Answer:
a. P.E = 3430Joules.
b. Workdone = 3430Nm
Explanation:
Given the following data;
Mass = 70kg
Distance = 5m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy = mgh
P.E = 70*9.8*5
P.E = 3430J
b. To find the workdone;
Workdone = force * distance
But force = mass * acceleration
Force = 70*9.8
Force = 686 Newton.
Workdone = 686 * 5
Workdone = 3430Nm
What is unusual about the material that Emily invented?
Answer:
The material that Emily invented can be easily repaired by shining ultraviolet light on it.
Explanation:
Hope it helps! Please mark brainliest.
Answer:
The material that Emily invented can be easily repaired by shining ultraviolet light on it.
Explanation:
Agnes makes a round trip at a constant speed to a star that is 16 light-years distant from Earth, while twin brother Bert remains on Earth. When Agnes returns to Earth, she reports that she has celebrated 20 birthdays during her journey. (a) What was her speed during her journey
Answer:
Speed of Agnes during her journey was 0.848c
Explanation:
Given that;
Age of Agnes t₀ = 20 years
distance d = 2 × distance of star from Earth = 2 × 16 light-years
= 32 light-years
so get her speed speed; we use the following expression
Yvt₀ = d
( v / √(1 - ([tex]\frac{v}{c}[/tex])²) )² = ( 32 light-years / 20 yrs )²
v² / (1 - ( v²/c²)) = ( 32 × c / 20 )²
v² / (1 - ( v²/c²)) = 2.56 × c²
v² / c²-v²/c² = 2.56 × c²
v²c² / c² - v² = 2.56 × c²
v² / c² - v² = 2.56
v² = 2.56 (c² - v²)
v² = 2.56 (c² - v²)
v² = 2.56c² - 2.56v²
v² + 2.56v² = 2.56c²
3.56v² = 2.56c²
v² = (2.56/3.56)c²
v = √((2.56/3.56)c²)
so v = 0.848c
Therefore, Speed of Agnes during her journey was 0.848c
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the mean value of your results to three significant digits. ________
Answer: The mean value = 9.85m/s².
Explanation:
Mean = [tex]\dfrac{\text{Sum of n observations}}{n}[/tex]
The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.
Number of measurements =9
Sum of measurements = 88.69
Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]
Hence, the mean value = 9.85m/s².
Now complete your visual overview by identifying the known variables and the variables you must find. Assume charge 1 is located at the origin of the x axis and the positive x axis points to the right. Let x1, x2, and x3 denote the positions of charge 1, charge 2, and charge 3, respectively. Determine which of the following quantities are known and which are unknown.
Answer:
Explanation:
From the given information;
If we assume that charge 1 is located at the origin;
Then, using the visual overview for identification, we will realize that the known quantities are:
[tex]\mathbf{= q_1, \ q_2 , \ q_3, \ x_1 \ and \ x_2}[/tex]
However, provided that we do not know the exact location of [tex]x_3[/tex],
Then, the unknown quantity is [tex]\mathbf{ x_3}[/tex]
Current Attempt in Progress The atomic radii of a divalent cation and a monovalent anion are 0.52 nm and 0.125 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance
Answer:
a) F = 1.70 10⁻⁹N, F = 1.47 10⁻⁸ N,
b) * the electronegative repulsion, from the repulsion by quantum effects
Explanation:
a) The atraicione force comes from the electric force given by Coulomb's law,
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
divalent atoms
In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given
F = k q² / r²
F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]
F = 1.70 10⁻⁹N
monovalent atoms
in this case the load is q = q₀
F = 2 \ 10^9 \ \frac{ (1.6 \ 10^{-19} )^2}{ (0.125 10^{-9} )^2 }
F = 1.47 10⁻⁸ N
b) repulsive forces come from various sources
* the electronegative repulsion of positive nuclei
* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together
* from the repulsion of electron clouds, by quantum effects
3N
3 N
What is the net force of the box?
A 10 kg remote control plane is flying at a height of 111 m. How much
potential energy does it have?
Answer:
10.88kJ
Explanation:
Given data
mass= 10kg
heigth= 111m
Applying
PE= mgh
assume g= 9.81m/s^2
substitute
PE= 10*9.81*111
PE=10889.1 Joules
PE=10.881kJ
Hence the potential energy is 10.88kJ