A child is twirling a 0.0154-kg ball on a string in a horizontal circle whose radius is 0.149 m. The ball travels once around the circle in 0.639 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase?

Answer:

Capacitive reactance is 132.6 Ω.

Explanation:

It is given that,

Capacitance, [tex]C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F[/tex]

Voltage source, V = 20 volt

Frequency of source, f = 60 Hz

We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :

[tex]X_C=\dfrac{1}{2\pi fC}[/tex]

[tex]X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}[/tex]

[tex]X_C=132.6\ \Omega[/tex]

So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.

Answer:

135°.

Explanation:

R = 75 ohm, L = 0.01 H, C = 4 micro F = 4 x 10^-6 F

Frequency is equal to the half of resonant frequency.

Let f0 be the resonant frequency.

[tex]f_{0}=\frac{1}{2\pi \sqrt{LC}}[/tex]

[tex]f_{0}=\frac{1}{2\times 3.14 \sqrt{0.01\times 4\times 10^{-6}}}[/tex]

f0 = 796.2 Hz

f = f0 / 2 = 398.1 Hz

So, XL = 2 x 3.14 x f x L = 2 x 3.14 x 398.1 x 0.01 = 25 ohm

[tex]X_{c}=\frac{1}{2\pi fC}[/tex]

Xc = 100 ohm

[tex]tan\phi = \frac{X_{L}-_{X_{C}}}{R}[/tex]

tan Ф = (25 - 100) / 75 = - 1

Ф = 135°

Thus, the phase difference is 135°.

The impulse delivered to the ball by the floor after rebounding with the same speed and angle is; Δp = 2.74 N.s

We are given;

Mass of basketball; m = 0.6 kg

Speed; v = 5.4 m/s

Angle to the vertical; θ = 65°

       We want to find the impulse if the ball rebounds with the same speed and angle above.

Now, as the x-component of the momentum remains constant, the impulse would be equal to the change in the y-component of the balls' momentum.

Thus;

Δp = m[(v_f * cos θ) - (v_i * cos θ)]

        Earlier we saw that we were given the rebound speed to be the same with the initial speed but however, as the y-axis is pointing upwards, it means that the initial velocity will be negative.

Thus;

v_f = 5.4 m/s

v_i = -5.4 m/s

        Plugging in the relevant values into the impulse equation gives;

Δp = 0.6[(5.4 * cos 65) - (-5.4 * cos 65)]

Δp = (0.6 × cos 65) × (5.4 + 5.4)

Δp = 2.74 N.s

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The impulse delivered to the ball by the floor after the rebound is 5.87 Ns.

The given parameters;

The impulse delivered to the ball is the change in the linear momentum of the ball in vertical direction.

The magnitude of the impulse delivered to the ball by the floor is calculated as;

[tex]J = \Delta P = m(v_f - v_0)\\\\J = m(vsin\theta - vsin\theta)\\\\the \ final \ speed \ occured \ in \ opposite \ direction;\\\\J = m(vsin\theta + vsin\theta)\\\\J = m(2vsin\theta)\\\\J = 2mv \times sin(\theta)\\\\J = 2\times 0.6 \times 5.4 \times sin(65)\\\\J = 5.87 \ Ns[/tex]

Thus, the impulse delivered to the ball by the floor after the rebound is 5.87 Ns.

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Answer:0.2 rad/s

Explanation:

Given data

Velocity of the bottom point of the ladder=1.2Ft/s

Length of ladder=10ft

distance of the bottom most point of ladder from origin=8ft

From the data the angle θ with ladder makes with horizontal surface is

Cosθ=[tex]\frac{8}{10}[/tex]

θ=36.86≈37°

We have to find rate of change of θ

From figure we can say that

[tex]x^{2}[/tex]+[tex]y^{2}[/tex]=[tex]AB^{2}[/tex]

