Answers
On the surface of the Moon, the free-fall acceleration is roughly one-sixth that of the Earth. Their average densities are 0.667 apart.
Given the radius of the Moon is (Rm) = 0.250 RE
The radius of the earth (RE) = 6.4 x 10^6 m
Let freefall acceleration of the moon = gm
Let freefall acceleration of the earth = ge
Given gm = ge/6
We know that g = Gm/r^2 where G is the gravitational constant
Let mass of moon = m1 and mass of earth = m2
Gm1/Rm^2 = Gm2/6xRE^2
mass = densityxvolume
let density of moon = d1 and density of earth = d2
d1 x (4/3Rm^3)/Rm^2 = d2x (4/3RE^3)/6xRE^2
d1xRm = d2xRE/6
d1/d2 = RE/6X0.250RE
d1/d2 = 0.667
Hence the ratio of densities of moon and earth is 0.667
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Related Questions
In a compound microscope the objective and the eyepiece have focal lengths of 0.95 cm and 5 cm respectively, and are kept at a distance of 20 cm. The last image is formed at a distance of 25 cm from the eyepiece.The position of object and the total magnification are
a. 95/94 cm in front of field lens 94
b. 80/79 cm in front of the field lense, 93
c. 70/65 cm in front of the field lense, 65
d. 60/55cm in front of the field lense, 50
Answers
A) In a compound microscope, the focal lengths of the objective and eyepiece are 2.0 cm and 3.0 cm, respectively.
The eyepiece's final picture is formed at infinity and the objective and eyepiece are separated by 15.0 cm. Both lenses are quite narrow.
f \s0 \s \s =0.95cm
f \se \s \s =5cm
L=20cm
v \se \s \s =25cm
U \s0 \s \s =? M=?
L D =v 0 + D+fe Dfe
20=v \s0 \s \s + \s \s30 \s6
25× \s \s5
6 \s95 \s \s =v \s0 \s
95 \s6 \s \s − \sv \s0
1 \s \s = \s95 \s100
− \sv \s0 \s
1 \s \s = \s95 \s94
v \s0 \s \s =− \s94 \s95 \s \s cm
M \sD \s \s = \sv \s0 \s
v \s0 \s
\s (1+ \sfe \sD \s \s )
M \sD \s \s = \s94 \s95
6 \s95
\s (1+ \s5 \s25 \s \s )
= \s6×98 \s95×94 \s \s ×6
=94
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A cat runs and jumps from one roof top to another which is 5 meters away and 3 meters below
Answers
Answer: 6.45 m/s
Explanation:
Find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given as 5 m
thus speed,= distance/time.
Putting values,
speed = 5/0.77
= 6.45 m/s
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A car travels 400 km in the first 4.5 hours of a trip. It stops for an hour and then travels final 300 km in 2.5 hours. Find the average speed of the car.
Answers
Given data:
Distance traveled by car in t_1=4.5 hr is s_1=400 km.
Distance traveled by car in t_2=1 hr is s_2=0 km (as the car was stopped).
Distance traveled by car in t_3=2.5 hr is s_3=300 km.
The average speed is given as,
\(\begin{gathered} v_{avg}=\frac{\text{ total distance traveled}}{\text{total time taken}} \\ =\frac{s_1+s_2+s_3}{t_1+t_2+t_3} \end{gathered}\)Substitute all known values,
\(\begin{gathered} v_{avg}=\frac{(400\text{ km})+(0\text{ km})+(300\text{ km})}{(4.5\text{ hr})+(1\text{ hr})+(2.5\text{ hr})} \\ =87.5\text{ km/h} \end{gathered}\)Therefore, the average speed of the car is 87.5 km/h.
Describe the unique characteristics of fluids.
Answers
Answer:
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension
A mass m at the end of a spring of spring constant k is undergoing simple harmonic oscillations with amplitude A.
Part (a) At what positive value of displacement x in terms of A is the potential energy 1/9 of the total mechanical energy?
Part (b) What fraction of the total mechanical energy is kinetic if the displacement is 1/2 the amplitude?
