Why is it important to exclude onlockers from a crime scene
1 answer:
You might be interested in
Answer:
The total time it is in the air for the ball is 1.6326 s
Given:
Initial velocity = 8
To find:
the total time it is in the air = ?
Formula used:
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
Solution:
A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.
The time taken by the ball to reach the maximum height is given by,
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
t =
t = 0.8163 s
Thus, time taken by the ball to reach the ground again = time taken to reach maximum height
So, Total time required for ball to reach ground = 2t = 2 × 0.8163
Total time required for ball to reach ground = 1.6326 s
The total time it is in the air for the ball is 1.6326 s
Answer:
The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s
= 16.125 Pa
Explanation:
The Bernoulli's equation is essentially a law of conservation of energy.
It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.
For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.
We also assume that the initial velocity of wind is 0 m/s.
This calculation is presented in the attached images to this solution.
Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.
The density is obtained to be 1.29 kg/m³.
Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.
We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.
Hope this Helps!!!
Answer:
= 14.88 N
Explanation:
Let's begin by listing out the given variables:
M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,
g = 9.8 m/s²
At equilibrium, the sum of all external torque acting on an object equals zero
τ(net) = 0
Taking moment about we have:
(M + m) g * 0.5L - (L - d) = 0
⇒ = [(M + m) g * 0.5L] ÷ (L - d)
= [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)
= 59.535 ÷ 2.4
= 24.80625 N ≈ 24.81 N
Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N
Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N
Using sum of equilibrium in the vertical direction, we have:
+
= W + w ------- Eqn 1
Substituting T2, W & w into the Eqn 1
+ 24.81 = 26.46 + 13.23
= <u>14.88</u> N