Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current I1 through the first resistor and the potential difference ΔV2 between the ends of the second resistor.]

Answer:

Option D is the correct answer.

Explanation:

The specific gravity of copper is 8.91.

We have

            [tex]\frac{\texttt{Density of copper}}{\texttt{Density of water}}=8.91\\\\\texttt{Density of copper}=8.91\times 1000kg/m^3=8910kg/m^3[/tex]

We also have

           1 kg = 2.205 lb

            1 m = 3.28 ft

Substituting

            [tex]\texttt{Density of copper}=8910kg/m^3=\frac{8910\times 2.205}{3.28^3}=556.76lb/ft^3[/tex]

Option D is the correct answer.

Answer:

Option D is the correct answer.

Answer:

B). Kelvin Scale

Explanation:

Kelvin scale is the absolute scale which is used to express temperature in English System

We have different temperature scales

1) Fahrenheit Scale

Generally use in British system of units

2) Celcius Scale

It is used to given temperature of different scales and its relation with kelvin

[tex]^0 C = K - 273[/tex]

3) Rankine Scale

It is used in thermodynamic scales with large temperature range

Answer:1200

Explanation:

Given data

Upper Temprature[tex]\left ( T_H\right )=240^{\circ}\approx 513[/tex]

Lower Temprature [tex]\left ( T_L\right )=20^{\circ}\approx 293[/tex]

Engine power ouput[tex]\left ( W\right )=910 W[/tex]

Efficiency of carnot cycle is given by

[tex]\eta =1-\frac{T_L}{T_H}[/tex]

[tex]\eta =\frac{W_s}{Q_s}[/tex]

[tex]1-\frac{293}{513}=\frac{910}{Q_s}[/tex]

[tex]Q_s=2121.954 W[/tex]

[tex]Q_r=1211.954 W[/tex]

rounding off to two significant figures

[tex]Q_r=1200 W[/tex]

Answer:

1200

Explanation:

Answer:

The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.

Explanation:

Tnp= 4 sec

gnp= 9.83 m/sec²

Lnp= 3.97m

Tequ= 4 sec

gequ= 9.78 m/sec²

Lequ=?

Lequ= (Lnp* gequ) / gnp

Lequ= 3.949 m

Answer:

a) V = 195.70 m/s

b) f=3.02 × 10⁻⁴ Hz

c) T = 3311.25 seconds

Explanation:

Given:

Wavelength, λ = 646 Km = 646000 m

Distance traveled = 3410 Km = 3410000 m

Time = 4.84 h = 4.84 × 3600 s = 17424 seconds

a) The speed (V) of the wave is given as

V = distance / time

V = 3410000 m/ 17424 seconds

or

V = 195.70 m/s

b) The frequency (f) of the wave is given as:

f = V / λ

f= 195.70 / 646000

f=3.02 × 10⁻⁴ Hz

c) The time period (T)  is given as:

T = 1/ f

T = 1/ (3.02 × 10⁻⁴) Hz

T = 3311.25 seconds

Answer:

a) Tmax=827.12N

b) Tmin = 626.22N

c) 776.895 N

Explanation:

Given:

Mass of wrecking ball M1=63.9 Kg

Mass of the chain M2=20.5 Kg

acceleration due to gravity, g=9.8m/s²

Now,

(a)The Maximum Tension generated in the chain,

    Tmax=(M1+M2)×g)

    Tmax=(M1+M2)×(9.8 m/s²)

    Tmax=(63.9+ 20.5)×(9.8 m/s²)  

    Tmax=827.12N

(b) The Minimum Tension Tmin will be due to the weight of the wrecking ball only

Mathematically,

Tmin=weight of the wrecking ball

Tmin = 63.9kg×9.8m/s²

Tmin = 626.22N

(c)Now. the tension at 3/4 from the bottom of the chain

In this part we have to use only 75% of the chain i.e the weight acting below the point of consideration

thus, the tension will be produced by the weight of the 3/4 part of the chain and the wrecking ball

Therefore, the weight of the 3/4 part of the chain = [tex]\frac{3}{4}\times 20.5\times 9.8 N[/tex]

= 150.675 N

Hence, the tension at a point 3/4 of the way up from the bottom of the chain will be  = 150.675 N + (63.9×9.8) N = 776.895 N

Answer:The sled slides 16.875m before rest.

