How much more light does one collect with a 10-meter diameter telescope in 1 second than with a 0.1-meter diameter telescope in 10,000 seconds (nearly 3 hours)?

A. 1/100th as much

B. 1/10th as much

C. Same amount

D. 10 times as much

E. 100 times as much

Answer and Explanation:

Since black holes deal with dark matter and dark energy and are dark enough not to allow even light to escape them, so their detection is much more difficult.

There are two basic methods to detect a black hole in space and these are listed below:

Answer:

a)1.13×10³

b)1.6×10³

Explanation:

Given:

Boltzmann's constant (K)=1.38×10^-23 J/K

atmoic mass of helium = 4 AMU or 4×1.66×10^-27kg

a)The formula for RMS speed (Vrms) is given as

[tex]Vrms=\sqrt{\frac{3KT}{m} }[/tex]

where

K= Boltzmann's constant

T= temperature

m=mass of the gas

[tex]Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 206}{6.64\times 10^{-27}}}[/tex]

[tex]Vrms=1.13\times 10^{3}m/s[/tex]

b) RMS speed of helium when the temperature is doubled

[tex]Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 2\times 206}{6.64\times 10^{-27}}}[/tex]

[tex]Vrms=1.598\times 10^{3}m/s[/tex]

Answer: a)1.13×10³ b)1.6×10³

Explanation:

Answer:

1.84 x 10^7 N/C

Explanation:

Let q 1 = - 3 C, q2 = 2 C, q3 = 1 C

Electric field at P due to q1 is E1, due to q2 is E2 and due to q3 is E3.

E1 = k (3) / (50)^2 = 9 x 10^9 x 3  / 2500 = 108 x 10^5 N/C

E2 = k (2) / (50)^2 = 9 x 10^9 x 2 / 2500 = 72 x 10^5 N/C

E3 = k (1) / (150)^2 = 9 x 10^9 x 1 / 22500 = 4 x 10^5 N/C

Resultant electric field at P is given by

E = E1 + E2 + E3 = (108 + 72 + 4) x 10^5 = 184 x 10^5 = 1.84 x 10^7 N/C

Answer:

[tex]L = 20 + 37 = 57 cm[/tex]

Explanation:

As we know that two charges connected with spring is at equilibrium

so here force due to repulsion between two charges is counter balanced by the spring force between them

so here we have

[tex]F_e = F_{spring}[/tex]

here we have

[tex]\frac{kq_1q_2}{r^2} = kx[/tex]

[tex]\frac{(9 \times 10^9)(40 \mu C)(40 \mu C)}{(0.20 + x)^2} = 120 x[/tex]

[tex]14.4 = (0.20 + x)^2 ( 120 x)[/tex]

by solving above equation we have

[tex]x = 0.37 m[/tex]

so the distance between two charges is

[tex]L = 20 + 37 = 57 cm[/tex]

Answer:

Speed of air = 1106.38 ft/s

Explanation:

Speed of sound in air with temperature

[tex]v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\ [/tex]

Here speed is in m/s and T is in celcius scale.

T = 50°F

[tex]T=(50-32)\times \frac{5}{9}=10^0C \\ [/tex]

Substituting

[tex]v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=

337.31m/s \\ [/tex]

Now we need to convert m/s in to ft/s.

1 m = 3.28 ft

Substituting

[tex]v_{air}=337.31\times 3.28=1106.38ft/s \\ [/tex]

Speed of air = 1106.38 ft/s

Answer:

[tex]\dot{m_{2}}=0.865 kg/s[/tex]

Explanation:

[tex]\dot{m_1}= 0.5kg/s[/tex]

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

[tex]\dot{m_{in}}=\dot{m_{out}}[/tex]

[tex]\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}[/tex]

according to energy balance equation

[tex]\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}[/tex]

[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]

[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s[/tex]

The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s

Given data :

m₁ = 0.5 kg/s

m₂ = ?

