The speed with which Melinda’s car is bumped across the floor is 14 ms-1.
Total mass of the twins and the car = 300 Kg
Initial velocity of the twins car = 10 m/s
Total mass of Melinda and her car = 125 Kg
Initial velocity of Melinda's car = 0 m/s (at rest)
Final velocity of the twins car = 4.12 m/s
Final velocity of Melinda's car = ?
Initial momentum of the system = (300 × 10) + (125 × 0) = 3000 Kgms-1
Final momentum of the system = (300 × 4.12) + (125 × x) = (1236 + 125x) Kgms-1
Using the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m1u1 + m2u2 = m2v1 + m2v2
(300 × 10) + (125 × 0) = (300 × 4.12) + (125 × x)
3000 = 1236 + 125x
x = 3000 - 1236 /125
x = 14 ms-1
The speed with which Melinda’s car is bumped across the floor is 14 ms-1.
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Answer:
The distance traveled by Tina before passing David is 900 m
Given:
Initial speed of David,
Acceleration of Tina,
Solution:
Now, as per the question, we use 2nd eqn of motion for the position of David after time t:
where
s = distance covered by David after time 't'
a = acceleration of David = 0
Thus
Now, Tina's position, s' after time 't':
where
, initially at rest
(1)
At the instant, when Tina passes David, their distances are same, thus:
s = s'
t = 30 s
Now,
The distance covered by Tina before she passes David can be calculated by substituting the value t = 30 s in eqn (1):
= 900 m
It will be cloudy and there will be rain.
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Answer:
<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q =
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( = 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
Answer:
10 hours earlier than regular train
Explanation:
In this case you are already giving the expression to be used which is:
S = D/t (1)
The problem is giving us the data of the speed of both trains, and we also know the distance between City A and B, which is 4000 km, therefore, we just need to solve for t in the above expression for both trains, and then, do the difference between their times and see how much earlier the express train arrives.
Solving for t, we have:
t = D/S (2)
For Train 1 (The regular):
t₁ = 4000 / 80
t₁ = 50 h
For Train 2 (Express):
t₂ = 4000 / 100
t₂ = 40 h
Now, as expected express train arrives earlier, now let's see how much:
T = t₁ - t₂
T = 50 - 40
<h2>
T = 10 h</h2><h2>
</h2>
Therefore, Express train arrives 10 hours earlier than regular train.
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