A 50-kg copper block initially at 140°C is dropped into an insulated tank that contains 90 L of water at 10°C. Determine the final equilibrium tempera

Explanation:

acceleration is weight*gravity

tension is the weight In Newtons

Please delete my answer. I made a mistake

Answer:

The mechanical energy lost due to friction is 2,424.5 J

Explanation:

Given;

mass of the child, m = 25 kg

intial velocity of the child, u = 0

final velocity of the child, v = 7.8 m/s

initial position of the child, h₁ = 21 m

final position of the child, h₂ = 8 m

Let the energy lost due to heat = ΔE

ΔE + ΔK.E  + ΔP.E = 0

ΔE  +   ¹/₂m(v² - u²)  +  mg(h₂ - h₁) = 0

ΔE   +   ¹/₂ x 25(7.8² - 0)    +   25 x 9.8(8 - 21) = 0

ΔE   +  760.5 J   - 3185 J =

ΔE   -  2,424.5 J = 0

ΔE = 2,424.5 J

Therefore, the mechanical energy lost due to friction is 2,424.5 J

Answer:

In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.

Explanation:

Answer:

[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]

Answer:

108 km/hr or 0.03 km/s

Explanation:

conversion factor for m/s to km/hr is 5/18

conversion factor for m/s to km/s is 1/1000

Explanation:

Glasses or Contacts. You might not realize it, but if you wear glasses or contact lenses, this is light refraction at play. ...

Human Eyes. Human eyes have a lens. ...

Prism. ...

Pickle Jar. ...

Ice Crystals. ...

Glass. ...

Twinkling Stars. ...

Microscope or Telescope.

Explanation:

the answer is in the above image

Answer:

v = 0.064 m/s

Explanation:

Given that,

The mass of two skaters = 75 kg

The mass of a ball = 0.4 kg

The speed of the ball = 6 m/s

The speed of skater = 6 m/s

We need to find the velocity of each of the two skaters.

Under the values given the moment with respect to the ball and which is subsequently transmitted to people it would be given by:

[tex]P=0.4(6)+0.4(6)\\\\=4.8\ kg-m/s[/tex]

We know that,

P = mv

Where

v is the velocity of each skater.

[tex]v=\dfrac{p}{m}\\\\v=\dfrac{4.8}{75}\\\\=0.064\ m/s[/tex]

So, the velocity of each of the skaters is 0.064 m/s.

Answer:

When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12

Explanation:

Given;

initial area of the artery, A₁ = 1.3 cm²

Area of each smaller 18 arteries, a₂ = 0.6 cm²

Total area of the smaller 18 arteries, A₂ = 18 x 0.6 cm²

Apply flow rate equation;

Q = AV

where;

Q is the flow rate of the blood

V is the average velocity of the blood

If the flow rate is constant, then;

A₁V₁ = A₂V₂

[tex]V_2 = \frac{A_1V_1}{A_2} = \frac{1.3\times V_1}{18\times 0.6} \\\\V_2 = 0.12 \ V_1[/tex]

When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12

Complete Question

A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.

(a) What is the moment of inertia of the system

(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm

Answer:

a) [tex]M.I=0.32kg.m^2[/tex]

b) [tex]K.E=621.8J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=2.0kg[/tex]

Disk Radius [tex]r_d=50cm=0.5m[/tex]

Mass of annular cylinder [tex]M_c=1.0kg[/tex]

Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]

Outer Radius of cylinder [tex]R_o=0.3m[/tex]

Angular  Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]

Generally the equation for moment of inertia is mathematically given by

[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]

[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]

[tex]M.I=0.32kg.m^2[/tex]

Generally the equation for Rotational Kinetic Energy  is mathematically given by

[tex]K.E=0.5M.I \omega^2[/tex]

[tex]K.E=0.5 *0.32*62.83[/tex]

[tex]K.E=621.8J[/tex]

Answer:

Work done applied = 12 newton-meter

Explanation:

Given examples:

Force applied = 6 newton

Distance of book = 2 meter

Find from the given data:

Work done

Computation:

The equation can be used to compute work.