Explanation:

[tex]a=\frac{F}{m} =\frac{12N}{20kg}[/tex]

a=0.6 m/s²

[tex]Vf=0=Vi-a.t[/tex]

[tex]t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg[/tex]

[tex]d= Vi.t - \frac{a.t^{2}}{2}[/tex]

[tex]d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m[/tex]

Explanation:

The mass of freight car, m₁ = 30,000 kg

Velocity of freight car, u₁ = 0.85 m/s

Mass of hopper, m₂ = 110,000 kg

(a) Let v is the final velocity of the loaded freight car. Initial momentum of the car before the dump, [tex]p_i=30000\ kg\times 0.850\ m/s=25500\ kg-m/s[/tex]

Final momentum, [tex]p_f=(30000\ kg+110000\ kg)v=140000\ v[/tex]

According to the conservation of momentum,

initial momentum = final momentum

25500 kg-m/s = 140000 v

v = 0.182 m/s

So, the  final velocity of the loaded freight car is 0.182 m/s.

(b) Initial kinetic energy, [tex]k_i=\dfrac{1}{2}\times 30000\ kg\times (0.850\ m/s)^2=10837.5\ J[/tex]

Final kinetic energy, [tex]k_f=\dfrac{1}{2}(m_1+m_2)v^2[/tex]

[tex]k_f=\dfrac{1}{2}\times (30000\ kg+110000\ kg)\times (0.182\ m/s)^2=2318.68\ J[/tex]

So, loss in kinetic energy, [tex]\Delta k=k_f-k_i[/tex]

[tex]\Delta k=2318.68\ J-10837.5\ J=-8518.82\ J[/tex]

So, 8518.82 J of kinetic energy is lost after collision. Hence, this is the required solution.

Using the conservation of momentum, we find the final velocity of the loaded freight car to be 0.182 m/s. By comparing the initial and final kinetic energies, we find that 8,459.89 Joules of kinetic energy is lost during the process.

This problem is a classic example of conservation of momentum, and to a lesser extent, conservation of kinetic energy. Firstly, we calculate the initial momentum of the system. Momentum is mass multiplied by velocity, so the initial momentum of the freight car is 30,000 kg * 0.850 m/s = 25,500 kg*m/s. After the scrap metal is dumped into the freight car, the total mass of the system is now 30,000 kg + 110,000 kg = 140,000 kg. Because momentum must be conserved, we calculate the final velocity of the freight car as the total momentum divided by the total mass, 25,500 kg*m/s / 140,000 kg = 0.182 m/s.

Secondly, before the scrap metal is added, the initial kinetic energy of the freight car is 0.5 * 30,000 kg * (0.850 m/s)² = 10,781.25 Joules. After the metal is added, the final kinetic energy is 0.5 * 140,000 kg * (0.182 m/s)² = 2,321.36 Joules. The change in kinetic energy can then be obtained by subtracting the final kinetic energy from the initial one, which gives us 8,459.89 Joules as the amount of kinetic energy that is lost.

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Answer:

The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]

Explanation:

Given that,

Mass of object = 6.22 kg

Distance = 1.02 m

We need to calculate the number of electron

Using formula of electric force

[tex]F_{e}=\dfrac{k(q)^2}{r^2}[/tex]....(I)

We know that,

[tex]q = Ne[/tex]

Put the value of q in equation (I)

[tex]F_{e}=\dfrac{k(Ne)^2}{r^2}[/tex].....(II)

Using gravitational force

[tex]F_{G}=\dfrac{Gm^2}{r^2}[/tex].....(III)

Equating equation (II) and (III)

[tex]F_{e}=F_{G}[/tex]

[tex]\dfrac{k(Ne)^2}{r^2}=\dfrac{Gm^2}{r^2}[/tex]

[tex]N=\sqrt{\dfrac{G}{k}}\times\dfrac{m}{e}[/tex]....(IV)

Put the value in the equation(IV)

[tex]N=\sqrt{\dfrac{6.67\times10^{-11}}{9\times10^{9}}}\times\dfrac{6.22}{1.6\times10^{-19}}[/tex]

[tex]N=3.35\times10^{9}[/tex]

Hence, The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]

Answer:

The answers are: a)Fcp=0,23N b)As Fcp=0,93N, it increases 4 times when the speedis doubled

Explanation:

Let´s explain what´s the centripetal force about: It´s the force applied over an object moving on a curvilinear path. This force is directed to the rotation center.

