What is the magnitude of the electric field at a distance 60 cm from the center of the sphere? The radius of the sphere 30 cm, the charge on the sphere is 1.56 × 10−5 C and the permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in units of N/C.

Answer:

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

[tex]m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}[/tex]

[tex]\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s[/tex]

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Answer:

42.4 Npa

Explanation:

Explanation is attached in the picture below

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

Answer:

10ns

Explanation:

Suppose the EM wave travels at light speed [tex]c = 3\times10^8 m/s[/tex]. A change in travel distance of the order of 3 m would result of a change in travel time of

[tex] \Delta t = \frac{\Delta s}{c} = \frac{3}{3\times10^8} = 10^{-8} s = 10 ns[/tex]

To detect a change in travel distance of about 3 m, a GPS receiver must measure a change in travel time of approximately 10 nanoseconds. This is calculated using the speed of light and the time = distance/speed formula.

To determine the associated travel time that must be measured for a change in travel distance of 3 meters, we need to know the speed of electromagnetic waves, which is the speed of light, typically denoted 'c'. The speed of light is approximately 3.00 x 108 meters per second (m/s).

Now we can calculate the time it takes for light to travel a given distance using the formula time (t) = distance (d) / speed (s). Substituting for our given distance (3 m) and the speed of light, we get t = 3 m / 3.00 x 108 m/s. This results in a time of approximately 1 x 10-8 seconds or 10 nanoseconds (ns).

Therefore, to detect a change in travel distance of about 3 m, the GPS receiver must be able to measure a change in travel time of roughly 10 nanoseconds.

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Answer:

The time is [tex]110.16\times10^{-3}\ sec[/tex]

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

[tex]i_{0}=\dfrac{V_{0}}{R}[/tex]

Put the value into the formula

[tex]i_{0}=\dfrac{150}{1.8\times10^{3}}[/tex]

[tex]i_{0}=83.3\times10^{-3}\ A[/tex]

We need to calculate the time

Using formula of current

[tex]i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}[/tex]

Put the value into the formula

[tex]50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}[/tex]

[tex]\dfrac{50}{83.3}=e^{\frac{-t}{RC}}[/tex]

[tex]\dfrac{-t}{RC}=ln(0.600)[/tex]

[tex]t=0.51\times1.8\times10^{3}\times120\times10^{-6}[/tex]

[tex]t=110.16\times10^{-3}\ sec[/tex]

Hence, The time is [tex]110.16\times10^{-3}\ sec[/tex]

The current across the student's chest will exceed the danger level for approximately 216 seconds.

The current that flows through the student's chest when he accidentally touches the terminals of the 120 mF capacitor can be calculated using Ohm's law: I = V/R, where I is the current, V is the voltage, and R is the resistance. In this case, the voltage is 150 V and the resistance of the student's body is 1.8 kΩ. Plugging in these values, we can find the current.

I = (150 V)/(1.8 kΩ) = 0.0833 A

To find how long the current will exceed the danger level of 50 mA, we need to calculate the time it takes for the charge on the capacitor to dissipate through the student's body. Since the current is constant, we can use the formula Q = I*t, where Q is the charge and t is the time. Rearranging the formula, we can solve for t.

t = Q/I = (120 mF * 150 V) / (0.0833 A) = 216 s

Therefore, the current will exceed the danger level of 50 mA for approximately 216 seconds.

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Answer:

0.65 A.

Explanation:

Given:

Pmax = 10 W

R = 23.9 Ω

Formula for calculating power,

P = I × V

= I^2 × R

I^2 = 10/23.9

I = 0.65 A.

Answer:

Therefore, the force exerted by the magnetic field on the particle goes along the positive y- direction

Explanation:

The equation for the force exerted on a particle due to the magnetic field is equal to:

F = q*(v * B)

If we replace vi for v, +q for q and -Bk for B, we have:

F = +q*(vi*(-Bk)) = -q*v*B*(i * k)

From this equation we have that the vector i is in direction -x and k in direction -z

We compute the cross product of two unit vectors:

i * k = -j, where j is the vector along -y direction

Replacing we have:

F = -q*v*B*(-j) = -q*v*B*j

Therefore, the force exerted by the magnetic field on the particle goes along the positive y- direction

Answer:

The lethal voltage for the electrician under those conditions is 126.5 V.

