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3 years ago

What condition is necessary for two bodies that are in physical contact with each other to be in thermal equilibrium?

1 answer:

3 0

<span>for two bodies that are in physical contact with each other to be in thermal equilibrium the condition is that they require the same temperature and they no longer effect each other temperature
hope it helps</span>

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Use the drop down menu to complete the sentence. Resistance is measured in units called...

liubo4ka [24]

Ohms is the correct answer

7 0

4 years ago

Read 2 more answers

Artemon [7]

Answer:

constant volicty of the pumper when they hit ground 7.03-/s

8 0

3 years ago

A block of mass M1 = 288.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and

AlekseyPX

Answer:

M2 = 278.06 kg

Explanation:

We calculate the weight of M1

W=m*g

Where

m: mass (kg)

g: acceleration due to gravity (m/s²)

W₁=288* 9.8= 2822.4 N

Look at the attached graphic

We calculate the x-y components of the weight :

W₁x= 2822.4*sin41° N =1851.66 N

W₁y= 2822.4 *cos41° N = 2130.09 N

We apply Newton's first law for the balance in M1:

Σ Fy=0

Fn-W₁y=0 , Fn: normal force

Fn=W₁y=2130.09N

Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N

Σ Fx=0

T- W₁x- Ff=0

T= 1851.66 + 873.34

T=2725 N

We apply Newton's first law for the balance in M2:

Σ Fy=0

T- W₂ =0

W₂ = T = 2725 N

W₂ = M2*g

M2 = W₂/g

M2 = 2725/9.8

M2 = 278.06 kg

6 0

3 years ago

An 30-turn coil has square loops measuring 0.341 m along a side and a resistance of 3.61 Ω. It is placed in a magnetic field tha

Verdich [7]

Answer:

Thus, the induced current in the coil at $t =1.73s$ is 9.98 A.

Explanation:

Faraday's law says

$\varepsilon = N \frac{d \Phi_B}{dt}$

where $N$ is the number of turns and $\Phi_B$ is the magnetic flux through the square coil:

Now,

$N = 30$

$\theta = 37.5^o$

$A = (0.341m)^2= 0.11623m^2$

$B = 1.45t^3$ ;

therefore,

$\varepsilon = N \frac{d \Phi_B}{dt} = N\frac{d ( BA\:cos(\theta))}{dt} = 30*\frac{d ( (1.45t^2)(0.1163)\:cos(37.5^o))}{dt}$

$=30*(0.1163)\:cos(37.5^o)*1.45*\dfrac{d ( t^3)}{dt} = 12.04t^2$

$\boxed{\varepsilon = 12.04t^2}$

is the emf induced in the coil.

Now, the loop is connected to $R = 3.61\Omega$ resistance; therefore, at $t = 1.73s$

$\varepsilon = RI = 12.04t^2$

$RI = 12.04(1.73)^2$

$RI = 36.03$

$I = \dfrac{36.03}{3.61\Omega }$

$\boxed{I = 9.98A }$

Thus, the current in the coil at $t =1.73s$ is 9.98 A.

8 0

4 years ago

A car moving at 10 m/s crashes into a large bush and stops in 1.3 m. Using the Work-Energy theorem, calculate the average force

ad-work [718]

Answer:

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

It takes 0.26 s to bring the passenger in the car to a halt.

Explanation:

Hi there!

The negative work (W) needed to bring a moving object to stop is equal to its kinetic energy (KE):

W = KE

F · s = 1/2 · m · v²

Where:

F = applied force on the passenger.

s = displacement of the passenger.

m = mass of the passenger.

v = velocity of the passenger.

Solving the equation for F:

F = 1/2 · m · v² / s

Replacing with the data:

F = 1/2 · 70 kg · (10 m/s)² / 1.3 m

F = 2692.3 N

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

According to the second law of Newton:

F = m · a

Where "a" is the acceleration of the passenger.

We also know from kinematics that the velocity of an object can ve calculated as follows:

v = v0 + a · t

Where:

v = velocity of the passenger at time t.

v0 = initial velocity.

t = time.

a = acceleration.

When the passenger stops, its velocity is zero. So replacing a = F/m, let´s solve the equation for the time it takes the passenger to stop:

v = v0 + a · t

0 = 10 m/s + (-2692.3 N/ 70 kg) · t

-10 m/s / (-2692.3 N/ 70 kg) = t

t =0.26 s

It takes 0.26 s to bring the passenger in the car to a halt.

5 0

3 years ago