<h2>
Average speed of transit train is 60 mph</h2>
Explanation:
Average speed of passenger train = 45 mph
Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours
Distance between station A to station B = 45 x 4 = 180 miles.
Time taken from station A to station B for transit train = 4 - 1 = 3 hours
Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train
180 = Average speed of transit train x 3
Average speed of transit train = 60 mph
Average speed of transit train is 60 mph
Newton’s law of gravity is considered “universal” because it is believed to be applicable to the entire Universe (to a good approximation). Gravity does not just pull an apple from a tree, but also pulls the Moon to the Earth, the Earth to the Sun, and so on. Gravity is a fundamental force of nature.
Answer:
A) 10 m/s
Explanation:
We know that according to conservation of momentum,
m1v1 + m2v2 = m1u1 + m2u2 ..............(equation 1)
where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.
From the given data
If truck and car are two bodies
truck : m1 = 2000 Kg v1 = 5 m/s u1 = 0
car : m2 = 1000 kg v2 = 0 u2 = ?
final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.
substituting the values in equation 1, we get
(2000 x 5) + 0 = 0 + (1000 x u2)
u2 =
x 5
= 10 m/s
Hence after collision, car moves at a velocity of 10 m/s
Answer:
Given that
The earth spins on its axis once a day and orbits the sun once a year (365 1/4 days)
a)
When earth spins on its axis
We know that earth take 1 day to complete one revolution around its own axis.
T= 1 day = 24 hr = 24 x 3600 s
T=86400 s
We know that
T=2π/ω
ω= 2π/T
ω= 2π/86400
ω=7.27 x 10⁻5 rad/s
b)
When earth revolve around earth
T =365 1/4 days = 365.25 days
T= 365.24 x 86400 s
T=31557600
We know that
T=2π/ω
ω= 2π/T
ω= 2π/31557600
ω=1.99 x 10⁻⁷ rad/s
Answer:
The time for final 15 cm of the jump equals 0.1423 seconds.
Explanation:
The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

where
'v' is the final velocity of the player
'u' is the initial velocity of the player
'a' is acceleration due to gravity
's' is the height the player jumps
Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

where symbols have the usual meaning
Applying the given values we get