Explanation:

To discover what is the lethal voltage to the electrician we need to find out what is the voltage that produces 55 mA = 0.055 A when across a resistance of 2300 Ohms (Electrician's body resistancy). For that we'll use Ohm's Law wich is expressed by the following equation:

V = i*R

Where V is the voltage we want to find out, i is the current wich is lethal to the electrician and R is his body resistance. By applying the given values we have:

V = 0.055*2300 = 126.5 V.

The lethal voltage for the electrician under those conditions is 126.5 V.

Answer:

126.5 V

Explanation:

Using Ohm's Law,

V = IR............................. Equation 1

Where V = Voltage, I = current, R = resistance.

Note: The current needed to bring about the fatal voltage is equal to the current that will cause human being to be electrocuted.

Given: I = 55 mA = 55/1000 = 0.055 A, R = 2300 Ω

Substitute into equation 1

V = 0.055×2300

V = 126.5 V.

Hence the fatal voltage = 126.5 V

Answer:

a) [tex]-22.5 rad/s^2[/tex]

b) 30.6 revolutions

c) 4.13 s

d) 52.9 m

e) 25.6 m/s

Explanation:

a)

The relationship between linear acceleration and angular acceleration for an object in circular motion is given by

[tex]a=\alpha r[/tex]

where

[tex]a[/tex] is the linear acceleration

[tex]\alpha[/tex] is the angular acceleration

r is the radius of the motion of the object

For the tires of the  car in this problem, we have:

[tex]a=-6.2 m/s^2[/tex] is the linear acceleration (the car is slowing  down, so it is a deceleration, therefore the negative sign)

r = 0.275 m is the radius of the tires

Solving for [tex]\alpha[/tex], we find the angular acceleration:

[tex]\alpha = \frac{a}{r}=\frac{-6.2}{0.275}=-22.5 rad/s^2[/tex]

b)

To solve this part of the problem, we can use the suvat equation for the rotational motion, in particular:

[tex]\omega^2 - \omega_0^2 = 2\alpha \theta[/tex]

where:

[tex]\omega[/tex] is the final angular velocity

[tex]\omega_0[/tex] is the initial angular velocity

[tex]\alpha[/tex] is the angular acceleration

[tex]\theta[/tex] is the angular displacement

Here we have:

[tex]\omega=0[/tex] (the tires come to a stop)

[tex]\omega_0 = 93 rad/s[/tex]

[tex]\alpha = -22.5 rad/s^2[/tex]

Solving for [tex]\theta[/tex], we find the angular displacement:

[tex]\theta=\frac{\omega^2-\omega_0^2}{2\alpha}=\frac{0^2-(93)^2}{2(-22.5)}=192.2 rad[/tex]

And since 1 revolution = [tex]2\pi rad[/tex],

[tex]\theta=\frac{192.2}{2\pi}=30.6 rev[/tex]

c)

To solve this part, we can use another suvat equation:

[tex]\omega=\omega_0 + \alpha t[/tex]

where in this case, we have:

[tex]\omega=0[/tex] is the final angular velocity, since the tires come to a stop

[tex]\omega_0 = 93 rad/s[/tex] is the initial angular velocity

[tex]\alpha=-22.5 rad/s^2[/tex] is the angular acceleration

t is the time

Solving for t, we can find the time required for the tires (and the car) to sopt:

[tex]t=\frac{\omega-\omega_0}{\alpha}=\frac{0-93}{-22.5}=4.13 s[/tex]

d)

The car travels with a uniformly accelerated motion, so we can find the distance it covers by using the suvat equations for linear motion:

[tex]s=vt-\frac{1}{2}at^2[/tex]

where:

v = 0 is the final velocity of the car (zero since it comes to a stop)

t = 4.13 s is the time taken for the car to stop

[tex]a=-6.2 m/s^2[/tex] is the deceleration for the car

s is the distance covered during this motion

Therefore, substituting all values and calculating s, we find the distance covered:

[tex]s=0-\frac{1}{2}(-6.2)(4.13)^2=52.9 m[/tex]

e)

The relationship between angular velocity and linear velocity for a rotational motion is given by

[tex]v=\omega r[/tex]

where

v is the linear velocity

[tex]\omega[/tex] is the angular speed

r is the radius of the circular motion

In this problem:

[tex]\omega_0 = 93 rad/s[/tex] is the initial angular speed of the tires

r = 0.275 m is the radius of the tires

Therefore, the initial velocity of the car is:

[tex]u=\omega_0 r = (93)(0.275)=25.6 m/s[/tex] is the initial velocity of the car

Answer:

The electric force increases by a factor of 4.

