The average speed is 20.8 m/s
Explanation:
The average speed for the trip is given by:

where
d is the distance covered
t is the time elapsed
For the trip in this problem, we have:
d = 187 km = 187,000 m is the distance travelled
The initial time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,

Therefore, the average speed for the trip is

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Answer:
0.438kg/ms-¹
Explanation:
Momentum, denoted by p, can be calculated by using the formula;
p = mv
Where;
m = mass (kg)
v = velocity (m/s)
Momentum (p) of bird = 0.216 kg × 5.87 m/s = 1.268kg/ms-¹
Momentum (p) of crawling baby = 7.29 kg kg × 0.234 m/s = 1.706kg/ms-¹
Having calculated the momentum of the bird to be 1.268kg/ms-¹, and the momentum of the baby to be 1.706kg/ms-¹, the difference in momentum between the flying bird and the crawling baby is:
{1.706kg/ms-¹ - 1.268kg/ms-¹} = 0.438kg/ms-¹
Answer:

Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So
and
(Since
). Then, we get:

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).
Answer:
Gamma radiation or Cathode rays
Explanation:
by striking incident gamma or cathode rays onto the solid when placed on a photographic plate
Answer:
The canon B hits the ground fast.
Explanation:
Given that,
Speed of cannon A = 85 m/s
Speed of cannon B= 100 m/s
Speed of cannon C = 75 m/s
We need to calculate the cannonballs will hit the ground with the greatest speed
Using conservation of energy
The final kinetic energy of canon depends on initial kinetic energy and potential energy.
The final velocity depends upon initial velocity and initial height.
So, the initial velocity of canon B is high.
Hence, The canon B hits the ground fast.