From the geometry of the problem, the 20 m-long cable creates
the hypotenuse of a right triangle, with the extended of the other two sides of
size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased
by 20 m - 17.3 m = 2.7 m.
The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 *
1.6 m , or about 37044 joules.
When you bring two objects of different temperature together, energy will always be transferred from the hotter to the cooler object. The objects will exchange thermal energy, until thermal equilibrium is reached, i.e. until their temperatures are equal. We say that heat flows from the hotter to the cooler object. Heat is energy on the move.
Units of heat are units of energy. The SI unit of energy is Joule. Other often encountered units of energy are 1 Cal = 1 kcal = 4186 J, 1 cal = 4.186 J, 1 Btu = 1054 J.
Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.
Answer:
In most cases, the sources of information for both levels are same as follows:
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Different media: Media is one of the best sources of information in these days.
Explanation:
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
