PHYSICS COLLEGE

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?
3) For the conditions, how long does it take for her to reach the opposite bank?

Answers

Answer 1

Answer:

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^(-1)$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^(-1)$

speed of the swimmer with respect to water, $v=4\ ft.s^(-1)$

Now the resultant of the two velocities perpendicular to each other:

$v_r=√(v^2+v_s^2)$

$v_r=√(4^2+8^2)$

$v_r=8.9442\ ft.s^(-1)$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=(v_s)/(v)$

$\tan\beta=(8)/(4)$

$\beta=63.43^(\circ)$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d* \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=√(d^2-w^2)$

$\Delta s=√(1118.034^2-500^2)$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4* \cos37$

$v'=3.1945\ ft.s^(-1)$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^(-1)$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^(-1)$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=(v_n)/(v'_r)$

$\tan\phi=(4.8055)/(2.4073)$

$\phi=63.39^(\circ)$

Now let the distance swam in this direction be d'.

$d'* \cos\phi=w$

$d'=(500)/(\cos63.39)$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=√(d'^2-w^2)$

$\Delta s'=√(1116.344^2-500^2)$

$\Delta s'=998.11\ ft.s^(-1)$

Time taken in crossing the rive in case 1:

$t=(d)/(v_r)$

$t=(1118.034)/(8.9442)$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=(d')/(v'_r)$

$t'=(1116.344)/(2.4073)$

$t'\approx463.733\ s$

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Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$

$\lambda = 280nm . (1m)/(1x10^(9)nm)$ ⇒ $2.8x10^(-7)m$

$\nu = (3x10^(8)m/s)/(3.2x10^(-7)m)$

$\nu = 9.375x10^(14)s^(-1)$

$\nu = 9.375x10^(14)Hz$

$\nu = (3x10^(8)m/s)/(2.8x10^(-7)m)$

$\nu = 1.071x10^(15)Hz$

Hence, the UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

(b) In which of these three categories does radiation with a frequency of $7.9x10^(14)Hz$ belong.

The same approach followed in part A will be used to answer part B.

Case for UV-A:

$\lambda = 400nm . (1m)/(1x10^(9)nm)$ ⇒ $4x10^(-7)m$

$\nu = (3x10^(8)m/s)/(4x10^(-7)m)$

$\nu = 7.5x10^(14)s^(-1)$

$\nu = 7.5x10^(14)Hz$

Hence, the UV-A has frequencies between $7.5x10^(14)Hz$ and $9.375x10^(14)Hz$ .

Therefore, the radiation with a frequency of $7.9x10^(14)Hz$ belongs to UV-A category.

Select True or False for the following statements about Heisenberg's Uncertainty Principle. True False It is not possible to measure simultaneously the x and y positions of a particle exactly.True False It is not possible to measure simultaneously the x and y momentum components of a particle exactly.
True False It is not possible to measure simultaneously the z position and the z momentum component of a particle exactly.

Answers

Answer:

Statement 1) False

Statement 2) False

Statement 3) True

Explanation:

The uncertainty principle states that " in a physical system certain quantities cannot be measured with random precision no matter whatever the least count of the instrument is" or we can say while measuring simultaneously the position and momentum of a particle the error involved is

$P\cdot\delta x\geq (h)/(4\pi )$

Thus if we measure x component of momentum of a particle with 100% precision we cannot measure it's position 100% accurately as the error will be always there.

Statement 1 is false since measurement of x and y positions has no relation to uncertainty.

Statement 2 is false as both the momentum components can be measured with 100% precision.

Statement 3 is true as as demanded by uncertainty principle since they are along same co-ordinates.

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.50 m from the mirror. The filament is 6.00 mm tall, and the image is to be 37.5 cm tall. Part A: How far in front of the vertex of the mirror should the filament be placed?Part B: to what radius of curvature should you grind the mirror?