From steam tables

At 250 kPa

at T1 = 80℃ , h₁ = 335.02 kJ/kg

at T2 = 20℃, h₂ = 83.915 kJ/kg

at T3 = 42℃, h₃ = 175.90 kJ/kg

Given that :

Min = Mout    also m₁ + m₂ = m₃

applying the principle of energy balance

M₁h₁ + M₂ h₂  = ( m₁ + m₂ ) h₃ ---- ( 1 )

insert values into equation ( 1 )

m₂ = 0.865 kg/s

Hence In conclusion The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s

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Answer:

The elevator has moved 54 meters upwards.

Explanation:

m= 50kg

F= 600 N

t= 3 sec

a= [F-(m*g)]/m

a= (600N-490N)/50kg

a= 2.2 m/s²

h= (a*t²)/2

h= (2.2m/s²*(3 sec)²)/2

h= 9.9 m

Final answer:

The elevator has moved upward a distance of 7 meters at the end of 3 seconds by considering the motion and acceleration of the elevator and the student.

Explanation:

The elevator has moved upward a distance of 7 meters at the end of 3 seconds.

To calculate this, we need to consider the motion of the elevator and the student. The apparent weight of the student is the normal force exerted by the elevator floor on the student, which is 600 N when moving up. This value is different from the student's actual weight due to the acceleration of the elevator.

Using the equation Net Force = Mass x Acceleration, and knowing that the student's weight (mg) - normal force = ma, we can find the acceleration, and subsequently, the distance the elevator has moved 7 meters after 3 seconds.

Answer:

Heat energy needed = 1243.45 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 0.500 kg water from 50°C to vapor at 110°C

First the water changes to 100°C from 50°C , then it changes to steam and then its temperature increases from 100°C to 110°C.

Mass of water = 500 g

Heat energy required to change water temperature from 50°C to 100°C

                   [tex]H_1=mC\Delta T=500\times 4.18\times (100-50)=104.5kJ[/tex]

Heat energy required to change water from 100°C to steam at 100°C

                   [tex]H_2=mL=500\times 2257=1128.5kJ[/tex]

Heat energy required to change steam temperature from 100°C to 110°C

                   [tex]H_3=mC\Delta T=500\times 2.09\times (110-100)=10.45kJ[/tex]

Total heat energy required

                   [tex]H=H_1+H_2+H_3=104.5+1128.5+10.45=1243.45kJ[/tex]

Heat energy needed = 1243.45 kJ

Answer:

Angle of refraction.

Explanation:

Refractive index is given as the ratio of the angle of incidence to the angle of refraction.

If angle of incidence is i and the refractive index of the incident medium is n₁ , and if angle of refraction is r and refractive index of the refracting medium is n₂,

according to Snell's law, n₁ sin i = n₂ sin r,

relative refractive index = sin i / sin r

The distance from where the ball hits the ground is approximately 7.53 m.

To find the distance from where the ball hits the ground, we need to analyze the horizontal and vertical motion separately. First, we can find the time it takes for the ball to hit the ground using the vertical motion equation. The final position in the y-axis is 0, the initial position is 1 m, the initial vertical velocity is 8*sin(40°) m/s, and the acceleration is -9.8 m/s² (due to gravity). By solving the equation, we find that the time of flight is 1.15 s. Using this time, we can find the horizontal distance traveled by the ball using the horizontal motion equation. The initial horizontal velocity is 8*cos(40°) m/s, and multiplying it by the time of flight gives us the answer: approximately 7.53 m.

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Answer:

Part a)

Magnitude = 5.06 unit

Part b)

[tex]\theta = 36.2 ^0[/tex]

Explanation:

Part a)

Vector is given as

[tex]\vec A = 4.08 \hat x + 3.0 \hat y[/tex]

now from above we can say that

x component of the vector is 4.08

y component of the vector is given as 3.0

so the magnitude of the vector is given as

[tex]|A| = \sqrt{4.08^2 + 3^2}[/tex]

[tex]|A| = 5.06 unit[/tex]

Part b)

Now the angle made by the vector is given as

[tex]\theta = tan^{-1}(\frac{y}{x})[/tex]

[tex]\theta = tan^{-1}(\frac{3}{4.08})[/tex]

[tex]\theta = 36.3 degree[/tex]

Explanation:

The weight of the car is equal to, [tex]W_c=m\times g[/tex]...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, [tex]F_N=mg\ cos\theta[/tex]

or

[tex]F_N=mg\ cos(13)[/tex].............(2)

The horizontal component of the force is, [tex]F_H=mg\ sin\theta[/tex]

Taking ratio of equation (1) and (2) as :

[tex]\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}[/tex]

[tex]\dfrac{F_N}{W_c}=cos(13)[/tex]

[tex]\dfrac{F_N}{W_c}=0.97[/tex]

or

[tex]\dfrac{F_N}{W_c}=\dfrac{97}{100}[/tex]

Hence, this is the required solution.

