A hockey puck of mass m traveling along the x axis at 6.0 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 4.8 m/s at an angle of 30° above the x axis, what is the final velocity of the first puck

Answers

Answer 1

Answer:

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.

Explanation:

Let unknown velocity be v.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 6i + m x 0i = 6m i

Final momentum = m x (4.8cos 30 i + 4.8sin 30 j) + m x v = 4.16 m i + 2.4 m j + m v

Comparing

4.16 m i + 2.4 m j + m v = 6m i

v = 1.84 i - 2.4 j

Magnitude of velocity

       [tex]v=\sqrt{1.84^2+(-2.4)^2}=3.02m/s[/tex]

Direction,  

        [tex]\theta =tan^{-1}\left ( \frac{-2.4}{1.8}\right )=-53.13^0[/tex]     

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.

Answer 2
Final answer:

To find the final velocity of the first puck, apply conservation of momentum and solve for v'. The final velocity of the first puck is 4.8 m/s.

Explanation:

To find the final velocity of the first puck, we can apply the principle of conservation of linear momentum. The initial momentum of the system is given by m1v1 + m2v2, where m1 is the mass of the first puck, m2 is the mass of the second puck, v1 is the initial velocity of the first puck, and v2 is the initial velocity of the second puck.

Since the collision is elastic, the total momentum before and after the collision is conserved. So, m1v1 + m2v2 = (m1 + m2)v', where v' is the final velocity of both pucks after the collision. We can plug in the given values to find the final velocity of the first puck.

Let's solve for v': m1v1 + m2v2 = (m1 + m2)v' => (m1)(6.0 m/s) + (m2)(0 m/s) = (m1 + m2)(4.8 m/s).

From the given information, we know that the two pucks are identical, so m1 = m2. Substituting this into the equation, we get (m)(6.0 m/s) = 2(m)(4.8 m/s), where m is the mass of each puck. Simplifying this equation, we find the mass cancels out, leaving 6.0 m/s = 2(4.8 m/s). Solving for v', we find v' = 4.8 m/s.

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Related Questions

13. A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and the secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.

Answers

Answer:

The output power is 2 kW

Explanation:

It is given that,

Number of turns in primary coil, [tex]N_p=250[/tex]

Number of turns in secondary coil, [tex]N_s=500[/tex]

Voltage of primary coil, [tex]V_p=200\ V[/tex]

Current drawn from secondary coil, [tex]I_s=5\ A[/tex]

We need to find the power output. It is equal to the product of voltage and current. Firstly, we will find the voltage of secondary coil as :

[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}[/tex]

[tex]\dfrac{250}{500}=\dfrac{200}{V_s}[/tex]

[tex]V_s=400\ V[/tex]

So, the power output is :

[tex]P_s=V_s\times I_s[/tex]

[tex]P_s=400\ V\times 5\ A[/tex]

[tex]P_s=2000\ watts[/tex]

or

[tex]P_s=2\ kW[/tex]

So, the output power is 2 kW. Hence, this is the required solution.

Rearrange each of the following combinations of units to obtain a units that involves Joule ) Show kg m s intermediate steps, ( ).

Answers

Answer:

Explanation:

Joule is SI unit of work

Work = force x distance

Work = mass x acceleration x distance

Unit of mass is kg

Unit of acceleration is m/s^2

Unit of distance is m

So, unit of work = kg x m x m /s^2

So, Joule = kg m^2 / s^2

Part A: A charge +Q is located at the origin and a second charge, +9Q, is located at x= 15.8 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x -axis.

Part B: A charge +Q is located at the origin and a negative charge, -7Q, is located at a distance x= 19.6 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x-axis.

Answers

Answer:

Part a)

x = 3.95 cm

Part b)

x = - 11.9 cm

Explanation:

Part a)

Since both charges are of same sign

so the position at which net force is zero between two charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(15.8 - r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = 9Q[/tex]

[tex]\frac{Q}{r^2} = \frac{9Q}{(15.8 - r)^2}[/tex]

square root both sides

[tex](15.8 - r) = 3r[/tex]

[tex]r = 3.95 cm[/tex]

Part b)

Since both charges are of opposite sign

so the position at which net force is zero will lie on the other side of smaller charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(19.6 + r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = -7Q[/tex]

[tex]\frac{Q}{r^2} = \frac{7Q}{(19.6 + r)^2}[/tex]

square root both sides

[tex](19.6 + r) = 2.64r[/tex]

[tex]r = 11.9 cm[/tex]

so on x axis it will be at x = - 11.9 cm

The total solar irradiance (TSI) at Earth orbit is 1400 watts/m2 . Assuming this value can be represented as a single Poynting flux. Find the corresponding flux at the solar visible-light surface. Explain your methods.

