Em inglês, esse ponto é 'perihelion' ... 'periélio'. É o mais próximo que a Terra ou o planeta chega do sol.
C real,inverted and smaller than the object
Answer:
22.2 m/s
Explanation:
First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.
Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.
The average speed can be found by using the equation
. After substitution, this gives the fraction
, which reduces to 22
m/s, or about 22.2 m/s.
Explanation:
Given that,
Initial speed of the sports car, u = 80 km/h = 22.22 m/s
Final speed of the runner, v = 0
Distance covered by the sports car, d = 80 km = 80000 m
Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :




Value of g, 


Hence, this is required solution.
Answer:
rpm
Explanation:
Given that rotational kinetic energy = 
Mass of the fly wheel (m) = 19.7 kg
Radius of the fly wheel (r) = 0.351 m
Moment of inertia (I) = 
Rotational K.E is illustrated as 





Since 1 rpm = 


