A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance r1 (r1 < R) from the center of the sphere, the electric field has magnitude E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R, the magnitude of the electric field at the same distance r1 from the center would be equal to:

Answer:

175  [tex]\frac{mi}{h}[/tex]

Explanation:

v = speedometer reading of a european sportscar = [tex]282 \frac{km}{h}[/tex]

we know that,

1 km = 0.621 miles

So the speedometer reading of a european sportscar in  mi/h is given as

v = [tex]282 \frac{km}{h}[/tex] = [tex]282\left ( \frac{km}{h} \right ) \left ( \frac{0.621 mi}{1 km} \right )[/tex]

v = (282 x 0.621) [tex]\frac{mi}{h}[/tex]

v = 175  [tex]\frac{mi}{h}[/tex]

Answer:

The height of father = 178 cm.

Explanation:

The height of your father is 70 inches.

There are 2.54 centimeters in an inch

1 inch = 2.54 cm

70 inches = 70 x 2.54 = 177.8 cm

Rounding to the nearest centimeter.

70 inches = 177.8 cm = 178 cm

The height of father = 178 cm.

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

   Horizontal displacement = 40.5 m

   Time  

         [tex]t=\frac{40.5}{28.64}=1.41s[/tex]          

   Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

    Time to reach the goal posts 40.5 m away = 1.41 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

    Height of throw = 1.4 m

    Height traveled by ball = 2.83 m

    Total height = 2.83 + 1.4 = 4.23 m

    When the ball goes to first base it will be 4.23 m high.

Answer:

[tex]5.32\cdot 10^4 g[/tex]

Explanation:

First of all, we need to find the volume of the room, which is given by

[tex]V=2.50 m \cdot 5.50 m \cdot 3.00 m =41.3 m^3[/tex]

Now we  can find the mass of the air by using

[tex]m=dV[/tex]

where

[tex]d=1.29 g/dm^3[/tex] is the density of the air

[tex]V=41.3 m^3 = 41,300 dm^3[/tex] is the volume of the room

Substituting,

[tex]m=(1.29)(41300)=5.32\cdot 10^4 g[/tex]

Final answer:

The mass of the air in a room with dimensions 2.50 m x 5.50 m x 3.00 m and an air density of 1.29 g/dm³ is calculated to be 53.2 kg.

Explanation:

The mass of the air in the room can be calculated by using the formula for density, which is mass (mass) equals density (density) times volume (volume), or m = ρV. Given that the density of air is 1.29 g/dm³, first we need to convert the measurements of the room to dm³ (decimeters cubed) as the given room dimensions are in meters. The volume of the room is 2.50 m x 5.50 m x 3.00 m which equals 41.25 m³. Converting from cubic meters to cubic decimeters results in 41,250 dm³ (1 m³ = 1,000 dm³). Therefore, the mass of air is calculated as 1.29 g/dm³ * 41,250 dm³, which equals 53,212.5 grams or 53.2 kg.

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

[tex]B=\dfrac{\mu_oI}{2\pi r}[/tex]

[tex]B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}[/tex]

B = 0.000002 T

[tex]B=10^{-5}\ T[/tex]

(2) Exact formula:

[tex]B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }[/tex]

[tex]B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }[/tex]

B = 0.00000199 T

or

B = 0.000002 T

Hence, this is the required solution.

Answer:

Mass of the stick = 139.04 gram

Explanation:

Let the mass of the meter stick be = M grams

given:

Attached mass, m = 50.0 gram

position of the fulcrum = 39.2 cm

The meter stick is balanced at 49.7 cm, therefore the center of mass of the stick will be at 49.7 cm

Now for the system to be balanced the moment due to all the masses about the fulcrum must be equal.

thus,

moment = Force × perpendicular distance from the point of movement

Force = mass × acceleration due to gravity(g)

therefore,

(refer figure for the distances)

50g × (39.2-10) = Mg × (49.7 - 39.2)

⇒[tex]M=\frac{50\times 29.2}{10.5}[/tex]

⇒M = 139.04 gram

Answer:

applied force on piston is 302.308163 N

Explanation:

Given data

mass = 1510 kg

weight = 1510 × 9.81 = 14813.1 N

radius = 8 cm

piston =  1.0 cm

to find out

force

solution

we will apply here pascal principal i.e

weight of automobile / area 1  = applied force on piston / area 2

so applied force on piston = weight of automobile × area 2  / area 1

and we know that areas is proportional to square of diameter

so area 2  / area 1 = (0.01/0.08)²

so  applied force on piston = weight of automobile × (0.01/0.08)²

applied force on piston = 14813.1  × (0.01/0.07)²

applied force on piston is 302.308163 N

The force necessary to raise the automobile using the hydraulic lift can be calculated using Pascal's Principle and the areas of the piston and the shaft. After calculating these, it is found that approximately 231 Newtons of force is needed to lift the car.