(Distance between Sun & Earth is 150 million km)

Answers

Answer:

3.958 × 10²⁶ watt

Explanation:

Given:

Distance between earth and sun, [tex]R_e[/tex] = 150 ×10⁸ m

Total solar irradiance at earth orbit  = 1400 watt/m²

Now,

Area irradiated ([tex]A_e[/tex]) will be = [tex]4\pi R_e^2[/tex]

⇒ [tex]A_e[/tex] =  [tex]4\pi \times (150\times 10^{8})^2[/tex]

⇒ [tex]A_e[/tex] =  [tex]2.827\times 10^{23}m^2[/tex]

Therefore, the flux = Total solar irradiance at earth orbit × [tex]A_e[/tex]

the flux =   [tex]1400watt/m^2\times 2.827\times 10^{23}m^2[/tex]

⇒the flux = 3.958 × 10²⁶ watt

The 400 kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F= (3200) N, where t is in seconds. If the car has an initial velocity V1= 2m/s at s 0 and t= 0, determine the distance it moves the plane when (a) t 1 s and (b) f-5 s.

Answers

Answer:

(a) 110 m/s

(b) 42 m/s

Explanation:

mass, m = 400 kg, F = 3200 N, V1 = 2 m/s,

acceleration, a = Force / mass = 3200 / 400 = 8 m/s^2

(a) Use first equation of motion

v = V1 + a t

v = 2 + 8 x 1 = 10 m/s

(b) Again using first equation of motion

v = V1 + a t

v = 2 + 8 x 5 = 42 m/s

Thus, the velocity of plane after 1 second is 10 m/s and after 5 second the velocity is 42 m/s.

Final answer:

The 400 kg mine car moves a distance of 6 m after one second and 110 m after five seconds given an initial velocity of 2 m/s and a force of 3200 N.

Explanation:

The problem describes a physics scenario where a 400 kg mine car is pulled up an incline by a cable and motor. The force on the cable is given as F = 3200N and the initial velocity, V1, is given as 2 m/s. We can calculate the distance the car moves on the plane at different times using the physics equations of motion.

Let's use the equation of motion: s = ut + 1/2 at², where 's' is the distance moved, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.

Given that the net force is equal to mass times acceleration (F = ma), we can calculate the acceleration, 'a', as F/m. So, a = 3200N/400kg = 8 m/s².

(a) t = 1s: The distance moved is s = 2m/s * 1s + 1/2 * 8 m/s² * (1s)² = 2m + 4m = 6m.(b) t = 5s: The distance moved is s = 2m/s * 5s + 1/2 * 8 m/s² * (5s)² = 10m + 100m = 110m.

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The output of a generator is 440 V at 20 A. It is to be transmitted on a line with resistance of 0.60 Ω. To what voltage must the generator output be stepped up with a transformer if the power loss in transmission is not to exceed 0.010% of the original power?

Answers

Answer:

The voltage of the generator is 7.27 kV.

Explanation:

Given that,

Output of generator = 440 V

Current = 20 A

Resistance = 0.60 Ω

Power loss =0.010%

We need to calculate the total power of the generator

Using formula of power

[tex]P=VI[/tex]

Where, V = voltage

I = current

Put the value into the formula

[tex]P=440\times20[/tex]

[tex]P=8800\ W[/tex]

Th power lost on the transmission lines

[tex]P_{L}=0.010\% P[/tex]

[tex]P_{L} = 0.010\%\times8800[/tex]

[tex]P_{L}=0.88\ W[/tex]

The current passing through the transmission line

[tex]I'=\sqrt{\dfrac{P_{L}}{R}}[/tex]

[tex]I'=\sqrt{\dfrac{0.88}{0.60}}[/tex]

[tex]I'=1.211\ A[/tex]

We need to calculate the voltage of the generator

Using formula of voltage

[tex]V_{g}=\dfrac{P}{I'}[/tex]

Put the value into the formula

[tex]V_{g}=\dfrac{8800}{1.211}[/tex]

[tex]V_{g}=7.27\times10^{3}\ V[/tex]

[tex]V_{g}=7.27\ kV[/tex]

Hence, The voltage of the generator is 7.27 kV.

