Answer:
a)
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
the required distance is 40.98 m
Explanation:
Given that;
velocity of the river u = 1.70 m/s
velocity of boat v = 14.0 m/s
Now to get the velocity of the boat relative to shore;
( north of east), we say
a² + b² = c²
(1.70)² + (14.0)² = c²
2.89 + 196 = c²
198.89 = c²
c = √198.89
c = 14.1028 m/s
tan∅ = v/u = 14 / 1.7 = 8.23529
∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east
Therefore, the velocity of the boat relative to shore is;
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
width of river = 340 m,
ow far downstream has the boat moved by the time it reaches the north shore in meters = ?
we say;
340sin( 90° - 83.0765°)
⇒ 340sin( 6.9235°)
= 40.98 m
Therefore, the required distance is 40.98 m
Answer:
it's important because it shows how thermal energy transforms or continues to be all around us in everything
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
<span>Even in space, there is still presence of gravity. The
cause of weightlessness is not how far above the earth the space shuttle is but
rather how fast it is travelling. The shuttle is in free fall causing
weightlessness, but it is travelling fast enough to miss the earth as it falls.
Similarly, the airplane could also provide weightlessness if it went free fall
as well. However, that ends as the plane hits the ground. </span>
Answer:
4. 10.0 m/s²
Explanation:
I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:


II) using these two equations after substitution v₀=0; v=30 and L=45:

