Thomson effect is the combined effect of Seebeck effect
and Peltier effect. Justify your answer.plz help me​

Answer:

Centripetal force is defined as, "the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation,"

Explanation:

so if you put a small ball por grape and put it on a table and you take a wine glass and spin it around the grape the grape will continue spinning even if you stop spinning the glass hope this helped

;)

Answer:

A force acting on a moving body at an angle to the direction of motion, tending to make the body follow a circular or curved path. The force of gravity acting on a satellite in orbit is an example of a centripetal force; the friction of the tires of a car making a turn similarly provides centripetal force on the car.

Explanation:

Answer:

v = 3.951 m/s

Explanation:

Given that,

Mass of a ball, m = 6.5 kg

Radius of the circle, r = 0.9 m

Angular speed of the ball, [tex]\omega=0.7\ rev/s=4.39\ rad/s[/tex]

Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :

[tex]v=r\omega\\\\v=0.9\times 4.39\\\\v=3.951\ m/s[/tex]

So, the tangential speed of the ball is 3.951 m/s.

Answer:

kinetic

Explanation:

Answer:

kinetic energy.

Explanation:

Q= Q1 +Q2 +Q3+Q4+ Q5

3066 J

Answer:

If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.

Explanation:

Based on the Principle of Energy Conservation we find that the square of the initial speed of the object ([tex]v[/tex]), measured in meters per second, is directly proportional to its change in height ([tex]\Delta h[/tex]), measured in meters. That is:

[tex]v^{2} \propto \Delta h[/tex]

[tex]v^{2} = k\cdot \Delta h[/tex] (Eq. 1)

Where [tex]k[/tex] is the proportionality constant, measured in meters per square second.

Such constant is eliminated by using the following relationship:

[tex]\left(\frac{v_{2}}{v_{1}} \right)^{2} = \frac{\Delta h_{2}}{\Delta h_{1}}[/tex]  

[tex]\Delta h_{2} = \left(\frac{v_{2}}{v_{1}} \right)^{2}\cdot \Delta h_{1}[/tex] (Eq. 2)

Where:

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the object, measured in meters per second.

[tex]\Delta h_{1}[/tex], [tex]\Delta h_{2}[/tex] - Initial and final changes in height, measured in meters.

If we know that [tex]\frac{v_{2}}{v_{1}} = 2[/tex] and [tex]\Delta h_{1} = 10\,m[/tex], then the change in height is:

[tex]\Delta h_{2} = 2^{2}\cdot (10\,m)[/tex]

[tex]\Delta h_{2} = 40\,m[/tex]

If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.

When the object is thrown with double the speed, it reaches a height of 40m.

Given that the height attained by the object is h = 10m.

Let the initial speed of the object be u

and its final speed will be v = 0 at the highest point.

Then from the third equation of motion, we get,

[tex]v^2=u^2-2gh\\\\0=u^2-2gh\\\\h=\frac{u^2}{2g}[/tex]

if the object is thrown at double the speed, that is u' = 2u, then the height attained will be:

[tex]h'=\frac{u'^2}{2g}\\\\h'=\frac{(2u)^2}{2g}\\\\h'=\frac{4u^2}{2g}\\\\h'=4h[/tex]

h' = 4 × 10m

h' = 40m

So, the height attained will be 40 m.

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Answer:

9.77 m/s

Explanation:

Since the cylinder is rolling, it will have both translational and rotational kinetic energy.

From conservation of energy, its initial mechanical energy equals its final mechanical energy.

So, K' + U' = K + U where K and U are the initial kinetic and potential energies respectively and K' and U' are the final kinetic and potential energies respectively.

So, Since the cylinder starts from rest, its initial kinetic energy K = 0, its initial potential energy = mgh where m = mass of cylinder, g = acceleration due to gravity = 9.8 m/s² and h = height of incline = 7.30 mm = 7.3 × 10⁻³ m . Its final kinetic energy K' = 1/2Iω² + 1/2mv² where I = moment of inertia of cylinder = 1/2mr² where r = radius of cylinder and v = translational velocity of cylinder. Its final potential energy U' = 0. Substituting these into the equation, we have  

K' + U' = K + U

1/2Iω² + 1/2mv² + 0 = 0 + mgh

1/2(1/2mr²)ω² + 1/2mv² = mgh

1/4mr²ω² + 1/2mv² = mgh          

1/4m(rω)² + 1/2mv² = mgh    v = rω

1/4mv² + 1/2mv² = mgh

3/4mv² = mgh

v² = 4gh/3

v = √(4gh/3)

v = √(4 × 9.8 m/s² × 7.3 × 10⁻³ m/3)

v = (√286.16/3) m/s

v = (√95.39) m/s

v = 9.77 m/s

Answer:

Explanation:

a) The total distance traveled by the balloon = (100 + 60) m

                                                          = 160 metes

b) speed = [tex]\frac{distance covered}{time taken}[/tex]

The average speed for the first flight = [tex]\frac{100}{10}[/tex]

                                                    = 10 m/s

c) The speed during its next flight = [tex]\frac{60}{6}[/tex]

                                                    = 10 m/s

d) Average speed for the entire trip = [tex]\frac{(10 + 10)}{2}[/tex]

                                                    = 10 m/s

e) Total displacement = [tex]\sqrt{(100)^{2} + (60)^{2} }[/tex]

                                   = [tex]\sqrt{13600}[/tex]

                                   = 116.62 m

f) velocity = [tex]\frac{displacement}{time}[/tex]

Average velocity during the first 10 seconds = [tex]\frac{100}{10}[/tex]

                              = 10 m/s

g) Average velocity during the next 6 seconds = [tex]\frac{60}{6}[/tex]

                               = 10 m/s

h) Average velocity for the entire trip = [tex]\sqrt{(10)^{2} + (10)^{2} }[/tex]

                                                              = [tex]\sqrt{200}[/tex]

                                                              = 14.14 m/s

Answer:

Energy lost = 55.28Joules

Explanation:

Using the law of momentum:

m1u1+m2u2 = m1v1+m2v2

m1 and m2 are the masses of the objects

u1 and u2 are the initial velocities.

v1 and v2 are final velocities

Given

m1 = 5kg

m2 = 4kg

u1 = 12m/s

u2 = 2m/s

v1 = 10m/s

v2 = ?

Substitute

5(12)+4(2)=4(10)+5v2

60+8 = 40+5v2

68-40 = 5v2

28 = 5v2

v2 = 28/5

v2 = 5.6m/sec

Hence the speed of the 5kg block is 5.6m/s

Energy lost = Kinetic energy after collision - kinetic energy before collision

KE after collision = 1/2m1v1²+1/2m2v2²

KE after collision = 1/2(5)10²+1/2(4)(5.6)²

= 250+62.72

= 312.72Joules

KE before collision = 1/2m1u1²+1/2m2u2²

KE before collision = 1/2(5)12²+1/2(4)(2)²

= 360+8

= 368Joules

Energy lost = 368-312.72

Energy lost = 55.28Joules

Answer:

Net force = 419.5N

Explanation:

Given the following data;

Mass = 50kg

Radius = 15m

Time = 2.1 secs

Turns = 1/4 = 0.25

In order to find the net force, we would first of all solve for the speed and acceleration of the halfback.

To find speed;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the equation;

[tex]Speed = \frac{distance}{time}[/tex]

But the distance traveled is given by the circumference of a circle = [tex] 2\pi r[/tex]

Since he covered 1/4th of a turn;

[tex] Distance = 0.25 * 2 \pi *r[/tex]

Substituting into the equation;

[tex] Speed, v = \frac {(0.25*2*3.142 * 15)}{2.1}[/tex]

[tex] Speed, v = \frac {23.565}{2.1}[/tex]

Speed, v = 11.22m/s

To find acceleration;

[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]

Where, v = 11.22m/s and r = 15m

Substituting into the equation, we have;

[tex] Acceleration, a = \frac {11.22^{2}}{15}[/tex]

[tex] Acceleration, a = \frac {125.8884}{15}[/tex]

Acceleration, a = 8.39m/s²

To find the net force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

Substituting into the equation, we have;

[tex] F = 50 * 8.39[/tex]

F = 419.5N

Therefore, the net force acting upon the halfback is 419.5 Newton.

Answer:

a = 15.68 [m/s²]; v = 112.92 [m/s]

Explanation:

To solve this problem we must use the combination of two interesting topics of physics, the principle of energy conservation and then the application of kinematics.

The principle of Energy Conservation tells us that kinetic energy is transformed into potential energy or vice versa. For this particular problem, we must imagine the rocket at a height above 650 [m], suddenly this rocket falls to the ground. We will propose the reference point of potential energy at ground level, at this point the potential energy is equal to zero.

[tex]E_{pot}=E_{kin}[/tex]

[tex]E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}[/tex]

where:

m = mass of the rocket [kg]

g = gravity acceleration = 9.81[m/s²]

v = velocity of the rocket [m/s]

m*g*h = 0.5*m*v²

9.81*650 = 0.5*v²

v = √((6376.5)/0.5)

v = 112.92 [m/s]

Now using the following kinematic equation we have:

[tex]v_{f} = v_{o} + (a*t)[/tex]

where:

Vf = final velocity = 112.92 [m/s]

Vo = initial velocity = 0 (at the begining the rocket is at rest)

a = rocket acceleration [m/s²]

t = time = 7.2 [s]

Note: in the kinematics equation the positive sign of acceleration means that the rocket accelerates in the direction of motion, i.e. activates its thrusters to descend and descend to 650 [m] in 7.2 [s].

