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3 years ago

Diabatic processes occur when:

Physics

1 answer:

Svetach [21]3 years ago

4 0

Answer:

The correct answer is d. add or remove energy from a system.

Explanation:

A diabatic process can be defined as one where a system exchanges energy with the surroundings resulting in a thermodynamic change of the system's state.

Diabatic processes can be grouped into five categories viz

1. Terrestrial radiation.

2. Solar radiation

3. latent heat flux

4. sensible heat flux

5. friction process

In the same vein, an adiabatic process is one in which there is zero heat, mass or momentum transfer across the bounary of the thermodynamic system

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The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom

BabaBlast [244]

Answer:

The difference between ice and steam in Celsius (Centigrade) is 100 deg.

So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.

So 8 cm is 4 cm greater than the ice point

4 cm / 20 cm = 1/5 since the steam point and the ice point are 20 cm apart

Then 1/5 * 100 deg C = 20 deg C the requested temperature

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3 years ago

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notka56 [123]

but I think it's a and f

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3 years ago

a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.

PtichkaEL [24]

During that final period of time,
his acceleration is
(9 m/s - 5 m/s) / (4 sec) = 1 m/s² .

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8 0

3 years ago

Need help asap; put you as brainliest

Vera_Pavlovna [14]

Answer:

unmmmmmmmm I think the answerA

8 0

3 years ago

-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a

bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

$\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1$

The velocity of the carts after the event is 1 m/s

3 0

2 years ago