Two distinct systems have the same amount of stored internal energy. 500 J are added by heat to the first system and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy?

Answer:

Explanation:

Bhjb

Explanation:

its because a concrete surface is a hard surface which doesn't absorb the energy of gravitation when we fall down so we get hurt more badly..

hope this helps

Answer:

T₁ = 232.5 ºC

Explanation:

For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law

           V = i R

            R = V / i

i₀ = 1.28 A

            R₀ = 120 / 1.28

            R₀ = 93.75 ohm

i₁ = 1.229 A

             R₁ = 120 / 1.229

             R₁ = 97.64 or

Resistance in a metal is linear with temperature

            ΔR = α R₀ ΔT

where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹

            ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]

            ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]

            ΔT = 2,075 10² C

            ΔT = T₁-T₀ = 2,075 10²

            T₁ = T₀ + 207.5

             T₁ = 25+ 207.5

             T₁ = 232.5 ºC

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.

The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:

∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0

(right is positive, left is negative)

∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0

(up is positive, down is negative)

Solve the system of equations. I use elimination here:

• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):

sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0

cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0

T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0

T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)

• Subtract the first equation from the second to eliminate T₁ :

T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

• Solve for T₂ :

T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

T₂ sin(74.0°) = (215 N) cos(29.5°)

… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))

T₂ = (215 N) cos(29.5°) / sin(74.0°)

T₂ ≈ 195 N

• Solve for T₁ :

T₁ cos(29.5°) - T₂ cos(44.5°) = 0

T₁ cos(29.5°) = T₂ cos(44.5°)

T₁ = T₂ cos(44.5°) / cos(29.5°)

T₁ ≈ 160. N

Answer:

The energy stored in a body due to either it's position or change in shape is called gravitational potential energy.

Answer:

I think it is of science is it true na i knew it bro dont take tension

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

Answer:

Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.

Explanation:

Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.

Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.

Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.

Answer:

v(7) = 52.915 m/s

Explanation:

First, find the value for acceleration.

F = ma

100 = .5 * a

a = 200 m/s²

Next find the velocity at x = 7 using kinematic equations.

v² = v₀² + 2a(Δx)

v² = (0)² + 2(200)(7)

v = [tex]\sqrt{2800}[/tex]

v = 52.915 m/s

Answer:

a) [tex]F=940.8N[/tex]

b)  [tex]S=234.14m[/tex]

c)  [tex]W=2.2*10^5J[/tex]

d)  [tex]W=0[/tex]

Explanation:

Mass [tex]m=160kg[/tex]

Velocity [tex]v=53m/s[/tex]

Time [tex]t=9seconds[/tex]

a)

Generally the Newton's equation for motion is mathematically given by

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

[tex]a=\frac{53}{9}[/tex]

[tex]a=5.9m/s^2[/tex]

Therefore

F=ma

[tex]F=160*5.88[/tex]

[tex]F=940.8N[/tex]

b)

Generally the Newton's equation for motion is mathematically given by

[tex]S=0.5at^2[/tex]

[tex]S=0.5*5.9*9^2[/tex]

[tex]S=234.14m[/tex]

c)

Generally the Newton's equation for  work done is mathematically given by

[tex]W=Fd[/tex]

[tex]W=940.8*238.14[/tex]

[tex]W=2.2*10^5J[/tex]

d)

Generally the Newton's equation for  work done  by the force of the road on the wheels is mathematically given by

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

[tex]W=0[/tex]

Answer:

u can ask it to the person who give ot to u i dont no

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

Answer:

Force A newton Law first law

F = M.A which Force in 0 N as you Questions Above

Force B

Newton Law 3

Action = -Reaction

Hope you can explain this formula as you want to scribe to explaining

Answer:

[tex]\omega=109.67\ rad/s[/tex]

Explanation:

Given that,

The speed of the ball, u = 34 m/s

The ball is 0.310 m from the elbow joint.

We need to find the angular velocity (in rad/s) of the forearm.

We know that,

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]

So, the required angular velocity of the forearm is 109.67 rad/s.

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

Nearly four hours every day, doing little to no physical activity.