The ratio of the magnitude of the normal force to the weight of the car can be determined by taking the cosine of the angle of inclination.

When a car is traveling up a hill inclined at an angle θ above the horizontal, the ratio of the magnitude of the normal force to the weight of the car can be determined. The vertical component of the normal force is N cos θ, while the weight of the car is mg. Since the net vertical force must be zero, these two forces must be equal in magnitude:

N cos θ = mg

To find the ratio, divide both sides of the equation by mg:

N/mg = cos θ

So, the ratio of the magnitude of the normal force to the weight of the car is cos θ.

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Answer:

Is dissipated E) P= 48 W in the 12 Ω resistor.

Explanation:

V= 36v

Req= (12 + 6)Ω

Req= 18Ω

I = V/ Req

I= 2 A

P= I² * R(12Ω)

P= 48 W

Answer:

6900 m/s

Explanation:

The mass of the rocket is:

m = 330000 − 280000 (t / 250)

m = 330000 − 1120 t

Force is mass times acceleration:

F = ma

a = F / m

a = F / (330000 − 1120 t)

Acceleration is the derivative of velocity:

dv/dt = F / (330000 − 1120 t)

dv = F dt / (330000 − 1120 t)

Multiply both sides by -1120:

-1120 dv = -1120 F dt / (330000 − 1120 t)

Integrate both sides.  Assuming the rocket starts at rest:

-1120 (v − 0) = F [ ln(330000 − 1120 t) − ln(330000 − 0) ]

-1120 v = F [ ln(330000 − 1120 t) − ln(330000) ]

1120 v = F [ ln(330000) − ln(330000 − 1120 t) ]

1120 v = F ln(330000 / (330000 − 1120 t))

v = (F / 1120) ln(330000 / (330000 − 1120 t))

Given t = 250 s and F = 4.1×10⁶ N:

v = (4.1×10⁶ / 1120) ln(330000 / (330000 − 1120×250))

v = 6900 m/s

Answer:

so specific heat capacity of unknown metal is 656.8 J/degree C kg

Explanation:

Here by energy balance we can say that energy given by metal alloy at 100 degree = heat absorbed by water at 18.7 degree C

now we have

[tex]Q_{in} = Q_{out}[/tex]

[tex]390 s (100 - 26.8) = 553 (4186 ) ( 26.8 - 18.7)[/tex]

[tex] 28548 s = 18750349.8[/tex]

now we have

[tex]s = \frac{18750349.8}{28548}[/tex]

[tex]s = 656.8 J/kg ^o C[/tex]

so specific heat capacity of unknown metal is 656.8 J/degree C kg

Answer:

The final speed of the rocket once the engine has fired is 29.16 m/s.

Explanation:

By the principle of momentum and amount of movement, we match the momentum data to the amount of movement and find the value of speed.

I=3.5 kg*m/s

m=120g= 0.12kg

V=?

I=F*t

F*t=m*V

I/m=V

(3.5 kg*m/s) / 0.12kg = V

V=29.16 m/s

Explanation:

The given data is as follows.

   J = 3.5 kg-m/s,    m = 120 g = 0.120 kg      (as 1 g = 0.001 kg)

It is known that formula to calculate impulse is as follows.

        J = [tex]m(v_{2} - v_{1})[/tex]

          [tex]0.120(v_{2} - v_{1})[/tex] = 3.5

      [tex]v_2 - v_1[/tex] = 29 m/s

So, [tex]v_{2} = 29 + v_{1}[/tex]

                = 29 + 0 = 29 m/s

Thus, we can conclude that final speed ([tex]v_{2}[/tex]) is 29 m/s.