This definition is described this way:

[tex]Fcp*r=m*V^2[/tex] where:

Fcp is the Centripetal Force

r is the horizontal circle radius

m is the ball mass

V is the tangencial speed, same as the rotation speed

w is the angular speed

Here we need to note that the information we have talks about 1cycle/0,639s (One cycle per 0,639s). We need to express this in terms of radians/seconds. To do it we define that 1cycle is equal to 2pi, so we can find the angular speed this way:

[tex]w=(1cycle/0,639s)*(2pi/1cycle)[/tex]

So the angular speed is [tex]w=9,83rad/s[/tex]

Now that we have this information, we can find the tangencial speed, which will be the relation between the angular speed and the circle radius, this way:

[tex]V=w*r[/tex] so the tangencial speed is:

[tex]V=(9,83rad/s)*(0,149m)[/tex]

[tex]V=1,5m/s[/tex]

Now we have all the information to find the Centripetal Force:

[tex]Fcp=(m*V^2)/(r)[/tex]

[tex]Fcp=((0,0154kg)*(1,5m/s)^2)/(0,149m)[/tex]

a) So the Centripetal Force is: [tex]Fcp=0,23N[/tex]  

b) If the tangencial speed is doubled, its new value will be 3m/s. replacing this information we will get the new Centripetal Force is:

[tex]Fcp=((0,0154kg)*(3m/s)^2)/(0,149m)[/tex]

The Centripetal Force is: [tex]Fcp=0,93N[/tex]

Here we can see that if the speed is doubled, the Centripetal Force will increase four times.

Answer:

The top of the ladder go down by 0.15 m.

Explanation:

Here we have right angled triangle.

Hypotenuse = 4.6 m

Bottom angle = 66º

Length from ladder bottom to wall bottom = 4.6 cos66 = 1.87 m

Length from ladder top to wall bottom = 4.6 sin66 = 4.20 m

New length from ladder bottom to wall bottom =  1.87 + 0.31 = 2.18 m

By Pythagoras theorem

New length from ladder top to wall bottom is given by

          [tex]\sqrt{4.6^2-2.18^2}=4.05m[/tex]

Distance  the top of the ladder go down = 4.20 - 4.05 = 0.15 m

Answer:

5100 eV

Explanation:

q = 3.2 x 10^-19 C

Va = 2.4 x 10^3 J/C

Vb = 4.95 x 10^3 J/C

Work done = q (Vb - Va)

W = 3.2 x 10^-19 x (4.95 - 2.4 ) x 10^3 = 8.16 x 10^-16 J

A we know that 1 eV = 1.6 x 10^-19 J

So, W = (8.16 x 10^-16) / (1.6 x 10^-19) = 5100 eV

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.

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.

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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Using the acceleration due to gravity on the planet, we calculate its mass to be 8.97 x 10²⁴ kg. For the mass of the star, using Kepler's third law, we find it to be approximately 1.99 x 10³⁰ kg, assuming the planet's orbit is about 1 AU.

To find the mass of the planet and the star, we can use Newton's form of Kepler's third law and Newton's universal law of gravitation. Let's denote the mass of the star as M and the mass of the planet as m.

Mass of the Planet

The acceleration due to gravity (g) on the surface of the planet is given by:

g = Gm / r²

Where:

G is the gravitational constant (approximately 6.674 x 10⁻¹¹ N m2 kg-²).

m is the mass of the planet.

r is the radius of the planet, which is half of its diameter.

Rearranging the formula to solve for m:

m = g r² / G

Substituting the given values (g = 59.7 m/s² and r = 9 x 106 m), we get:

m = 59.7 x (9 x 106)² / 6.674 x 10⁻¹¹
= 8.97 x 10²⁴ kg

Mass of the Star

Using Newton's version of Kepler's third law, we get:

M = 4π² a3 / G T²

Where: a is the semi-major axis of the planet's orbit, which is the average distance from the planet to the star.

T is the orbital period of the planet around the star.

Assuming that the planet's orbit is approximately 1 Astronomical Unit (AU) since the period is close to 1 Earth year, and converting 402 Earth days to seconds, we find M.

T = 402
days x 24 hours/day x 3600 seconds/hour = 3.47 x 10⁷ seconds

a = 1 AU = 1.496 x 10¹¹ m

Then the mass of the star M is:

M = 4π² (1.496 x 1011)3 / (6.674 x 10⁻¹¹ x (3.47 x 10⁷)²) = 1.99 x 10³⁰ kg

Answer:

The mass of neon is 13.534 g.

Explanation:

Given that,

Volume = 3.8 L

Pressure = 6.8 atm

Temperature = 470 K

Atomic mass of neon =20.2 g/mol

Gas constant R = 8.314 = 0.082057 L atm/mol K

We need to calculate the mass of neon

Using equation of gas

[tex]PV=nRT[/tex]

[tex]6.8\times3.8=n\times0.082057\times470[/tex]

[tex]n=\dfrac{6.8\times3.8}{0.082057\times470}[/tex]

[tex]n= 0.670[/tex]

We know that,

[tex]n = \dfrac{m}{molar\ mass}[/tex]

[tex]m = n\times molar\ mass[/tex]

[tex]m=0.670\times20.2[/tex]

[tex]m=13.534\ g[/tex]

Hence, The mass of neon is 13.534 g.

