The height of the ball when lifted is given by 7sin(25)=2.96
the gravitational energy is mgh, the kinetic is (1/2)mv². We can set these equal since the pendulum doesn't lose much energy
mgh = (1/2)mv²
we can divide by m (since we don't have it anyways)
gh = v²/2
v=√(gh/2) = √(9.81*2.96/2)=3.8m/s.
Not exactly one of your choices, but the right one none the less
The lower limit that could be used to determine the position of each object along the direction of the velocity is 1.136 x 10⁻³ m.
<h3>
Uncertainty in the position of the electron</h3>
The uncertainty in the position of the electron is calculated as follows;
where;
- h is Planck's constant = 6.63 x 10⁻³⁴ Js
- m is mass of electron
- Δv uncertainty in velocity = (0.01 x 10⁻²) x 510 = 0.051 m/s
Thus, the lower limit that could be used to determine the position of each object along the direction of the velocity is 1.136 x 10⁻³ m.
Learn more about uncertainty in position here: brainly.com/question/1970053
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Answer:
57 N
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling down.
Normal force N pushing up.
Friction force F pushing horizontally.
Sum of the forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Friction force is the product of normal force and coefficient of friction:
F = Nμ
F = mgμ
F = (65 kg) (9.8 m/s²) (0.09)
F = 57.3 N
Rounded, the friction force is 57 N.
Answer:
if you are asking for static friction the answer is <u>372N</u>
Explanation
friction static=normal force x coefficient of static friction
and normal force is equal to weight of a body
so static friction = 600N x 0.62= 372N
Answer:
a) ΔV = 2,118 10⁻⁸ m³ b) ΔR= 0.0143 cm
Explanation:
a) For this part we use the concept of density
ρ = m / V
As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats
m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg
V = m / ρ
V = 0.56 / 3.52
V = 0.159 cm3
We use the relation of the bulk module
B = P / (Δv/V)
ΔV = V P / B
ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹
ΔV = 2,118 10⁻⁸ m³
b) indicates that we approximate the diamond to a sphere
V = 4/3 π R³
For this part let's look for the initial radius
R₀ = ∛ ¾ V /π
R₀ = ∛ (¾ 0.159 /π)
R₀ = 0.3361 cm
Now we look for the final volume and with this the final radius
= V + ΔV
= 0.159 + 2.118 10⁻²
= 0.18018 cm3
= ∛ (¾ 0.18018 /π)
= 0.3504 cm
The radius increment is
ΔR = - R₀
ΔR = 0.3504 - 0.3361
ΔR= 0.0143 cm