Differentiating above equation we get

[tex]\frac{dx}{dt}[/tex]=-[tex]\frac{dy}{dt}[/tex]

i.e [tex]{V_A}=-{V_B}=1.2ft/s[/tex]

[tex]{at\theta}={37}[/tex]

[tex]Y=6ft[/tex]

[tex]and\ about\ Instantaneous\ centre\ of\ rotation[/tex]

[tex]{\omega r_A}={V_A}[/tex]

[tex]{\omega=\frac{1.2}{6}[/tex]

ω=0.2rad/s

i.e.Rate of change of angle=0.2 rad/s

Answer:

18.4

Explanation:

Q1 = 4390 J

Q2 = 3582.2 J

The efficiency of heat engine is given by

n = 1 - Q2 / Q1

n = 1 - 3582.2 / 4390

n = 0.184

n = 18.4 %

Answer:

14.112 mV

Explanation:

L = 16 m, v = 21 m/s, B = 42 μ T = 42 x 10^-6 T

The formula for the induced emf is given by

e = B x v x L

e = 42 x 10^-6 x 21 x 16 = 14.112 x 10^-3 V = 14.112 mV

Thus, the induce emf is 14.112 mV.

Answer:

Work done, W = 401 kJ

Explanation:

It is given that,

Mass of the car, m = 1100 kg

Initial, velocity of the car, u = 27 m/s

Finally it stops, v = 0

According to work energy theorem, the change in kinetic energy is equal to the work done i.e.

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]W=\dfrac{1}{2}\times 1100\ kg\times (0-(27\ m/s)^2)[/tex]

W = -400950 J

or

[tex]W=-401\ kJ[/tex]

-ve sign shows the direction of force and displacement are in opposite direction.

So, 401 kJ work is required to stop a car. Hence, this is the required solution.

To stop an 1100 kg car traveling at 27 m/s, the work required is calculated using the kinetic energy formula KE = ½ mv² and is 401 kJ.

The question asks about the amount of work required to stop a car with a mass of 1100 kg that is traveling at a velocity of 27 m/s. To solve this, we can use the equation for kinetic energy (KE), which is the work needed to accelerate a body of a given mass from rest to its stated velocity. Using the equation KE = ½ mv², we find that KE = ½ (1100 kg) * (27 m/s)² = 399,150 J, which is approximately 401 kJ. Thus, the correct amount of work required to stop the car is 401 kJ.

Answer:

222.22 N

Explanation:

r = 5 cm = 0.05 m, R = 1.5 m, f = ? , F = 200,000 N

Use pascal's law

F / A = f / a

F / 3.14 x R^2 = f / 3.14 x r^2

F / R^2 = f / r^2

f = F x r^2 / R^2

f = 200000 x 0.05 x 0.05 / (1.5 x 1.5)

f = 222.22 N

Answer:

a)

125.6 rad/s

b)

25.12 rad/s²

Explanation:

a)

t = time required by the fan to get up to final operating speed = 5 sec

w = final operating rotational speed = 1200 rpm

we know that :

1 revolution = 2π rad

1 min = 60 sec

w = [tex]1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}[/tex]

w = [tex]\frac{1200\times 2\pi }{60}\frac{rad}{s}[/tex]

w = 125.6 rad/s

b)

w₀ = initial angular speed = 0 rad/s

α = angular acceleration

using the equation

w = w₀ + α t

125.6 = 0 + α (5)

α = 25.12 rad/s²

Answer:

[tex]3.6\cdot 10^{-8} N[/tex]

Explanation:

The electrostatic force between the proton and the electron is given by:

[tex]F=k\frac{q_p q_e}{r^2}[/tex]

where

[tex]k=9.00\cdot 10^9 Nm^2 C^{-2}[/tex] is the Coulomb constant

[tex]q_p = 1.6\cdot 10^{-19} C[/tex] is the magnitude of the charge of the proton

[tex]q_e = 1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the electron

[tex]r=8\cdot 10^{-11}m[/tex] is the distance between the proton and the electrons

Substituting the values into the formula, we find

[tex]F=(9\cdot 10^9 ) \frac{(1.6\cdot 10^{-19})^2}{(8\cdot 10^{-11})^2}=3.6\cdot 10^{-8} N[/tex]

Answer:

Density of unknown material = 3.86 g/cm³

Explanation:

Mass of unknown material, M = 54 gm

The object occupied 14 gm of water.

