An object is placed in material a at point P, as shown in the diagram. The light is refracted when it strikes the interface with material b. When viewed from material b, at which point will the image appear?

Answer:

Explanation:

Force between two charged conducting sphere

= k x Q₁ x Q₂ / r² ,  k is a constant  Q₁ and Q₂ are charges and   r is distance between them .

= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²

= 21600 x 10⁻⁹

= 2.16 x 10⁻⁵ N .

b )

After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .

Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2

= - 3 x 10⁻⁹ C .

Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²

= 900 x 10⁻⁹ N .

Answer:

B

Explanation:

The crest of all three waves are of equal height

Answer:

[tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]

Explanation:

[tex]u_1[/tex] = Velocity of one lump = [tex]3x+3y-3z[/tex]

[tex]u_2[/tex] = Velocity of the other lump = [tex]-4x+0y-4z[/tex]

m = Mass of each lump = [tex]30\ \text{g}[/tex]

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

[tex]mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]

The velocity of the stuck-together lump just after the collision is [tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex].

Answer:

is there a. b.  c  or d?

Explanation:

Answer:

hello your question is incomplete attached below is the complete and the required circuit diagrams

answer :

Ai) This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

Explanation:

Ai) The voltage across the resistor will double when you double the current through the resistor

Given that : V = I*R.  

lets assume : I = 2 amperes , R = 3 ohms

V = 2*3 = 6 v

secondly lets assume double the value of  (I)   i.e. I = 4 amperes

hence : V = 4*3 = 12 volts

This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

Aii) Showing the two data points from simulation

I1                    V1            I2= 2I1         V2=2V1       V1/ I1 =V2/I2

0.9*10^3     9 * 10^3     1.8*10^3       18*10^3          10 ohms

1.6 * 10^3    16 * 10^3    3.2*10^3     32*10^3         10 ohms

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

Answer:

Part A

Newton's second law of motion states that the force applied to an object is directly proportional to the rate of change of momentum that is produced

Mathematically, we have;

F = m·v - m·u/Δt

Where;

m = The mass of the object

v = The final velocity of the object

u = The initial velocity of the object

Δt = The duration of motion of the object during change in velocity

Therefore, given that the mass, 'M', of the truck is larger than the mass, 'm', of the minivan, where the time of change in velocity Δt, and the initial and final velocities of both automobiles are the same such as in a sudden stop, the garbage ruck will exert more force than the minivan, and therefore, the garbage truck has a greater initial momentum before the automobiles are brought to a stop

Part B;

According to Newton's first law of motion, we have;

The use of a seat belt (and airbag for front seated passengers) will prevent dashboard or windscreen for the front passengers or the front seat for the passengers in the back, from being the item that stops the continued forward motion of the passengers in the car, which can lead to injury

Part C; The Newton's law of motion that act on a body at the point of impact is Newton's third law of motion, which states that the action and reaction are equal and opposite

Therefore, the action of the garbage truck on the minivan upon impact is equal to the reaction of the minivan to the force the garbage truck exerts on the minivan

Explanation:

Answer:

1.93×10²⁸ s

Explanation:

From the question given above, the following data were obtained:

Number of electron (e) = 2×10²⁴

Current (I) = 10 A

Time (t) =?

Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:

1 e = 96500 C

Therefore,

2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e

2×10²⁴ e = 1.93×10²⁹ C

Thus, 1.93×10²⁹ C of electricity is passing through the point.

Finally, we shall determine the time. This can be obtained as follow:

Current (I) = 10 A

Quantity of electricity = 1.93×10²⁹ C

Time (t) =?

Q = it

1.93×10²⁹ = 10 × t

Divide both side by 10

t = 1.93×10²⁹ / 10

t = 1.93×10²⁸ s

Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point

Answer:

The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.

Explanation:

Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.

Answer:

air

Explanation:

Answer:

2.66×10⁻⁹ N.

Explanation:

From the question,

Applying newton's law of universal gravitation,

Fg = GMm/r²............................... Equation 1

Where Fg = gravitational force, G = universal constant, M = mass of the mercury, m = mass of the human, r = radius of Mercury

Given: M = 3.31023 kg, M = 70 kg, r = 2.4106

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

Fg = 6.67×10⁻¹¹(70×3.31023)/(2.4106²)

Fg = 2.66×10⁻⁹ N.

Answer:

a. Power = 1000 Watts or 1 Kilowatts.

b. Current = 4 Amperes.

Explanation:

Given the following data;

Energy consumed = 1.8MJ = 1.8 × 10^6 = 1800000 Joules

Voltage = 250V

Time = 30 minutes to seconds = 30 * 60 = 1800 seconds

To find the power rating;

Power = energy/time

Substituting into the equation, we have;

Power = 1800000/1800

Power = 1000 Watts or 1 Kilowatts.

b. To find the current taken from the supply;

Power = current * voltage

1000 = current * 250

Current = 1000/250

Current = 4 Amperes.

Answer:

Work required to empty the tank by pumping all of the water to the top of the tank = 1674700 Kgm/s^2

Explanation:

Volume of Circular cone = V = (1/3)πr2h

where r is the radius in meters

and h is the height in meters

Substituting the given values in above equation, we get -

V = [tex]\frac{1}{3} * 3.14 * 4^2 * 10 = 167.47[/tex] cubic  meters.