Answer:

6.97 m/s, 344 degree

Explanation:

mass of car, m = 900 kg, uc = 15 m/s, vc = 5 m/s, θ = 40 degree

mass of truck, M = 1500 kg, uT = 0, vT = ?, Φ = ?

Here, vT be the velocity of truck after collision and Φ its direction above x axis.

Use conservation of momentum in X axis

900 x 15 + 1500 x 0 = 900 x 5 Cos 40 + 1500 x vT Cos Φ

13500 - 3447.2 = 1500 vT CosΦ

vT CosΦ = 6.7 .....(1)

Use conservation of momentum in y axis

0 + 0 = 900 x 5 Sin 40 + 1500 vT SinΦ

vT SinΦ = - 1.928 .....(2)

Squarring both the equations and then add

vT^2 = 6.7^2 + (-1.928)^2

vT = 6.97 m/s

Dividing equation 2 by 1

tan Φ = - 1.928 / 6.7

Φ = - 16 degree

Angle from + X axis = 360 - 16 = 344 degree

Answer:

[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]

Explanation:

Let the speed of light in vacuum is c and the speed of light in medium is v. Let the angle of incidence is i.

By using the definition of refractive index

refractive index of the medium is given by

n = speed of light in vacuum / speed of light in medium

n = c / v  ..... (1)

Use Snell's law

n = Sin i / Sin r

Where, r be the angle of refraction

From equation (1)

c / v = Sin i / Sin r

Sin r = v Sin i / c

[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]

Answer:

5.3 cm

Explanation:

Let the charges on the spheres be q and Q.

r = 9.20 cm

F = k Q q / (9.20)^2 ..... (1)

Let the new distance be r' and force F'

F' = 3 F

F' = k Q q / r'^2

3 F = k Q q / r'^2 ...... (2)

Divide equation (1) by (2)

F / 3 F = r'^2 / 84.64

1 / 3 = r'^2 / 84.64

r' = 5.3 cm

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

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

[tex]H = 4.905 m[/tex]

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

[tex]a = g = 9.81 m/s^2[/tex]

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

[tex]v_f = 0[/tex]

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

[tex]\Delta y = 0 = v_y t + \frac{1}{2} at^2[/tex]

[tex]0 = v_y (2) - \frac{1}{2}(9.81)(2^2)[/tex]

[tex]v_y = 9.81 m/s[/tex]

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we have

[tex]0 - (9.81^2) = 2(-9.81)H[/tex]

[tex]H = 4.905 m[/tex]

The acceleration of the ball is -9.8 m/s², the velocity of the ball at max height is 0 m/s, the initial velocity is 9.8 m/s and the max height is 4.9 m.

The questions refer to the motion of a tennis ball under gravity. (a) The acceleration of the ball while it is in flight is -9.8 m/s², due to Earth's gravity. This will be the case throughout the ball's flight, regardless of whether it's moving up or down. (b) The velocity of the ball when it reaches its maximum height is 0 m/s, as the ball stops moving vertically for an instant before it starts coming down. (c) We can calculate the initial velocity using the equation v = u + at. Here, v is final velocity, u is initial velocity, a is acceleration and t is time. As v = 0 m/s at max height, a = -9.8 m/s² and t = 1 s (time taken to reach max height, half of total time), we find u to be 9.8 m/s. (d) The maximum height can be calculated using the equation h = ut + 0.5at². Here, h is height, u is initial velocity, t is time and a is acceleration. Subsituting u = 9.8 m/s, a = -9.8 m/s², t = 1 s, we get h to be 4.9 m.

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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

Final answer:

The heat flow through the lead brick can be calculated using Fourier's law of heat conduction. The thermal conductivity of lead is used along with the provided dimensions and temperature difference to find the heat flow, which is 2.2 W to two significant figures.

Explanation:

To calculate the heat flow through a lead brick, we can use Fourier's law of heat conduction, which states that the heat transfer rate (Q) through a material is proportional to the thermal conductivity of the material (k), the area (A) through which the heat is being transferred, the temperature difference (ΔT) across the material, and inversely proportional to the thickness (d) of the material.