Volume of unknown material = Volume of 14 gm of water.

Density of water = 1 gm/cm³

Volume of 14 gm of water  

                 [tex]V=\frac{14}{1}=14cm^3[/tex]

Volume of unknown material, V = Volume of 14 gm of water = 14 cm³

Density of unknown material    

                              [tex]\rho =\frac{M}{V}=\frac{54}{14}=3.86g/cm^3[/tex]    

Answer:

1119.1 K

Explanation:

From Clausius-Clapeyron equation:

[tex]\frac{dP}{dT}=[/tex]Δ[tex]\frac{h_{v} }{R} (\frac{1}{T^{2} } )dT[/tex]

The equation may be integrated considering the enthalpy of vaporization constant, and its result is:

[tex]ln(\frac{P_{2} }{P_{1} } )=-[/tex]Δ[tex]\frac{h_{v} }{R}*(\frac{1}{T_{2} }-\frac{1}{T_{1} })[/tex]

Isolating the temperature [tex]T_{2}[/tex]

[tex]T_{2}=\frac{1}{\frac{-R}{dhv}*ln(\frac{P_{2} }{P_{1}}) +\frac{1}{T_{1}} }[/tex]

[tex]T_{2}=\frac{1}{\frac{-8.314}{1.00*10^5}*ln(\frac{500}{40}) +\frac{1}{906.15}}[/tex]

[tex]T_{2}=1119.1K[/tex]

Note: Remember to change the units of the enthalpy vaporization to J/mol; and the temperatures must be in Kelvin units.

There is a format mistake with the enthalpy of vaporization, each 'Δ' correspond to that.

There is no illustration of the problem provided but I'll attempt to provide an answer.

The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:

║E║ = ΔV/d

║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.

We are given the following values:

║E║= 10N/C

d = 3m

Plug these values in and solve for ΔV

10 = ΔV/3

ΔV = 30V

Answer:

speed of block is 0.665 m/s

Kinetic energy loss is 237.16 J

Explanation:

Here we can use momentum conservation as there is no external force on the system in horizontal direction

so here we will have

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

[tex]0.022(210) + 2(0) = 0.022(150) + 2 v[/tex]

[tex]4.62 = 3.3 = 2 v[/tex]

[tex]v = \frac{4.63 - 3.3}{2}[/tex]

[tex]v = 0.665 m/s[/tex]

Now kinetic energy loss in the system is given as

[tex]Loss = KE_i - KE_f[/tex]

[tex]Loss = \frac{1}{2}(0.022)(210^2) - (\frac{1}{2}(0.022)(150^2) + \frac{1}{2}(2)(0.665)^2)[/tex]

[tex]Loss = 237.16 J[/tex]

The block's final velocity after the bullet passes through it is about 1.68 m/s. The kinetic energy loss of the system is about 602 Joules. These conclusions are drawn using the conservation of momentum and energy.

This problem can be approached using the principles of conservation of momentum and energy. Let's first focus on momentum. Before the bullet hits, the momentum of the system is only contributed by the bullet. After the bullet passes through the block, both the bullet and the block contribute to the total momentum, which should be the same as before. This gives us the equation: 0.022 kg * 210 m/s (bullet's initial momentum) = 0.022 kg * 150 m/s (bullet's final momentum) + 2.0 kg * v (final momentum of the block). Solving this for the block's final velocity, v, will yield about 1.68 m/s.

