Answer:
the answer is 2000Nm
Explanation:
wprk done = force × distance moved
w.d = 200N × 10m
w.d = 2000Nm
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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.
Best Answer:<span> </span><span>hf = work function + KE
However if you are looking at the max wavelength (or threshold frequency) then there will only be just enough energy for the photoelectrons to be liberated, hence their KE will be 0.
So hf = work function
convert eV to joules, 2.4 x (1.6 x 10^-19) = 3.84 x 10^-19
therefore, hf = 3.84 x 10^-19
f = 3.84 x 10^-19 / planck's constant which is 6.63 x 10^-34
f = 5.79 x 10^14 Hz
c = frequency x wavelength,
wavelength = speed of light/frequency
= (3x10^8)/(5.79x10^14)
=5.18 x 10^-7 metres</span>