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion [tex]\alpha= 10.5\times10^{-6}\ K^{-1}[/tex]

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

[tex]\Delta L=L\alpha\Delta T[/tex]

Where, [tex]\Delta T [/tex]= temperature difference

[tex]\alpha[/tex]=Coefficient of thermal expansion

L= length

Put the value into the formula

[tex]\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})[/tex]

[tex]\Delta L=0.42\ m[/tex]

Hence, The difference in the length of the bridge is 0.42 m.

Answer:

The distance the ball moves up the incline before reversing its direction is 3.2653 m.

The total time required for the ball to return to the child’s hand is 3.2654 s.

Explanation:

When the girl is moving up:

The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t  to reach the top.

Using

v = u + a×t

0 = 4 - 2.45*t

t = 1.6327 s

Since, this is the same time the ball will come back. So,

Total time to go and come back = 2* 1.6327 = 3.2654 s

To find the distance, using:

v² = u² + 2×a×s

0² = 4² + 2×(-2.45)×s

s = 3.2653 m

Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.

To determine the distance and the total time for the ball to return to the girl, we apply kinematic equations with the initial velocity of 4 m/s and a constant acceleration of 0.25g down the incline. The ball travels 3.27 meters up the incline before stopping, and the total time for the round trip is 3.26 seconds.

To determine the distance s that a ball moves up an incline before reversing its direction and the total time t required for it to return to the girl, we can use kinematics equations. The acceleration a is given as 0.25g, which means the acceleration is 0.25 times the acceleration due to gravity, with g being 9.8 m/s². The initial velocity u of the ball is 4 m/s up the incline.

The final velocity v when the ball stops at the highest point is 0 m/s. By using the kinematic equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can solve for s:

The ball travels a distance of s = 3.27 m up the incline before stopping and reversing its direction.

To find the total time t, we know that the time taken to go up and down the incline is the same. We use the equation v = u + at, and by rearranging for t, we get t = (v - u) / a.

The ball takes 1.63 seconds to reach the highest point, so the total time for the round trip is t = 1.63 s * 2 = 3.26 s. Thus, it takes 3.26 seconds for the ball to return to the girl's hand.

Answer:

A large elliptical

Explanation:

A S0 type galaxy is a large elliptical.

A S0 type galaxy is a large elliptical.

Answer is C.

Answer:

The power required to move each bale is 30 lbf.ft/sec.

Explanation:

F= 40lbf * sin (30º)

F=20lbf

P= F*v

P=20 lbf * 1.5 ft/sec

P= 30 lbf.ft/sec

Given:

heat generated by John's cooling system,  [tex]H = \rho A v^{3}[/tex]  = 45 W    (1)

If ρ, A, and v corresponds to John's cooling system then let [tex]\rho_{1}, A_{1}, v_{1}[/tex] be the variables for Mike's system then:

[tex]\rho  = 9.5\rho_{1}[/tex]

[tex]\rho_{1}  = \frac{\rho}{9.5}[/tex]

[tex]v_{1} =3.5 v[/tex]

Formula use:

Heat generated, [tex]H = \rho A v^{3}[/tex]

where,

[tex]\rho[/tex] = density

A = area

v = velocity

Solution:

for Mike's cooling system:

[tex]H_{2}[/tex] = [tex]v_{1}^{3}{1}A_{1}\rho_{1}[/tex]

⇒ [tex]H_{2}[/tex] = [tex](3.5v)^{3}[/tex] × A × [tex]\frac{\rho}{9.5}[/tex]

[tex]H_{2}[/tex] = 4.513[tex]v^{3}[/tex] A  [tex]\rho[/tex]

Using eqn (1) in the above eqn, we get:

[tex]H_{2}[/tex] = 4.513 × 45 = 203.09 W

We have that The Heat loss H2 is given as

From the question we are told

Generally the equation for the   is mathematically given as

H=PAV^3

Therefore

[tex]\frac{110}{h_2}=\frac{P_1}{p1/3.5}*(\frac{V_1}{3.5*v_1})^3[/tex]