In the hydraulic lift, the force applied on the smaller piston is transmitted to the larger piston. This is based on Pascal's Principle that states that a change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. The force necessary to lift the car can be calculated using this principle and the formula F1/F2 = A1/A2, where F is the force and A is the area of the pistons.

First, we need to calculate the areas of the pistons. The area of a circle is given by the formula A = πr^2 where r is the radius of the circle. Plugging in the given values, A1 (piston) = π*(1cm)^2 = π cm^2 and A2 (shaft) = π*(8 cm)^2 = 64π cm^2.

The force on the larger piston (F2), which is the weight of the car, can be calculated by multiplying the mass of the car by gravitational acceleration F2 = m*g, with m = 1510 kg and g = 9.8 m/s^2. Hence, F2 is roughly 14798 Newtons. Therefore, using the formula F1/F2 = A1/A2 we can solve for F1: F1 = F2 * (A1/A2) = 14798N * (1/64) = approx 231N.

So, to lift the automobile using the hydraulic lift, a force of approximately 231 Newtons must be applied to the piston.

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Answer:

a) [tex]\frac{m_1}{m_2}=\frac{1}{3}[/tex]

b) Acceleration = 0.75 m/s²

Explanation:

a) We have force , F = mass x acceleration.

[tex]\texttt{Value of force}=F_s[/tex]

[tex]\texttt{Acceleration of }m_1=3m/s^2\\\\\texttt{Acceleration of }m_2=1m/s^2[/tex]

We have force value is same

       [tex]m_1\times 3=m_2\times 1\\\\\frac{m_1}{m_2}=\frac{1}{3}[/tex]

b) We have

         [tex]m_1\times 3=m_2\times 1\\\\m_2=3m_1[/tex]

Combined mass

       [tex]m=m_1+m_2=m_1+3m_1=4m_1[/tex]

Force

         [tex]F_s=4m_1\times a\\\\a=\frac{F_s}{4m_1}=\frac{1}{4}\times \frac{F_s}{m_1}=\frac{1}{4}\times 3=0.75m/s^2[/tex]

Answer:

[tex]a_c = 120.4 m/s^2[/tex]

Explanation:

As we know that stone is revolving in horizontal circle

so here height of the stone is H = 1.9 m

now by kinematics we can find the time to reach it at the ground

[tex]H = \frac{1}[2}gt^2[/tex]

now we have

[tex]1.9 = \frac{1}{2}(9.81)t^2[/tex]

now we have

[tex]t = 0.622 s[/tex]

now the speed of the stone is given as

[tex]v = \frac{x}{t}[/tex]

[tex]v = \frac{8.9}{0.622} = 14.3 m/s[/tex]

now for finding centripetal acceleration we have

[tex]a_c = \frac{v^2}{R}[/tex]

we know that radius of circle is

R = 1.7 m

now we have

[tex]a_c = \frac{14.3^2}{1.7}[/tex]

[tex]a_c = 120.4 m/s^2[/tex]

Answer:

L = 30 m

Explanation:

As per mechanical energy conservation law we can say that total kinetic energy of the hoop is equal to the total gravitational potential energy at the top

So here we can say for initial total kinetic energy as

[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]

for pure rolling we will have

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

also for large hoop we will have

[tex]I = mR^2[/tex]

now we have

[tex]KE = \frac{1}{2}mR^2\omega^2 + \frac{1}{2}(mR^2)\omega^2[/tex]

[tex]KE = mR^2\omega^2[/tex]

[tex]KE = (3.8)(1.3)^2(6.7)^2[/tex]

[tex]KE = 288.3 J[/tex]

now for gravitational potential energy we can say

[tex]U = mg(Lsin\theta)[/tex]

now by energy conservation we have

[tex]288.3 = (3.8)(9.81)(L sin15)[/tex]

[tex]L = 30 m[/tex]

The horizontal distance traveled by the hoop is approximately 30 m.

