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Anika [276]
3 years ago
5

A power station with an efficiency e generates W watts of electric power and dissipates D J of heat energy each second to the co

oling water that flows through it. The cooling water increases its temperature by T degrees of Celsius. Find the mass of cooling water which flows through the plant each second. Specific heat capacity of water cw = 4,184 J/kgC e = 0.5 W = 4.2 x 108 watts D = 3.5 x 108 J T = 6 C Enter value in tons (1 ton = 1000 kg). Enter two digits after decimal point.
Physics
1 answer:
Andrews [41]3 years ago
3 0

Answer: 13.94 tons/s

Explanation:

On adding heat energy to a substance, the temperature would be changed by a particular amount. This relationship between heat energy and temperature is often different for each material. The specific heat, is a value that describes how they relate.

Heat energy = mass flow rate * specific heat * Δ T

Q = MC (ΔΦ)

Heat energy, Q= 3.5*10^8J

Mass flow rate, M= ?

Specific heat, C= 4184j/KgC

Change in temperature, ΔΦ= 6°C

M = Q/CΔΦ

M = (3.5*10^8)/4184*6

M = 13942kg/s

M = 13.94 tons/s

You might be interested in
PLEASE HELP LOTTA POINTS
sergiy2304 [10]

There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question.  There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems.   That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

That tells us that the yellow bar resting on the pivot is actually a meter stick, but the pictures don't show the centimeter marks on the stick.  The left end of the stick is "0 cm", the right end of the stick is "100 cm", and the pivot is under the "50 cm" mark.  

When the question talks about hanging a weight, it tells the <em>centimeter mark on the stick</em> where the weight is tied.  To solve the problem, we have to first figure out <em>how far that is from the pivot</em>, then calculate how far from the pivot to put the weight on the other side, and finally <u><em>what centimeter mark that is</em></u> on the stick.      

How to solve the problems:

-- The "moment" of a weight is (the weight) x (its distance from the pivot) .

-- To balance the stick, (the sum of the moments on one side) = (the sum of the moments on the other side).

= = = = = = = = = =  

#1).  Only one moment on the left side.  

(160 gm) x (30 cm from pivot) = 4,800 gm-cm

To balance, we need 4,800 gm-cm of moment on the right side.

(500 gm) x (distance from pivot) = 4,800 gm-cm

Distance from pivot = (4,800 gm-cm) / (500 gm)  =  9.6 cm

The 500 gm has to hang 9.6 cm to the right of the pivot.  But that's not the answer to the problem.  They want to know what mark on the stick to hang it from.  The pivot is at the 50cm mark.  The 500gm has to hang 9.6 cm to the right of the pivot.  That's the <em>59.6 cm</em> mark on the stick.

= = = = =

#2).  There are 2 weights hanging from the left side. We have to find the moment of each weight, add them up, then create the same amount of moment on the right side.

one weight:  120gm, hanging from the 25cm mark.

That's 25cm from the pivot.  Moment = (120gm) (25cm) = 3,000 gm-cm

the other weight:  20gm, hanging from the 10cm mark;

That's 40cm from the pivot.  Moment = (20gm) (40cm) = 800 gm-cm

Add up the moments on the left side:

(3,000 gm-cm) + (800 gm-cm) = 3,800 gm-cm.

To balance, we need 3,800 gm-cm of moment on the right side.

(500 gm) x (its distance from the pivot) = 3,800 gm-cm

Distance from the pivot = (3,800 gm-cm) / (500 gm) = 7.6 cm

The pivot is at the 50cm mark on the stick.  You have to hang the 500gm from 7.6cm to the right of that.  The mark at that spot on the stick is                (50cm + 7.6cm) = <em>57.6 cm </em>.

3 0
3 years ago
Read 2 more answers
Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1
harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum p_{i} (before the elastic collision) must be equal to the final momentum p_{f} (after the elastic collision):

p_{i}=p_{f} (1)

Being:

p_{i}=m_{1}V_{i} + m_{2}U_{i}

p_{f}=m_{1}V_{f} + m_{2}U_{f}

Where:

m_{1}=200 kg +100 kg=300 kg is the combined mass of Tubby and Libby with the car

V_{i}=10 m/s is the velocity of Tubby and Libby with the car before the collision

m_{2}=25 kg + 100 kg=125 kg is the combined mass of Flubby with its car

U_{i}=0 m/s is the velocity of Flubby with the car before the collision

V_{f}=4.12 m/s is the velocity of Tubby and Libby with the car after the collision

U_{f} is the velocity of Flubby with the car after the collision

So, we have the following:

m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f} (2)

Finding U_{f}:

U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}} (3)

U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg} (4)

Finally:

U_{f}=14.1 m/s

8 0
3 years ago
(e)<br> Give an example of a process when electrons behave as:<br><br> A.particles.<br> B.a wave.
borishaifa [10]

Answer:

particles

Explanation:

6 0
3 years ago
3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000
cestrela7 [59]

Answer:

d)    α = 1693.5 rad / s² , a = 392.7 m / s² ,   a_total = α √(R² +1) ,

e)   tan θ = a / α

Explanation:

This is an exercise in linear and angular kinematics.

We initialize reduction of all the magnitudes to the SI system

   w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s

   w = 6000 rev / mi = 628.32 rad / s

   θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad

d) ask for centripetal, tangential and total acceleration.

Let's start by looking for centripetal acceleration, let's use the formula

          w² = w₀² + 2 α θ

          α = (w²- w₀²) / 2θ

we calculate

           α = (628.32²2 - 314.16²) / 2 75.398

           α = 1693.5 rad / s²

the quantity is linear and angular are related

         

the linear or tangential acceleration is

            a =    α  R

where R is the radius of the drum

            a = 1693.5 R

Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m

           a = 1693.5 0.20

           a = 392.7 m / s²

the total acceleration is

           a_total = √(a² + α²)

           a_total = √ (α² R² + α²)

           a_total = α √(R² +1)

e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant

Tangential acceleration is tangency to radius and its value varies proportionally radius

the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry

            tan θ = a / α

the angular velocity increases linearly when with centripetal acceleration

8 0
3 years ago
An air-gap, parallel plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potentia
sergejj [24]

Answer:

The new potential energy decreases by the factor of 2 to the old potential energy.

Explanation:

Capacitance of a parallel plate capacitor is given by the relation :

C = (ε₀A)/d

Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.

Potential energy of the capacitor, U = \frac{1}{2}CV^{2}

Here V is the potential difference between the plates.

According to the problem, the distance between the plates get double but area remains same. So,

d₁ = 2d

Here d₁ is new distance between the plates.

Hence, new capacitance is :

C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2

The capacitor have same potential difference that is V. Hence, the new potential energy is :

U₁ = \frac{1}{2}C_{1} V^{2} = \frac{1}{2}\frac{C}{2} V^{2}

U₁ = U/2

\frac{U_{1} }{U} = \frac{1}{2}

7 0
3 years ago
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