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Aleonysh [2.5K]

2 years ago

6

An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th

is process, in joules?

1 answer:

pogonyaev2 years ago

5 0

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, $c=345J/^oC$

Initial temperature, $T_i=88^{\circ}C$

Final temperature, $T_f=45^{\circ}C$

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

$Q=mc\Delta T$

Let the mass of the object is 10 g or 0.01 kg

So,

$Q=0.01\times 345\times (88-45)$

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.

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schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0

2 years ago

A free-falling golf ball strikes the ground and exerts a force on it. Which sentences are true about this situation? A golf ball

Harlamova29_29 [7]

Answer:

The ground exerts an equal force on the golf ball

Explanation:

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

In this problem, object A is the golf ball while object B is the ground, so we can say that:

- the golf ball exerts a force on the ground

- the ground exerts an equal and opposite force on the golf ball

8 0

2 years ago

Read 2 more answers

the millersburg ferry (m=13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. if the speed before brakin

kenny6666 [7]

The braking force is -400 N

Explanation:

We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

$F \Delta t = m(v-u)$

where in this problem, we have:

F is the force applied by the brakes

$\Delta t = 65 s$ is the time interval

m = 13,000 kg is the mass of the ferry

u = 2.0 m/s is the initial velocity

v = 0 is the final velocity

And solving for F, we find the force applied by the brakes:

$F=\frac{m(v-u)}{\Delta t}=\frac{(13000)(0-2.0)}{65}=-400 N$

where the negative sign indicates that the direction is backward.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

4 0

2 years ago

Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller

jeka94

Answer:

Explanation:

Torque on smaller wheel

= F x r

50 x .30

= 15 Nm

Torque on larger wheel

= F x .5

For equilibrium

F x .5 = 15

F = 15 / .5

= 30 N

8 0

2 years ago

Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m

OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

$E_m = \frac{1}{2} m\omega_1^2A_1^2$

The energy of the system having mass 2m is,

$E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2$

For the two expressions mentioned above remember that the variables mean

m = mass

$\omega =$ Angular velocity

A = Amplitude

The energies of the two system are same then,

$E_m = E_{2m}$

$\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2$

$\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}$

Remember that

$k = m\omega^2 \rightarrow \omega^2 = k/m$

Replacing this value we have then

$\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}$

But the value of the mass was previously given, then

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}$

$\frac{A_1}{A_2} = 1$

Therefore the ratio of the oscillation amplitudes it is the same.

5 0

2 years ago