Two forces are applied to a 5.0-kg crate; one is 3.0 N to the north and the other is 4.0 N to the east. The magnitude of the acceleration of the crate is: a. 1.0 m/s^2 b. 2.8 m/s^2 c.7.5 m/s^2 d. 10.0 m/s^2

Answer:

Electric force between the charges, [tex]F=1.35\times 10^{10}\ N[/tex]

Explanation:

It is given that,

Charge 1, q₁ = 3 C

Charge 2, q₂ = 2 C

Distance between them, r = 2 m

We need to find the electric force between them. The formula for electric force is given by :

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

k is the electrostatic constant

[tex]F=9\times 10^9\times \dfrac{3\ C\times 2\ C}{(2\ m)^2}[/tex]

[tex]F=1.35\times 10^{10}\ N[/tex]

So, the force between the charges is [tex]1.35\times 10^{10}\ N[/tex]. Hence, this is the required solution.

Answer:

The radius of the sphere is 3.6 m.

Explanation:

Given that,

Potential of first sphere = 450 V

Radial distance = 7.2 m

If the potential of sphere =150 V

We need to calculate the radius

Using formula for potential

For 450 V

[tex]V=\dfrac{kQ}{r}[/tex]

[tex]450=\dfrac{kQ}{r}[/tex]....(I)

For 150 V

[tex]150=\dfrac{kQ}{r+7.2}[/tex]....(II)

Divided equation (I) by equation (II)

[tex]\dfrac{450}{150}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{r+7.2}}[/tex]

[tex]3=\dfrac{(r+7.2)}{r}[/tex]

[tex]3r=r+7.2[/tex]

[tex]r=\dfrac{7.2}{2}[/tex]

[tex]r=3.6\ m[/tex]

Hence, The radius of the sphere is 3.6 m.

The radius of the sphere whose surface has a potential difference of  450 V is 3.6 m.

We know that the potential difference can be written as,

[tex]V = k\dfrac{Q}{R}[/tex]

We know that at  R= R, Potential difference= 450 V,

[tex]450 = k\dfrac{Q}{R}[/tex]

Also, at R = (R+7.2), Potential difference = 150 V,

[tex]150 = k\dfrac{Q}{(R+7.2)}[/tex]

Taking the ratio of the two,

[tex]\dfrac{450}{150} = \dfrac{kQ}{R} \times \dfrac{(R+7.2)}{kQ}\\\\\dfrac{450}{150} = \dfrac{(R+7.2)}{R}\\\\R = 3.6\ m[/tex]

Hence, the radius of the sphere whose surface has a potential difference of  450 V is 3.6 m.

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To calculate the diffusion flux at 1000˚C for the same concentration gradient, use the Arrhenius equation.

J ≈ 2.4 × 10-12 kg/m2-s

To calculate the diffusion flux at 1000˚C (1273 K) for the same concentration gradient, we can use the Arrhenius equation:

J = J0 * exp(-Q/RT)

Where J is the diffusion flux, J0 is the pre-exponential factor, Q is the activation energy for diffusion, R is the gas constant, and T is the absolute temperature.

Given that the diffusion flux at 1220˚C (1493 K) is 7.8 × 10-8 kg/m2-s and the activation energy for diffusion is 145,000 J/mol, we can calculate the diffusion flux at 1000˚C as:

J = (7.8 × 10-8) * exp(-145000/(8.314*1273))

J ≈ 2.4 × 10-12 kg/m2-s

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To calculate the steady-state diffusion flux at a different temperature, use the Arrhenius equation to find the diffusion coefficients at the two temperatures, and find the ratio based on the fact that the diffusion flux is proportional to the diffusion coefficient when the concentration gradient is constant. Input the known values into the equation to solve for the unknown diffusion flux.

The steady-state diffusion through a metal plate can be calculated using the Arrhenius equation, which relates the diffusion coefficient (D) to temperature. The equation is D = D0e^-(Q/RT), where D0 is the pre-exponential factor, Q is the activation energy for diffusion, R is the gas constant and T is the temperature in K.

