The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 350 km above the surface of the Earth, where the International Space Station (ISS) flies? (The mass of the Earth is 5.97×1024 kg, and the radius of the Earth is 6370 km.) It is somewhat greater than 9.81 m/s2. It is zero, since the ISS is in the state of weightlessness. It is half of 9.81 m/s2. It is somewhat less than 9.81 m/s2. It is twice of 9.81 m/s2. It is 9.81 m/s2, the same.

Answer:

1.8 x 10⁻⁷ N/m

Explanation:

[tex]F_{max}[/tex] = maximum force with which the optical trap can pull = 11 x 10⁻¹⁵ N

x = stretch caused in DNA molecule due to the force = 0.06 x 10⁻⁶ m

k = spring constant of the spring

Maximum force is given as

[tex]F_{max}= k x[/tex]

[tex]11\times 10^{-15}= k (0.06\times 10^{-6})[/tex]

k = 1.8 x 10⁻⁷ N/m

The spring constant of the DNA molecule is calculated using Hooke's Law with the given force of 11 fN and stretch distance of 0.06 μm, resulting in a spring constant of approximately 183.33 N/m.

To calculate the spring constant (k) of a DNA molecule modeled as a spring, we can use Hooke's Law, which states that the force (F) applied to stretch or compress a spring is directly proportional to the displacement (x) it causes, as represented by the equation F = kx. The optical trap pulls with a maximum force of 11 fN and stretches the DNA molecule by 0.06 μm. Using the given values, the spring constant (k) can be calculated as:

k = F / x

Therefore, k = 11 fN / 0.06 μm, and to ensure the units are consistent, we convert 0.06 μm to meters (0.06 μm = 0.06 x 10-6 m).

k = 11 x 10-15 N / 0.06 x 10-6 m

k = (11 / 0.06) x 10-9 N/m

k ≈ 183.33 N/m

The spring constant of the DNA molecule is therefore approximately 183.33 N/m.

Answer:

The speed of the ball is 0.8 m/s.

(D) is correct option.

Explanation:

Given that,

mass of ball = 3.0 kg

length = 50 cm

We need to calculate the speed of the ball

Using relation between the centripetal force and tension

[tex]\dfrac{mv^2}{r}=T[/tex]

[tex]v^2=\dfrac{T\times r}{m}[/tex]

Where, m = mass

T = tension

r = radius

Put the value into the formula

[tex]v=\sqrt{\dfrac{3.8\times50\times10^{-2}}{3.0}}[/tex]

[tex]v=0.8\ m/s[/tex]

Hence, The speed of the ball is 0.8 m/s.

Answer:

Period of oscillation, T = 2 sec

Explanation:

It is given that,

Mass of the object, m = 5 kg

Spring constant of the spring, k = 50 N/m

This object is hanging from a coiled spring. We need to find the period of oscillation of the spring. The time period of oscillation of the spring is given by :

[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]

[tex]T=2\pi\sqrt{\dfrac{5\ kg}{50\ N/m}}[/tex]

T = 1.98 sec

or

T = 2 sec

So, the period of oscillation is about 2 seconds. Hence, this is the required solution.

Answer:

a) [tex]a_c= 1.33 m/s^2 [/tex]

b) a= 1.79 m/s²

   θ = 41.98⁰

Explanation:

arc radius  = 12 m

constant speed = 4.00 m/s

(a) centripetal acceleration

     [tex]a_c=\frac{v^2}{R}[/tex]

     [tex]a_c=\frac{4^2}{12} [/tex]

                  = 1.33 m/s²

(b) now we have given

        [tex]a_t= \ 1.20 m/s^2 [/tex]

        now,

         [tex]a=\sqrt{a^2_c+ a^2_t}[/tex]

         [tex]a=\sqrt{1.33^2+ 1.20^2}[/tex]

            a= 1.79 m/s²

 direction

[tex]\theta = tan^{-1}(\frac{a_t}{a_r} )[/tex]

[tex]\theta = tan^{-1}(\frac{1.2}{1.33} )[/tex]

     θ = 41.98⁰

The centripetal acceleration of the hawk is 1.33 m/s².

The resultant acceleration  of the hawk at the given moment is 1.79 m/s².

The direction resultant acceleration of the hawk is 48⁰.