Explanation:

The electric force between two charges [tex]q_1[/tex] and [tex]q_2[/tex] separated a distance d can be calculated using Coulomb's Law:

[tex]F=\frac{kq_1q_2}{d^2}[/tex]

where [tex]k=9\times10^9Nm^2/C^2[/tex] is the Coulomb constant.

If the value of each charge is doubled, then we will have a force between them which is:

[tex]F'=\frac{k(2q_1)(2q_2)}{d^2}=4\frac{kq_1q_2}{d^2}=4F[/tex]

So the new force is 4 times larger than the original force.

Doubling the charge on each particle increases the electric force between them by a factor of 4.

The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.

Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.

So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.

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Answer:

a) [tex]||\vec F_{B}||=62.664\,N[/tex], b) From east to west.

Explanation:

Vectorially, the magnetic force can be calculed by the following formula:

[tex]\vec F_{B} = i\cdot \vec l\, \times \, \vec B[/tex]

The cross product is:

[tex]\vec F_{B} = \left|\begin{array}{ccc}i&j&k\\1015000\,A\cdot m&0\,A\cdot m&0\,A\cdot m\\22.471\times 10^{-6}\,T&-61.738\times 10^{-6}\,T&0\,T\end{array}\right|[/tex]

[tex]\vec F_{B} = - 62.664\,N\cdot k[/tex]

a) The magnitude of the magnetic force is:

[tex]||\vec F_{B}||=62.664\,N[/tex]

b) The direction of the magnetic force is:

From east to west.

Answer:

a. Perpendicular.

Explanation:

The relation between angular velocity vector and linear velocity vector is described by a cross product:

[tex]\vec v = \vec \omega \times \vec r[/tex]

Where [tex]\vec r[/tex] is perpendicular to the rotation axis and [tex]\vec \omega[/tex] is parallel to the same axis. By definition of cross product, [tex]\vec v[/tex] is a vector which is perpendicular to both vectors. Therefore, linear velocity vector is perpendicular to angular velocity vector. The correct answer is A.

The angular velocity vector, which points along the rotation axis, is perpendicular (Option A) to the linear velocity vector, which is tangent to the rotation path.

The angular velocity of a point on a rotating object is a vector that points along the axis of rotation. This vector's direction is determined with respect to the right-hand rule: if the fingers on your right-hand curl from the x-axis toward the y-axis, your thumb points in the direction of the positive z-axis. For an angular velocity pointing along the positive z-axis, the rotation is counterclockwise, whereas for an angular velocity pointing along the negative z-axis, the rotation is clockwise.

On the other hand, the linear velocity of the same point is a vector that is tangent to the path of rotation. It represents the instantaneous linear speed of the point as it moves along the path.

Therefore, given the definitions and directions of both vectors, the angular velocity vector of a point on a rotating object is perpendicular to the linear velocity vector of that point.

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Answer: Average kinetic energy is greater than gravitational potential energy.

Explanation: The average kinetic energy formulae for a molecule of a gas relative to temperature (in Kelvin) is given below as

E = 3/2 (KT).

Where E = average kinetic energy =?

K = boltzman constant = 1.381×10^-23 m²kg/s²k

T = temperature = 300 k

By substituting the parameters in the formulae, we have that

E = 3/2 ×( 1.381×10^-23 × 300)

E = 1.5 × 4.143×10^-21

E = 6.2145×10^-21 J

To get the gravitational potential energy, we use the fact that

Gravitational potential energy = gravitational energy at the top - gravitational energy at bottom.