Given that the diffusion flux (J) is defined as J = -D×(dc/dx), where dc/dx is the concentration gradient. We can find that when the concentration gradient remains the same, the ratio of the two diffusion fluxes at different temperatures can be represented as J1/J2 = D1/D2.

Substitute the Arrhenius equation into the ratio, we get J1/J2 = e^(Q/R)×(1/T1-1/T2). Then you can use the given values, namely Q = 145,000 J/mol, R = 8.314 J/(mol×K), and temperatures T1 = 1493K , T2 = 1273K, as well as the known J1, to calculate J2.

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Answer:

a) 7250.5 N

b) 4.6 m/s²

Explanation:

a)

F = applied force = 8000 N

θ = angle with the horizontal = 65 deg

Consider the motion along the vertical direction :

[tex]F_{y}[/tex] = Applied force in vertical direction in upward direction = F Sinθ = 8000 Sin65 = 7250.5 N

[tex]F_{g}[/tex] = weight of the plane in vertical direction in downward direction = ?

[tex]a_{y}[/tex] = Acceleration in vertical direction = 0 m/s²

Taking the force in upward direction as positive and in downward direction as negative, the force equation along the vertical direction can be written as

[tex]F_{y}-F_{g} = m a_{y}[/tex]

[tex]7250.5 -F_{g} = m (0)[/tex]

[tex]F_{g}[/tex] = 7250.5 N

b)

m = mass of the plane

force of gravity is given as

[tex]F_{g} = mg [/tex]

[tex]7250.5 = m(9.8) [/tex]

m = 739.85 kg

Consider the motion along the horizontal direction

[tex]F_{x}[/tex] = Applied force in horizontal direction = F Cosθ = 8000 Cos65 = 3381 N

[tex]a_{x}[/tex] = Acceleration in horizontal direction

Acceleration in horizontal direction is given as

[tex]a_{x}=\frac{F_{x}}{m}[/tex]

[tex]a_{x}=\frac{3381}{739.85}[/tex]

[tex]a_{x}[/tex] = 4.6 m/s²

Answer:

Part a)

the two frequencies are in ratio of odd numbers so it must be closed at one end

Part b)

L = 34.3 cm

Explanation:

Part a)

Since the shortest frequency is known as fundamental frequency

It is given as

[tex]f_o = 250 Hz[/tex]

next higher frequency is given as

[tex]f_1 = 750 Hz[/tex]

since the two frequencies here are in ratio of

[tex]\frac{f_1}{f_o} = \frac{750}{250} = 3 : 1[/tex]

since the two frequencies are in ratio of odd numbers so it must be closed at one end

Part b)

For the length of the pipe we can say that fundamental frequency is given as

[tex]f_o = \frac{v}{4L}[/tex]

here we have

[tex]250 = \frac{343}{4(L)}[/tex]

now we will have

[tex]L = \frac{343}{4\times 250}[/tex]

[tex]L = 34.3 cm[/tex]

Answer:

The acceleration of the ball is 666.67 m/s²

Explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Applied force to it, F = 100 N

We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :

F = ma

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{100\ N}{0.15\ kg}[/tex]

[tex]a=666.67\ m/s^2[/tex]

So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.

Answer:

(a): The bird speed after swallowing the insect is V= 4.83 m/s

(b): The impulse on the bird is I= 0.3 kg m/s

(c): The force between the bird and the insect is F= 20 N

Explanation:

ma= 0.3 kg

va= 6 m/s

mb= 0.01kg

vb= 30 m/s

(ma*va - mb*vb) / (ma+mb) = V

V= 4.83 m/s (a)

I= mb * vb

I= 0.3 kg m/s  (b)

F*t= I

F= I/t

F= 20 N (c)

This physics problem uses the principle of conservation of momentum to calculate the bird's speed after swallowing the insect, the impulse experienced by the bird, and the force between the bird and the insect.