Answers

Answer:13.6 cm

Explanation:

Given

v(image distance)=-8.5 m

height of object $(h_1)$ =6 mm

height of image $(h_2)$ =37.5 cm

and magnification of concave mirror is given by $m=(-v)/(u)=(-h_2)/(h_1)$

$m=(-37.5* 10)/(6)=-62.5$

$-62.5=(8.5* 100)/(u)$

u=13.6 cm

so object is at a distance of 13.6 cm from mirror.

for focal length

$(1)/(f)=(1)/(v)+(1)/(u)$

$(1)/(f)=(-1)/(850)+(-1)/(13.6)$

$(1)/(f)=-0.00117-0.0735$

f=-13.4 cm

thus radius of curvature of mirror is R=2f=26.8 cm

Final answer:

The filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror. The radius of curvature for the concave mirror should be approximately 0.85 m.

Explanation:

To determine how far in front of the vertex of the mirror the filament should be placed, we can use the mirror equation:

1/f = 1/do + 1/di

Where f is the focal length of the concave mirror, do is the object distance, and di is the image distance.

With the given information, we have:

do = ?

di = 8.50 m

Using the magnification formula:

magnification = -di/do

By substituting the values we know, we can solve for do:

37.5 cm / 6.00 mm = -8.50 m / do

Solving for do, we find that do ≈ - 0.85 m.

Since the object distance cannot be negative, we conclude that the filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror.

To find the radius of curvature for the concave mirror, we use the mirror formula:

1/f = 1/do + 1/di

With do = -0.85 m and di = 8.50 m, we can rearrange the formula to solve for f:

1/f = 1/-0.85 + 1/8.50

1/f ≈ -1.1765

Solving for f, we find that the focal length is approximately 0.85 m.

Learn more about Concave mirror here:

brainly.com/question/3555871

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he magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.9 AA , how many turns of wire would you need? Express your answer using two significant figures.

Answers

Answer: 3400

Explanation:

Given

Magnetic field, B = 0.1 T

Diameter of magnet, d = 2 cm = 0.02 m

Length of magnet, l = 8 cm = 0.08 m

Current of the magnet, I = 1.9 A

Number of turns needed, N = ?

To solve this problem, we would use the formula,

N = (LB) / (μI), where

μ = 1.257*10^-6 Tm/A, so that

N = (0.08 * 0.1) / (1.257*10^-6 * 1.9)

N = 0.008 / 2.388*10^-6

N = 3350

N ~ 3400

Therefore, the number of turns of wire needed is 3400

At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)

Answers

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

$R_t = R_o[1 + \alpha(T-T_o)]\n\n$

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

$R_t = R_o[1 + \alpha(T-T_o)]\n\n\R_t = 1.59*10^(-8)[1 + 0.0038(T-20)]$

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

$R_t,_(silver) = 2R_o,_(iron)\n\n1.59*10^(-8)[1 + 0.0038(T-20)] =(2 *9.71*10^(-8))\n\n\ \ (divide \ through \ by \ 1.59*10^(-8))\n\n1 + 0.0038(T-20) = 12.214\n\n1 + 0.0038T - 0.076 = 12.214\n\n0.0038T +0.924 = 12.214\n\n0.0038T = 12.214 - 0.924\n\n0.0038T = 11.29\n\nT = (11.29)/(0.0038) \n\nT = 2971.1 \ ^0C$

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

While on a hike, a pair of friends get caught in a thunderstorm. Four seconds after seeing the flash of a distant lightning strike, they hear the thunder. How far away was this lightning strike in miles? Note: sound, in air, travels at 340 m/s.

Answers

Answer:

1360 m

Explanation:

Time taken for the thunder to travel the distance to the hikers = 4 seconds

Speed of the thunder = 340 m/s

Speed of light = 3×10⁸ m/s

It can be seen that the speed of light is substantially faster than the speed of sound. This is the reason why there is a delay in seeing the lightning and hearing the thunder.

Distance = Speed × Time

$\text{Distance}=340* 4\n\Rightarrow \text{Distance}=1360\ m$

Hence, the lightning strike was 1360 m away from the hikers

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