Part (c) By what factor does the maximum kinetic energy change if the amplitude is increased by a factor of 3?
Answers
Answer:
a) The potential energy of the system is 1/9 of the total mechanical energy, when \(x= \frac{1}{3}\cdot A\).
b) The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) The maximum kinetic energy is increased by a factor of 9.
Explanation:
a) From Mechanical Physics, we remember that the mechanical energy of mass-spring system (\(E\)), measured in joules, is the sum of the translational kinetic energy (\(K\)), measured in joules, and elastic potential energy (\(U\)), measured in joules. That is:
\(E = K + U\) (1)
By definitions of translational kinetic energy and elastic potential energy, we have the following expressions:
\(K = \frac{1}{2}\cdot m \cdot v^{2}\) (2)
\(U = \frac{1}{2}\cdot k\cdot x^{2}\) (3)
Where:
\(m\) - Mass, measured in kilograms.
\(v\) - Velocity of the mass, measured in meters per second.
\(k\) - Spring constant, measured in newtons per meter.
\(x\) - Elongation of the spring, measured in meters.
If we know that \(U = \frac{1}{9}\cdot E\), \(k = k\) and \(E = \frac{1}{2}\cdot k \cdot A^{2}\), then:
\(\frac{1}{18}\cdot k\cdot A^{2} = \frac{1}{2}\cdot k\cdot x^{2}\)
\(\frac{1}{9}\cdot A^{2} = x^{2}\)
\(x= \frac{1}{3}\cdot A\)
The potential energy of the system is 1/9 of the total mechanical energy, when \(x= \frac{1}{3}\cdot A\).
b) If we know that \(k = k\), \(x = \frac{1}{2}\cdot A\) and \(E = \frac{1}{2}\cdot k \cdot A^{2}\), then the equation of energy conservation associated with the system is:
\(\frac{1}{2}\cdot k\cdot A^{2} = \frac{1}{4}\cdot k\cdot A^{2}+K\)
\(K = \frac{1}{4}\cdot k\cdot A^{2}\)
The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) From the Energy Conservation equation associated with the system, we know that increasing the amplitude by a factor of 3 represents an increase in the elastic potential energy by a factor of 9. Then, the maximum kinetic energy is increased by a factor of 9.
Please help, these questions are actually irritating me
**If the simple machine reduces the amount of force, what must be increased? (Think of using an inclined plane)**
A.) the speed of the input force
B.) the distance over which the force is applied
C.) the work the simple machine performs
D.) the size of the simple machine
**What is an input force?**
A.) the force a simple machine applies to an object
B.) the force a person applies to a simple machine
C.) the applied force on a system
D.) the net force on a system
**A pair of scissors is which category of lever?**
A.) 1st class
B.) 2nd class
C.) 3rd class
D.) 4th class
Answers
1.The distance over which the force is applied
2. The force a person applies to a simple machine
3. 1st Class
A student designs an experiment where he takes a small wood car with 1.0 inch diameter wheels and rolls it down a ramp. The student measures the speed of the car when it reaches the bottom of the ramp. The student then takes a small plastic car with 1.5 inch diameter wheels and repeats the process. Finally, the student takes a small metal car with 2.0 inch diameter wheels and repeats the process a final time. The student notes that the metal car had the fastest speed at the bottom of the ramp
Answers
suppose a hiker is on a mountain ridge 1200 meters above sea level. approximately what air pressure will she experience
Answers
Assuming standard atmospheric conditions, the air pressure at sea level is approximately 1013.25 hectopascals (hPa) or 1 atmosphere (atm). The air pressure decreases with altitude following the barometric formula, which states that pressure decreases by about 1 hPa for every 8 meters of ascent.
Using this formula, we can estimate the air pressure at 1200 meters above sea level as follows:
1200 m / 8 m per hPa = 150 hPa
Therefore, the hiker on a mountain ridge 1200 meters above sea level would experience an air pressure of approximately 863.25 hPa (1013.25 hPa - 150 hPa) or about 0.85 atm.
