A). Both the energy and the wave travel in the same direction.
If they didn't, they'd wind up in different cities almost instantly.
Answer:
The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Explanation:
Given that,
Mass of object = 5 kg
Speed = 3 m/s
Mass of stationary object = 3 kg
Moving object deflected = 30°
Stationary object deflected = 31°
We need to calculate the velocity of each ball after collision
Using conservation of momentum
Along x-axis
Put the value into the fomrula
....(I)
Along y -axis
Put the value into the formula
...(II)
From equation (I) and (II)
Put the value of v₁ in equation (I)
Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Answer: Your answer is 1250J
Explanation:
K
E
=
1/2
m
v
2
The mass is
m
=
25
k
g
The velocity is v
=
10
m
s
−
1
So,
K
E
=
1
/2
x25
x
10
2^2=
1250
J
pls mark brainiest answer
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
The correct statements are that the speed decreases as the distance decreases and speed increases as the distance increases for the same time.
Answer:
Option A and Option B.
Explanation:
Speed is defined as the ratio of distance covered to the time taken to cover that distance. So Speed = Distance/Time. In other words, we can also state that speed is directly proportional to the distance for a constant time. Thus, the speed will be decreasing as there is decrease in distance for the same time. As well as there will be increase in speed as the distance increases for the same time. So option A and option B are the true options. So if there is decrease in the distance due to direct proportionality the speed will also be decreasing. Similarly, if the distance increases, the speed will also be increasing.