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Answer:

Explanation:

For frequencies n generated in a string , the expression is as follows

n = 1 /2L√ ( T/m )

n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.

If T and m are constant , then

n x L = constant , hence n is inversely proportional to L or length of string.

Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times

Initial length of string is 65 cm .

x1 = 65 x 1 / 1.75 = 37.14 cm

x2 = 37.14 x 1/ 1.75 = 21.22 cm

x3 = 21.22 x 1 / 1.75 = 12.12 cm

x4= 12.12 x 1 / 1.75 = 6.92 cm

x5 = 6.92 x 1 / 1.75 = 3.95 cm

x6 = 3.95 x 1 / 1.75 = 2.25 cm

Answer:

The bullet will rise 320 meters above the point of projection.

Explanation:

Assuming that air friction is negligent we can use the kinematic equation:

[tex]v_{2} ^2=v_{1} ^2+2(-a)d\\0\frac{m^2}{s^2} =6400\frac{m^2}{s^2} +2(-10\frac{m}{s^2} )d\\-6400\frac{m^2}{s^2} =(-20\frac{m}{s^2}) d\\320m=d[/tex]

*acceleration is negative (-a) as it is acting in the opposite direction of the motion of the bullet.*

The bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.

Assume the air friction is negligible, the kinematic equation:

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

Where,

[tex]v_i^2[/tex] - iinitial velocity = 80 m/s

[tex]v_f^2[/tex]- final velocity = 0

[tex]d[/tex]- distance= ?

[tex]a[/tex]- gravitational acceleration = 9.8 m/s² = 10 m/s²

Put the values in the formula,

[tex]\begin {aligne} 0 = (80)^2 + 2 (10)^2 d\\\\d = \dfrac {6400}{ 200}\\\\d &= 3600 \rm \ m\end {aligne}\\[/tex]

Therefore, the bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.

To know more about kinematic equation:

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Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

Answer:

b. approaches zero.

Explanation:

The phenomenon is known as length contraction.

Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.

let the length of the train = L

Let the length observed when the train is in motion = L₀

Apply Einstein's special theory of relativity;

[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]

from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)

As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L).  (L₀ < L)

Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero.  (L₀ = 0)

Thus, the length of this object, as measured by a stationary observer approaches zero

Answer:

49 N

Explanation:

Given,

Mass ( m ) = 50 kg

To find : Weight ( W ) = ?

Take the value of acceleration due to gravity as 9.8 m/s^2

Formula : -

W = mg

W = 50 x 9.8

W = 49 N

Answer:

Option (2) is correct.

The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Explanation:

An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.

So, the correct option is (2).

The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.

The given problem is based on the concept and fundamentals of electromagnetic waves.  The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.

In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.

Learn more about the electromagnetic waves here:

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Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

Explanation:

[tex] \implies v_{av} = \dfrac{total \: displacement}{total \: time} [/tex]

[tex] \implies v_{av} = \dfrac{100}{10} [/tex]

[tex]\implies v_{av} =10 \: {ms}^{ - 1} [/tex]

Answer:

256 ohms

Explanation:

Applying,

R = R'[1+α(T-T')]............. Equation 1

Where R = Final resistance of the wire, R' = Initial resistance of the wire, T = Final temperature, T' = Initial temperature, α = Temperature coefficient of resistance

From the question,

Given: R' = 200 ohms, T = 90 degrees, T' = 20 degrees, α = 0.004/degree

Substitute these values into equation 1

R = 200[1+0.004(90-20)]

R = 200[1+0.28]

R = 200(1.28)

R = 256 ohms

The resistance of the wire at 90 °C correct to the nearest ohm assuming the coefficient of resistance is 0.004 °C¯¹ is 256 ohm

α = R₂ – R₁ / R₁(T₂ – T₁)

0.004 = R₂ – 200 / 200(90 – 20)

0.004 = R₂ – 200 / 200(70)

0.004 = R₂ – 200 / 14000

Cross multiply

R₂ – 200 = 0.004 × 14000

R₂ – 200 = 56

Collect like terms

R₂ = 56 + 200

R₂ = 256 ohm

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