The total kinetic energy of the ball is the sum of the translational and rotational kinetic of the ball.

K.E(total) = K.E(trans) + K.E(rotational)

[tex]K.E(t) = \frac{1}{2}mv^2 + \frac{1}{2} I \omega ^2\\\\K.E(t) = \frac{1}{2}mv^2 + \frac{1}{2} (mR^2) \omega ^2\\\\K.E(t) = \frac{1}{2}m(\omega R)^2 + \frac{1}{2} (mR^2) \omega ^2\\\\K.E(t) = m\omega ^2R^2\\\\K.E(t) = 3.8 \times (6.7)^2 \times (1.3)^2\\\\K.E(t) = 288.28 \ J[/tex]

The horizontal distance traveled by the hoop is calculated as follows;

K.E = P.E

288.28  = mg(Lsinθ)

288.28 = (3.8 x 9.8) x (L) x sin(15)

288.28 = 9.63 L

L = 30 m

Thus, the horizontal distance traveled by the hoop is approximately 30 m.

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Answer:

Part a)

[tex]\phi' = 9943.5 Nm^2/C[/tex]

Part b)

[tex]\phi' = 4971.75 Nm^2/C[/tex]

Explanation:

Part a)

Total electric flux due to charge contained in a closed surface is given as

[tex]\phi = \frac{Q}{\epsilon_0}[/tex]

now we have

[tex]Q = 88 nC[/tex]

now from above equation total flux is given as

[tex]\phi = \frac{88 \times 10^{-9}}{8.85 \times 10^{-12}}[/tex]

[tex]\phi = 9943.5 Nm^2/C[/tex]

Part b)

Now we need to find the flux through the hemispherical surface

so we will have

[tex]\phi' = \frac{\phi}{2}[/tex]

here we have

[tex]\phi' = \frac{9943.5}{2}[/tex]

[tex]\phi' = 4971.75 Nm^2/C[/tex]

(a) The total electric flux through the surface of the shell is 9943.5[tex]\frac{Nm^2}{C}[/tex]  

(b) The total electric flux through any hemispherical portion of the      shell's surface is 4971.75 [tex]\frac{Nm^2}{C}[/tex]

Electric flux is the measure of the electric field through a given surface. It helps us to describe the strength of an electric field at any distance from the charge.

(a) The formula for the electric flux is given by,

[tex]\rm{\phi=\frac{Q}{\varepsilon_0}}[/tex]      

Where , Q= 88.0 nC = [tex]88\times10^{-9}[/tex]

             [tex]\varepsilon_0[/tex] =[tex]{8.85\times10^{-12} }[/tex]

[tex]\phi=\frac{88\times10^{-9} }{8.85\times10^{-12} }[/tex]

[tex]\phi=9943.5\frac{Nm^{2} }{C}[/tex]

The total electric flux through the surface of the shell is 9943.5[tex]\frac{Nm^2}{C}[/tex]

(b) For the spherical surface electric flux is taken as [tex]\phi_2[/tex]

[tex]\rm{\phi_2=\frac{\phi_1}{2}}[/tex]

[tex]\rm{\phi_2=\frac{9943.5}{2}}[/tex]

[tex]\phi_2 =4979.5\frac{Nm^2}{C}[/tex]

The total electric flux through any hemispherical portion of the shell's surface is 4971.75 [tex]\frac{Nm^2}{C}[/tex]

To learn more about the electric flux refer to the link.

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Answer:

[tex]L_f = 71.25 kg m^2/s[/tex]

Explanation:

As we know by Newton's law of rotational motion that Rate of change in angular momentum is total torque on the system

So here we have

[tex]\tau = \frac{\Delta L}{\Delta t}[/tex]

here we can say it

[tex]L_f - L_i = \tau \Delta t[/tex]

so final angular momentum of the disc is given by the equation

[tex]L_f = L_i + \tau \Delta t[/tex]

now we know that

[tex]\tau = 7.50 Nm[/tex]

time interval is given as

[tex]\Delta t = 9.50 s[/tex]

since it is initially at rest so initial angular momentum is ZERO

so we have

[tex]L_f = 0 + (7.50)(9.50)[/tex]

[tex]L_f = 71.25 kg m^2/s[/tex]

Answer:

3.36m or 1680 N

Explanation:

Given:

Mass of the satellite =m

Mass of the moon Ganymede, M = 1.48 × 10²³ kg

Radius of Ganymede, R = 2631 km

Distance of satellite above the surface of the moon, d = 300 km

According to Universal Gravitational law:

[tex]F=\frac{GMm}{(R+d)^2}[/tex]

where, G is the gravitational constant, M and m are mass of the objects and (R+d) is the distance between the centers of the objects.

