(-4)cm)^2=7.85*10^(-7)cm^2$

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

$n=(A_1v_1)/(A_2v_2)=((3.1415cm^2)(40cm/s))/((7.85*10^(-7)cm^2)(0.10cm/s))=1.6*10^9$

Then, the number of capillaries is 1.6*10^9

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?

Answers

The value of final angular speed of the uniform rod which rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

What is angular speed of a body?

The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,

$\omega= (\Delta \theta)/(\Delta t)$

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it.

The mass of the bullet is one-fourth the mass of the rod. The diagram for the above condition is attached below.

In the attached image the angular momentum about the point A is constant just before and after the collision. Thus,

$L_i=L_f\nL_i=(m)/(4)*(vL)/(2)=I\omega\n(m)/(4)*(vL)/(2)=I\omega$

Put the value of inertia as,

$(m)/(4)*(vL)/(2)=\left((mL^2)/(3)+(mL^2)/(4*4)\right)\n(mvL)/(8)*(vL)/(2)=\left((19mL^2)/(48)\right)$

Solving it further we get,

$(v)/(8)=(19)/(48)L\omega\n\omega=(6v)/(19L)$

Hence, the value of final angular speed of the uniform rodwhich rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

Learn more about the angular speed here;

brainly.com/question/540174

Answer: a) 0.315 (V/L)

Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

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Answers

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

Final answer:

The interactions of a TV remote and the TV involve short-range infrared communication, while the TV receives signals from long-range electromagnetic waves broadcasted for channels in frequency ranges for VHF and UHF.

Explanation:

When you use a TV remote control to change the channel, two main interactions are involved. The first interaction is the infrared communication between the remote and the TV, which is a form of electromagnetic radiation. Infrared signals require a direct line of sight, operating over a relatively short range. On the other hand, the TV itself receives broadcast signals through antennas that capture electromagnetic waves broadcasted over a long range - these signals can be VHF or UHF TV channels.

Additionally, the TV channels are broadcasted on frequencies ranging from 54 to 88 MHz and 174 to 222 MHz for VHF, while UHF channels utilize frequencies from 470 to 1000 MHz. These signals are sent over a significant distance to your TV’s antenna, showing that television broadcast interaction is long range. These broadcast signals are part of electromagnetic spectrum and carry a large range of frequencies due to the variety of content (audio and visual information) that needs to be transmitted.

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Answers

Answer and Explanation:

Doppler effect : According to Doppler effect whenever there is a relative motion between the source and observer then there is an increase or decrease in frequency of sound light or waves.

REASON OF DOPPLER EFFECT : Doppler effect is mainly due to the sudden change in pitch of the sound

EXAMPLE OF DOPPLER EFFECT : The best example of Doppler effect is when an ambulance passes and when it comes closer then the frequency of the siren increases and when it goes away its frequency decreases.

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