Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect. If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be

Answer:

The resolving power remains same.

Explanation:

The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.

As the diameter is same but the focal length is doubled so the resolving power remains same.

Answer:

Microwave is a types of a electromagnetic radiation

Answer:

Transvers

Explanation:

Because microwave is electromagnetic  waves and all electromagnetic waves are transvers.

Answer:

1.37 mm²

Explanation:

From the image attached below:

Let's take a look at the two rays r and r' hitting the same mirror from two different positions.

Let x be the distance between these rays.

[tex]d_o =[/tex] distance between object as well as the mirror

[tex]d_{eye}[/tex] = distance between mirror as well as the eye

Thus, the formula for determining the distance between these rays can be expressed as:

[tex]x = 2d_o tan \theta[/tex]

where; the distance between the eye of the observer and the image is:

[tex]s = d_o + d_{eye}[/tex]

Then, the tangent of the angle θ is:

[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]

replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:

[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]

[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]

[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]

x = (0.13157 × 10) mm

x = 1.32 mm

Finally, the area A = π r²

[tex]A = \pi(\frac{x}{2})^2[/tex]

[tex]A = \pi(\frac{1.32}{2})^2[/tex]

A = 1.37 mm²

The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.

Momentum is conserved, so the total momentum of the system is the same before and after the collision:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

==>

(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'

==>

-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'

where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.

Kinetic energy is also conserved, so that

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²

or

m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²

==>

(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²

==>

1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²

Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is

v₁' ≈ -3.11 m/s

v₂' ≈ -0.167 m/s

and take the absolute values to get the magnitudes.

If you want to instead use the masses from the "Required" section, you would end up with

v₁' ≈ -3.18 m/s

v₂' ≈ -0.236 m/s

Answer:

[tex]f=9.55Hz[/tex]

Explanation:

From the question we are told that:

Number of Turns [tex]N=200[/tex]

Length [tex]l=5cm to 10cm[/tex]

Voltage [tex]V=18V[/tex]

Magnetic field [tex]B=300mT[/tex]

Generally, the equation for Frequncy of an amarture is mathematically given by

[tex]f =\frac{ V}{(N B A * 2 pi )}[/tex]

[tex]f =\frac{ 18}{(200 300*10^{-3} (10*10^-2)(5*10^{-2}) * 2 *3.142 )}[/tex]

[tex]f=9.55Hz[/tex]

THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-

• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.

• FPS System- (Foot–Pound–Second system).

The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.

• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."

Answer:

It will travel 350 meters each second.

Explanation:

The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.

Answer:

5.83 seconds

Explanation:

60 seconds in 1 minute

350 meters per second

350/60

=5.83

Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.

Hope it helps you

Mark my answer as brainlist

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Answer:

zzyibgsdwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwe

Explanation:

Answer:

|x| = √53

Explanation:

We are told that the vector starts at the point (0.0) and ends at (2,-7) .

Thus, magnitude of displacement is;

|x| = √(((-7) - 0)² + (2 - 0)²)

|x| = √(49 + 4)

|x| = √53

Answer:

3.1 kg

Explanation:

Applying,

R = m(g-a)..................... Equation 1

Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.

From the question,

Given: m = 5 kg, a = 3.8 m/s²

Constant: g = 9.8 m/s²

Substitute these values into equation 1

R = 5(9.8-3.8)

R = 5(6)

R = 30 N

Hence the spring scale is

m' = R/g

m' = 30/9.8

m' = 3.1 kg

Answer:

b = frequency

Answer:

opinion a

Explanation:

the si units of distance is metre (m)

Answer:

A

Explanation:

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

[tex]R = \frac{\rho L}{A}[/tex]

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]

L = 169.5 m

Answer:

the speed of the block at the given position is 21.33 m/s.

Explanation:

Given;

spring constant, k = 3500 N/m

mass of the block, m = 4 kg

extension of the spring, x = 0.2 m

initial velocity of the block, u = 0

displacement of the block, d =1.3 m

The force applied to the block by the spring is calculated as;

F = ma = kx

where;

a is the acceleration of the block

[tex]a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2[/tex]

The final velocity of the block at 1.3 m is calculated as;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2ad

v = √2ad

v = √(2 x 175 x 1.3)

v = 21.33 m/s

Therefore, the speed of the block at the given position is 21.33 m/s.

The speed of the block at a height of 1.3 m above the starting position is 21.33 m/s

To solve this question, we'll begin by calculating the acceleration of the block.

F = Ke

Also,

F = ma

Thus,

ma = Ke

Divide both side by m

a = Ke / m

a = (3500 × 0.2) / 4

a = 175 m/s²

v² = u² + 2as

v² = 0² + (2 × 175 × 1.3)

v² = 455

Take the square root of both side

v = √455

v = 21.33 m/s

Learn more about spring constant:

https://brainly.com/question/9199238

Answer:

The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.

