Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
Answer:
The ball stops instantaneously at the topmost point of the motion.
Explanation:
Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.
The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.
The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.
Answer:
68 °F, 293.15 K
Explanation:
Fahrenheit, Kelvin and Celsius are the different scales of temperature in which temperature is measured.
Given : T = 20°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
<u>T = (20 + 273.15) K = 293.15 K </u>
The conversion of T( °C) to T(F) is shown below:
T (°F) = (T (°C) × 9/5) + 32
So,
<u>T (°F) = (20 × 9/5) + 32 = 68 °F</u>
Newton's 3rd law: The ball exerts a force on the bat that is equal and opposite to the force exerted by the bat on the ball.