This is a physics problem relating to the conservation of momentum. Let's start by defining some facts, where m bird = 0.3 kg and v bird = 6.0 m/s are the mass and speed of the bird before the incident and m insect = 0.01 kg and v insect = 30 m/s are the mass and speed of the insect.

a. To find the bird's speed immediately after swallowing the insect, we need to apply the conservation of momentum principle: initial total momentum = final total momentum, which can be written as m bird * v bird + m insect * v insect = (m bird + m insect) * v final.

b. The impulse on the bird equals the change in momentum of the bird, thus equals to the final momentum of the bird - initial momentum of the bird.

c. The force between the bird and the insect is obtained from the definition of impulse: Force * time = impulse, or Force = Impulse/time.

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Answer:

0.84 Ω

Explanation:

r = mean radius of the turn = 6.5 mm

n = number of turns of copper wire = 150

Total length of wire containing all the turns is given as

L = 2πnr

L =  2 (3.14)(150) (6.5)

L = 6123 mm

L = 6.123 m

d = diameter of the wire = 0.4 mm = 0.4 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.4 x 10⁻³)²

A = 1.256 x 10⁻⁷ m²

ρ = resistivity of copper = 1.72 x 10⁻⁸ Ω-m

Resistance of the coil is given as

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

[tex]R = \frac{(1.72\times 10^{-8}) (6.123))}{(1.256\times 10^{-7}))}[/tex]

R = 0.84 Ω

Answer: The mass of the object will be 11250 kg.

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given :

Density of the object= [tex]12500kg/m^3[/tex]

Mass of object = ?

Volume of the object = [tex]0.9m^3[/tex]

Putting in the values we get:

[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]

[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]

[tex]mass=11250kg[/tex]

Thus the mass of the object is 11250 kg.

Answer:

37.86 kg

Explanation:

The weight of sign board is equally divided on each rope. It means the tension in all the ropes is equal to the weight of the sign board in equilibrium condition.

Tension in each rope = 53 N

Tension in 7 ropes = 7 x 53 N = 371 N

Thus, The weight of sign = 371 N

Now, weight = m g

where m is the mass of sign.

m = 371 / 9.8 = 37.86 kg

Answer:

The average total energy density of this electromagnetic wave is [tex]72.5\ \mu\ J/m^3[/tex].

Explanation:

Given that,

Magnetic field [tex]B = 13.5\mu T[/tex]

We need to calculate the average total energy density

Using formula of energy density

[tex]Energy\ density =\dfrac{S}{c}[/tex]....(I)

Where, S = intensity

c = speed of light

We know that,

The intensity is given by

[tex]S = \dfrac{B^2c}{2\mu_{0}}[/tex]

Put the value of S in equation (I)

[tex]Energy\ density =\dfrac{\dfrac{B^2c}{2\mu_{0}}}{c}[/tex]

[tex]Energy\ density = \dfrac{(13.5\times10^{-6})^2}{2\times4\pi\times10^{-7}}[/tex]

[tex]Energy\ density = 0.0000725\ J/m^3[/tex]

[tex]Energy\ density = 72.5\times10^{-6}\ J/m^3[/tex]

[tex]Energy\ density = 72.5\ \mu\ J/m^3[/tex]

Hence, The average total energy density of this electromagnetic wave is [tex]72.5\ \mu\ J/m^3[/tex].

