Answer:
Explanation:
This is a simple gravitational force problem using the equation:
where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
I'm going to do the math on the top and then on the bottom and divide at the end.
and now when I divide I will express my answer to the correct number of sig dig's:
6.45 × 10¹⁶ N
Answer:
If one end of a metal bar is heated, the atoms at that end vibrate more than the atoms at the cold end. The vibration spreads along the bar from atom to atom.
Explanation:
The spread of heat in this way is called conduction. Metals are good conductors of heat.
Answer:
2.75 m/s^2
Explanation:
The airplane's acceleration on the runway was 2.75 m/s^2
We can find the acceleration by using the equation: a = (v-u)/t
where a is acceleration, v is final velocity, u is initial velocity, and t is time.
In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1
a = 2.75 m/s^2
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :
Horizontal component of force(Normal reaction) :
Since, box is on the verge of slipping :
Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
Answer:
Explanation:
Heat capacity A = 3 x heat capacity of B
initial temperature of A = 2 x initial temperature of B
TA = 2 TB
Let T be the final temperature of the system
Heat lost by A is equal to the heat gained by B
mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)
heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)
3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB)
3 TA - 3 T = T - TB
6 TB + TB = 4 T
T = 1.75 TB
