Answer:
This would be traveling at the lower reaches.
Explanation:
A river would be traveling the fastest at the upper reaches and it becomes slower at the middle reaches and the slowest at the lower reaches. A place where water flows fast in a river is where the width is narrow and the bottom is steep. (This is just examples incase you would like to keep notes).
Answer:
The neutron can be found in the nucleus of the atom with the proton.
Answer:
The frog's horizontal velocity is 0.2 m/s.
Explanation:
To solve this problem, we must first remember what velocity is and how we solve for it. Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance. If we plug in the given numbers for these variables and solve, we get the following:
v = x/t
v = 0.8m/4s
v = 0.2 m/s
Therefore, the correct answer is 0.2 m/s. We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.
Hope this helps!
Answer:
a) 4.2m/s
b) 5.0m/s
Explanation:
This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.
The problem is also an illustration of elastic collision where there is no loss in kinetic energy.
Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses
and
whose respective velocities before collision are
and
;

where
and
are their respective velocities after collision.
Given;

Note that
=0 because the second mass
was at rest before the collision.
Also, since the two masses are equal, we can say that
so that equation (1) is reduced as follows;

m cancels out of both sides of equation (2), and we obtain the following;

a) When
, we obtain the following by equation(3)

b) As
stops moving
, therefore,