Given:

mass of satellite, m = 1710 kg

altitude, h = [tex]2.6\times 10^{6} m[/tex]

G =  [tex]6.67\times 10^{-11} [/tex]

we know

mass of earth, [tex]M_{E}[/tex] =  [tex]5.972\times 10^{24} kg[/tex]

Here, according to question we will consider

[tex]2M_{E}[/tex] =  [tex]11.944\times 10^{24} kg[/tex]

radius of earth,  [tex]R_{E}[/tex] =  [tex]6.371\times 10^{6} m[/tex]

Formulae Used and replacing [tex]M_{E}[/tex] by  [tex]2M_{E}[/tex] :

1). [tex]v = \sqrt{\frac{2GM_{E}}{R_{E} + h}}[/tex]

2). [tex]T = \sqrt{\frac{4\pi ^{2}(R_{E} + h)^{3}}{2GM_{E}}}[/tex]

3). [tex]KE = \frac{1}{2}mv^{2}[/tex]

4). [tex]Total Energy, E = -\frac{2GM_E\times m}{2(R_{E} + h)}[/tex]

where,

v = orbital velocity of satellite

T = time period

KE = kinetic energy

Solution:

Now, Using Formula (1), for orbital velocity:

 [tex]v = \sqrt{\frac{6.67 \times 10^{-11} \times 11.944 \times 10^{24}}{6.371 \times 10^{6} + 2.6 \times 10^{6}}[/tex]

v =  [tex]9.423 \times 10^{3}[/tex]  m/s

Using Formula (2) for time period:

[tex]T = \sqrt{\frac{4\pi ^{2}(6.371\times 10^{6} + 2\times 10^{6})^{3}}{6.67\times 10^{-11}\times 9.44\times 10^{24}}}[/tex]

[tex]T = 6.728\times 10^{3} s[/tex]

Now, Using Formula(3) for kinetic energy:

[tex]KE = \frac{1}{2}(9.44\times 10^{24})(9.42\times 10^{3})^{2}[/tex]

[tex]KE = \frac{1}{2}(1710)(9.42\times 10^{3})^{2} = 7.586\times 10^{10} J[/tex]

Now, Using Formula(4) for Total energy:

[tex]E = -\frac{6.67\times 10^{-11}\times 9.44\times 10^{24}\times 1710}{2( 6.371\times 10^{6} + 2.6\times 10^{6})}[/tex]

[tex]E = - 7.59\times 10^{10} J[/tex]

Answer:

(a) 6.22 x 10^-6 C

(b) 3.8 x 10^13

Explanation:

Let the second charge is q2 = q

q1 = 1.6 x 10^-19 C

F = 3.5 x 10^9 N

d = 1.6 mm = 1.6 x 10^-3 m

(a) Use the formula of Coulomb's law

F = K q1 x q2 / d^2

3.5 x 10^-9 = 9 x 10^9 x 1.6 x 10^-19 x q / (1.6 x 10^-3)^2

q = 6.22 x 10^-6 C

(b)

Let the number of electrons be n

n = total charge / charge of one electron

n = 6.22 x 10^-6 / (1.6 x 10^-19) = 3.8 x 10^13

Answer:

[tex]6.35\cdot 10^{-9} W[/tex]

Explanation:

The relationship between power and intensity of a sound is given by:

[tex]I=\frac{P}{A}[/tex]

where

I is the intensity

P is the power

A is the area considered

In this problem, we know

[tex]A=2.90\cdot 10^{-3}m^2[/tex] is the surface area of the ear

[tex]I = 2.19\cdot 10^{-6} W/m^2[/tex] is the intensity of the sound

Re-arranging the equation, we can find the power intercepted by the ear:

[tex]P=IA=(2.19\cdot 10^{-6} W/m^2)(2.90\cdot 10^{-3} m^2)=6.35\cdot 10^{-9} W[/tex]

Answer:

1.176m

Explanation:

Local ambient pressure(P1) = 100 kPa

Absolute pressure(P2)=260kPa

Net pressure=absolute pressure-local ambient absolute pressure

Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)

Net pressure=260-100=160kPa

Pressure= ρgh

160kPa=13600*10*h

h=[tex]\frac{160000}{136000}[/tex]

h=1.176m

Answer:

0.5 Ns

Explanation:

When a large force acting on a body for a very small time it is called impulsive force.

Impulse = force × small time

Impulse = 25 × 0.02 = 0.5 Ns

It is a vector quantity

Answer:

A positive charge experiences a force in the direction of an electric field.