Final answer:

The voltage of the generator output needs to be stepped up with a transformer to limit power losses in transmission.

Explanation:

To limit power losses in transmission, the voltage of the generator output needs to be stepped up with a transformer. The formula to calculate power loss is P_loss = I^2 x R. Given that the power loss should not exceed 0.010% of the original power, we can calculate the maximum allowable power loss. By rearranging the formula, we can find the voltage required for the generator output with a transformer.

First, we calculate the original power using P = V x I. Then, we calculate the maximum allowable power loss as a percentage of the original power. Next, we substitute the given values into the formula and solve for the new voltage using the rearranged formula.

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Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnected and the capacitors are connected to a 33 ohm resistor, how long should it take for the voltage to cross the capacitors to drop to 2.2 volts (36.8% of the original 6.0 volts)?

Answers

Answer:

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

C = 1.0 + 1.0

C = 2.0 F

R = resistance = 33 Ω

Time constant is given as

T = RC

T = 33 x 2

T = 66 sec

V₀ = initial potential difference across the combination = 6.0 Volts

V = final potential difference = 2.2 volts

Using the equation

[tex]V = V_{o} e^{\frac{-t}{T}}[/tex]

[tex]2.2 = 6 e^{\frac{-t}{66}}[/tex]

t = 66.2 sec

wo kids are on a seesaw. The one on the left has a mass of 75 kg and is sitting 1.5 m from the pivot point. The one on the right has a mass of 60 kg and is sitting 1.8 m from the pivot point. What is the net torque on the system?

Answers

Answer:

4.5 Nm (Anticlockwise)

Explanation:

Let the 75 kg kid is sitting at the left end and the 60 kg kid is sitting on the right end.

Anticlockwise Torque = 75 x 1.5 = 112.5 Nm

clockwise Torque = 60 x 1.8 = 108 Nm

Net torque = Anticlockwise torque - clockwise torque

Net Torque = 112.5 - 108 = 4.5 Nm (Anticlockwise)

What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m.s^{-1} ( i ) 9.5 × 10 5 m . s − 1 (ii)\:\:\:\:2.2\:\times\:10^9\:m.s^{-1} ( i i ) 2.2 × 10 9 m . s − 1 (iii)\:\:\:3.9\:\times\:10^8\:m.s^{-1} ( i i i ) 3.9 × 10 8 m . s − 1 (iv)\:\:\:\:1.7\:\times\:10^6\:m.s^{-1}

Answers

Answer:

Velocity of a proton, [tex]v=1.7\times 10^6\ m/s[/tex]    

Explanation:

It is given that,

Potential difference, [tex]V=15\ kV=15\times 10^3\ V[/tex]

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=qV[/tex]

q is the charge of proton

m is the mass of proton

[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}[/tex]

[tex]v=1695361.75\ m/s[/tex]

[tex]v=1.69\times 10^6\ m/s[/tex]

or

[tex]v=1.7\times 10^6\ m/s[/tex]

So, the velocity of a proton is [tex]1.7\times 10^6\ m/s[/tex]. Hence, this is the required solution.

A string of length L= 1.2 m and mass m = 20g is under 400 N of tension, its two ends are fixed. a. How many nodes will you see in the 5th harmonic.
i. 4
ii. 5
iii. 6
iv. 7
v. None of the above.

Answers

Answer:

(iii) 6

Explanation:

Part a)

Since it is given that both ends are fixed and it is vibrating in 5th harmonic

So here it will have 5 number of loops in it

so we can draw it in following way

each loop will have 1 antinode and two nodes which means number of nodes is one more than number of anti-nodes

So there are 5 loops which means it will have 5 antinodes

and hence there will be 6 nodes in it

so correct answer will be

(iii) 6

Write a balanced half-reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:  [tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Explanation:

Reduction is a process where electrons are gained and acidic solution means presence of [tex]H^+[/tex] ions.

Reduction of [tex]MnO_2[/tex] to [tex]Mn^{2+}[/tex]

Mn is in +4 oxidation state in [tex]MnO_2[/tex] which goes to +2 state in [tex]Mn^{2+}[/tex] by gain of 2 electrons.

