Answer:
The handrails must be approximately 10.63 meters long
Explanation:
The given parameters are;
The height of the bleachers, h = 8 m
The depth of the bleachers, d = 7 m
The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;
The length of the hand rail = √(d² + h²)
∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63
In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.
Answer:
1. 2.5s
Explanation:
1. For time, divide Distance / speed
25m / 10
=2.5s
This is an insidious question. Quite frankly, I would not have
expected to see it here on Brainly. But I'm ready to play the
cards that you have dealt me.
None of the choices offered is a correct solution.
If the output of the AC generator is nice and sinusoidal, and
its maximum (peak) emf is 150 volts, then its RMS emf is
(1/2) (150) (√2) = 106.07 volts.
The resistor's dissipation is
Power = (current) x (voltage) .
If the resistor is dissipating its full rated 35W, then
35W = (current) x (106.07 V)
Divide each side by 106.07 V:
RMS Current = (35W) / (106.07 V) = 0.33 Ampere .
_________________________________________
Looking over the choices offered . . .
The largest choice ... 3.1 A ... is the current in a resistor
that is dissipating 35W if the voltage is
(35W / 3.1A) = 11.29 volts .
The smallest choice ... 1.2 A ... is the current in a resistor
that is dissipating 35W if the voltage is
(35W / 1.2A) = 29.17 volts .
Whatever you meant the so-called "150 V" of the generator
to represent ... whether the RMS sinusoidal, peak sinusoidal,
peak square-wave, RMS square-wave, DC, average, etc. ...
none of the choices for current, in combination with any of these
generators, would dissipate 35W.
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.