Substitute the values:

[tex]F=\frac{6.67\times 10^{-11}\times 1.48 \times 10^{23} m}{(2.931\times 10^6)^2}=3.36m[/tex]

If we consider mass of a satellite to be about 500 kg, the gravitational force between the moon and the satellite would be:

[tex] F = 3.36\times 500 = 1680 N[/tex]

Answer:

Torque = -1799.52 Nm or 1799.52 Nm (clockwise direction)

Explanation:

Torque =  Force x Perpendicular distance

Here force is given as 18.4 N and Perpendicular distance = 97.8 m

Substituting

Torque =  Force x Perpendicular distance = 18.4 x 97.8 = 1799.52 Nm

Torque = 1799.52 Nm (clockwise direction)

Usually anticlockwise is positive

So torque = -1799.52 Nm

Answer:

The torque is 1799.52 N-m

Explanation:

Given that,

Force = 18.4 N

Length = 97.8 m

We calculate the torque,

Torque is the product of the force and perpendicular distance.

[tex]\tau =Fl\sin\theta[/tex]

Where, F = force

l = length

Put the value into the formula

[tex]\tau =18.4\times97.8\sin90^{\circ}[/tex]

[tex]\tau =1799.52\ N-m[/tex]

Hence, The torque is 1799.52 N-m

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       [tex]v=\sqrt{2.14^2+0.34^2}=2.17m/s[/tex]

Direction,  

       [tex]\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0[/tex]

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Answer:

flow rate in pipe B is 16 times the flow in pipe A

Explanation:

According to the poiseuille's law, flow rate  is is given as

[tex]Q = \frac{ \pi Pr^{4}}{8\eta*L}[/tex]

Flow rate  in the pipe will remain same as above i.e,

[tex]Q_{A} = \frac{\pi Pr^{4}}{8\eta*L}[/tex]

Flow in the pipe be will be

As diameter OF PIPE B is doubled

AND length of both pipes remained same

[tex]Q_{B} = \frac{\pi P(2r)^{4}}{8\eta*L}[/tex]

          [tex]= \frac {16\pi P(r)^{4}} {8\eta*L}}[/tex]

          so we have

flow rate in pipe B is 16 times the flow rate  in pipe A

The flow rate in pipe B is four times larger than the flow rate in pipe A.

The flow rate in pipe B can be determined using the continuity equation, which states that the flow rate must be the same at all points along the pipe. The equation is given as Q = Av, where Q is the flow rate, A is the cross-sectional area, and v is the average velocity.

Since pipe B has twice the diameter of pipe A, its cross-sectional area is four times larger. Therefore, the flow rate in pipe B will be four times larger than the flow rate in pipe A.

So, if the flow rate in pipe A is Q, then the flow rate in pipe B is 4Q.

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Answer:

The frequency of such a radio wave is 1 MHz.

(d) is correct option

Explanation:

Given that,

Wave length = 300 m

We know that,

Speed of light [tex]c= 3\times10^{8}\ m/s[/tex]

We calculate the frequency

Using formula of frequency

[tex]c = f\times\lambda[/tex]

[tex]f = \dfrac{c}{\lambda}[/tex]

Put the value into the formula

[tex]f=\dfrac{3\times10^{8}}{300}[/tex]

[tex]f =1\times10^{6}\ Hz[/tex]

[tex]f =1\ MHz[/tex]

Hence, The frequency of such a radio wave is 1 MHz.

The frequency of an AM radio wave with a wavelength of 300 meters is 1.00 MHz, using the formula c = λf, and knowing that the speed of light c is 3×10^8 m/s, the calculation results in a frequency of 1.00 MHz, which is 1×10^6 Hz.

