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2 years ago

Assuming that a2 is the magnitude of the horizontal acceleration of the block of mass m2, what is T, the tension in the string

Physics

1 answer:

cestrela7 [59]2 years ago

6 0

Answer:

T= (m2a2)/2

Explanation:

Using Newton's second law

F21+F22+ F33.....= M2a2

Where F21 and F22 are forces acting on M2

Thus T = ( M2a2)/2 this is tension on the string

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A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

4 0

3 years ago

why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level

Genrish500 [490]

Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.

5 0

2 years ago

Help, anyone?? please:/

NeTakaya

A,E,C,B,D is the order

5 0

3 years ago

Kinetic energy of an object is equal to

Gnom [1K]

See
K.E=1/2(mass*velocity²)
so option B is the correct answer.
Brainliest pls :-)

8 0

2 years ago

An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extra

WARRIOR [948]

Answer:

The inventors claim is not real

a) No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

The power input is $P_i = 0.25 kW = 0.25 *10^{3} \ W$

The rate of heat transfer $J = 3050 J/s$

The temperature of the freezer content is $T = 270 \ K$

The ambient temperature is $T_a = 293 \ K$

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

$COP = \frac{T }{Ta - T}$

substituting values

$COP = \frac{270 }{293 - 270}$

$COP =11.7$

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

$COP = \frac{J}{P_i}$

substituting values

$COP = \frac{3050}{0.25 *10^{3}}$

$COP = 12.2$

Now given that the COP of an ideal refrigerator is less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

4 0

2 years ago