<h3>
<u>Provided</u><u>:</u><u>-</u></h3>
- Initial velocity = 15 m/s
- Final velocity = 10 m/s
- Time taken = 2 s
<h3><u>To FinD:-</u></h3>
- Accleration of the particle....?
<h3>
<u>How</u><u> </u><u>to</u><u> </u><u>solve</u><u>?</u></h3>
We will solve the above Question by using equations of motion that are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
Here,
- v = Final velocity
- u = Initial velocity
- a = acceleration
- t = time taken
- s = distance travelled
<h3>
<u>Work</u><u> </u><u>out</u><u>:</u></h3>
By using first equation of motion,
⇛ v = u + at
⇛ 10 = 15 + a(2)
⇛ -5 = 2a
Flipping it,
⇛ 2a = -5
⇛ a = -2.5 m/s² [ANSWER]
❍ Acclearation is negative because final velocity is less than Initial velocity.
<u>━━━━━━━━━━━━━━━━━━━━</u>
The area of a square is given by:
A = s²
A is the square's area
s is the length of one of the square's sides
Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:
d[A]/dt = d[s²]/dt
The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt
dA/dt = 2s(ds/dt)
A) Given values:
s = 14m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(14)(3)
dA/dt = 84m²/s
B) Given values:
s = 25m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(25)(3)
dA/dt = 150m²/s
10 newton east is the answer because the two force vector form 0 degree angle with each other
Answer:
The angle is
Solution:
As per the question:
Mass of the objects are same, say m Kg each
Let their initial speed be 'u'
Both the objects move apart from each other at m/s
Now, the angle between the objects' initial velocities is given by using the law of conservation of momentum:
Now, applying the principle of momentum conservation along the horizontal axis:
Now, momentum conservation along vertical axis:
Now, angle between the initial velocities of the objects:
Answer:
(A) The maximum height of the ball is 40.57 m
(B) Time spent by the ball on air is 5.76 s
(C) at 33.23 m the speed will be 12 m/s
Explanation:
Given;
initial velocity of the ball, u = 28.2 m/s
(A) The maximum height
At maximum height, the final velocity, v = 0
v² = u² -2gh
u² = 2gh
(B) Time spent by the ball on air
Time of flight = Time to reach maximum height + time to hit ground.
Time to reach maximum height = time to hit ground.
Time to reach maximum height is given by;
v = u - gt
u = gt
Time of flight, T = 2t
(C) the position of the ball at 12 m/s
As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.
v² = u² - 2gh
12² = 28.2² - 2(9.8)h
12² - 28.2² = - 2(9.8)h
-651.24 = -19.6h
h = 651.24 / 19.6
h = 33.23 m
Thus, at 33.23 m the speed will be 12 m/s