At the top, the height of cube is h= 15cm = 0.15m, g = acceleration dude to gravity = 9.8m/s², m = mass of molecule of oxygen = 1.661×10^-27 kg

Gravitational potential energy = mgh = 1.661×10^-27× 9.8 × 0.15 = 2.442×10^-27 J

At the bottom, height is zero, hence gravitational potential energy is also zero.

Hence the final gravitational potential energy = potential energy at top - potential energy at bottom =

2.442×10^-27 - 0 = 2.442×10^-27 J.

Gravitational potential energy = 2.442×10^-27 J

Average kinetic energy = 6.2145×10^-21 J

As we can see that the average kinetic energy is bigger than the gravitational potential energy.

Compare the average kinetic energy of oxygen molecules at 300 K to the change in their gravitational potential energy if they fall from the top to the bottom of a container.

Oxygen (O2) molecules at 300 K have an average kinetic energy of 3kT, where k is Boltzmann's constant. If an oxygen molecule falls from the top of a container to the bottom, its change in gravitational potential energy is mgΔh, where m is the mass of the molecule, g is the acceleration due to gravity, and Δh is the height change.

Answer:

The net force is 300.8 N

Explanation:

∆H = Cp(T1 - T2) = 1/2(V2^2 - V1^2)

Cp is the heat capacity of air at constant pressure = 1005 J/kg

T1 is initial temperature of air = 500 K

T2 is the exit temperature of air = 440 K

V1 is the initial velocity of air = 50 m/s

V2 is the exit velocity of air

1005(500 - 440) = 1/2(V2^2 - 50^2)

60,300×2 = V2^2 - 2,500

V2^2 = 120,600 + 2,500 = 123,100

V2 = sqrt(123,100) = 350.8 m/s

Net force (F) = mass flow rate × change in velocity = 1 kg/s × (350.8 - 50)m/s = 1 kg/s × 300.8 m/s = 300.8 kgm/s^2 = 300.8 N

The net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].

Given:

- Inlet conditions (state 1):

 - [tex]\( P_1 = 12 \) bar = \( 12 \times 10^5 \) Pa[/tex]

 - [tex]\( T_1 = 500 \) K[/tex]

 - [tex]\( v_1 = 50 \) m/s[/tex]

- Exit conditions (state 2):

 - [tex]\( P_2 = 7 \) bar = \( 7 \times 10^5 \) Pa[/tex]

 - [tex]\( T_2 = 440 \) K[/tex]

- Mass flow rate, [tex]\( \dot{m} = 1 \) kg/s[/tex]

1. Calculate specific volumes:

 [tex]\( v_1 = \frac{RT_1}{P_1} = \frac{287 \times 500}{12 \times 10^5} \approx 0.02379 \) m^3/kg[/tex]

  [tex]\( v_2 = \frac{RT_2}{P_2} = \frac{287 \times 440}{7 \times 10^5} \approx 0.02086 \) m^3/kg[/tex]

2. Calculate the change in specific volume:

 [tex]\( \Delta v = v_2 - v_1 = 0.02086 - 0.02379 = -0.00293 \) m^3/kg[/tex]

3. Calculate the net force in the direction of flow:

  [tex]\( F = \dot{m} \cdot \Delta v = 1 \cdot (-0.00293) = -0.00293 \) N[/tex]

Since the force is negative, it indicates that the force is acting opposite to the direction of flow.

Therefore, the net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].

Answer with Explanation:

We are given that

Distance between two parallel long wires=r=3.3 cm=[tex]\frac{3.3}{100}=0.033m[/tex]

1 m=100 cm

[tex]\frac{F}{l}=5\times 10^{-5} N/m[/tex]

[tex]I_1=0.62 A[/tex]

a.We have to find the current in the second wire.

We know that

[tex]\frac{F}{l}=\frac{2\mu_0I_1I_2}{4\pi r}[/tex]

Using the formula

[tex]5\times 10^{-5}=\frac{2\times 10^{-7}\times 0.62\times I_2}{0.033}[/tex]

Where [tex]\frac{\mu_0}{4\pi}=10^{-7}[/tex]

[tex]I_2=\frac{5\times 10^{-5}\times 0.033}{2\times 10^{-7}\times 0.62}[/tex]

[tex]I_2=13.3 A[/tex]

Hence, the current in the second wire=13.3 A

b.We are given that the wires repel each other.When the current carrying in the wires in opposite direction then, the wires repel to each other.