Next, let's examine the kinetic energy loss. Before the collision, kinetic energy is only contributed by the bullet: KE_initial = 0.5 * 0.022 kg * (210 m/s)². After, it comes from both the bullet and the block: KE_final = 0.5 * 0.022 kg * (150 m/s)² + 0.5 * 2.0 kg * (1.68 m/s)². The kinetic energy loss is KE_initial - KE_final, which should be about 602 Joules. It's worth mentioning that this energy loss is transferred to other forms like heat and sound, illustrating the law of conservation of energy.

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Answer:

47.17 degree C

Explanation:

mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C

mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C

Let T be the temperature at equilibrium.

Heat given by the gold = Heat taken by water

mg x sg x (T1 - T) = mw x sw x (T - T2)

3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)

42441 - 451.5 T = 840 T - 18480

60921 = 1291.5 T

T = 47.17 degree C

Answer:

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Explanation:

Let the linear charge density of the charged wire is given as

[tex]\frac{q}{L} = \lambda[/tex]

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]

here we have

[tex]E \int dA = \frac{\lambda L}{\epsilon_0}[/tex]

[tex]E. 2\pi r L = \frac{\lambda L}{\epsilon_0}[/tex]

so we have

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Answer:

Change in length is 8.3 cm

Explanation:

Coefficient of thermal expansion for iron is given as

[tex]\alpha = 12 \times 10^{-6} per ^0C[/tex]

now as we know that the change in the length due to thermal expansion depends of change in temperature and initial length of the object

so here we will have

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

here we know that

[tex]L_o = 300 m[/tex]

[tex]\Delta T = 23^o C[/tex]

now we will have

[tex]\Delta L = (300)(12 \times 10^{-6})(23)[/tex]

[tex]\Delta L = 0.083 m[/tex]

so change in length is approx 8.3 cm

Final answer:

To determine the rate at which the sphere's volume is dissolving, differentiate the volume formula with respect to time and substitute the given rate of the radius change. The dissolution rate at the instant when the radius is 1 meter is 4π*0.08 m^3/hour.

Explanation:

To find the rate at which the sphere's volume is being dissolved with respect to time, we can apply the formula for the volume of a sphere, V = &frac43;πr^3, and differentiate it with respect to time (t). Since the radius (r) of the sphere is changing, we have:

dV/dt = d(&frac43;πr^3)/dt = 4πr^2*(dr/dt)

Given that the radius is shrinking at a rate of 8 cm/hour (which we need to convert to meters since our volume will be in cubic meters), the radius change in m/hour is:

dr/dt = 0.08 m/hour

At the instant when r is 1 meter, plugging the values into the equation gives us:

dV/dt = 4π(1 m)^2*(0.08 m/hour) = 4π*0.08 m^3/hour

Therefore, the rate at which the sphere's volume is being dissolved is 4π*0.08 m^3/hour.

Answer:

b = 1084.5 kg/s

Explanation:

As we know that the amplitude of damped oscillation is given as

[tex]A = A_o e^{-\frac{bt}{2m}}[/tex]

here we know that

[tex]A = \frac{A_o}{2}[/tex]

after time t = 4.9 minutes

also we know that

[tex]m = 2.30 \times 10^5 kg[/tex]

now we will have

[tex]\frac{A_o}{2} = A_o e^{-\frac{bt}{2m}}[/tex]

[tex]\frac{bt}{2m} = ln2[/tex]

[tex]b = \frac{ln2 (2m)}{t}[/tex]

[tex]b = \frac{2(ln2)(2.30 \times 10^5)}{4.9 \times 60}[/tex]

[tex]b = 1084.5 kg/s[/tex]

C = 0.274μF.

The capacitance when a dielectric is present is given by the equation:

C = kε₀(A/d)

Where k is the dielectric constant of the material, ε₀ is a universal constant, A is the area of the capacitor's plates, and d is the distance that separate the plates.

A parallel plates capacitor with plates of area 1.0cm² that are separated by a distance of 0.1mm, the dielectric constant of the stronium titanate is 310 at a temperature of 20 ° C.

