Answers
Answer:
probably won't help now but the answer is the 4th one lol
Explanation:
Related Questions
Suppose a 118kg watermelon is held 5m above the ground before sliding along a frictionless ramp to the ground. How high above the ground is the watermelon at the moment it's kinetic energy is 4,610J
Answers
This question involves the concepts of kinetic energy, potential energy, and the law of conservation of energy.
The watermelon is "1.02 m" above the ground.
The total energy of the watermelon can be found by its potential energy at the highest point:
Total Energy = mgh
where,
m = mass = 118 kgg = 9.81 m/s²h = height = 5 mTherefore,
Total Energy = (118 kg)(9.81 m/s²)(5 m)
Total Energy = 5787.9 J
LAW OF CONSERVATION OF ENERGYNow, according to the law of conservation of energy, at the given point:
Total Energy = Kinetic Energy + Potential Energy
5787.9 J = 4610 J + mgh'
[tex]h'=\frac{5787.9\ J-4610\ J}{(118\ kg)(9.81\ m/s^2)}[/tex]
h' = 1.02 m
Learn more about the law of conservation of energy here:
https://brainly.com/question/2264339
Suppose that you make a series RC circuit with a capacitor and a known resistor that has a 5% tolerance: R= 5.20 ± 0.26kΩ. You plan is to measure the time constant of the circuit in order to determine the value of the capacitor. You do ten trials of your experiment and measure the time constant each time, and then you calculate the average and standard error of the ten results to determine that τ= 2.150 ± 0.002s.
Which of the following is the best estimate of the uncertainty on your determination of the value of your capacitor?
a. 1%
b. 0.1%
c. 5%
Answers
Answer:
correct answer is C
Explanation:
The time constant of an RC circuit is
τ = RC
so to find the capacitance
C = τ/ R
C = 2.150 / 5.20 10³
C = 4.13 10⁻⁴ F
to find the error we use the worst case
ΔC = | [tex]|\frac{dC}{d \tau }| \ \Delta \tau + | \frac{dC}{dR} | \ \Delta R[/tex]
the absolute value guarantees that we find the worst case, we evaluate the derivatives
ΔC = 1 /R Δτ + τ/R² ΔR
the absolute values of the errors are
Δτ = 0.002 s
ΔR = 0.3 kΩ
we substitute
ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)² 0.3 10³
ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵
ΔC = 1.77 10⁻⁵ F
the uncertainty or error must be expressed with a significant figure
ΔC = 2 10⁻⁵ F
the percentage error is
Er% =[tex]\frac{\Delta C}{C} \ 100[/tex]
Er% = [tex]\frac{2 \ 10^{-5} }{ 4.13 \ 10^{-4} } \ 100[/tex]
Er% = 4.8%
the correct answer is C
Which of the following does NOT create a mechanical wave? Text to speech
A. a shaking rope
B. a guitar being played
C. dropping a rock in a pond
D. light coming through a window
Answers
Answer:
light coming through a window
A ball is kicked at 10.4 m/s at an angle of 32 degrees to the horizontal
how long (time) is the ball in the air?
find the horizontal displacement (range) of the ball
Answers
Answer:
3M/S
Explanation:
The amount of energy for a coulomb of charge at one end of a conductor is 40 joules. The amount of energy for a coulomb of charge at the other end of the conductor is 20 joules. The potential difference across the ends of the conductor is ______.
A.) 60 joules per coulomb
B.) 60 joules
C.) 20 joules per coulomb
D.) 20 joules
Answers
Which statement best describes covalent bonding?
A. two nonmetal atoms share electrons between them
B. many atoms give up electrons that can move among the atoms
C. two metal atoms share electrons between them
D. a metal atom transfers electrons to a nonmetal atom
Answers
Identify and describe one past geologic process that formed copper ore deposits.
Answers
Answer:
tornado and land
Explanation:
In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright fringe on a flat screen is 0.0176 m, when the light has a wavelength of 425 nm. Assume that the angles are small enough so that is approximately equal to . Find the separation y when the light has a wavelength of 614 nm.
Answers
Answer:
Y = 0.0254 m = 25.4 mm
Explanation:
The formula for the fringe spacing in Young's Double-slit experiment is given by the following formula:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = fringe spacing = 0.0176 m
λ = wavelength = 425 nm = 4.25 x 10⁻⁷ m
L = Distance between screen and slits
d = slit separation
Therefore,
[tex]\frac{0.0176\ m}{4.25\ x\ 10^{-7}\ m} = \frac{L}{d}\\\\\frac{L}{d} = 41411.76[/tex]
Now, for:
λ = 614 nm = 6.14 x 10⁻⁷ m
[tex]Y = (6.14\ x\ 10^{-7}\ m)(41411.76)\\[/tex]
Y = 0.0254 m = 25.4 mm