The force the box is exerting on Manuel is the weight of the box, downward:
and this force is perfectly balanced by the constraint reaction applied by Manuel's hand, pushing upward.
Answer: 20 kgm/s
Explanation:
Given that M1 = M2 = 10kg
V1 = 5 m/s , V2 = 3 m/s
Since momentum is a vector quantity, the direction of the two object will be taken into consideration.
The magnitude of their combined
momentum before the crash will be:
M1V1 - M2V2
Substitute all the parameters into the formula
10 × 5 - 10 × 3
50 - 30
20 kgm/s
Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s
Answer:
The mass of the massive object at the center of the Milky Way galaxy is 
Explanation:
Given that,
Diameter = 10 light year
Orbital speed = 180 km/s
Suppose determine the mass of the massive object at the center of the Milky Way galaxy.
Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.
We need to calculate the radius of the orbit
Using formula of radius
We need to calculate the mass of the massive object at the center of the Milky Way galaxy
Using formula of mass
Put the value into the formula
Hence, The mass of the massive object at the center of the Milky Way galaxy is
Answer:
Explanation:
potential energy of compressed spring
= 1/2 k d²
= 1/2 x 730 d²
= 365 d²
This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .
Kinetic energy after crossing the rough patch
= 1/2 x 1.2 x 2.3²
= 3.174 J
Loss of energy
= 365 d² - 3.174
This loss is due to negative work done by frictional force
work done by friction = friction force x width of patch
= μmg d , μ = coefficient of friction , m is mass of block , d is width of patch
= .44 x 1.2 x 9.8 x .05
= .2587 J
365 d² - 3.174 = .2587
365 d² = 3.4327
d² = 3.4327 / 365
= .0094
d = .097 m
= 9.7 cm
If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .
I can guarantee you that it is not
C.<span>the angle that the incident ray makes with a line drawn perpendicular to the reflecting surface I hope this somewhat helps</span>
