A student mixes two clear liquids together. After a few minutes, a white powdery solid can be seen settling on the bottom of the
2 answers:
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Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.
In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:
Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:
Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)
To solve for "t" in this quadratic equation, we can factor it out as shown:
Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:
Answer:
a)1.37 s
b)∞ ( Infinite)
Explanation:
Given that
L= 47 cm ( 1 m =100 cm)
L= 0.47 m
a)
On the earth :
Acceleration due to gravity = g
We know that time period of the simple pendulum given as
Here
Now by putting the values
T=1.37 s
b)
Free falling elevator :
When elevator is falling freely then
( This is case of weightless motion)
Therefore
T=∞ (Infinite)
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,