Hence,the two  currents in opposite directions.

Answer:

2y

Explanation:

Electric field in terms of Electric potential is given as:

E = dV/dr(x, y, z)

Where r(x, y, z) = position in x, y, z plane

The y component of the Electric field will be:

Ey = -dV/dy

Given that

V = 2x - y² - cos(z)

dV/dy = -2y

=> E = - (-2y)

E = 2y

Answer: elastic (e = 2.43)

Explanation:

The price elasticity formulae is given below as

Elasticy of price = change in quantity demanded / change in price.

P1 =$2.60, P2 = $6.30, q1 =17 and q2 = 8

e = q2 - q1/ P2 - P1

e = 8 - 17/ 6.30 - 2.60

e = - 9 /3.7

e = - 2.43

We take the modulus of e to have a positive value. Hence e = 2.43

Since e is greater than 1, then the elasticity of demand is elastic

Final answer:

The elasticity of demand between the given prices is calculated using the midpoint formula. It results in a value of -0.865 (ignoring the minus sign), indicating that the demand is elastic.

Explanation:

To calculate the elasticity of demand, we can use the midpoint formula, which is defined as the percentage change in quantity demanded divided by the percentage change in price. The formula for computing elasticity is:

Price Elasticity of Demand (Ed) = ([(Q2 - Q1) / ((Q2 + Q1)/2)] / [(P2 - P1) / ((P2 + P1)/2)]) * 100

Using the information provided: At a price of $2.60, quantity demanded is 17, and at a price of $6.30, quantity demanded is 8. We can fill in the values:

(Q2 - Q1) is (8 - 17) = -9

(Q2 + Q1)/2 is (8 + 17)/2 = 12.5

(P2 - P1) is ($6.30 - $2.60) = $3.70

(P2 + P1)/2 is ($6.30 + $2.60)/2 = $4.45

Now putting all these into the formula:

Ed = [(-9 / 12.5) / (3.70 / 4.45)] * 100

The percentage change in quantity is -72%, and the percentage change in price is 83.15%. Therefore:

Ed = (-0.72 / 0.8315) * 100

Ed = -0.865 (ignoring the minus sign)

This value is greater than 1, which indicates that the demand is elastic between these two prices; meaning, the quantity demanded is quite responsive to price changes.

Remember, we ignore the negative sign and use the absolute value when discussing elasticity, even though it is understood that price and quantity demanded move in opposite directions.

Answer:

V = 144mi/h

Explanation:

Please see the attachment below.

Answer:

Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.

The conserved form of the equation is given by

E = K + Ug + Wother.

In this case Wother = work done by the tension force.

In that form the total mechanical energy is conserved.

Final answer:

The correct answer is:  An external force is doing work on the system.

Explanation:

When the object is lowered at constant velocity, an external force (in this case, the force exerted by the person lowering the object) is doing work against gravity to keep the velocity constant. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh).

Since work is being done on the system externally, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved.

However, in the scenario described, an external force is doing work on the system. Work is being done to counteract gravity and maintain a constant velocity. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh), where "m" is the mass of the object, "g" is the acceleration due to gravity, and "h" is the vertical distance through which the object is lowered.

As a result, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved. The work done by the external force manifests as a change in the object's gravitational potential energy, causing a decrease in potential energy as the object is lowered. This decrease in potential energy is exactly balanced by the work done by the external force, so the total mechanical energy of the system remains constant, despite the fact that kinetic energy remains constant throughout the process.

Answer:

37.5 N

Explanation:

Horizontal component is represented by Fx and  is given as

Fx = F cos θ

Here,

θ = 60 degrees

F= 75 N

So,

Fx= 75 cos (60°)

==> Fx = 75 ×  0.5 =37.5 N

Final answer:

The horizontal component of the force acting parallel to the floor when a 150 N force is applied at a 60-degree angle to the horizontal is 75 N, calculated using the cosine of the angle.

Explanation:

The question involves resolving a force into its components, which is a common problem in physics. When a force is applied at an angle to the horizontal, it has both horizontal and vertical components. The horizontal (parallel) component ([tex]F_{parallel}[/tex]) is found using the cosine function of the angle Θ , which in this case is 60 degrees. The formula is [tex]F_{parallel}[/tex] = F * cos(Θ), where F is the magnitude of the force.