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)[/tex]

In order to balance oxygen atoms:

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)+2H_2O[/tex]

In order to balance hydrogen atoms:

[tex]MnO_2(s)+4H^+(aq)\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

In order to balance charges:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l[/tex]

Thus the net balanced half reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Final answer:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, the balanced half-reaction is MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l).

Explanation:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, we can follow these steps:

1. Write the half-reaction for reduction, adjusting the physical states of the reactants and products:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

2. Balance the number of Mn and O atoms on each side of the equation:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

3. Balance the H atoms by adding H+ ions:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

4. Balance the charges by adding electrons:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

Therefore, the balanced half-reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

A rhinoceros beetle rides the rín of a small disk that rotates like a merry-go-round. If the beetle crawls toward the center of the disk, do the following (each relative to the central axis) increase, decrease, or remain the same for the beetle-disk system: (a) rotational inertia, (b) angular momentum, and (c) angular speed?

Answers

Answer:

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

Explanation:

Here the Beetle is initially moving along the rim of the disc

So here during the motion of beetle there is no external force on the system of beetle and the disc.

So here we can also say that there is no torque acting on the system

so angular momentum of the disc + beetle system will remain conserved

so here we have

[tex]I_1\omega_1 = I_2 \omega_2[/tex]

here as the beetle crawls towards the centre of the disk then

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 250 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 30 centimeters to 60 centimeters? N-cm

Answers

Answer:

112.5 J

Explanation:

F = 250 N, x = 30 cm = 0.3 m

Let the spring constant be K.

By using the Hooke's law

F = k x

250 = k x 0.3

k = 833.3 N / m

xi = 30 cm = 0.3 m, xf  60 cm = 0.6 m

Work done = 1/2 k (xf^2 - xi^2)

Work done = 0.5 x 833.33 x (0.6^2 - 0.3^2)

Work done = 112.5 J

Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29×1011 m.

Answers

Answer:

(a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

Explanation:

Given that,

Mass of baby = 4.20 kg

Mass of father = 100 kg

Distance = 0.200 m

We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times100}{0.200^2}[/tex]

[tex]F=7\times10^{-7}\ N[/tex]

(b). We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times1.9\times10^{27}}{(6.29\times10^{11})^2}[/tex]

[tex]F=1.35\times10^{-6}\ N[/tex]

Hence, (a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

The magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N

What is gravitational potential energy?

Gravitational potential energy is the energy which a body posses because of its position.

The gravitational potential energy of a body is given as,

[tex]G=\dfrac{Fr^2}{Mm}[/tex]

Here, (m) is the mass of the body, (F) is the gravitational force and (r) is the height of the body.

The mass of baby is 4.20 kg and mass of father is 100 kg. The distance between them is 0.200 m. Put the values in the above formula as,

[tex]6.67\times10^{-11}=\dfrac{F(0.200)^2}{100(4.20)}\\F=7\times10^{-7}\rm \; N[/tex]

(b) The magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29×1011 m.

Put the values in the above formula again as,

[tex]6.67\times10^{-11}=\dfrac{F(6.29\times10^{11})^2}{(1.9\times10^{27})(4.20)}\\F=1.35\times10^{-6}\rm \; N[/tex]

Thus, the magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N.

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lectric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads. What is the resistance of the heater wire?

Answers

Answer:

The resistance of the heater wire is of R= 0.68 Ω.

Explanation:

1 cal/s = 4.184 W

P= 50 cal/s = 209.2 W

V= 12V

P= V* I

I= P/V

I= 17.43 A

P= I² * R

R= P / I²

R= 0.68 Ω

A 0.42 kg football is thrown with a velocity of 17 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.017 S. What is the force exerted on the receiver? Answer in units of N

Answers

Answer:

420 N

Explanation:

m = 0.42 Kg, u = 17 m/s, v = 0 m/s, t = 0.017 s

By first law of Newtons' laws of motion, the rate of change in momentum is force, F = m (v - u) / t

F = 0.42 x ( 0 - 17) / 0.017

F = - 420 N

Negative sign shows hat the force is resistive that means the ball finally comes to rest.

A current of 0.2 A flows through a 3 m long wire that is perpendicular to a 0.3 T magnetic field. What is the magnitude of the force on the wire in units of newtons?