112.92 = 0 + (a*7.2)

a = 15.68 [m/s²]

Answer:

595,680,000 miles

(= 5.96 x 10⁸ miles)

Explanation:

we are given the speed of travel as 68,000 mi/hr

if we assume one common year

= 365 days

= 365 days x 24 hours/day

= 8760 hours

Then the distance travelled

= speed x time

= 68,000 x 8760

= 595,680,000 miles

= 5.96 x 10⁸ miles   (rounded to 3 sig. fig and expressed in scientific form)

Answer:

Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).

Hope this helps!

Answer:

3.33 m/s

Explanation:

Just add up all of the speeds and divide it by 3 (since there are 3 different speeds)

(4 m/s + 2.5 m/s + 3.5 m/s) / 3 = 3.33 m/s

Answer:

1.2 seconds

Explanation:

Formula for total time of flight in projectile is; T = 2u/g

Now, for the time to reach maximum height, it is gotten from Newton's first equation of motion;

v = u + gt

t is time taken to reach maximum height.

But in this case, g will be negative since motion is against gravity. Final velocity will be 0 m/s at max height.

Thus;

0 = u - gt

u = gt

t = u/g

Putting t for u/g in the equation of total time of flight, we have;

T = 2t

We are given t = 1.2

Thus, T = 2 × 1.2 = 2.4

Since total time of flight is 2.4 s and time to get to maximum height is 1.2 s, then it means time to fall from maximum height back to the students hand is; 2.4 - 1.2 = 1.2 s

Answer:

a. 50 m/s b. 0 m/s c. 50 m/s d. 4.04 s e. -39.6 m/s f. 63.78 m/s g. 202 m

Explanation:

a. What is the initial horizontal velocity?

Since the object is launched horizontally, it initial horizontal velocity is 50 m/s

b. What is the initial vertical velocity?

Since the object is launched horizontally, it has no initial vertical component. So, its initial vertical velocity is 0 m/s

c. What is the final horizontal velocity?

Its final horizontal velocity is 50 m/s since no force acts on it in the horizontal direction to change its value.

d. How much time did it take for the object to hit the ground?

We use the equation s = ut - 1/gt² since the object is falling under gravity where u = initial vertical velocity = 0 m/s, s = height of cliff = 80 m, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the object to hit the ground.

s = ut - 1/2gt²

80 m = 0 × t - 1/2 × -9.8 m/s² × t²

80 m = 4.9 m/s² × t²

t² = 80 m ÷ 4.9 m/s²

t² = 16.33 s²

t = √(16.33 s²)

t = 4.04 s

e. What is the final vertical velocity?

Using v = u + at where u = initial vertical velocity = 0 m/s, v = final vertical velocity, a = acceleration = -g =  -9.8 m/s²and t = time it takes object to reach the ground = 4.04 s.

Substituting these values into the equation, we have

v = u + at

v = 0 m/s + (-9.8 m/s²) × 4.04 s

v = -39.6 m/s

f. What is the final resultant speed?

The final resultant speed v' is the resultant of the final horizontal velocity and the final vertical velocity. Let u' = final vertical velocity = 50 m/s.

v' = √(u'² + v²)

v' = √((50 m/s)² + (-39.6 m/s)²)

v' = √(2500 m²/s² + 1568.16 m²/s²)

v' = √(4068.16 m²/s²)

v' = 63.78 m/s

g. How far from the base of the cliff will the projectile land?

The distance from the base of the cliff, d where the projectile lands is

d = u't where u' = horizontal velocity = 50 m/s and t = time it takes object to land = 4.04 s

d = 50 m/s × 4.04 s

d = 202 m

Answer:

Kinetic Energy

Explanation:

I can't really explain it, but that's the correct answer.

Answer:

Simple enough that it seems like a trick question: gravity.

Answer:

The speed of the particle is proportional to:

c. t²

Explanation:

Given;

initial velocity of the particle, u = 0

let the net force on the object = F

Work done on the particle is given by;

W = F x d

W ∝ t

[tex]Fd \ \alpha \ t\\\\\frac{mv}{t}d \ \ \alpha \ t\\\\mvd \ \ \alpha \ t^2\\\\v \ \ \alpha \ t^2[/tex]

Therefore, the speed of the particle is proportional to t²

The speed of the particle is proportional to t.

It should be noted that work is a change in kinetic energy. Therefore, power will be calculated as:

Power = Work / Time = force × velocity

Power is proportional to t. Since force is a constant, then the velocity will be proportional to t. Therefore, the speed of the particle is proportional to t.

