Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
E=mc² where c is speed of the light
3 m/s more andmore less than speed of the light. So mass of the person still 100 kg
Answer:
About 66 miles per hour
Explanation:
Based on the information given we can assume the car traveled the same number of miles every hour meaning all we need to do is divide.
400/6 ≈ 66 miles per hour
Answer:
6.8 m/s2
Explanation:
Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is
W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N
For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of
a = F/m = 13.6 / 2 = 6.8 m/s2
So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail
Answer:
a ) 11.1 *10^3 m/s = 39.96 Km/h
b) T_{o2} =1.58*10^5 K
Explanation:
a)
= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h
b)
M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol
gas constant R = 8.31 j/mol.K
So, 
multiply each side by M_{o2}, so we have
solving for temperature T_{o2}
In the question given,
T_{o2} =1.58*10^5 K