Answers

Answer:

Magnetic force, F =  0.18 N

Explanation:

It is given that,

Current flowing in the wire, I = 0.2 A

Length of the wire, L = 3 m

Magnetic field, B = 0.3 T

It is placed perpendicular to the magnetic field. We need to find the magnitude of force on the wire. It is given by :

[tex]F=ILB\ sin\theta[/tex]

[tex]F=0.2\ A\times 3\ m\times 0.3\ T\ sin(90)[/tex]

F = 0.18 N

So, the magnitude of force on the wire is 0.18 N. Hence, this is the required solution.

g Complete the following statements. (a) A substance that conducts electricity but whose conduction is not temperature dependent is called Correct: Your answer is correct. . (b) An example of a conductor is . (c) An example of a semiconductor is . (d) An example of an insulator is .

Answers

Answer:

Explanation:

(a) It is called conductors.

The conductors are the materials which can allow the current to pass through it.

(b) The example of conductor is copper, iron, etc.

The best conductor of electricity is silver.

(c) The example of semiconductor is silicon, germanium, etc.

The semiconductors are the materials which are insulators at normal temperature, but if the temperature increases, the conductivity of semiconductor increases.

(d) An example of insulator is wood.

Insulators are the materials which do not allow the current to pass through them.

A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time interval?

Answers

Answer:

Number of electrons, [tex]n=5.62\times 10^{21}[/tex]

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, [tex]I=\dfrac{q}{t}[/tex]

[tex]I\times t=n\times e[/tex]

[tex]n=\dfrac{It}{e}[/tex]

e is the charge of an electron

[tex]n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}[/tex]

[tex]n=5.62\times 10^{21}[/tex]

So, the number of electrons pass through the resistor is [tex]5.62\times 10^{21}[/tex]. Hence, this is the required solution.

A 15-uF capacitor is connected to a 50-V battery and becomes fullycharged. The battery is
removed and a slab of dielectric that completely fills the spacebetween the plates is inserted.
If the dielectric has a dielectric constant of 5.0:

A. what is the capacitance of the capacitor after the
slab is inserted?


B. what is the voltage across the capacitor's plates
after the slab is inserted?

Answers

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0)(15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.

What is capacitance of capacitor?

The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference.

The capacitance of the capacitor is 15-uF and it is connected to  50-V battery. The value of the dielectric constant is 5.0.

A. The capacitance of the capacitor after the slab is inserted

The dielectric constant of the slab is the ratio of capacitance of the capacitor after the slab and the capacitance of the before the slab inserted.

As the value of the dielectric constant is 5.0. Thus the capacitance of the capacitor after the slab is inserted given as,

[tex]0.5=\dfrac{C}{15\times10^{-6}}\\C=75\rm \mu F[/tex]

Thus, the capacitance of the capacitor after the slab is inserted is 75-uF.

B. The voltage across the capacitor's plates after the slab is inserted

The dielectric constant of the slab is the ratio of voltage across the capacitor's of the capacitor before the slab inserted and the voltage across the capacitor after the slab inserted.

As the value of the dielectric constant is 5.0. Thus the voltage across the capacitor's plates after the slab is inserted can be given as,

[tex]V=\dfrac{50}{5.0}\\C=10\rm V[/tex]

Thus, the voltage across the capacitor's plates after the slab is inserted is 10 volts.

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.

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Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B. B) three times smaller than the flux of electric field through Gaussian surface B. C) unrelated to the flux of electric field through Gaussian surface B. D) equal to the flux of electric field through Gaussian surface B. E) three times larger than the flux of electric field through Gaussian surface B

Answers

The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Therefore, the flux through surface A is three times larger than the flux through surface B.

The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Since the flux of electric field is proportional to the area of the surface, the flux through surface A is three times larger than the flux through surface B. Therefore, the correct answer is E) three times larger than the flux of electric field through Gaussian surface B.

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Final answer:

The correct answer is D) equal to the flux of the electric field through Gaussian surface B. The electric flux depends on the charge enclosed by the Gaussian surface, not on its size or shape, according to Gauss's law.

Explanation:

The question asks whether Gaussian surfaces A and B, enclosing the same positive charge +Q, with surface A having three times the area of surface B, have different electric fluxes. According to Gauss's law, the electric flux (Φ) through a closed surface is proportional to the charge enclosed (Φ = Q/ε0). Since both surfaces enclose the same charge, the flux through each surface must be the same, regardless of their respective areas.

Thus, the correct answer is D) equal to the flux of the electric field through Gaussian surface B because the flux depends only on the amount of enclosed charge, not on the size or shape of the Gaussian surface.

Identify the least dense and most dense planets and provide an explanation as to why they are the least and most dense.

Answers

Answer:

Explained

Explanation:

The most dense planet in  our solar system is Earth. Earth is most dense planet because

Earth is made of dense material like rocks, core is made of metals, soil and water. The average density of earth is 5.5 gram/cm^3.The overall mass of Earth is higher than those of other rocky planets. This heaviness accounts for higher density of earth.

The least dense planet of the solar system is Saturn because Saturn is mostly made of gases and its size is smaller than the Jupiter. Jupiter has more gravity hence its density is higher to Saturn. Moreover, Uranus and Neptune are ice giants. Although they are also made of gases but due their distance from sun most of these gases have solidified. Hence making them more dense than Saturn and Jupiter.

Final answer:

Saturn is the least dense planet, while Earth is the densest planet in our solar system.

Explanation:

The least dense planet in our solar system is Saturn, while the densest planet is Earth. Saturn is a gas giant composed mostly of hydrogen and helium, which have lower densities compared to the rocks and metals found on rocky planets like Earth. Earth, on the other hand, has a higher density due to its solid composition and a core made of iron and nickel.

The return-air ventilation duct in a home has a cross-sectional area of 900 cm^2. The air in a room that has dimensions 7.0 m ×× 11.0 m ×× 2.4 m is to be completely circulated in a 40-min cycle. What is the speed of the air in the duct? (Express your answer to two significant figures.) (m/s)

Answers

Answer:

0.86 m/s

Explanation:

A = cross-sectional area of the duct = 900 cm² = 900 x 10⁻⁴ m²

v = speed of air in the duct

t = time period of circulation = 40 min = 40 x 60 sec = 2400 sec

V = Volume of the air in the room = volume of room = 7 x 11 x 2.4 = 184.8 m³

Volume of air in the room is given as

V = A v t

inserting the values

184.8 = (900 x 10⁻⁴) (2400) v

v = 0.86 m/s

An electric coffee maker has a heating element that operates at 120 V and with a current of 2.00 A. Assuming the water absorbs all the energy delivered to the heating element, calculate the time interval (in s) during which the temperature of 0.489 kg of water rises from room temperature (23.0°C) to the boiling point. (The specific heat of water is 4,186 J/(kg · °C).)

Answers

Answer:

10.95 minute

Explanation:

V = 120 V, I = 2 A, m = 0.489 kg, c = 4186 J/kgC, T1 = 23 C , T2 = 100 C

Let time be the t.

Heat energy is equal to electrical energy

m x c (T2 - T1) = V x I x t

0.489 x 4186 x (100 - 23) = 120 x 2 x t

t = 656.73 second

t = 10.95 minute

A 10 kg mass starts from rest at the top of a frictionless incline and slides down the ramp. The ramp makes an angle of 0=30" . Take the positive x axis to be down the ramp. What is the acceleration of the mass and it moves down the ramp? m a) a 0 -4.9 m b) a=-4.9- m S m c) a+4.9- a, 0 2. 2 d) None of the above.

Answers

Answer:

Explanation:

When a body starts sliding on an inclined plane, the acceleration of body is due to its impotent 9f weight. The component of weight is mg Sin theta along the plane.

Thus, the acceleration is g Sin theta

= 9.8 × Sin 30 = 4.9 m/s^2

A bolt comes loose from underneath an elevator that is moving upward at a speed of 5 m/s. The bolt reaches the bottom of the elevator shaft in 3.1 s. (a) How high up was the elevator when the bolt came loose? (In m)
(b) What is the speed of the bolt when it hits the bottom of the shaft? (In m/s)

Answers

Answer:

a)  The elevator was  31.64 m high up when the bolt came loose.

b)  Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

Explanation:

a) Considering motion of bolt:-

Initial velocity, u =  5 m/s

Acceleration , a = -9.81 m/s²

Time = 3.1 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= 5 x 3.1 - 0.5 x 9.81 x 3.1²

    s = 0 x t + 0.5 x 9.81 x t²

    s = -31.64 m

The elevator was  31.64 m high up when the bolt came loose.

b) We have equation of motion v = u + at

  Initial velocity, u =  5 m/s

 Acceleration , a = -9.81 m/s²

 Time = 3.1 s  

Substituting

  v = u + at

  v  = 5 - 9.81 x 3.1 = -25.41 m/s

Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is perpendicular to the door. What is the magnitude of the torque?