K = 310.00, ε₀ = 8.85x10⁻¹²F/m, A = 1.00x10⁻²m², and d = 0.10x10⁻³m. Substituing the values in the equation C = kε₀(A/d):

C = (310.00)(8.85x10⁻¹²F/m)(1.00x10⁻²m²/0.10x10⁻³m)

C = (2.74x10⁻⁹F/m)(100m)

C = 2.74x10⁻⁷F

C = 0.274x10⁻⁶ = 0.274μF

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

     4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

So Given values of Planck Constant are equivalent in English system and metric system.

Answer:

0.1029 Kg

Explanation:

[tex]Given\ data[/tex]

[tex]mass\of \ice\ cube =50gm[/tex]

[tex]Latent\ heat \of\ vapourization\ of nitrogen(L) =200kJ/kg[/tex]

[tex]specific\ heat\ of\ ice=2100 J/(kgc)[/tex]

[tex]boiling\ point\ of\ nitrogen=77K\approx-169^{\circ}c[/tex]

[tex]now\ Equating\ heat \ absorb\ by \ ice\ from\ liquid\ nitrogen [/tex]

[tex]{m_{ice}}c[{\Delta T}]={m_{nitrogen}}L[/tex]

[tex]{m_{ice}}2100[0-(-196)}=m_{nitrogen}{200\times 1000}[/tex]

[tex]m_{nitrogen}=0.1029Kg[/tex]

Explanation:

Electric force is the force acting between two charged particles. Electric force between electron and proton is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]

[tex]F=9\times 10^9\times \dfrac{-(1.6\times 10^{-19})^2}{(5.3\times 10^{-11}\ m)^2}[/tex]

[tex]F=-8.2\times 10^{-8}\ N[/tex]

Gravitational force is the force acting between two masses. The gravitational force between the electron and proton is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\ kg\times 1.67\times 10^{-27}\ kg}{(5.3\times 10^{-11}\ m)^2}[/tex]

[tex]F=3.61\times 10^{-47}\ N[/tex]

Hence, this is the required solution.

Answer:

P = 34034.2 Watt

Explanation:

As we know that the slope angle is given as

[tex]\theta = 15.6^0[/tex]

now the weight of the rider along the slope is given as

[tex]W = mgsin\theta[/tex]

[tex]W = 69(9.81)sin15.6[/tex]

[tex]W = 182 N[/tex]

now total weight of all 53 riders along the slope is given as

[tex]F = 53 W[/tex]

[tex]F = 9647.5 N[/tex]

now the speed of the rope is given as

[tex]v = 12.7 km/h = 3.53 m/s[/tex]

now the power required is given as

[tex]P = F.v[/tex]

[tex]P = 9647.5(3.53) [/tex]

[tex]P = 34034.2 Watt[/tex]

The power required to operate the tow is 34020.45 W.

The weight of a single rider is calculated as follows;

[tex]W = mgsin(\theta)\\\\W = (69 \times 9.8) sin(15.6)\\\\W = 181.84 \ N[/tex]

W = 53 x 181.84

W = 9637.52 N

The power required to operate the tow is calculated as follows;

P = Fv

where;

P = 9637.52 x 3.53 = 34020.45 W

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Answer:

a) 35.5

b)25.1

Explanation:

Torque=distance* (perpendicular component of the force to the distance)

a) force=50 N distance=0,71m(we need the distance in meters to have the same units with the Newton)

T=50*0.75

b) only one component of the force makes torque, so we multiply all the 50N with sin(45) to find the perpendicular component to the distance