Work done applied = Force applied x Distance of book

Work done applied = Force x Distance

Work done applied = 6 x 2

Work done applied = 12 newton-meter

Answer:

a)  V = 45 10³ V, b) U = 4.59 J

Explanation:

a) The electric potential for a series of point charges is

         V = k ∑ [tex]\frac{q_i}{r_i}[/tex]

in this case point P is at a distance of 1 m from each charge, so the point is located perpendicular to the charges at its midpoint

         V = k ( [tex]\frac{q_1}{r} + \frac{q_2}{r}[/tex])

         V = 9 10⁹ (10 - 5/ 1) 10⁻⁶

         V = 45 10³ V

b) the potential energy is

           U = k (  [tex]\frac{q_1q}{r} + \frac{q_2q}{r} + \frac{q_1q_2}{r_2}[/tex] )

where r = 1m and r₂ is the distance between the two charges r₂ = 0.10 m

           U = 9 10⁹ (10 2 / 1 - 5 2/1 - 10 5 /0.10) 10⁻¹²

           U = 9 10⁻³  510

           U = 4.59 J

Explanation:

Energy input = F×d = (150 N)(5.5 m) = 825 J

Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J

efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%

Answer:

So the answer is B. A is wrong because negative answer = deceleration

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -  

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

For safety reasons, this spring constant is increased by 13 % So the new spring constant is  

 k = 19600 * 1.13 = 22148 N/m = 22.15 N/m

Answer:

30

Explanation:

since  [tex]\frac{30}{5}[/tex]=6

         [tex]\frac{30}{6}[/tex]=5

then both 5 and 6 are factors of 30

Have a nice day

Answer:

B

Explanation:

The seasons are measured in how far or close the earth is to the sun.

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)

[tex] v_{s}[/tex]: is the speed of the source =?

The frequency of the train before slowing down is given by:

[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex]  (1)                  

Now, the frequency of the train after slowing down is:

[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex]   (2)  

Dividing equation (1) by (2) we have:

[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]

[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex]   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

[tex] v_{s_{b}} = 2v_{s_{a}} [/tex]     (4)

Now, by entering equation (4) into (3) we have:

[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]  

[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]

By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:

[tex] v_{s_{a}} = 11.06 m/s [/tex]

Finally, the speed of the train before slowing down is:

[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

Answer:

Answer:

Ba đặc trưng sinh lí của âm là độ cao, độ to và âm sắc.

Explanation:

Answer:

focal length=12cm

Explanation:

object size is equal to 1.00cm

object distance = 18cm

heigh of image = 2.00cm

image distance = ??

but magnification is given by;

M = 2.00/1.00 = 2

but u/v = M

u/18 = 2

u = 36

1/f = 1/u+1/v

1/f = 1/18+ 1/36

1/f = 1/12

f = 12cm

Answer:

The output electric power is 1338876 W.

Explanation:

Volume, V = 690 cubic meter

height, h = 220 mm = 0.22 m

efficiency = 90 %

time , t = 1 s

Let the mass is m.

m = volume x density  

m = 690 x 1000 = 690000 kg

The input power is

P = m g h = 690000 x 9.8 x 0.22 = 1497640 W

The electric power out put is

[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

Answer:

Option B. 3.25×10¯¹⁰ mm.

Explanation:

Measurement (cm) = 0.00325×10⁻⁸ cm

Measurement (mm) =?

The measurement in mm can be obtained as follow:

1 cm = 10 mm

Therefore,

0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm

0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm

Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.

The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm

The number given is in standard form and can be written as 3.25*10^-11 cm.

To convert this from centimeter to millimeter, we have to multiply this value by 10.

So, let's 3.25*10^-11 by 10 and get our value in mm

[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]

From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]

Learn more about conversion of units here;

https://brainly.com/question/8426032

Answer:

Answer:

A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.

Explanation:

Edge.

Answer:

The motion of the paper airplane  is best explained by horizontal inertia and vertical pull of gravity.

Explanation:

Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .

While vertical pull is due to the earth .

Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.

Also read it;

https://brainly.com/question/11049671

#SPJ2

Answer:

[tex]v=6.65m/sec[/tex]

Explanation:

From the Question we are told that:

Mass [tex]m=97.6[/tex]

Coefficient of kinetic friction  [tex]\mu k=0.555[/tex]

Generally the equation for Frictional force is mathematically given by

 [tex]F=\mu mg[/tex]

 [tex]F=0.555*97.6*9.8[/tex]

 [tex]F=531.388N[/tex]

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 [tex]a_f=-\mu g[/tex]

 [tex]a_f=-0.555 *9.81[/tex]

 [tex]a_f=-54455m/sec^2[/tex]

Therefore

 [tex]v=u-at[/tex]

 [tex]v=0+5.45*1.22[/tex]

 [tex]v=6.65m/sec[/tex]