All categories

3 years ago

Answer plz also give reason

Physics

1 answer:

user100 [1]3 years ago

4 0

See this suggested solution.

1. Let a force F' is the vector sum of the forces P and Q, then it is shown on the attached picture and marked with red color.

2. according to the condition the force F holds the object, then F should have the same length as the force F' and the opposite direction.

3. using the conditions described in 2. the answer is C.

You might be interested in

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

3 years ago

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s

ad-work [718]

Answer:

$\theta = 0.0022 rad$

$I = 0.000304 I_o$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

The distance of the slit separation is $d = 0.500 \ mm = 5.0 *10^{-4} \ m$

Generally the condition for two slit interference is

$dsin \theta = m \lambda$

Where m is the order which is given from the question as m = 2

=> $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

substituting values

$\theta = 0.0022 rad$

Now on the second question

The distance of separation of the slit is

$d = 0.300 \ mm = 3.0 *10^{-4} \ m$

The intensity at the the angular position in part "a" is mathematically evaluated as

$I = I_o [\frac{sin \beta}{\beta} ]^2$

Where $\beta$ is mathematically evaluated as

$\beta = \frac{\pi * d * sin(\theta )}{\lambda }$

substituting values

$\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }$

$\beta = 0.06581$

So the intensity is

$I = I_o [\frac{sin (0.06581)}{0.06581} ]^2$

$I = 0.000304 I_o$

3 0

3 years ago

Help me, please with this queastion

sergejj [24]

Answer:

Kinetic energy is the energy due to motion. Potential energy is energy stored in matter. The joule (J) is the SI unit of energy and equals (kg×m2s2) ( kg × m 2 s 2 ) .

please mark me brainliest and follow me my friend.

7 0

2 years ago

The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum

Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.

5 0

3 years ago

A milk truck carries milk with density 64.6 lb/ft3 in a horizontal cylindrical tank with diameter 12 ft. (a) Find the force exer

7nadin3 [17]

Answer:

F = 351×10³lb

Explanation:

Given the density

ρg = 64.6lb/ft³

Diameter d = 12ft

The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft

The pressure in the tank is

P = ρgh = 64.6 × 12 = 775.2lb/ft²

The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²

F = 351×10³lb.

4 0

3 years ago