The given parameters;

The centripetal acceleration of the hawk is calculated as follows;

[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(4)^2}{12} \\\\a_c = 1.33 \ m/s^2[/tex]

The resultant acceleration is calculated as;

[tex]a = \sqrt{a_c^2 + a_t} \\\\a = \sqrt{(1.33)^2 + (1.2)^2} \\\\a = 1.79 \ m/s^2[/tex]

The direction of the acceleration is calculated as follows;

[tex]tan(\theta) = \frac{a_c}{a_t} \\\\\theta = tan^{-1} ( \frac{a_c}{a_t} )\\\\\theta = tan^{-1} ( \frac{1.33}{1.2} )\\\\\theta = 48^0[/tex]

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Answer:

2.17 x 10^8 m/s

Explanation:

Angle of incidence, i = 30 degree, refractive index of mineral oil, n = 1.38

Let r be the angle of refraction.

By use of Snell's law

n = Sin i / Sin r

Sin r = Sin i / n

Sin r = Sin 30 / 1.38

Sin r = 0.3623

r = 21.25 degree

Let the speed of light in oil be v.

By the definition of refractive index

n = c / v

Where c be the speed of light

v = c / n

v = ( 3 x 10^8) / 1.38

v = 2.17 x 10^8 m/s

Mario's velocity vector relative to the ice is

[tex]\vec v_{M/I}=-4.79\,\vec\jmath[/tex]

(note that all velocities mentioned here are given in m/s)

The puck is moving in a direction of 17.8 degrees west of south, or 252.2 degrees counterclockwise relative to east. Its velocity vector relative to the ice is then

[tex]\vec v_{P/I}=9.13(\cos252.2^\circ\,\vec\imath+\sin252.2^\circ\,\vec\jmath)=-2.79\,\vec\imath-8.69\,\vec\jmath[/tex]

The velocity of the puch relative to Mario is

[tex]\vec v_{P/M}=\vec v_{P/I}+\vec v_{I/M}=\vec v_{P/I}-\vec v_{M/I}[/tex]

[tex]\vec v_{P/M}=-2.79\,\vec\imath-3.90\,\vec\jmath[/tex]

Then, relative to Mario,

a. the puck is traveling at a speed of [tex]\boxed{\|\vec v_{P/M}\|=4.80}[/tex], and

b. is moving in a direction [tex]\theta[/tex] such that

[tex]\tan\theta=\dfrac{-3.90}{4.80}\implies\theta=-126^\circ[/tex]

which is about [tex]\boxed{35.6^\circ}[/tex] west of south.

The magnitude of the puck's velocity as observed by Mario is 4.9 m/s and the direction is 35.6 degrees west of south.

When addressing this problem, we're dealing with relative velocity in two dimensions. Since we need to find the puck's velocity as seen by Mario, we should use the principle that the relative velocity of the puck with respect to Mario is the vector difference between the velocity of the puck and Mario.

Given that Mario is travelling south at 4.79 m/s and the puck is travelling at an angle of 17.8 degrees west of south at a speed of 9.13 m/s, we can break the puck's velocity into southward and westward components. The southward component is 9.13 m/s × cos(17.8°) = 8.74 m/s and the westward component is 9.13 m/s × sin(17.8°) = 2.84 m/s.

The relative southward velocity of the puck with respect to Mario is 8.74 m/s - 4.79 m/s = 3.95 m/s. The westward component is unchanged since Mario has no westward velocity. So, the puck appears to Mario to be moving southwest at an angle of arctan(2.84 m/s / 3.95 m/s) = 35.6 degrees west of south.

The magnitude is given by the Pythagorean theorem, sqrt((2.84 m/s)² + (3.95 m/s)²) = 4.9 m/s.

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Answer:

C. 150 N to the left

Explanation:

If we take right to be positive and left to be negative, then:

∑F = -450 N + 300 N

∑F = -150 N

The net force is 150 N to the left.

Answer:

(C) 150 N to the left

Explanation:

It is given that,

Force acting in left side, F = 450 N

Force acting in right side, F' = 300 N

Let left side is taken to be negative while right side is taken to be positive. So,

F = -450 N

F' = +300 N

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

[tex]F_{net}=-450\ N+300\ N[/tex]

[tex]F_{net}=-150\ N[/tex]    

So, the net force on the table is 150 N and it is acting to the left side. Hence, the correct option is (c).

