When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.
We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.

where,
- c: specific heat capacity of water
- m: mass of water
- ΔT: change in the temperature
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
Learn more: brainly.com/question/16104165
Thermal equilibrium is a state in which all parts of a system are at the same temperature
Mitochondria help take energy from sugar or glucose and convert it into a simpler form, called ATP, that the cell can more easily use. This process is called cellular respiration, and the mitochondria plays a central role in it.Mitochondria are composed of two membranes.An outer membrane covers the organelle like a skin, protecting it.An inner membrane, that is folded over again and again to create a layered structure called crista, which is studded with proteins. The fluid inside these folds is called the matrix.Chloroplasts are very similar to mitochondria, but are found only in the cells of plants and some algae. Like mitochondria, chloroplasts produce food for their cells. Chloroplasts help turn sunlight into food that can be used by the cell, a process known as photosynthesis.
Answer:


Explanation:
As the disc is unrolling from the thread then at any moment of the time
We have force equation as

also by torque equation we can say



Now we have



Also from above equation the tension force in the string is


Answer:
The time it will take for the object to hit the ground will be 4.
Explanation:
You have:
h(t)=−16t²+v0*t+h0
Being v0 the initial velocity (54 ft/s) and h0 the initial height (40 ft) and replacing you get:
h(t)=−16t²+54*t+40
To know how long it will take for the object to touch the ground, the height h(t) must be zero. So:
0=−16t²+54*t+40
Being a quadratic function or parabola: f (x) = a*x² + b*x + c, the roots or zeros of the quadratic function are those values of x for which the expression is 0. Graphically, the roots correspond to the points where the parabola intersects the x axis. To calculate the roots the expression is used:

In this case you have that:
Replacing in the expression of the calculation of roots you get:
Expresion (A)
and
Expresion (B)
Solving the Expresion (A):

Solving the Expresion (B):

These results indicate the time it will take for the object to hit the ground can be -5/8 and 4. Since the time cannot be negative, then <u><em>the time it will take for the object to hit the ground will be 4.</em></u>