H_2=1347.5w

Therefore

The Heat loss H2 is given as

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Answer:

The acceleration of the car is 8.50 m/s²

Explanation:

Given that,

Initial velocity = 10 m/s

Final velocity = 61 m/s

Time = 6 s

We need to calculate the acceleration of the car

Using equation of motion

[tex]v=u+at[/tex].....(I)

[tex]a=\dfrac{v-u}{t}[/tex]

Where, u = initial velocity

v = final velocity

t = time

Put the value in the equation (I)

[tex]a=\dfrac{61-10}{6}[/tex]

[tex]a=8.50\ m/s^2[/tex]

Hence, The acceleration of the car is 8.50 m/s²

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.27 s

c) Total distance traveled = 402.43 m

Explanation:

a) Maximum speed is obtained after the end of acceleration

        v = u + at

        v = 13.5 + 1.9 x 6.2 = 25.28 m/s

    Maximum speed = 25.28 m/s

b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.

         v = u + at  

         0 = 25.28 - 1.2 t

         t = 21.07 s

    Total time = 6.2 + 21.07 = 27.27 s

c) Distance traveled for the first 6.2 s

          s = ut + 0.5 at²

          s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m

   Distance traveled for the second 21.07 s

          s = ut + 0.5 at²

          s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m

   Total distance traveled = 120.22 + 282.21 = 402.43 m

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.26 s

c) Total distance traveled = 390,5537

Explanation:

In order to solve the first proble we just have to use the next formula:

[tex]Vf= Vo+ Acc-t\\Vf= 13,5 + 6,2*1,9\\Vf=25,28 m/s\\[/tex]

So the maximum speed would be 25,28 m/s.

THe total time of the trip will be given by adding the inital time plus the velocity divided by the acceleration rate:

[tex]Time= 6,2 s+ \frac{25,28}{-1,2} \\Time= 6,2 +21,06\\Time= 27,26[/tex]

Remember that when dealing with time in physics you will always use positive numbers since there is no negative time.

To calculate the total distance covered we use the next formula:

[tex]D=Vo*t+ 1/2a*t^2\\D= 13,5*6,2 + 1/2(1,9)(6,2)^2\\D=87,75+36,518\\D=124,268m[/tex]

This is the first part now we calculate it with the stopping of the car:

[tex]D=Vo*t+ 1/2a*t^2\\D=25,28*21,06+ 1/2(-1,2)(21,06)^2\\D=532,3968-266,1141\\D= 266,2857meters[/tex]

No we just add the two distances to discover the whole distance:

Total distance= 124,268+266,2857= 390,5537

Answer:

[tex]r = 5.13 \times 10^6[/tex]

Explanation:

As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed

so here we will have

[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]

here we know that

v = 19800 mi/h

[tex]v = 19800 \frac{1602 m}{3600 s}[/tex]

[tex]v = 8811 m/s[/tex]

also we know that

[tex]M = 5.97 \times 10^{24} kg[/tex]

now we will have

[tex]r = \frac{GM}{v^2}[/tex]

[tex]r = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(8811)^2}[/tex]

[tex]r = 5.13 \times 10^6[/tex]

Answer:

The momentum of the two-block system is 18 kg-m/s

Explanation:

It is given that,

Mass of block A, [tex]m_A=4\ kg[/tex]  

Mass of block B, [tex]m_B=6\ kg[/tex]  

Velocity of block A, [tex]v_A=3\ m/s[/tex]  

Velocity of block B, [tex]v_B=-5\ m/s[/tex] (it is moving in opposite direction)

We need to find the momentum of two block system. It is given by the product of mass and velocity for both blocks i.e.

[tex]p=m_Av_A+m_Bv_B[/tex]

[tex]p=4\ kg\times 3\ m/s+6\ kg\times (-5\ m/s)[/tex]

p = -18 kg-m/s

So, the momentum of two block system is 18 kg-m/s. Hence, this is the required solution.                                            

Final answer:

The total momentum of the two-block system, involving block A with a mass of 4.0 kg moving at 3.0 m/s and block B with a mass of 6.0 kg moving at 5.0 m/s in the opposite direction, is computed as -18 kg·m/s, which is in the -x-direction.

