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murzikaleks [220]
3 years ago
7

A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0

Physics
1 answer:
ValentinkaMS [17]3 years ago
7 0

M = mass of aluminium = 1.11 kg

c_{a} = specific heat of aluminium = 900

T_{ai} = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

c_{w} = specific heat of water = 4186

T_{wi} = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M c_{a} (T_{ai} - T) = m c_{w} (T - T_{wi} )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

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Which of the following best describes the circuit shown below?
Rudiy27

Answer:

Short circuit

Explanation:

The given figure shows a short circuit. It is defined as the circuit which allows the flow of electric current when there is no resistance. It shows a battery, bulb and connecting wires.

The wire across the bulb is connected from one terminal to another without any resistance in between them.

So, the correct option is (d) " short circuit ".

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3 years ago
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A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

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8 0
2 years ago
block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally
Alexeev081 [22]

Answer:

v_{2} =\frac{1}{2}

Explanation:

From the second law of Newton movement laws, we have:

F=m*a, and we know that a is the acceleration, which definition is:

a=\frac{dv}{dt}, so:

F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

dv=\frac{1}{8}(t+1)dt\\\int\limits^{v_{2}}_0 \, dv=\int\limits^{2}_{0} {\frac{1}{8}(t+1)} \, dt

v_{2}=\frac{1}{8}*(\frac{t^{2}}{2}+t) (This is the indefinite integral), the definite one is:

v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}

3 0
3 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
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The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

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When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

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Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

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Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

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As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

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