Explanation:

Electric field is defined as the electric force acting per unit positive charge. Mathematically, it is given by :

[tex]E=\dfrac{F}{q}[/tex]

We know that like charges repel each other while unlike charges attract each other. The direction of electric field is in the direction of electric force. For a positive charge the field lines are outwards and for a negative charge the electric field lines are inwards.

So, the correct option is (b) "A positive charge experiences a force in the direction of an electric field".

Answer:

[tex]\mu_k = 0.15[/tex]

Explanation:

according to the kinematic equation

[tex]v^{2} - u^{2} = 2aS[/tex]

Where

u is initial velocity  = 0 m/s

a = acceleration

S is distance = 8.00 m

final velocity = 1.0 m/s

[tex]a = \frac {v^{2}}{2S}[/tex]

[tex]a = \frac {1{2}}{2*8.6}[/tex]

a = 0.058 m/s^2

from newton second law

Net force = ma

[tex]f_{net} = ma[/tex]

F - f = ma

2[tex]5 - \mu_kN = ma[/tex]

[tex]25 - \mu_kmg = ma[/tex]

[tex]\frac {25 - ma}{mg} =\mu_k[/tex]

[tex]\frac {25 - 16*0.058}{16*9.81} = 0.15[/tex]

[tex]\mu_k = 0.15[/tex]

Answer:

To the nearest integer, the penny's speed in m/s be right as it reaches the ground if it was dropped from rest = 91 m/s

Explanation:

We have equation of motion

S = ut + 0.5at²

Here u = 0, a = g  and t = 9.3 s

We have equation of motion v = u +at

Substituting

       v = u +at

       v = 0 + 9.8 x 9.3 = 91.14 m/s

To the nearest integer, the penny's speed in m/s be right as it reaches the ground if it was dropped from rest = 91 m/s

Answer:

a) 0.0693 m

b) Work done = 8.644 J

Explanation:

Given:

Spring constant, k = 3600 N/m

Radius of the piston, r = 0.028 m

Now, we know that the atmospheric pressure at STP = 1.01325 × 10⁵ Pa  = 101325 Pa

Now,

The force ([tex]F_P[/tex]) due to the atmospheric pressure on the piston will be:

[tex]F_P[/tex] = Pressure × Area of the piston

on substituting the values we get,

[tex]F_P[/tex] = 101325 × πr²

F = 101325 × π × (0.028)² = 249.56 N

also,

Force on spring is given as:

F = kx

where,

x is the displacement in the spring

 on substituting the values we get,

 249.56 N = 3600N/m × x

or

x = 0.0693 m

thus, the compression in the spring will be = 0.0693 m

b) Applying the concept of conservation of energy

we have,

Work done by the atmospheric pressure in compressing the spring = Potential energy gained  by the spring

mathematically,

[tex]W = \frac{1}{2}kx^2[/tex]

 on substituting the values we get,

[tex]W = \frac{1}{2}\times 3600\times (0.0693)^2[/tex]

W = 8.644 J

a) x = 0.0693 m

b) W = 8.644 J

Given :

Spring constant, K = 3600 N/m

Radius of the piston, r = 0.028 m

Solution :

Now the atmospheric pressure at STP = 1.01325 × 10⁵ Pa  = 101325 Pa

Force due to the atmospheric pressure on the piston is,

Force = Pressure × Area of the piston

on substituting the values we get,

[tex]\rm F_P = 101325\times \pi r^2[/tex]

[tex]\rm F_P = 249.56\;N[/tex]

a) We know that the force on spring is given by,

F = Kx

where, k is spring constant and x is the displacement in the spring.