A ceramic capacitor has an effective plate area of 4 cm2 separated by 0.1 mm of ceramic of relative
permittivity 100.
a) Calculate the capacitance of the capacitor in picofarads.
b) If the capacitor in part (a) is given a charge of 1.2 μC what will be the pd between the plates?
Answers
a) The capacitance of the ceramic capacitor is approximately 354.16 pF.
b) The potential difference between the plates of the capacitor will be approximately 3.39 x 10^6 volts.
To calculate the capacitance of the ceramic capacitor, we can use the formula:
C = (ε₀ * εᵣ * A) / d
where:
C is the capacitance,
ε₀ is the vacuum permittivity (approximately 8.854 x 10^(-12) F/m),
εᵣ is the relative permittivity of the ceramic (given as 100),
A is the effective plate area (given as 4 cm², which is equal to 4 x 10^(-4) m²),
d is the separation between the plates (given as 0.1 mm, which is equal to 0.1 x 10^(-3) m).
Let's calculate the capacitance in picofarads (pF):
a) Calculation of capacitance (C):
C = (ε₀ * εᵣ * A) / d
= (8.854 x 10^(-12) F/m * 100 * 4 x 10^(-4) m²) / (0.1 x 10^(-3) m)
= (8.854 x 100 x 4) / 0.1
= 354.16 pF
Therefore, the capacitance of the ceramic capacitor is approximately 354.16 pF.
b) To find the potential difference (PD) between the plates when the capacitor is given a charge of 1.2 μC (microcoulombs), we can use the formula:
PD = Q / C
where:
PD is the potential difference,
Q is the charge (given as 1.2 μC, which is equal to 1.2 x 10^(-6) C),
C is the capacitance (calculated in part a) as 354.16 pF, which is equal to 354.16 x 10^(-12) F).
Let's calculate the potential difference (PD):
b) Calculation of potential difference (PD):
PD = Q / C
= (1.2 x 10^(-6) C) / (354.16 x 10^(-12) F)
= 3.39 x 10^6 V
Therefore, the potential difference between the plates of the capacitor will be approximately 3.39 x 10^6 volts.
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Assume that two cars have the same kinetic energy, but that the red car has twice the speed of the blue car. We then know that the red car has ____ mass as the blue car.
Answers
Answer:
equal
Explanation:
Multiple part question Here are the needed details:Five rotations took 5.15 seconds 1 rotation took 1.07s Distance from shoulder to elbow is 29 cm distance from shoulder to middle of the hand is 57cm.7. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the elbow?B. What was the average centripetal acceleration (m/s squared) of the elbow?C. How do these compare to the hand? Why are they the same and or/ different?
Answers
Given data
One rotation means 360 degree
The angular speed for five rotations is
\(\omega_1=\frac{5\times2\pi}{5.15}=6.09\text{ rad/s}\)The angular speed for one rotation is
\(\omega_2=\frac{1\times2\pi^{}}{1.07}=5.86\text{ rad/s}\)(A)
The average angular acceleration is zero here as there is no change in the angular velocity. But, if you consider that first case (A) where the angular velocity is 6.09 rad/s and the second case where the angular velocity is 5.86 rad/s. You can say that there is a change in angular velocity but there is no particular time interval. Hence, you can not calculate the average angular acceleration. For individual cases, there will be no average acceleration.
But, there will be centripetal acceleration which will act towards the center of the circle.
(B)
The length of the elbow is given as
\(r=29\text{ cm=0.29 m}\)The centripetal acceleration of the elbow is calculated as
\(\begin{gathered} a_e=\omega^2_2r \\ =(5.86)^2(0.29) \\ =9.99m/s^2 \end{gathered}\)(C)
The length of the hand is given as,
\(\begin{gathered} L=2\times57\operatorname{cm} \\ =114\text{ cm} \\ =1.14\text{ m} \end{gathered}\)Therefore, the centripetal acceleration on the hand is calculated as
\(\begin{gathered} a_h=\omega^2L \\ =(5.86)^2(1.14) \\ =39.14m/s^2 \end{gathered}\)Thus, the centripetal acceleration for the hand is different from the elbow
(b) Calculate the force required to topple a
person of mass 70 kg, standing with his feet
spread 0.9 m apart as shown in figure. Assume
the person does not slide and the weight of
the person is equally distributed on both feet.