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Answer:

Vf = Vi + at

t = Vf - Vi = 0.0m/s - 1.75 m/s

    --------- = ------------------------- = 8.8s

         a              -0.20 m/s(squared)

Explanation:

This is what I got I am pretty sure it's correct but correct me if I'm wrong : )

Answer:

10 m/h^2

Explanation:

a=v/t

a=60/6

a=10 m/h^2

Answer:

0.71 N

Explanation:

The inertia of a body, I, is the product of the force applied on it and the time which it acts. Also, it is equal to the change in  the momentum, P, of a body.

So that:

I = Ft

I = ΔP

Ft = m(v - u)

Where F is the force, t is the time, m is the mass of object, v is the final velocity and u is the initial velocity.

Given that: m = 4.33 kg, u = -2.74 m/s, v = 6.35 m/s and t = 55.33 s

F x 55.33 = 4.33(6.35 - (-2.74))

F x 55.33 = 4.33(6.35 + 2.74)

55.33 F = 4.33 x 9.09

55.33 F = 39.3597

F = [tex]\frac{39.3597}{55.33}[/tex]

  = 0.711363

The average force on the ball by the player is 0.71 N.

Answer:

[tex]\alpha =240.79\ rad/s^2[/tex]

Explanation:

Given that,

Initial angular velocity, [tex]\omega_i=0[/tex]

Final angular velocity, [tex]\omega_f=16740\ rev/min = 1753\ rad/s[/tex]

We need to find the angular acceleration of the drill. It is given by the formula as follows :

[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{1753-0}{7.28}\\\\\alpha =240.79\ rad/s^2[/tex]

So, the angular acceleration of the drill is [tex]240.79\ rad/s^2[/tex].

Answer:

Explanation:

a ) magnitude of force = centripetal force = m v² / R , m is mass , v is velocity of skater , R is radius of circular path .

= 54 x 3.94² / .9

= 931.41 N

= .93141 kN.

b ) Her weight = mg = 54 x 9.8 = 529.2 N

= .5292 kN .

Ratio = .93141 / .5292 = 1.76

Answer:

Larger in magnitude

Explanation:

As the vectors get closer reducing the 90 degree angle, based on the "parallelogram rule" for vector addition, the magnitude of the resultant vector gets larger than when they are forming a 90 degree angle.

Answer:8

Explanation:

Given :

If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F.

To Find :

The volume of air in ft³ exiting the evaporator coil.

Solution :

We know, relation between initial and final temperature and volume is given by :

[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]

Here, temperature is in Kelvin .

We know, Kelvin temperature is given by :

[tex]K=(F-32)\times \dfrac{5}{9} + 273[/tex]

Putting F = 60° F,

[tex]K=(60-32)\times \dfrac{5}{9} + 273\\\\K = 288.56\ K[/tex]

Putting F = 80° F,

[tex]K=(80-32)\times \dfrac{5}{9} + 273\\\\K = 299.67\ K[/tex]

Putting all values, we get :

[tex]\dfrac{V_1}{2000}=\dfrac{60}{80}\\\\V_1=\dfrac{60\times 2000}{80}\ ft^3\\\\V_1=1500\ ft^3[/tex]

Hence, this is the required solution.

The dog starts from rest and has a speed of 2.10 m/s at the bottom. The height of the slope will be 0.225 m

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

The friction force prevents any two surfaces of objects from easily sliding over each other or slipping across one another. It depends upon the force applied to the object.

As there is almost zero friction between the total potential energy height is equal to the kinetic energy

m*g*h = 1/2* m*v²

Where m is the mass of the dog

g is the acceleration due to gravity,h is the height of the slope

v is the velocity of the sliding dog

g*h = 1/2* v²

9.81*h = 0.5* 2.1²

           =0.225 m

The height of the slope would be 0.225 m.        

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Answer:

Explanation:

Mitochondria is the powerhouse of the cell. And Chroloplast helps plants capture energy

Answer:

The tension in the rope is 588 N

Explanation:

I have attached a free body diagram of the problem.

First we calculate the force applied downwards by the person's weight:

[tex]F_{weight} =m*a = 80 kg * 9.8 \frac{m}{s^{2}} = 784 N[/tex]

If the scale reads a 25% of the person' weight:

[tex]F_{Scale} = F_{weight}*0.25=784N*0.25=196N[/tex]

which is the same value (but opposite) to the effective weight of the persons ([tex]F_{Person}=-F_{Scale}[/tex])

The other 75% of the total weight is the force of the rope pulling the person upwards.

[tex]F_{Pull}= F_{weight}*0.75=784N*0.75=588N[/tex]

To verify, we can use 1st Newton's law

∑F = 0

[tex]F_{Rope}-F_{Pull}+F_{Person}-F_{Scale}=0 \\588N-588N+196N-196N=0[/tex]