Answer:
The normal force will be "122.8 N".
Explanation:
The given values are:
Weight,
W = 100 N
Force,
F = 40 N
Angle,
θ = 35.0°
As we know,
⇒
On substituting the given values, we get
⇒
⇒
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If you put a object in front of the mirror why does it appear the opposite way
1 answer:
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Answer:
The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles
Explanation:
The parameters of the motion of the driver are;
The upgrade of the road, θ = 10°
The rate of constant acceleration of the bus driver = 5 ft./s²
The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s
The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s
The acceleration due to gravity, g ≈ 32.1740 ft./s²
Therefore, we have;
The acceleration due to gravity down the incline plane, gₓ = g·sinθ
∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²
The net acceleration of the bus, on the incline plane, = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²
The vertical component of the velocity, = v × sin(θ)
∴ = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s
vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s
The velocity of the car, v₂, on the inclined plane is given as follows;
v₂ = v₁ - × t
∴ t = (v₁ - v₂)/ = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s
The distance covered, 's', is given as follows;
s = v₁·t - 1/2··t²
∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.
The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles
Answer:
Explanation:
For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:
v₀ = 48 ft/s
a = -32 ft/s/s
v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:
v² = v₀² + 2aΔx and filling in:
Δx and
0 = 2300 - 64Δx and
-2300 = -64Δs so
Δx = 36 feet.
Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:
v = v₀ + at and filling in:
0 = 48 - 32t and
-48 = -32t so
t = 1.5 sec. That's part a. Onto part b:
The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:
Δx = v₀t + and filling in:
and
and
0 = 16t(3 - t) so
t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:
We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:
and get everything on one side and factor it again:
and we find that
1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.