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AlexFokin [52]
3 years ago
10

Object A is placed on top of object B. Object A is the same temperature as object B. How will heat flow between object A and obj

ect B?
A.Heat will flow from object A to object B.
B.No heat will flow between object A and object B.
C.Heat will flow from object B to object A.
Physics
2 answers:
Alex17521 [72]3 years ago
8 0

Answer:

B.No heat will flow between object A and object B.

Explanation:

As we know that heat will flow due to temperature gradient so in all cases the flow of thermal energy is from high temperature to low temperature

So whenever thermal energy is transferred then it has tendency to make the two objects into thermal equilibrium

At the condition of thermal equilibrium the net flow of heat from one object to other object is always zero

So here when object A is placed on the top of object B then in that case it is given that the temperature of two objects are same here.

So these objects are already in thermal equilibrium so there is no net heat flow.

so correct answer will be

B.No heat will flow between object A and object B.

mash [69]3 years ago
6 0
"No heat will flow between object A and object B" is the one among the following choices given in the question that describes how heat will <span>flow between object A and object B. The correct option among all the options that are given in the question is the second option or option "B". I hope it helps you.</span>
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Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle
tangare [24]

Answer:

F=1.159

Explanation:

From the question we are told that:

Mass of pulley M=1kg

Radius r=12cm

Mass of block A M_a=2.1kg

Mass of block B m_b=4.1kg

Spring constant\mu= 358 J/m2

Generally the equation for Torque is mathematically given by

Since \sumF=ma

At mass A

 T_2-f_3=2.1a

At mass B

 4.8-T_1=4.1a

At  Pulley

 R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}

 R(T_1-T_2)=0.55a

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a

 -f_3+4.1=6.75a

 -f_3=6.75a+4.8

Since From above equation

M_{eff}=6.7kg

Therefore

T=2\pi \sqrt{{\frac{M_{eff}}{k}}

T=2\pi \sqrt{{\frac{6.75}{\mu}}

T=0.862s

Generally the equation for frequency is mathematically given by

F=\frac{1}{T} \\F=\frac{1}{0.862}

F=1.159

3 0
3 years ago
A body of mass 1kg has a 25kgsmsqure moment of inertia about its center of mass. if it rotated about an axis which has a displac
ASHA 777 [7]

Answer:

194 kg m²

Explanation:

Use parallel axis theorem:

I = I₀ + mr²

I = 25 kg m² + (1 kg) ((5 m)² + (12 m)²)

I = 25 kg m² + 169 kg m²

I = 194 kg m²

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3 years ago
if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice
Firlakuza [10]

Answer:

2F_{1}

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}

F_{1}= \frac{F_{2}}{2}\\

F_{2}= 2F_{1}

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4 years ago
Compared to microwaves, ultraviolet waves have ___________________________ frequency.
vampirchik [111]
I think its a higher frequency
4 0
3 years ago
An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with
Ket [755]

Answer:

Part a)

I = 1879.7 kg m^2

Part b)

\alpha = 0.70 rad/s^2

Part c)

I = 153.8 kg m^2

Part 4)

angular acceleration will be ZERO

Part 5)

I = 345.6 kg m^2

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)

So we have

I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)

I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)

I = (\frac{16}{3} + 127)m_r R^2

I = (\frac{16}{3} + 127)(6.85)(1.44)^2

I = 1879.7 kg m^2

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

\tau = F\frac{L}{2}

\tau = 460(2R)

\tau = 460 \times 2 \times 1.44

\tau = 1324.8 Nm

now angular acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{1324.8}{1879.7}

\alpha = 0.70 rad/s^2

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)

x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}

so moment of inertia about this position is given as

I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)

so we have

I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})

I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})

I = 6.85(1.44)^2\times 10.83

I = 153.8 kg m^2

Part 4)

If force is applied parallel to the length of rod

then we have

\tau = \vec r \times \vec F

\tau = 0

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)

so we have

I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)

I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)

I = 6.85(1.44)^2\times 24.33

I = 345.6 kg m^2

6 0
3 years ago
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