<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.
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Answer
given,
B = 6.40 mm , L = 26 cm , τ = 3.90 × 10⁻² s
general wave equation
y = A cos (k x - ωt)
where A is the amplitude of the
a) Amplitude of the given wave
B = 6.40 mm
b) Wavelength of the given wave
λ = L
λ = 26 cm
c) wave frequency
f = 25.64 Hz
d) speed of wave will be equal to
v = f λ
v = 25.64 x 0.26
v = 6.67 m/s
e) direction of propagation will be along +ve x direction because sign of k x and ωt is same as general equation.
Answer:before throwing and after catching the ball
Explanation:
When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.
After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.