the answer is d they are essential to all ecosystems
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
Answer:
The boat will be 74 .17 meters downstream by the time it reaches the shore.
Explanation:
Consider the vector diagrams for velocity and distance shown below.
converting 72 miles per hour to km/hr
we have 72 miles per hour 72 × 1.60934 = 115.83 km/hr
The velocity vectors form a right angled triangle, and can be solved using simple trigonometric laws


This is the vector angle with which the ship drifts away with respect to its northward direction.
<em>From the sketch of the displacement vectors, we can use trigonometric ratios to determine the distance the boat moves downstream.</em>
Let x be the distance the boat moves downstream.d



∴The boat will be 74 .17 meters downstream by the time it reaches the shore.
Refer to the diagram shown below.
The force, F, is applied at 5 cm from the elbow.
For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N
Answer: 1500 N