The force required will be equal to the mass of water in the cone

[tex]= 167.47 * 1000[/tex]

= 167470 Kg

Weight = Mass * g

= 167470 * 10

= 1674700 Kgm/s^2

Answer:

Explanation:

Given that,

The current in the light bulb, I = 0.5 A

(a) We know that,

Electric current = charge/time

or

Q = It

Put t = 2 hours = 7200 s

So,

Q = 0.5 × 7200

Q = 3600 C

(b) Charge on one electron, [tex]Q=1.6\times 10^{-19}\ C[/tex]

Let there are n electrons flow through the bulb in 2 hours.

I = Q/t

Since, Q = ne

So,

I = ne/t

[tex]n=\dfrac{I\times t}{e}\\\\n=\dfrac{0.5\times 7200}{1.6\times 10^{-19}}\\\\n=2.25\times 10^{22}[/tex]

Hence, this is the required solution.

Answer:

True

Explanation:

D. A hockey puck sliding across ice

Answer:

25 m/s

Explanation:

200/8 = 25

Answer:

Explanation:

When two charges of equal magnitude and sign approach each other, they interact through Coulomb's law

          F = [tex]k \frac{q_ 1q_2}{z^2 }[/tex]k q1 q2 / r2

In you case the house are of equal magnitude and sign

          q1 = q2 = q

          F = k q2 / r2

Let's analyze this expression, the charge is repulsive on each charge, when they are on a collision course as they approach they feel an electric field opposite to their direction of movement, this field decreases its speed, the closer they get, the greater the repulsive force. , up to the point where this force is equal to or greater than the impulse, therefore the point where the velocity reaches zero, for this reason the particles do not actually touch

Answer:

A) i) Dynamic error ≈ 3.1%

   ii) phase shift ≈ -12°

B) 79971.89 rad/s

Explanation:

Given data :

Damping ratio = 0.5

natural frequency = 18,000 Hz

a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz

i) Dynamic error

This can be calculated using magnitude ratio formula attached below is the solution

dynamic error ≈ 3.1%

ii)  phase shift

This phase shift can be calculated using frequency dependent phase shift formula

phase shift ≈ -12°

B) Determine resonance frequency

  Wr = 2[tex]\pi[/tex] ( 18000 [tex]\sqrt{0.5}[/tex] ) = 79971.89 rad/s

C) The maximum magnitude ratio that the system can achieve

Answer:

42,250

Explanation:

It goes inside=

Displacemt

It does work=

Work done

To find efficiency of jule we do=

Dicplacement × Work done

650 × 65

42,250

Please mark me as a brainlist

Answer:It will take about 3000 years

Explanation:

Answer:

okay I'm a bit confused but I like the little emoji dudw

Answer:

?

Explanation:

.

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

Answer:

7,812 J

Explanation:

Using the relation:

Q = mcΔθ

Q = quantity of heat

C = specific heat capacity of lead

Δθ = temperature change (T2 - T1)

M = mass of substance

Q = mass * specific heat * Δθ

Q = 0.125kg * 128 * (327 – 20)

Q = 0.125 * 128 * 307

Q = 4912 J

For melting:

Q = mass * Hf

0.125 * (2.32 * 10^4)

= 2,900 J

Total = 4,912 J + 2,900 J = 7,812 J

Answer:

a)  k = 701.8 N / m, b)  m_{ast} = 61.1 kg, c)  v ’= -1.3 10⁻⁴ m / s

Explanation:

a) For this exercise let's use the relationship of the angular velocity

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

          k = w² m

the angular velocity is related to the period

          w = 2π / T

we substitute

          k = 4 π²    [tex]\frac{m}{T^2}[/tex]

let's calculate

          k = 4 π²   10 /0.75²

          k = 701.8 N / m

b) now repeat the measurement with an astronaut on the chair

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

where the mass Month the mass of the chair plus the mass of the astronaut

        M = m + [tex]m_{ast}[/tex]

          M = k / w²

          w = 2π / T

let's calculate

           w = 2π / 2

            w = π rad / s

            M = 701.8 /π²

            M = 71,111 kg

now we use that

          M = m + m_{ast}

          m_{ast} = M - m

          m_{ast} = 71.111 - 10.0

          m_{ast} = 61.1 kg

c) if the astronaut's movement is simple harmonic

          x = A cos wt

therefore the speed is

         v = [tex]\frac{dx}{dt}[/tex]

         v = -Aw sin wt

maximum speed is

          v = - Aw

          v = 0.100 π

          v = 0.31416 m / s

we can suppose that the movement of the space station and the astronaut  is equivalent to division of the same

initial instant. Before the move

         p₀ = 0

final instant. When the astronaut is moving

        p_f = M_station v’+ m_{ast} v

the moment is preserved

         p₀ = pf

         0 = M__{station} v ’+ m_{ast} v

         v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]

we substitute

         v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]

         v ’= -1.3 10⁻⁴ m / s

the negative sign indicates that the station is moving in the opposite direction from the astronaut