Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
Explanation:
To find the volume of a box of Tissue (also known as a rectangular prism) just multiply the area of the base (length times width) by the height. The formula is V = l• w• h.
Answer:
If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast
Explanation:
When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:
Vr = velocity of the river = 6[km/h}
Vc = velocity of the canoe = -6 [km/h]
We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.
Vt = Vr + Vc = 6 - 6 = 0 [km/h]
For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.
So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).
Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:
![Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]](https://tex.z-dn.net/?f=Vt%20%3D%20%5Csqrt%7B%286%29%5E%7B2%7D%20%2B%28-6%29%5E%7B2%7D%20%7D%20%5C%5CVt%3D8.48%5Bkm%2Fh%5D)
Answer:
(7.8) x (9.8 m/s) = 76.44 m/s
during the time he spent falling.
Since his falling speed was zero when he 'stepped' off of the top,
he hit the ground at 76.44 m/s.
That's about 170 miles per hour.
I'll bet he left one serious crater!
I hope this helps too! :D
Explanation:
<span> </span>For any prism-shaped geometry, the volume
(V) is assumed by the product of cross-sectional area (A) and height (h).
<span> V = Ah </span>
<span>
Distinguishing with respect to time gives the
relationship between the rates.
dV/dt = A*dh/dt</span>
<span> in the meantime the area is not altering </span>
<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>
<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min
Water is draining from the tank at the rate of π/2
ft^3/min.</span>