Given that a force of 150 N is applied at an angle of 60 degrees to the horizontal, the horizontal component of the force can be calculated as follows:

[tex]F_{parallel}[/tex] = 150 N * cos(60°)
[tex]F_{parallel}[/tex] = 150 N * 0.5
[tex]F_{parallel}[/tex] = 75 N

The horizontal component of the force acting parallel to the floor is 75 N. To find the net force acting on the box and thus the acceleration, you would subtract the frictional force from the parallel component of the applied force and then use Newton's second law, F = m * a, where F is the net force, m is the mass of the object and a is the acceleration. However, the frictional force is not needed for this particular question about the horizontal component.

Answer:

Pithball electroscope is used to determine if any object has static charge.

It consists of one or two small balls of a lightweight non-conductive substance. When this material is moved near an object having static charge, polarization will be induced in the atoms of pithballs which will either attract or repel the object depending on the nature of the charge in it.

It will not move in case it is brought near to a neutral object.

Similarly, the pith ball will move when it will brought into the magnetic field of the magnet as it will also induce polarization within the atoms of the pithballs.

Answer:7.92 N

Explanation:

Given

mass of book [tex]m=1.83\ kg[/tex]

coefficient of static friction [tex]\mu _s=0.442[/tex]

coefficient of kinetic friction [tex]\mu _k=0.240[/tex]

To move the book, one need to overcome  the static friction  

Static friction [tex]F_s=\mu _sN[/tex]

[tex]F_s=\mu _s\times 1.83\times 9.8[/tex]

[tex]F_s=0.442\times 1.83\times 9.8[/tex]

[tex]F_s=7.92\ N[/tex]

After overcoming the Static friction , Force needed to move the block is

[tex]F_k=\mu _kN[/tex]

[tex]F_k=0.240\times 1.83\times 9.8[/tex]

[tex]F_k=4.30\ N[/tex]

Final answer:

The force needed to begin moving a 1.83 kg book on a flat desk, given a coefficient of static friction of 0.442, is approximately 7.90 N. This is calculated using the book's weight and the static friction formula fs(max) = μs * N.

Explanation:

To calculate the force needed to begin moving the book, we need to use the coefficient of static friction and the book's weight. The formula to find the maximum static frictional force (fs max) that must be overcome to start moving the object is fs(max) = μs * N, where μs is the coefficient of static friction and N is the normal force.

The book's weight is given by W = m * g, where m is the mass of the book (1.83 kg) and g is the acceleration due to gravity (9.81 m/s²). The weight of the book is equivalent to the normal force (N) exerted by the desk on the book since the book is resting on a flat surface and there are no other vertical forces acting on it. This simplifies the normal force to N = W = m * g.

Using the given coefficient of static friction (0.442) and the calculated normal force, the force needed to begin moving the book is fs(max) = 0.442 * (1.83 kg * 9.81 m/s²) ≈ 7.90 N. Therefore, a force slightly greater than 7.90 N is required to overcome static friction and start moving the book.

Answer:

58.6 N

Explanation:

We are given that

[tex]q_1=-7.97\mu C=-7.97\times 10^{-6} C[/tex]

[tex]q_2=3.55\mu C=3.55\times 10^{-6} C[/tex]

Using [tex]1\mu C=10^{-6} C[/tex]

[tex]r=6.59 cm=6.59\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

The magnitude of force that one particle exerts on the other

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

Where [tex]k=9\times 10^9[/tex]

Substitute the values

[tex]F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}[/tex]

F=58.6 N

Answer: [tex]||\vec A|| \approx 9.95 m,||\vec B|| \approx 8.19 m[/tex]

Explanation:

Magnitudes can be calculated by using Pythagorean theorem:

[tex]||\vec A|| =\sqrt{(5m)^{2}+(-7m)^{2}+(5m)^{2}}\\||\vec A|| \approx 9.95 m\\||\vec B|| =\sqrt{(3m)^{2}+(7m)^{2}+(-3m)^{2}}\\||\vec B|| \approx 8.19 m\\[/tex]

Answer:

EMF will induce in the conducting loop when

The magnetic field must be changing

Explanation:

As per Farady's law of EMI we know that the magnetic flux linked with the closed conducting loop must be changed with time

so we have

[tex]EMF = \frac{d\phi}{dt}[/tex]

so we know that

[tex]\phi = BAcos\omega t[/tex]

now if magnetic field is changing with time then we have

[tex]EMF = Acos\theta \frac{dB}{dt}[/tex]

so EMF will induce when magnetic field will change with time

The calculated heat Q is [tex]\( 1.22 \times 10^4 \mathrm{~J} \),[/tex] and the positive sign indicates that heat is being added to the system.

Let's break down the given answer step by step:

1. Given Values:

Temperature of gas, [tex]\( T = 22^{\circ} \mathrm{C} = 295 \mathrm{~K} \)[/tex]

Work done on the gas to compress it, [tex]\( W = -353 \mathrm{~J} \)[/tex](negative because work is done on the gas)

  Final temperature, [tex]\( T_f = 145^{\circ} \mathrm{C} = 418 \mathrm{~K} \)[/tex]

  Number of moles,[tex]\( n = 8.2 \) moles[/tex].

2. First Law of Thermodynamics:

The first law of thermodynamics is given by [tex]\( Q = \Delta U + W \),[/tex] where Q is the heat added to the system, [tex]\( \Delta U \)[/tex] is the change in internal energy, and W is the work done on the system.

For a monoatomic gas,[tex]\( C_v = \frac{3}{2} R \),[/tex]

where R is the universal gas constant [tex](\( R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \)). So, \( Q = n C_v \Delta T + W \).[/tex]

3. Calculation:

  Substituting the given values into the equation:

[tex]\[ \begin{aligned} Q & = \left(8.2 \times \frac{3}{2} \times 8.314 \times (418 - 295)\right) + (-353) \\ & = 12225 \mathrm{~J} = 1.22 \times 10^4 \mathrm{~J} \end{aligned} \][/tex]

4. Interpretation:

  Since Q is positive, it means heat is flowing inside the cylinder. A positive value of Q indicates that the system is gaining heat, and in this case, it's due to both the compression work done on the gas and the increase in internal energy.

In summary, the calculated heat Q is [tex]\( 1.22 \times 10^4 \mathrm{~J} \),[/tex] and the positive sign indicates that heat is being added to the system.

Answer: the mutual inductance is - 0.0978H

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

M = 0.1H

Explanation:

Please see attachment below.

Explanation:

Below is an attachment containing the solution.

To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.

To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for electric force is:

F = k * (q1 * q2) / r^2

Where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. In this case, the charge of the Moon is 27.3% of the charge of the Earth, and the radius of the Moon is also 27.3% of the radius of the Earth. Using these values, we can calculate the electric force.

Let's assume the charge of the Earth is q1 and the charge of the Moon is q2. Since the charge of the Moon is 27.3% as large as the charge of the Earth, we can write q2 = 0.273 * q1. Similarly, the radius of the Moon is 27.3% of the radius of the Earth, so we can write r = 0.273 * R, where R is the radius of the Earth. Plugging these values into Coulomb's law formula:

F = k * (q1 * (0.273 * q1)) / (0.273 * R)^2

Simplifying the equation, we get:

F = k * (q1^2 * 0.273) / (0.273^2 * R^2)

The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.

Learn more about Electric Force here:

https://brainly.com/question/21093704

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Kinetic energy lost in collision is 10 J.

Explanation:

Given,

Mass, [tex]m_{1}[/tex] = 4 kg

Speed, [tex]v_{1}[/tex] = 5 m/s

[tex]m_{2}[/tex] = 1 kg

[tex]v_{2}[/tex] = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

[tex]\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2[/tex]

By plugging in the values we get,

[tex]KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\[/tex]

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

[tex]KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)[/tex]

[tex]KE = 40J + KE(lost)[/tex]

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

Answer:

Explanation:

As there is no external torque is applied so the angular momentum remains constant.

L = I ω = constant

Where, I is the moment of inertia of the system and ω is the angular velocity

As we move towards the edge, the moment of inertia increases, hence the angular velocity decreases.