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

Answer:

21813.46 J

Explanation:

The Potential energy stored at the time it is hanging freely is mgl.

the potential energy stored at the time when it makes a angle 23 degree from the vertical is mglCos 23

Change in gravitational potential energy

U = mgl = mgl cos 23

U = mgl (1 - cos23)

U = 1400 x 9.8 x 20 ( 1 - Cos 23)

U = 21813.46 J

Answer:

2 kg m^2

Explanation:

M = 1 kg, R = 1 m

The moment of inertia of the hoop about its axis perpendicular to its plane is

I = M R^2

The moment of inertia of the hoop about its edge perpendicular to it splane is given by the use of parallel axis theorem

I' = I + M x (distance between two axes)^2

I' = I + M R^2

I' = M R^2 + M R^2

I' = 2 M R^2

I' = 2 x 1 x 1 x 1 = 2 kg m^2

The moment of inertia of a hoop about an axis perpendicular to its plane at an edge is calculated using the Parallel-Axis Theorem and for a hoop with mass M = 1kg and radius R = 1m, it is 2 kg*m².

Calculation of the Moment of Inertia for a Hoop

To find the moment of inertia of a hoop with mass M = 1kg and radius R = 1m about an axis perpendicular to the hoop's plane at an edge, we use the Parallel-Axis Theorem. The moment of inertia of a hoop about its central axis (through its center, perpendicular to the plane) is MR². According to the Parallel-Axis Theorem, the moment of inertia about an axis parallel to this but passing through the edge of the hoop is given by I = MR₂ + MR₂ (because the distance from the central axis to the outer edge of the hoop is R). Thus, the moment of inertia for the hoop about the edge is 2MR₂ which simplifies to 2 * 1kg * (1m)² = 2 kg*m².

Answer:

Distance between two adjacent wave crests = 24m

Explanation:

Distance= speed × time

Distance traveled by waves in 60 seconds (15 crests)= 15 × distance

15 × distance = 6,0 (meters/second) × 60 seconds

distance = (360 meters) / 15 = 24 meters (between two adyacent waves)

Answer:

1270.44 Hz

Explanation:

[tex]v_{L}[/tex]  = velocity of the our car = 35.0 m/s

[tex]v_{P}[/tex]  = velocity of the police car = ?

[tex]v_{S}[/tex]  = velocity of the sound = 343 m/s

[tex]f_{app}[/tex]  = frequency observed as police car approach = 1310 Hz

[tex]f_{rec}[/tex]  = frequency observed as police car go away = 1240 Hz

[tex]f[/tex]  = actual frequency of police siren

Frequency observed as police car approach is given as

[tex]f_{app}= \frac{(v_{s}-v_{L})f}{v_{s} -v_{P} }[/tex]

inserting the values

[tex]1310 = \frac{(343 - 35)f}{343 -v_{P} }[/tex]                          eq-1

Frequency observed as police car goes away is given as

[tex]f_{rec}= \frac{(v_{s} + v_{L})f}{v_{s} + v_{P} }[/tex]

inserting the values

[tex]1240 = \frac{(343 + 35)f}{343 + v_{P} }[/tex]                          eq-2

Dividing eq-1 by eq-2

[tex]\frac{1310}{1240} = \left ( \frac{343 - 35}{343 - v_{P} } \right )\frac{(343 + v_{P})}{343 + 35 }\\[/tex]

[tex]v_{P}[/tex]  = 44.3 m/s

Using eq-1

[tex]1310 = \frac{(343 - 35)f}{343 - 44.3 }[/tex]

f = 1270.44 Hz

In this situation, the neon gas should also have a temperature of 175 K, the same as the helium gas, given all the conditions and the principle of the ideal gas law.

The problem you're working on involves understanding the ideal gas law, which is PV = nRT. This equates the pressure (P), volume (V) of the gas, and temperature (T). There are two important points to consider. Firstly, the number of molecules or moles (n) and the ideal gas constant (R) aren't changing in this situation. Secondly, the volume and pressure are the same for both gases.

Given that the masses, volume, and pressure are the same for both helium and neon, we conclude that the number of moles (n) for helium and neon are equal (because the mass density is the same and for ideal gas mass density (ρ) = PM/RT, where M is molar mass). The temperature ratio should be the same as the ratio of Kelvin temperatures of two gases.

So, we can safely say that since the gases obey the ideal gas law, and all conditions are held constant aside from the identity of the gas and the temperature, the temperature of the neon gas must also be 175 Kelvin like the helium gas.

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Answer:

Product side

Explanation:

When water vapor reacts reversibly with solid carbon to yield a mixture of hydrogen gas and carbon monoxide and we continually add more water vapor to the reaction the equilibrium of the reaction shifts to the product side.

Because gaseous water is reactant that appears in the reaction quotient expression.