Answers
True or False: pH is a measure of how acidic or basic a solution is.
Answers
Answer:
Explanation:
True
There are very, very few times that scientists make estimations because they don’t have enough time or data to test all the possible answers to the question they have. TRUE OR FALSE
Answers
Answer:
FALSE
Explanation:
The Big Bang Theory is a hypothesis that states that all matter
and energy were at one time compressed into a small volume
and then
A
expanded outward
B.
imploded
C
stayed compressed
D
None of the above
Answers
Answer:
A expanded outward let me know if I'm right or wrong
One way to cool a gas is to let it expand. When a certain gas under a pressure of 5.00 x 10^6 Pa at 25.°C is allowed to expand to 3 times its original
volume, its final pressure is 1.07 x 10^6 Pa. What is its final temperature:
Answers
Answer:
The final temperature is T2= 5.35°C
Explanation:
Apply the Gay-lussacs's law we have
\(\frac{P1}{T1} = \frac{P2}{T2}\)
P1, initial pressure= 5.00 x 10^6 Pa
T1, initiation temperature= 25.°C
P2, final pressure= 1.07 x 10^6 Pa
T2, final temperature= ?
\(\frac{5.00 * 10^6}{25} =\frac{1.07 *10^6}{T2} \\\)
Cross multiplying and making T2 subject of formula we have
\(T2 =\frac{1.07 *10^6*25}{5.00 * 10^6} \\\\T2= \frac{26.75}{5} \\T2= 5.35\)
T2= 5.35°C
1) A race car travels east with a forward velocity of 40 m/s and slows at a constant rate
to 16 m/s over 9 S.
Answers
\(\\ \sf\longmapsto v=u+at\)
\(\\ \sf\longmapsto v=40+(16)(9)\)
\(\\ \sf\longmapsto v=40+144\)
\(\\ \sf\longmapsto v=184m/s\)
Explanation:
v=u+atv=40+16×9v=40+144v=184 m/s
PLEASE HELP AND SHOW WORK,THANK YOU!!
4) Suppose that two identical
mass planets are sitting
million miles apart. At that
distance the planets have a
gravitational force of 1,000,000 N.
If the planets are moved
to two million miles apart, what
is the new gravitational force
between them?
Answers
The new gravitational force between the two planets, when they are moved to two million miles apart, is 250,000 N
The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, which states that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Given:
Initial distance between the planets = 1 million miles
Initial gravitational force = 1,000,000 N
Final distance between the planets = 2 million miles
To determine the new gravitational force, we need to compare the ratios of the distances and apply the inverse square law.
Let's denote the initial distance as d1, the initial gravitational force as F1, the final distance as d2, and the unknown final gravitational force as F2.
According to the inverse square law, the ratio of the gravitational forces is the square of the ratio of the distances:
(F2/F1) = (d1/d2)²
Substituting the given values:
(F2/1,000,000 N) = (1 million miles / 2 million miles)²
Simplifying:
(F2/1,000,000 N) = (1/2)²
(F2/1,000,000 N) = 1/4
F2 = (1/4) * 1,000,000 N
F2 = 250,000 N
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Ideal mechanical advantage is equal to the displacement of the effort force divided by the displacement of the load.
True
False
Answers
Answer: False
Explanation:
Took the test.
Hide question 9 feedback
Rationale:
Ideal mechanical advantage is the ratio of input to output distance moved.
QUARTER 1 MC
DIRECTION: Read each question carefully and encircle
paper. (Bilugan and titik ng tamang sagot, maaring sula
MODULE 1: LAW OF ACCELERATION
1. If you want to accelerate an object with heavier
mass, what are you going to do?
a. apply a little force
b. apply a force unequal to the object's mass
C. apply a greater force
d. apply a little but faster force.
Answers
Answer:
i think its A
Explanation:
what's the name of first planet closest to the Sun?
