Answer:
P1 = 1.3 (500 + 60) = 728 kg-m total momentum to right at start
P2 = (v2 - 10) 60 + 500 v2
total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart
728 = 560 v2 - 600
v2 = 1328 / 560 = 2.37 m/s new speed of cart
Check:
After: p2 for cart = 500 * 2.37 = 1186
p1 for man = (2.37 - 10) * 60 = -458
P2 = p1 + p2 = 728 total momentum unchanged
Answer:
The frequency of vibration will be unchanged.
Explanation:
The frequency of oscillation of a simple pendulum is given by
where, 
is the acceleration due to gravity and 
is the length of the pendulum.
It can be seen from the formula that the frequency of the pendulum is independent of the mass.
Thus, the frequency of vibration will be same.
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-Brainly support team (Klara academic DEPT.)Spam words mean team is having trouble determining the question CONTACT For support.
What is horoscope?
A forecast of a person's future, typically including delineation of character and circumstances, based on the relative positions of the stars and planets at the time of that person's birth.
*A short forecast for people born under a particular sign, especially as published in a newspaper or magazine.
*A birth chart.
What is its uses?
It can also be calculated for an event, a question, and even a country. Symbols are used to represent planets, signs, and geometric connections called aspects. In most cases, the horoscope in Western astrology is drawn on a circular wheel.
Answer:
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms
2 ).
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
23 =20 3 N
While the uertical component of the force acting in upward direction=Fsin30=40× 21
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN = 528.64
=5.73m/s 2