[tex]249.56 = 3600\times x[/tex]

[tex]\rm x = 0.0693\;m[/tex]

b) We know that the Work Done is given by,

[tex]\rm W= \dfrac{1}{2} k x^2[/tex]

[tex]\rm W = 0.5\times 3600\times (0.0693)^2[/tex]

W = 8.644 J

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Answer:· Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed on a screen that is 2.74 m from the grating. In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.16 m. What is the difference between these wavelengths? . I know to apply the equation din(theta) = m*wavelength but I'm not sure how to find all the missing variables or to get the difference in wavelengths

Answer:

Force on eardrum

       [tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]

Explanation:

Force = Pressure x Area

Pressure = hρg

Height, h = 3 m

ρ = 1000 kg/m³

g = 9.8 m/s²

Pressure = hρg = 3 x 1000 x 9.8 = 29400 N/m²

Area = 1 cm²

Force on eardrum

       [tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]

Answer:

k = 212.55 newton per meter

Explanation:

A girl is standing on a trampoline. Her mass is 65 kg and she is able to jump 3 meters.

We have to find the spring constant.

Since by Hooke's law,

F = -kx

Where F = force applied by the spring

k = spring constant

x = displacement

And we know force applied by the spring will be equal to the weight of the girl.

So, F = mg

Therefore, (-mg) = -kx

65×(9.81) = k×(3)

k = [tex]\frac{(65)(9.81)}{3}[/tex]

k = 212.55 N per meter

Therefore, spring constant of the spring is 212.55 Newton per meter.

Answer:

b) Projectile MOTION

Explanation:

SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line

In this type of motion particle must be in straight line motion

So here we can say

a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM

b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM

d) Spring Motion : it is a straight line to and fro motion so it is also a SHM

So correct answer will be

b) Projectile MOTION

Final answer:

Projectile motion is not a simple harmonic motion because it does not meet the conditions for SHM.

Explanation:

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is proportional to the displacement. The three conditions that must be met to produce SHM are: a linear restoring force, a constant force constant, and no external damping forces. Based on these conditions, the answer to the question is (b) Projectile motion, as it does not meet the conditions for SHM. A projectile follows a parabolic path and does not have a linear restoring force.

Answer:

The distance of bob when Alice starts running is 50 m.

Explanation:

Given that,

Speed v = 10.0 m/s

Time t = 5 sec

We need to calculate the distance

Using formula of distance

[tex]D=v\times t[/tex]

[tex]D=10\times5[/tex]

[tex]D=50\ m[/tex]

Hence, The distance of bob when Alice starts running is 50 m.

The equation of motion of a pendulum is:

[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,[/tex]

where [tex]\ell[/tex] it its length and [tex]g[/tex] is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for small angles ([tex]\theta \ll 1[/tex]), we can use:

[tex]\sin\theta \simeq \theta.[/tex]

Additionally, let us define:

[tex]\omega^2\equiv\dfrac{g}{\ell}.[/tex]

We can now write:

[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.[/tex]

The solution to this differential equation is:

[tex]\theta(t) = A\sin(\omega t + \phi),[/tex]

where [tex]A[/tex] and [tex]\phi[/tex] are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

[tex]T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.[/tex]

This justifies that the period depends only on the pendulum's length.

•If a system gains 757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ. How much work is done is  - 581 kJ

•The correct statement is: Work was done by  the system

Let Change in internal energy ΔU = 176 kJ

Let Heat gained by the system (q) = 757 kJ

Using the  First law of thermodynamics

ΔU = q + w

Where:

ΔU  represent  change in internal energy

q represent  heat added to system and w is work done.

Let plug in the formula

176 kJ = 757 kJ + w

w = 176 kJ - 757 kJ

w= - 581 kJ

Based on the above calculation the negative sign means  that work is done by the system

Inconclusion:

•If a system gains 757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ. How much work is done is  - 581 kJ

•The correct statement is: Work was done by  the system

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The amount of work done by the system, based on given heat gain and change in internal energy is 581 kJ, meaning the work was done by the system.

The question asks about the amount of work done by or on a system in the field of thermodynamics. According to the first law of thermodynamics, the change in internal energy of a system (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W), or written as ΔU = Q - W. In this case, the heat added to your system was 757 kJ and the change in internal energy of the system was +176 kJ.