Answers
Answer:
Mercury
Explanation:
It is the smallest planet in our solar system and closest to the Sun
Mars is the closet to the sun
Can someone explain how to do the algebra for this question? I know everything else, I just don’t know how to rearrange the question to solve for v.
Answers
Answer:
Refer to the step-by-step Explanation.
Step-by-step Explanation:
Simplify the equation with given substitutions,
Given Equation:
\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)
Given Substitutions:
\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)
Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)
Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the left-hand side of the equation:
\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
Simplifying the third term.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)
"R²'s" cancel, we are left with:
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)
We have like terms, combine them.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)
Each term has an "m" in common, factor it out.
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)
Now we have the following equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the right-hand side of the equation:
\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)
\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)
Now we have the equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
\(\hrulefill\)
Now solving the equation for the variable "v":
\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
Dividing each side by "m," this will cancel the "m" variable on each side.
\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)
Subtract the term "gh" from either side of the equation.
\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)
Multiply each side of the equation by "10/7."
\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)
Now squaring both sides.
\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)
Thus, the simplified equation above matches the simplified equation that was given.
1 gram is 0.035 ounces.How many ounces is 200 grams
Answers
Answer:
7.05479
Explanation:
Answer:
200 grams: ounces:7.05479239 that's the answer
Explanation:
The stool is 120 cm tall. How tall is it in inches? Note: there are 2.54 cm in 1 inch
Answers
Work: 120 / 2.54
Answer:
47.24 in
Explanation:
120 / 2.54 = 47.24
In the figure, a red car and a green car, identical except for the colour, move toward each other in adjacent lanes and parallel to the xâaxis. At time t=0, the red car is at x r =0, and the green car is at x g =220m. If the red car has a constant velocity of 20km/h, the cars pass each other at x=44.5m, and if it has a constant velocity of 40km/h, they pass each other at x=76.6m. What is the constant acceleration of the green car?
Answers
The constant acceleration of the green car which passes red color car is 2.81 m/s^2.
In this problem, the initial positions of the red and green cars are xr = 0 and xg = 220 m, respectively, and their initial velocities are vr = 20 km/h and vg = 0. When the red car has a velocity of 20 km/h, the cars pass each other at x = 44.5 m, and when the red car has a velocity of 40 km/h, they pass each other at x = 76.6 m.
We can use these two equations to solve for the acceleration of the green car. First, we convert the velocities from kilometers per hour to meters per second by dividing by 3.6. This gives us vr = 5.56 m/s and vg = 0 m/s.
Then, we can substitute these values into the equation for position and solve for the acceleration. When the red car has a velocity of 20 km/h, the equation becomes:
x = 0 + 5.56t + 1/2at^2
x = 44.5
Solving for a, we find that the acceleration of the green car is a = 2.61 m/s^2.
When the red car has a velocity of 40 km/h, the equation becomes:
x = 220 + 11.11t + 1/2at^2
x = 76.6
Solving for a, we find that the acceleration of the green car is a = 2.61 m/s^2.
When the red car has a velocity of 40 km/h, the equation becomes:
x = 220 + 11.11t + 1/2at^2
x = 76.6
Solving for a, we find that the acceleration of the green car is a = 3.02 m/s^2.
Since the acceleration is the same in both cases, we can take the average of these two values to find the constant acceleration of the green car. The average of 2.61 m/s^2 and 3.02 m/s^2 is 2.81 m/s^2.
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Two closely spaced circular disks form a parallel-plate capacitor. Transferring 1,675,802,144.45 electrons from one disk to the other causes the electric field strength to be 926,331.13 N/C. What are the diameters, in mm, of the disks?
Answers
The electric field between the plates of a capacitor (assuming it is closely spaced) is given by:
\(\vec{E}=\frac{Q}{\epsilon_0A}\)We can replace our values, and we'll get the following:
\(926331.13=\frac{1675802144.45*1.6*10^{-19}}{8.8541878*10^{-12}A}\)By isolating the are, we get:
\(A=\frac{1675802144.45*1.6*10^{-19}}{8.8541878*10^{-12}*926331.13}=3.269*10^{-5}m^2\)Now, if we replace it on the area of a circle:
\(3.269*10^{-5}=\pi r^2\)Our value of r is:
\(r=\sqrt[\placeholder{⬚}]{\frac{3.269*10^{-5}}{\pi}}=3.2258mm\)Then, our final answer is d=6.4515mm
What is the mass, in kg, of a 136 pound gymnast on Earth?