[tex]H_{2}O+ C_{s}\leftrightharpoons H_{2}_{g}+ CO[/tex]

When we add more water vapor to the reaction the product formation is increased. The reaction goes in forward direction affecting the equilibrium.

Answer:

x = 0.95 m

y = 0.166 m

Explanation:

m = mass of the bullet = 24 g = 0.024 kg

v = speed of bullet before collision = 280 m/s

M = mass of the pendulum = 3.7 kg

L = length of the string = 2.8 m

h = height gained by the pendulum after collision

V = speed of the bullet and pendulum combination

Using conservation of momentum

m v = (m + M) V

(0.024) (280) = (0.024 + 3.7) V

V = 1.805 m/s

using conservation of energy

Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination

(m + M) g h = (0.5) (m + M) V²

(9.8) h = (0.5) (1.805)²

h = 0.166 m

y = vertical displacement = h = 0.166 m

x = horizontal displacement

horizontal displacement is given as

x = sqrt(L² - (L - h)²)

x =  sqrt(2.8² - (2.8 - 0.166)²)

x = 0.95 m

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of [tex]x= 5t^3+1[/tex]

Coordinate of position of [tex]y=3t^2+2[/tex]

Time = 2.00 s

We need to calculate the acceleration

[tex]a = \dfrac{d^2x}{dt^2}[/tex]

For x coordinates

[tex]x=5t^3+1[/tex]

On differentiate w.r.to t

[tex]\dfrac{dx}{dt}=15t^2+0[/tex]

On differentiate again w.r.to t

[tex]\dfrac{d^2x}{dt^2}=30t[/tex]

The acceleration in x axis at 2 sec

[tex]a = 60i[/tex]

For y coordinates

[tex]y=3t^2+2[/tex]

On differentiate w.r.to t

[tex]\dfrac{dy}{dt}=6t+0[/tex]

On differentiate again w.r.to t

[tex]\dfrac{d^2y}{dt^2}=6[/tex]

The acceleration in y axis at 2 sec

[tex]a = 6j[/tex]

The acceleration is

[tex]a=60i+6j[/tex]

We need to calculate the net force

[tex]F = ma[/tex]

[tex]F = 3.00\times(60i+6j)[/tex]

[tex]F=180i+18j[/tex]

The magnitude of the force

[tex]|F|=\sqrt{(180)^2+(18)^2}[/tex]

[tex]|F|=180.89\ N[/tex]

Hence, The net force acting on this object is 180.89 N.

To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.

To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. The x-component of the force can be found by taking the derivative of the x-coordinate equation with respect to time and multiplying it by the mass of the object. The same process can be applied to find the y-component of the force. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.

Starting with the x-coordinate equation, x = 5t³ - 1, taking its derivative gives dx/dt = 15t². Multiplying the derivative by the mass of the object (3.00 kg) gives the x-component of the force: Fx = 3.00 kg * 15t².

Similarly, for the y-coordinate equation y = 3t² + 2, taking its derivative gives dy/dt = 6t. Multiplying the derivative by the mass of the object (3.00 kg) gives the y-component of the force: Fy = 3.00 kg * 6t.

Using the Pythagorean theorem, the magnitude of the net force Fnet can be calculated as: Fnet = sqrt(Fx² + Fy²). Substituting the values into the equation will give you the magnitude of the net force at t = 2.00 s.

Answer:

Moessbauer Effect = eggy eggs

Explanation:

Answer:

Radius, r = 0.36 meters

Explanation:

It is given that,

Speed of cosmic ray electron, [tex]v=6.5\times 10^6\ m/s[/tex]

Magnetic field strength, [tex]B=10\times 10^{-5}\ T=10^{-4}\ T[/tex]

We need to find the radius of circular path the electron follows. It is given by :

[tex]qvB=\dfrac{mv^2}{r}[/tex]

[tex]r=\dfrac{mv}{qB}[/tex]

[tex]r=\dfrac{9.1\times 10^{-31}\ kg\times 6.5\times 10^6\ m/s}{1.6\times 10^{-19}\times 10^{-4}\ T}[/tex]

r = 0.36 meters

So, the radius of circular path is 0.36 meters. Hence, this is the required solution.

Answer:

(a):The time what it takes his voice reach the earth via radio waves is t1= 1.28 seconds.

(b): The time what it takes his voice reach from Mars to the earth via radio waves is t2= 186.66 seconds.