T=50*sin(45)*0.71

Answer:

a) 10462341.6×10⁻¹³ A/m²

b) 133709238.907 A

Explanation:

n = Density of the protons in the solar wind = 11.6 cm⁻³ = 11.6×10⁶ m⁻³

v = Velocity of the protons = 563 km/s = 563000 m/s

e = Charge of a proton = 1.602×10⁻¹⁹ coulombs

R = Radius of Earth = 6.3781×10⁶ m

A = Area of Earth = πR² = π(6.3781×10⁶)²=127.8×10¹² m²

a) Current density

J = nev

⇒J = 11.6×10⁶×1.602×10⁻¹⁹×563000

⇒J = 10462341.6×10⁻¹³ A/m²

∴ Current density of these protons is 10462341.6×10⁻¹³ A/m²

b) Current

I = JA

⇒I = 10462341.6×10⁻¹³×127.8×10¹²

⇒I = 1337092389.07×10⁻¹

⇒I = 133709238.907 A

∴ Total current Earth receives is 133709238.907 A

The angular acceleration α of the CD is 1.590 rev/min².

The angular acceleration α can be calculated using the formula:

α = [tex]\frac{\Delta \omega}{\Delta t}[/tex]

Where [tex]\(\Delta \omega\)[/tex] is the change in angular velocity and [tex]\(\Delta t\)[/tex] is the change in time. Angular velocity [tex](\(\omega\))[/tex] is related to angular displacement [tex](\(\theta\))[/tex] and time (\(t\)) by the equation:

[tex]\[ \omega = \frac{\theta}{t} \][/tex]

Given that the CD accelerates from rest [tex](\(\omega_0 = 0\))[/tex] to a constant angular velocity of 477 rev/min, and rotates through an angular displacement of 0.250 rev}, we can use the following relationships:

[tex]\[ \omega = \frac{\theta}{t} \][/tex]

[tex]\[ \omega_f = \frac{477}{1} \, \text{rev/min} \][/tex]

[tex]\[ \omega_0 = 0 \, \text{rev/min} \][/tex]

Substituting these values into the first equation, we get:

[tex]\[ \frac{477}{1} = \frac{0.250}{t} \][/tex]

Solving for t, we find t = 0.525 min.

Now, substitute these values into the angular acceleration formula:

α = [tex]\frac{\frac{477}{1} - 0}{0.525}[/tex]

α = 1.590 rev/min²

Therefore, the angular acceleration of the CD is 1.590 rev/min².

Answer:

The  bullet's initial speed 258.06 m/s.

Explanation:

It is given that,

It is given that,

Mass of the bullet, m₁ = 12 g = 0.012 kg

Mass of the pendulum, m₂ = 2.2 kg

The center of mass of the pendulum rises a vertical distance i.e. h = 10 cm = 0.1 m

We need to find the initial speed of the bullet. Here, the kinetic energy of the bullet gets converted to potential energy of the system as :

[tex]\dfrac{1}{2}(m_1+m_2)V^2=(m_1+m_2)gh[/tex]

[tex]V=\sqrt{2gh}[/tex]....................(1)

Where

V is the speed of bullet and pendulum at the time of collision.

Let v be the initial speed of the bullet. Using conservation of momentum as :

[tex]m_1v=(m_1+m_2)V[/tex]

[tex]V=(\dfrac{m_1+m_2}{m_1})\sqrt{2gh}[/tex]

[tex]V=(\dfrac{0.012\ kg+2.2\ kg}{0.012 kg})\sqrt{2\times 9.8\ m.s^2\times 0.1\ m}[/tex]

V = 258.06 m/s

So, the initial speed of the bullet is 258.06 m/s. Hence, this is the required solution.

Answer:

magnetic field will allow the electron to go through 2 x [tex]10^{4}[/tex]  T k

Explanation:

Given data in question

velocity = 5.0 × [tex]10^{7}[/tex]

electric filed = [tex]10^{4}[/tex]

To find out

what magnetic field will allow the electron to go through, undeflected

solution

we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions

so we can also say

F(net) = Fe + Fb i.e. = 0

q V B +  q E = 0

q will be cancel out

[tex]10^{4}j[/tex] + 5e + 7i × B =  0

B = 2 x [tex]10^{4}[/tex]  T k