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Romashka-Z-Leto [24]
3 years ago
10

3. What is the force of sliding friction

Physics
1 answer:
ale4655 [162]3 years ago
3 0

Answer:

the force of the friction is A-0.52

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A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p
anzhelika [568]

Answer:

The final temperature of the air is T_2= 605 K

Explanation:

We can start by doing an energy balance for the closed system

\Delta KE+\Delta PE+ \Delta U = Q - W

where

\Delta KE = the change in kinetic energy.

\Delta PE = the change in potential energy.

\Delta U = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so \Delta KE = 0 and \Delta PE=0

and our energy balance equation is \Delta U = Q - W

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change \Delta U=W+Q using the energy balance equation.

We convert the interval of time to seconds t = 5 \:min = 300\:s

\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t

\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ

We can use the change in specific internal energy \Delta U = m(u_2-u_1) to find the final temperature of the air.

We are given that T_1=300 \:K and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy u_1

u_1=214.07\:\frac{kJ}{kg}

Next, we will calculate the final specific internal energy u_2

\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1

\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}

u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}

With the value u_2=439.07 \:\frac{kJ}{kg} and the ideal gas tables for air we make a regression between the values u = 434.78 \:\frac{kJ}{kg},T=600 \:K and u = 442.42 \:\frac{kJ}{kg}, T=610 \:K and we find that the final temperature T_2 is 605 K.

3 0
2 years ago
A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the s
yan [13]

Answer:

The  force is F_1  = 400.8 \  N

Explanation:

From the question we are told that

   The first  diameter is  d_1 =  4.0 \ cm =  0.04 \ m

   The second diameter is  d_2  =  8.0 \ cm  = 0.08 \  m

   

Generally the first area is  

         A_1  =  \pi  * \frac{d^2_1 }{4}

=>      A_1  = 3.142  * \frac{0.04^2}{4}

=>       A_1  =  0.00126 \ m^2

The  second area is  

     A_2 =  \pi  * \frac{d^2_2 }{4}

     A_2  = 3.142  * \frac{0.08^2}{4}

     A_2  =  0.00503 \ m^2

For a hydraulic press the pressure at both end must be equal .

Generally  pressure is mathematically represented as

    P =  \frac{F}{A}

=>  

   \frac{F_1}{A_1 }  =  \frac{F_2}{A_2 }

=>   F_1  =  \frac{1600}{0.00503}  *  0.00126

=>    F_1  = 400.8 \  N

8 0
2 years ago
CAN SOMEONE TELL ME THE answer for this ?
Paladinen [302]

Answer:

b. light from earth is reflected by the moon surface

4 0
2 years ago
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When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart
mixer [17]

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

as we know that

m_1 = 5 kg

m_2 = 2 kg

v_{1i} = 2.8 m/s

v_{2i} = 0 m/s

now from above equation we have

5(2.8) + 2(0) = m_1v+ m_2v

14 = (5+ 2) v

v = 2 m/s

so the speed of combined system is 2 m/s

8 0
3 years ago
What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b
kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

8 0
2 years ago
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