Explanation:

d1= 3.85 *10⁸ m

d2= 5.6 *10¹⁰ m

C= 300,000,000 m/s

t1= d1/C

t1= 1.28 s

t2= d2/C

t2= 186.66 s

Answer:

286

Explanation:

p1v1/T1=p2v2/T2

then subtitude your values

T2=760*0.65*273/210*0.040

T2=135/8.4=16+273=289

The pressure-volume work done by the system during the expansion is 5.25 x 10³J.

The work that is done when a fluid is compressed or expanded is known as pressure-volume work. Pressure-volume work occurs whenever there is a change in volume but the outside pressure stays the same.

Here,

The pressure of the system of gas, P = 210 kPa

Initial volume of the system of gas, V₁ = 0.04 m³

Final volume of the system of gas, V₂ = 0.065 m³

The expression for the pressure-volume work done by a system of gas is given by,

Work done,

W = PΔV

W = P(V₂ - V₁)

Applying the values of P, V₁ and V₂,

W = 210 x 10³(0.065 - 0.04)

W = 210 x 10³x 0.025

W = 5.25 x 10³J

Hence,

The pressure-volume work done by the system during the expansion is 5.25 x 10³J.

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Answer:

diameter is 13.46 mm

Explanation:

length of rod = 200 mm  = 0.2 m

compression force = 80,000 N

factor of safety = 2.5

Sy = 496 MPa

E = 71 GPa

to find out

diameter

solution

first we calculate the allowable stress i.e.  = Sy/factor of safety

allowable stress = 496/ 2.5= 198.4 MPa  198.4 × [tex]10^{6}[/tex] Pa

now we calculate the diameter d by the Euler's equation i.e.

critical load = [tex]\pi ^{2}[/tex] E × moment of inertia / ( K × length )²   ..........1

now we calculate the critical load i.e.  allowable stress × area

here area = [tex]\pi[/tex] /4 × d²  

so critical load = 198.4 ×  [tex]\pi[/tex] /4 × d²

and K = 1 for pin ends  

and moment of inertia is =   [tex]\pi[/tex] / 64 × [tex]d^{4}[/tex]

put all value in equation 1 and we get d

198.4 ×[tex]10^{6}[/tex] ×  [tex]\pi[/tex] /4 × d²  = [tex]\pi ^{2}[/tex] 71 × [tex]10^{9}[/tex]  × [tex]\pi[/tex] / 64 × [tex]d^{4}[/tex]  / ( 1 × 0.2 )²

155.8229× [tex]10^{6}[/tex]  × d²  =   700.741912× [tex]10^{9}[/tex]× 0.049087×  [tex]d^{4}[/tex] / 0.04

d=0.01346118 m

d = 13.46 mm

diameter is 13.46 mm

Answer:83.17 cm/s

Explanation:

Let positive x be negative direction and negative x be positive direction in this question

General equation of motion of SHM is

x=[tex]Asin\left ( \omega_{n}t\right )[/tex] --------1

where [tex]\omega [/tex]is natural frequency of motion given by

[tex]\omega_n[/tex]=[tex]\sqrt{\frac{k}{m}}[/tex]

Where K is spring constant

here A=23.7cm

And it is given it is again at 23.7 from equilibrium position having passed through the equilibrium once.

i.e. it covers this distance in [tex]\frac{T}{2}[/tex] sec

where T is the time period of oscillation i.e. returning to same place after T sec

therefore T=0.634 sec

differentiating equation 1 we get

v=[tex]A\omega_n[/tex]cos[tex]\left ( \omega_{n}t\right )[/tex]

and [tex]T\times \omega_n[/tex]=[tex]2\pi[/tex]

[tex]\omega_n[/tex]=9.911rad/s

[tex]v[/tex]=[tex]23.7\times 9.911cos\left (789.261\degree\right)[/tex]

v=83.17 cm/s

Answer:

3.014 x 10⁻⁸ N

Explanation:

q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C

v = speed of the jet = 685 m/s

B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T

θ = angle between the magnetic field and direction of motion = 90

magnitude of the magnetic force is given as

F = q v B Sinθ

F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90

F = 3.014 x 10⁻⁸ N

Answer:

Electric charge, Q = 55.69 C

Explanation:

It is given that,

Electric current drawn by the compressor, I = 23.7 A

Time taken, t = 2.35 s

We need to find the electric charge passes through the circuit during this period. The definition of electric current is given by total charge divided by total time taken.

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

Where,

q is the electric charge

[tex]q=I\times t[/tex]

[tex]q=23.7\ A\times 2.35\ s[/tex]

q = 55.69 C

So, the electric charge passes through the circuit during this period is 55.69 C. Hence, this is the required solution.