Explanation:

The question involves finding the momentum of a two-block system where the blocks are moving in opposite directions. To compute the total momentum of the system, we apply the principle of conservation of momentum, which states that the momentum of a system remains constant if no external forces act on it.

Momentum is defined as the product of an object's mass and its velocity (p = mv). For block A, which has a mass of 4.0 kg and is moving at 3.0 m/s, its momentum is calculated as:
pA = (4.0 kg) × (3.0 m/s) = 12 kg·m/s in the +x-direction.
For block B, which has a mass of 6.0 kg and is moving at 5.0 m/s in the opposite direction, its momentum is calculated as:
pB = (6.0 kg) × (-5.0 m/s) = -30 kg·m/s in the -x-direction (note the negative sign because it is moving in the opposite direction).

To find the total momentum of the system, we sum the momenta of both blocks:
ptotal = pA + pB = 12 kg·m/s + (-30 kg·m/s) = -18 kg·m/s.

The total momentum of the two-block system is -18 kg·m/s, indicating that there is a net momentum in the -x-direction.

Answer:

a)

v = 31.15 m/s

t = 0.393 sec

h = 62.5 m

Explanation:

a)

v₀ = initially speed of the stone = 35.0 m/s

v = final speed of the stone = ?

y = vertical displacement of the stone = 13.0 m

a = acceleration due to gravity = - 9.8 m/s²

using the equation

v² = v₀² + 2 a y

v² = 35² + 2 (- 9.8) (13.0)

v = 31.15 m/s

t = time taken

using the equation

v = v₀ + a t

31.15 = 35 + (- 9.8) t

t = 0.393 sec

h = maximum height

v' = final speed at the maximum height = 0 m/s

using the equation

v'² = v₀² + 2 a h

0² = 35² + 2 (- 9.8) h

h = 62.5 m

Answer:

Answer:196 Joules

Explanation:

Hello

Note:  I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work  is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

[tex]F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules[/tex]

Answer:196 Joules

I hope it helps

Answer:

8.85 x 10⁴ Nm²/C

Explanation:

d = diameter of the conducting sphere = 0.10 m

r = radius of the conducting sphere = (0.5) d = (0.5) (0.10) = 0.05 m

Area of the sphere is given as

A = 4πr²

A = 4 (3.14) (0.05)²

A = 0.0314 m²

σ = Surface charge density = 150 x 10⁻⁶ C/m²

Q = total charge enclosed

Total charge enclosed is given as

Q = σA

Q = (150 x 10⁻⁶) (0.0314)

Q = 4.7 x 10⁻⁶ C

Electric flux through one of the side is given as

[tex]\phi = \frac{Q}{6\epsilon _{o}}[/tex]

[tex]\phi = \frac{4.7\times 10^{-6}}{6(8.85\times 10^{-12})}[/tex]

[tex]\phi [/tex] = 8.85 x 10⁴ Nm²/C

Answer:

Partial pressure of nitrogen gas,[tex]p^o_{N_2] =0.44 atm[/tex]

Explanation:

Pressure of the mixture of gases before adding nitrogen gas = 0.85 atm

Pressure of the mixture of gases after adding nitrogen gas = 988 mmHg

1 mmHg = 0.001315 atm

988 mmHg=[tex]988 mmHg\times 0.001315 atm = 1.29 atm[/tex]

Partial pressure of nitrogen gas,[tex]p^o_{N_2] = 1.29 atm - 0.85 atm = 0.44 atm[/tex]

According to Dalton's law of partial pressure , the total pressure of the mixture of gases is equal to sum of all the partial pressures of each gas present in the mixture.

[tex]P_{total}=\sum p^o_{i}[/tex]

So, on addition of nitrogen gas to the mixture the pressure of the mixture increases.

Answer:

The actual size is

69.33 nm.

Explanation:

It means that

3 cm is equivalent to 40 nm

So, 1 cm is equivalent to 40 / 3 nm

Thus, 5.2 cm is equivalent to

40 × 5.2 / 3 = 69.33 nm