Answer:

 emf = 312 V

Explanation:

In this exercise the electromotive force is asked, for which we must use Faraday's law

           emf =  [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt

           Ф = B. A = B A cos θ

bold type indicates vectors.

They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values ​​1

It also indicates that the area is reduced from  a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear

            emf = -N B [tex]\frac{dA}{dT}[/tex]

            emf = - N B (A_f - A₀) / Dt

we calculate

           emf = - 60 1.60 (0 - 0.325) /0.100

           emf = 312 V

The direction of this voltage is exiting the page

Answer:

endothermic

Explanation:

An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.

Answer:

The correct answer is - c.  Spaceship will have a velocity to the East and will be slowing down.

Explanation:

In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.

As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

         P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

        F = P A

We reduce the pressure to the SI system

       P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

       F = 150 10³ / 0.00133

       F = 1.128 10⁸ Pa

Answer:

A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.

A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.

Answer:

Here, m=9×10

−31

kg,

q=1.6×10

−19

C,v=3×10

7

ms

−1

,

b=6×10

−4

T

r=

qB

mv

=

(1.6×10

−19

)(6×10

−4

)

(9×10

−31

)×(3×10

7

)

=0.28m

v=

2πr

v

=

2πm

Bq

=

2×(22/7)×9×10

−31

(6×10

−4

)×(1.6×10

−19

)

=1.7×10

7

Hz

Ek=

2

1

mv

2

=

2

1

×(9×10

−31

)×(3×10

7

)

2

J

=40.5×10

−17

J=

1.6×10

−16

40.5×10

−17

keV

=2.53keV

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

         p₀ = m v₁₀ + 0

         p₀ = 0.6 2

         p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

        p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

         p₀ₓ = 1.2 kg m / s

         p_{fx} = m v1f cos 20 + m v2f cos θ

         p₀ = p_f

        1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

         1.2482 = v_{2f} cos θ

Y axis  

        p_{oy} = 0

        p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

        0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

        0.2736 = v_{2f} sin θ

we write our system of equations

         0.2736 = v_{2f} sin θ

         1.2482 = v_{2f} cos θ

divide to solve

         0.219 = tan θ

          θ = tan⁻¹ 0.21919

          θ = 12.36

let's look for speed

            0.2736 = v_{2f} sin θ

             v_{2f} = 0.2736 / sin 12.36

            v_{2f} = 1.278 m / s

Answer:

D) Quadruple.

Explanation:

We will use the second equation of motion to solve this problem:

[tex]s = v_it + \frac{1}{2}gt^2[/tex]

where,

s = distance travelled by the rock

vi = initial speed of rock = 0 m/s

t = time taken

g = acceleration due to gravity on the surface of the moon

Therefore,

[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)

Now, we double the time:

[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]

using equation (1):

s' = 4s

Hence, the correct option is:

D) Quadruple.

what do you mean about it

Solution :

We know,

Distance,

[tex]$S=ut+\frac{1}{2}at^2$[/tex]

[tex]$S=ut+0.5(a)(t)^2$[/tex]

For the first 20 ms,

[tex]$S=0+0.5(220)(0.020)^2$[/tex]

S = 0.044 m

In the remaining 30 ms, it has constant velocity.

[tex]$v=u+at$[/tex]

[tex]$v=0+(220)(0.020)[/tex]

v = 4.4 m/s

Therefore,

[tex]$S=ut+0.5(a)(t)^2$[/tex]

[tex]$S'=4.4 \times 0.030[/tex]

S' = 0.132 m

So, the required distance is = S + S'

                                              = 0.044 + 0.132

                                              = 0.176 m

Therefore, the tongue can reach = 0.176 m or 17.6 cm

Answer:

The total distance is 0.176 m.

Explanation:

For t = 0 s to t = 20 ms

initial velocity, u = 0

acceleration, a = 220 m/s^2

time, t = 20 ms

Let the final speed is v.

Use first equation of motion

v = u + at

v = 0 + 220 x 0.02 = 4.4 m/s

Let the distance is s.

Use second equation of motion

[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]

Now the distance is

s' = v x t

s' = 4.4 x 0.03 = 0.132 m

The total distance is

S = s + s' = 0.044 + 0.132 = 0.176 m

Explanation:

We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as

[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]

For the hanging mass [tex]m_2,[/tex] we can write NSL as

[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]

We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get

[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]

or

[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]

[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]

Using this value for the acceleration on Eqn(2), we find that the tension T is

[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:=24.7\:\text{N}[/tex]

Answer:

The correct answer is - 8.99N/C

Explanation:

[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.