Answers
Answer:
61.6886 kg
Explanation:
Answer:
61 kg
Explanation:
there ya go hope this was useful
What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each plate has a charge density of magnitude 5.0 pC/m^2?1.02 millivolts1.43 millivolts2.26 millivolts3.34 millivolts4.43 millivolts
Answers
In order to calculate the potential difference between the plates, we can use the following formula:
\(\begin{gathered} Q=\frac{\epsilon_0\cdot A}{d}\cdot\Delta V \\ \sigma\text{ (charge density)}=\frac{Q}{A} \end{gathered}\)Using d = 4 * 10^-3 m, sigma = σ * 10^-12 and ε0 = 8.85 * 10^-12, we have:
\(\begin{gathered} \Delta V=\frac{Q\cdot d}{\epsilon_0\cdot A} \\ \Delta V=\frac{\sigma\cdot d}{\epsilon_0} \\ \Delta V=\frac{5\cdot10^{-12}\cdot4\cdot10^{-3}}{8.85\cdot10^{-12}} \\ \Delta V=\frac{20\cdot10^{-3}}{8.85} \\ \Delta V=2.26\text{ mV} \end{gathered}\)Therefore the correct option is the third one.
A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.855 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.359 and he exerts a constant horizontal force of 289 N on the crate.
A man pushes a crate labeled m, which moves with a velocity vector v to the right, on a horizontal surface. The horizontal surface is textured from the right edge of the crate to a horizontal distance ℓ from the right edge of the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude
N
direction
Opposite Direction or Same Direction
(b) Find the net work done on the crate while it is on the rough surface.
J
(c) Find the speed of the crate when it reaches the end of the rough surface.
m/s
Answers
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 34.7 N in opposite direction.
(b) The net work done on the crate while it is on the rough surface is -22.6 J.
(c) The speed of the crate when it reaches the end is 0.5 m/s.
What is the net force on the crate while it is on the rough surface?
The magnitude and direction of the net force on the crate while it is on the rough surface is calculated as follows;
F (net) = F - Ff
where;
F is the applied forceFf is the frictional force on the crateF (net) = F - μmg
where;
μ is the coefficient of frictionm is massg is gravityF (net) = 289 N - (0.359 x 92 X 9.8)
F (net) = -34.7 N
The negative sign indicates opposite direction to the applied force.
The net work done on the crate while it is on the rough surface is calculated as follows;
W = F(net) x L
where;
L is the distance travelled by the crateW = -34.7 x 0.65
W = -22.6 J
The speed of the crate when it reaches the end is calculated as follows;
acceleration of the crate = F(net) / m
a = -34.7 N / 92 kg
a = -0.377 m/s²
v² = u² + 2aL
v² = ( 0.855)² + ( 2 x -0.377 x 0.65)
v² = ( 0.855)² - ( 2 x 0.377 x 0.65)
v² = 0.24
v = √ 0.24
v = 0.5 m/s
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Do you think a bicycle helps you develop more power?
Answers
Yes, it is possible that a bicycle can help you develop more power. Riding a bicycle involves using your muscles to pedal the bike, which can help to improve your muscle strength and endurance. As you ride the bike, your muscles will contract and relax repeatedly, which can help to build strength and power.
Additionally, biking can help to improve your cardiovascular fitness. As you ride, your heart will have to work harder to pump oxygen-rich blood to your muscles, which can help to improve the function of your heart and lungs. This can also help to increase your power and endurance, as your muscles will be able to work harder and for longer periods of time.
Overall, biking can be a great way to develop more power and improve your overall fitness. It can help to build strength and endurance in your muscles, and it can also improve the function of your heart and lungs.