Answers

Answer:

[tex]\tau = 7.63 Nm[/tex]

Explanation:

As we know that moment of force is given as

[tex]\tau = \vec r \times \vec F[/tex]

now we have

[tex]\vec r = 1.2 m[/tex]

[tex]\vec F = 14 N[/tex]

now from above formula we have

[tex]\tau = r F sin\theta[/tex]

here we know that

[tex]\theta = 27 degree[/tex]

so we have

[tex]\tau = (1.2)(14) sin27[/tex]

[tex]\tau = 7.63 Nm[/tex]

A batter hits a 0.140-kg baseball that was approaching him at 50.0 m/s and, as a result, the ball leaves the bat at 35.0 m/s in the direction of the pitcher. What is the magnitude of the impulse delivered to the baseball?

Answers

Answer:

Impulse, J = 2.1 kg-m/s

Explanation:

Given that,

Mass of baseball, m = 0.14 kg

It was approaching him at 50.0 m/s and, as a result, the ball leaves the bat at 35.0 m/s in the direction of the pitcher. We need to find the magnitude of Impulse delivered to the baseball.

The change in momentum is equal to the Impulse imparted to the ball i.e.

[tex]J=m(v-u)[/tex]

[tex]J=0.14(-35-50)[/tex]

J = -2.1 kg-m/s

So, the Impulse delivered to the baseball is 2.1 kg-m/s

A car starts from the origin and is driven 1.88 km south, then 9.05 km in a direction 47° north of east. Relative to the origin, what is the car's final location? a) Express your answer in terms of an angle (in degree)
b) and a distance.

Answers

Answer:

(a) θ = 55.85 degree

(b) 7.89 km

Explanation:

Using vector notations

A = 1.88 km south = 1.88 (- j) km = - 1.88 j km

B = 9.05 km 47 degree north of east

B = 9.05 ( Cos 47 i + Sin 47 j) km

B = (6.17 i + 6.62 j) km

Net displacement is

D = A + B

D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j

(a) Angle made with positive X axis

tanθ = 6.62 / 4.29 = 1.474

θ = 55.85 degree

(b) distance = [tex]Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}[/tex]

distance = 7.89 km

Final answer:

To find the car's final location, add the displacements in the north and east directions. The car's final location is 6.4 km north and 6.0 km east relative to the origin.

Explanation:

To find the car's final location, we need to add the displacements in both the north and east directions.

First, let's calculate the north displacement by using the distance formula:

North Displacement = 9.05 km * sin(47°) = 6.4 km

Next, let's calculate the east displacement by using the distance formula:

East Displacement = 9.05 km * cos(47°) = 6.0 km

Therefore, the car's final location relative to the origin is 6.4 km north and 6.0 km east.

An undiscovered planet, many light-years from Earth, has one moon which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.31500 x 10^5 km and the planet has a radius of 4.150 x10^3 km and a mass of 7.15 x10^22 kg, how long (in days) does it take the moon to make one revolution around the planet. The gravitational constant is 6.67 x10^-11 N m^2/kg^2.

Answers

Answer:

118.06 days

Explanation:

d = distance of the center of moon from surface of planet = 2.315 x 10⁵ km = 2.315 x 10⁸ m

R = radius of the planet = 4.15 x 10³ km = 4.15 x 10⁵ m

r = center to center distance between the planet and moon = R + d

M = mass of the planet = 7.15 x 10²² kg

T = Time period of revolution around the planet

Using Kepler's third law

[tex]T^{2}=\frac{4\pi ^{2}r^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4\pi ^{2}(R + d)^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4(3.14)^{2}((4.15\times 10^{5}) + (2.315\times 10^{8}))^{3}}{(6.67\times 10^{-11})(7.15\times 10^{22})}[/tex]

T = 1.02 x 10⁷ sec

we know that , 1 day = 24 h = 24 x 3600 sec = 86400 sec

T = [tex](1.02 \times 10^{7} sec)\frac{1 day}{86400 sec}[/tex]

T = 118.06 days

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