The formula to calculate heat flow is: Q = k × A × ΔT × t / d

First, we need to convert all measurements to SI units: the length from cm to m (13 cm = 0.13 m), the area from cm² to m² (13 cm² = 0.0013 m²), and the time from seconds to hours (1.0 s). The thermal conductivity of lead is approximately 35 W/m°C. Plugging in the values: Q = 35 W/m°C × 0.0013 m² × 8.0°C × 1.0 s / 0.13 m

By simplifying the equation, we find the heat flow through the lead brick:

Q = (35 × 0.0013 × 8.0 × 1.0) / 0.13 W = 2.215 W

To two significant figures, the heat flow is 2.2 W.

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:

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:

Option d is the correct option Kinetic energy is minimum while as potential energy is maximum

Explanation:

At the top most point of the flight since it cannot reach any further up in the vertical direction thus the potential energy at this position shall be maximum. Now since the total energy of the projectile is conserved so the remaining kinetic energy shall be minimum at that point so as the sum of the kinetic and potential energies remain constant.

Answer:

Answer to the question: 91.44m

Explanation:

L=100 yds = 300 ft = 91.44 m

To convert 100 yards to meters for an American football field, you multiply 100 yards by 3 to get 300 feet, then convert feet to meters using the given conversion factor, resulting in approximately 91.4 meters.

To convert the length of an American football field from yards to meters, we first need to understand the conversion factors between these units. We know that 1 yard equals 3 feet, and we are given that 1 meter equals 3.281 feet. Starting with the length of the field in yards, which is 100 yards, we convert yards to feet and then convert feet to meters.

Step-by-Step Conversion

Therefore, the length of an American football field is roughly 91.4 meters, excluding the end zones.

Answer:

2.20°

Explanation:

λ = wavelength of the coherent light = 500 x 10⁻⁹ m

d = slits separation = 1.30 x 10⁻⁵ m

n = order of the fringe = 1

θ = angle made by the first bright fringe with the center line = ?

For first bright fringe, Using the equation

d Sinθ = n λ

(1.30 x 10⁻⁵) Sinθ = (1) (500 x 10⁻⁹)

[tex]Sin\theta =\frac{500\times 10^{-9}}{1.30\times 10^{-5}}[/tex]

Sinθ = 0.0385

θ = 2.20°

Final answer:

To find the angle of the first bright fringe in a double-slit experiment, the formula for constructive interference is used. By substituting the provided measurements into the formula and calculating the inverse sine, the angle is found to be approximately 2.20°.

Explanation:

The question refers to a Young's double-slit experiment, where monochromatic light of a known wavelength is used to produce an interference pattern on a screen. To find the angle at which the first bright fringe occurs, we can use the formula for constructive interference in a double-slit setup, which is given by:

d sin(θ) = mλ

Where d is the separation between the slits, θ is the angle of the fringe from the centerline, m is the order number of the fringe (m=1 for the first bright fringe), and λ is the wavelength of the light.

Plugging in the values provided:

d = 1.30 x 10-5 m

λ = 500 x 10-9 m

m = 1

Now we solve for θ:

sin(θ) = mλ / d

sin(θ) = (1)(500 x 10-9 m) / (1.30 x 10-5 m)

sin(θ) ≈ 0.038462

Using a calculator to find the inverse sine, we get:

θ ≈ 2.20°

Answer:

Part A)

[tex]\alpha = 745 rad/s^2[/tex]

Part B)

[tex]\theta = 4563.1 rad[/tex]

Explanation:

Drill starts from rest so its initial angular speed will be

[tex]\omega_i = 0[/tex]

now after 3.50 s the final angular speed is given as

[tex]f = 2.49 \times 10^4 rev/min[/tex]

[tex]f = {2.49 \times 10^4}{60} = 415 rev/s[/tex]

so final angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega_f = 2607.5 rad/s[/tex]

now we have angular acceleration given as

[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]

[tex]\alpha = \frac{2607.5 - 0}{3.50}[/tex]

[tex]\alpha = 745 rad/s^2[/tex]

Part b)

The angle through which it is rotated is given by the formula

[tex]\theta = \frac{(\omega_f + \omega_i)}{2}\delta t[/tex]

now we have

[tex]\theta = \frac{(2607.5 + 0)}{2}(3.50)[/tex]

[tex]\theta = 4563.1 rad[/tex]

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:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, [tex]A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2[/tex]

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, [tex]\rho=11\times 10^{-8}\ \Omega-m[/tex]

We need to find the resistance of the iron rod. It is given by :

[tex]R=\rho\dfrac{L}{A}[/tex]

[tex]R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}[/tex]

[tex]R=0.000077 \Omega[/tex]

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.