To calculate the frequency of an AM radio wave given the wavelength, we can use the speed of light formula c = λf where c is the speed of light (approximately 3×108 meters per second), λ (lambda) is the wavelength, and f is the frequency. Since the typical wavelength for an AM radio wave is given as 300 meters, we plug this value into the equation to find the frequency:

f = c / λ

f = (3×108 m/s) / (300 m)

f = 1×106 Hz or 1.00 MHz

The frequency of an AM radio wave that has a wavelength of 300 meters is therefore 1.00 MHz, which corresponds to option (d).

The external resistance R is approximately 10.63 Ohms while the internal resistance of the battery is approximately 2.43 Ohms.

This problem pertains to the principles of electric circuits, namely Ohm's Law (V = IR) and power (P = VI). To solve this, we'll first compute the resistance R of the external load and then calculate the internal resistance (r) of the battery.

(a) We begin by using the power formula P = VI, rearranging to find resistance R: P/V = I. So, the current I flowing through the circuit is 14.0 W/12.2 V which equals to about 1.148 A. Then, we apply Ohm's Law to get R = V/I, which results in: R = 12.2 V /1.148 A = 10.63 Ω.

(b) The difference between the emf (ε) and the terminal voltage (Vt) gives the voltage drop across the internal resistance (i.e., ε - Vt = Ir). Thus, r = (ε - Vt) / I = (15V - 12.2V) / 1.148 A = 2.43 Ω.

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Final answer:

The argon temperature at the heater exit is determined to be 348.08 ℃, while the argon volume flow rate at the heater exit is calculated to be 3769.23 m³/s.

Explanation:

To determine the argon temperature and volume flow rate at the heater exit, we can use the principles of thermodynamics. The problem states that heat transfer in the rate of 150 kW is supplied to the argon. Since there is no information given about the heater efficiency, we can assume it to be 100%, meaning all the heat is transferred to the argon. Therefore, the heat transfer can be equated to the change in internal energy of the argon, which is given by:

Q = m * Cp * ΔT

Where Q is the heat transfer, m is the mass flow rate, Cp is the specific heat capacity of argon, and ΔT is the change in temperature.

Given that Q = 150 kW, m = 6.24 kg/s, and Cp = 0.52 kJ/kg·℃ (specific heat capacity of argon), we can rearrange the equation to solve for ΔT:

ΔT = Q / (m * Cp)

Substituting the values, we have ΔT = (150 * 10³ J/s) / (6.24 kg/s * 0.52 kJ/kg·℃), which equals 48.08 ℃.

Therefore, the argon temperature at the heater exit is 300 ℃ + 48.08 ℃, which equals 348.08 ℃.

To calculate the argon volume flow rate at the heater exit, we can use the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation to solve for V, we have:

V = (nRT) / P

Given that the initial conditions are 300 K and 100 kPa, and a mass flow rate of 6.24 kg/s, we can calculate the number of moles of argon:

n = m / M

Where M is the molar mass of argon. The molar mass of argon is 39.948 g/mol, so n = 6.24 / (39.948 * 10⁻³) = 156.172 mol.

Substituting the values into the ideal gas law equation, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100 kPa. Converting kPa to Pa, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100,000 Pa, which equals 3769.23 m³.

Therefore, the argon volume flow rate at the heater exit is 3769.23 m³/s.

Answer:

122.4 cm

Explanation:

[tex]d_{p}[/tex] = distance of phone from eye = 44 cm

[tex]d_{e}[/tex] = distance of eyeglasses from eye = 2.0 cm

[tex]d_{o}[/tex] = Object distance = [tex]d_{p}[/tex] - [tex]d_{e}[/tex] = 44 - 2 = 42 cm

P = Power of the eyeglasses = 1.55 diopter

focal length of eyeglass is given as

[tex]f = \frac{1}{P}[/tex]

[tex]f = \frac{100}{1.55}[/tex]

f = 64.5 cm

[tex]d_{i}[/tex] = image distance

using the lens equation

[tex]\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}[/tex]

[tex]\frac{1}{42} + \frac{1}{d_{i}} = \frac{1}{64.5}[/tex]

[tex]d_{i}[/tex] = - 120.4 cm

[tex]d_{n}[/tex] = distance of near-point

distance of near-point is given as

[tex]d_{n}[/tex] = |[tex]d_{i}[/tex]| + [tex]d_{e}[/tex]

[tex]d_{n}[/tex] = 120.4 + 2

[tex]d_{n}[/tex] = 122.4 cm

Answer:

- 5 °C

Explanation:

[tex]T_{s}[/tex] = Surrounding temperature = - 5 °C

[tex]T(t) [/tex] = temperature at any time "t"

As per newton's law

[tex]T(t) = T_{s} + C e^{kt}[/tex]

at t = 0

[tex]T(0) = - 5 + C e^{k(0)}[/tex]

[tex]21 = - 5 + C e^{k(0)}[/tex]

C = 26

at t = 1

[tex]T(1) = - 5 + (26) e^{k}[/tex]

[tex]10 = - 5 + (26) e^{k}[/tex]

k = - 0.55

at t = 34

[tex]T(t) = T_{s} + C e^{kt}[/tex]

[tex]T(34) = - 5 + (26) e^{34(- 0.55)}[/tex]

[tex]T(34)[/tex] = - 5 °C

Answer:

The magnitude of the centripetal acceleration of the car is 0.22 m/s².

Explanation:

t= 380 sec

r= 3.27km= 3270m

d=π*r=π*3270m

d= 10273m

d=Vt*t

Vt= d/t = 10273m/380s

Vt= 27.03 m/s

ac= Vt²/r

ac= (27.03 m/s)²/3270m

ac=0.22 m/s²

Answer:

Option (A)

Explanation:

The relation between degree Celsius and kelvin is

Degree C = K - 273

So change in 1 degree C is same as 1 kelvin.

Final answer:

The coefficient of static friction is determined by setting up equations based on the static equilibrium conditions of the ladder leaning against a wall, resulting in a coefficient of approximately 0.869 when using the ladder's angle of 41° with the horizontal.

Explanation:

To determine the coefficient of static friction between the ladder and the floor, we can analyze the forces acting on the ladder at the point of slipping. The forces involved in this problem are the weight of the ladder (W), the normal force exerted by the floor (N), the static frictional force (f), and the force exerted by the wall, which is horizontal due to the frictionless contact (H).

The ladder makes a 41.0° angle with the horizontal floor (θ), so we can use trigonometry and Newton's laws to set up equations. First, the sum of vertical forces must be zero, as the ladder is not moving vertically:

N = W * cos(θ)

Next, the sum of horizontal forces must be zero, as the ladder is not moving horizontally:

f = H

Now, since the wall is frictionless, the horizontal force the wall exerts on the ladder is equal to the horizontal component of the ladder's weight:

H = W * sin(θ)

At the point of slipping, the static frictional force (f) is at its maximum value and is given by:

f = μ * N

Using the fact that f = H at the point of slipping, we can solve for the coefficient of static friction (μ) using the following equation:

μ = W * sin(θ) / (W * cos(θ))

μ = tan(θ)

Substitute the value of θ = 41.0° into the equation:

μ = tan(41.0°) ≈ 0.869

Therefore, the coefficient of static friction is approximately 0.869.

Answer:

c) The magnitude of the magnification of a concave mirror is less than 1

Explanation:

we know

the mirror formula is given as:

[tex]\frac{1}{v}+\frac{1}{u}=\frac{2}{r}[/tex]   ..............(1)

also

magnification (m) of mirror is given as:

[tex]m=\frac{-v}{u}[/tex] . ..... (2)

where,

v = image distance

u = object distance

r = Radius of curvature of the mirror

Now

from (2)

[tex]v=-mu[/tex]

substituting in (1), we get

[tex]\frac{1}{u}(\frac{1}{-m}+1)=\frac{2}{r}[/tex]

or

[tex]\frac{1}{2}(1-\frac{1}{m})=\frac{u}{r}[/tex]     ..............(3)

now it is given that object is beyond the center of the the curvature

thus,

⇒[tex]\frac{u}{r}>1[/tex]     ................(4)

comparing the (3) and (4), we have

[tex]\frac{1}{2}(1-\frac{1}{m})>1[/tex]

or

[tex](1-\frac{1}{m})>2[/tex]

or

[tex]-\frac{1}{m}>2-1[/tex]

or

[tex]-\frac{1}{m}>1[/tex]

⇒[tex]m<-1[/tex]

Hence, the magnification is less than 1

The correct statement for the magnitude of magnification of a concave mirror, when the object is beyond the center of curvature, is that it is less than 1, indicating the image formed is real, inverted, and smaller than the object. So the correct option is C.