So we have: 176 kJ = 757 kJ - W. Subtracting 757 kJ from both sides of the equation would give us W = 757 kJ - 176 kJ. This results in the value of W = 581 kJ. Conclusively, since W is positive, we say that work was done by the system.

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Answer:

314 Joule

Explanation:

m = 10 kg, g = 10 m/s^2, d = 1 m, angle turn = 180 degree = π radian

work = torque x angle turn

torque = force x perpendicular distance

torque = m x g x d = 10 x 10 x 1 = 100 Nm

work = 100 x π

work = 100 x 3.14 = 314 Joule

Final answer:

The work done by a constant torque on a rigid body rotating through an angle of 180° can be found by multiplying the torque by the angle. The torque can be calculated using the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. Substituting the given values, we find that the work done by the torque on the rigid body is 18000 kg·m^2·rad/s^2.

Explanation:

The work done by a torque on a rigid body is given by the formula W = τθ, where τ is the torque and θ is the angle through which the body rotates. In this case, the torque is constant in both magnitude and direction, so we can use W = τθ. Given that the torque is constant and the body turns through an angle of 180°, we can calculate the work done as follows:

Since τ is constant, we can write W = τθ = τ (180° - 0°). The work done by the torque is equal to the torque multiplied by the change in angle. Substitute the given values into the formula: W = (τ) (180° - 0°) = (τ) (180°). The work done by the torque is equal to the torque multiplied by 180°.

To find the value of the torque, we need to use the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. In this case, the rigid body has a mass of 10 kg and a distance of 1 m from the pivot. The moment of inertia for a point mass rotating about a fixed axis is given by I = m(r^2), where m is the mass and r is the perpendicular distance from the axis. Substitute the given values into the formula: I = (10 kg)((1 m)^2) = 10 kg·m^2. Since α = a/r and the acceleration due to gravity is 10 m/s^2, we have α = (10 m/s^2)/(1 m) = 10 rad/s^2.

Substitute the values of τ and α into the equation τ = Iα: τ = (10 kg·m^2)(10 rad/s^2) = 100 kg·m^2·rad/s^2. Therefore, the torque is 100 kg·m^2·rad/s^2.

Finally, substitute the values of τ and θ into the equation W = τθ: W = (100 kg·m^2·rad/s^2 )(180°) = 18000 kg·m^2·rad/s^2.

Answer:

E =  ρ ( R1²) / 2 ∈o R

Explanation:

Given data

two cylinders are parallel

distance = d

radial distance = R

d < (R2−R1)

to find out

Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants

solution

we have two parallel cylinders

so area is 2 [tex]\pi[/tex] R × l

and we apply here gauss law that is

EA = Q(enclosed) / ∈o   ......1

so first we find  Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( [tex]\pi[/tex] R1² × l )

so put all value in equation 1

we get

EA = Q(enclosed) / ∈o

E(2 [tex]\pi[/tex] R × l)  = ρ ( [tex]\pi[/tex] R1² × l ) / ∈o

so

E =  ρ ( R1²) / 2 ∈o R

Final answer:

The resulting electric field in the specified region can be calculated using Gauss' Law. The equation for the electric field in that region is [tex]E = 2\pi R_1^2ho_E[/tex].

Explanation:

The resulting electric field in the region R>R3 is:

[tex]E = 2\pi R_1^2ho_E[/tex]

where R_1 is the radius of the inner cylinder, and ρ_E is the charge density. This expression is obtained by applying Gauss' Law for the region where R1 < r < R2.

C. [tex]E_{k}=2.03x10^{11}J[/tex]

The kinetic energy of a body is the ability to perform work due to its movement given by the equation [tex]E_{k}=\frac{1}{2}mv^{2}[/tex].

To calculate the kinetic energy of a rocket with mass 15000kg and speed of 5200m/s:

[tex]E_{k}=\frac{1}{2}(15000kg)(5200m/s)^{2}=202800000000J=2.03x10^{11}J[/tex]