Answer:

7.66 (litres); 523.66 (N).

Explanation:

all the details are provided in the attached picture. Note, P_i means 'Weight of iron', ρ_i means "Density of iron', ρ_w means 'Density of water', F_f means 'Arhimede force'.

The answers are marked with red colour.

Answer:

2.6 m/s²

Explanation:

m = mass of the object = 7.8 kg

Consider the right direction as positive and left direction as negative

F = horizontal applied force towards right = 50 N

f = frictional force acting towards left = - 30 N

a = acceleration

Force equation for the motion of the object is given as

F + f = ma

50 - 30 = (7.8) a

a = 2.6 m/s²

Answer:

0.324×10⁻³ m/s²

Explanation:

G=Gravitational constant=6.67408×10⁻⁸ m³/kg s

m₁=m₂=Mass of the ships=42000×10³ kg

r=Distance between the ships=93 m

The ships are considered particles

From Newtons Law of Universal Gravitation

[tex]F=G\frac {m_1m_2}{r^2}\\Here\ m_1=m_2\ hence\ m_1\times m_2=m_1^2\\\Rightarrow F=G\frac {m_1^2}{r^2}\\\Rightarrow F=\frac{6.67408\times 10^{-8}\times (42000\times 10^3)^2}{93^2}\\\Rightarrow F=13612.0674\ N\\[/tex]

[tex]F=ma\\\Rightarrow 13612.0674=42000\times 10^3a\\\Rightarrow a=0.324\times 10^{-3}\ m/s^2[/tex]

∴Acceleration of one of the liners toward the other due to their mutual gravitational attraction is 0.324×10⁻³

Answer:

0.5 A

Explanation:

N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm

The induced emf is given by

e = - N dФ/dt

Where, dФ/dt is the rate of change of magnetic flux.

Ф = B A

dФ/dt = A dB/dt

so,

e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V

negative sign shows the direction of magnetic field.

induced current, i = induced emf / resistance = 0.2 / 0.4 = 0.5 A

Answer:

[tex]log_28=3[/tex]

Explanation:

To calculate the value of [tex]log_28[/tex]

Also,Some relation for log values:

[tex]loga^m= mloga[/tex]  -----------------------------------------------------1

[tex]log_aa= 1[/tex]---------------------------------------------------------------2

Also, [tex]2^3= 8[/tex]

Applying in the question as:

[tex]log_28=log_2(2^3)[/tex]

Using Relation-1 mentioned above:

[tex]log_28=3\times log_2(2)[/tex]

Using Relation-2 mentioned above:

[tex]log_28=3\times 1[/tex]

Thus,

[tex]log_28=3[/tex]

Answer:

Power required, P = 0.49 kW

Explanation:

It is given that,

Mass of drum, m = 150 kg

Height, h = 20 m

Time, t = 1 min = 60 sec

Average power required is given by work done divided by total time taken. It is given by :

[tex]P=\dfrac{W}{t}[/tex]

W = F.d

Here, W = mgh

So, [tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{150\ kg\times 9.8\ m/s^2\times 20\ m}{60\ s}[/tex]

P = 490 watts

or P = 0.49 kW

So, the average power required to raise the drum is 0.49 kilo-watts. Hence, this is the required solution.

Final answer:

The average power required to lift a 150-kg drum to a height of 20 meters in 1 minute is calculated by first finding the work done and then dividing by time. The result is approximately 0.4905 kW.

Explanation:

Calculating Required Power to Lift a Drum

The problem involves finding the average power required to raise a certain weight to a specified height over a defined period of time. The weight in question is a 150-kg drum, the height is 20 meters, and the time is 1 minute (60 seconds).

To calculate power, you need to first calculate the work done, which is the product of force and displacement. Since we are lifting the object vertically, the force is the weight of the object, which is mass (m) times the acceleration due to gravity (g), and the displacement is the height (h). The formula for work (W) is:
W = m × g × h

The average power (P), then, is the work done over the time (t) it takes to do it:
P = W / t

Using the values given:

mass (m) = 150 kg

acceleration due to gravity (g) = 9.81 m/s²

height (h) = 20 m

time (t) = 60 s

First, calculate the work done:
W = 150 kg × 9.81 m/s² × 20 m = 29430 J

Now calculate the power required:
P = 29430 J / 60 s = 490.5 W

To convert this into kilowatts (kW), divide by 1000:
P = 0.4905 kW

Therefore, the average power required to lift the drum is approximately 0.4905 kW.