The correct statement about the magnitude of the magnification of a concave mirror when the object is beyond the center of curvature (do > r) is "The magnitude of the magnification of a concave mirror is less than 1." This is because when an object is located beyond the center of curvature, the image formed by a concave mirror is real, inverted, and smaller than the object. Hence, the magnification, which is the ratio of the image size to the object size, is positive but less than 1.

When dealing with concave mirrors, it is important to note that different object positions result in different types of images, depending on the object's distance relative to the mirror's focal length and center of curvature. If the object distance is greater than the radius of curvature, the image formed is smaller than the object, leading to a magnification of less than 1.

When a ray of light is incident on a flat reflecting surface and makes an angle of 28.7° with the normal, the reflected ray will also make an angle of 28.7° with the normal.

When light reflects off a flat surface, the angle of incidence at that time equals the angle of reflection. Given an incident angle of 28.7° with the normal, the reflected ray will also make an angle of 28.7° with the normal. This principle, known as the law of reflection.

In addition, this principle applies universally to all the flat surfaces and ensures that light behaves predictably when interacting with various reflective surfaces, thereby making it a fundamental concept in both optics and physics. Thus, in this given scenario, the angle of reflection is 28.7°, maintaining consistency with the law of reflection.

Answer:

No, the car will not make it to the top of the hill.

Explanation:

Let ΔX be how long the slope of the hill is, Δx be how far the car will travel along the slope of the hill, Ф be the angle the slope of the hill makes with the horizontal(bottom of the hill), ki be the kinetic energy of the car at the bottom of the hill and vi be the velocity of the car at the bottom of the hill and kf be the kinetic energy of the car when it stop moving at vf.

Since Ф is the angle between the horizontal and the slope, the relationship between the angle and the slope and the height of the hill is given by

sinФ = 12/ΔX

Which gives you the slope as

ΔX = 12/sinФ

Therefore for the car to reach the top of the hill it will have to travel ΔX.

Ignoring friction the total work done is given by

W = ΔK

W = (kf - ki)

Since the car will come to a stop, kf = 0 J

W = -ki

m×g×sinФ×Δx = 1/2×m×vi^2

(9.8)×sinФ×Δx = 1/2×(10)^2

sinФΔx = 5.1

Δx = 5.1/sinФ

ΔX>>Δx Ф ∈ (0° , 90°)

(Note that the maximum angle Ф is 90° because the slope of a hill can never be greater ≥ 90° because that would then mean the car cannot travel uphill.)

Since the car can never travel the distance of the slope, it can never make it to the top of the hill.

Answer:

d) None of the above

Explanation:

[tex]v_{o}[/tex] = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

[tex]v_{oy}[/tex] = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m  

t = time of travel

using the equation

[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.1 s

Using the kinematics equation

[tex]x =v_{ox} t+(0.5)a_{x} t^{2}[/tex]

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

[tex]v_{oy}[/tex] = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²

[tex]y_{o}[/tex] =initial vertical position at the time of launch = 20 m  

[tex]y[/tex] = vertical position at the maximum height = 20 m

[tex]v_{fy}[/tex] = final velocity along vertical direction at highest point = 0 m/s

using the equation

[tex]{v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})[/tex]

[tex]0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)[/tex]

[tex]y[/tex] = 20.02 m

h = height above the starting height

h = [tex]y[/tex] - [tex]y_{o}[/tex]

h = 20.02 - 20

h = 0.02 m

Answer:

1.7 × 10^8 N/C

Explanation:

Q = 7.6 nC, r = 2 cm = 0.02 m

E = KQ/r^2

E =

8.99 × 10^9 × 7.6 × 10^(-6)/(0.02 × 0.02)

E = 1.7 × 10^8 N/C

Answer:

Distance, x = 73.10 meters

Explanation:

It is given that,

Mass of the rock, m = 25 g = 0.025 kg

Acceleration due to gravity, a = 1.62 m/s²

We need to find the distance traveled by the rock when it falls from rest in 9.5 seconds if the only force acting on it was the gravitational force due to the Moon. It can be calculated using the second equation of motion i.e.

[tex]x=ut+\dfrac{1}{2}at^2[/tex]

[tex]x=0+\dfrac{1}{2}\times 1.62\ m/s^2\times (9.5\ s)^2[/tex]

x = 73.1025 meters

or

x = 73.10 meters

So, the distance traveled by the rock is 73.10 meters. Hence, this is the required solution.