Answer:
1800 J
Explanation:
GPE= mgh
60*3*10
=1800 J
A 10kg object is 15 meters up a hill. Find its potential energy
Answer:
Explanation:
Relative to an origin at the bottom of the hill,
PE = mgh = 10(9.8)(15) = 1470 J
The diagram shows the velocity-time graph for a car travelling in a straight line along a road. Calculate the acceleration between t = 2.0 s and t = 5.0 s.
Answer:
a = Δv/Δt = (0 - 20) / (5 - 2) = -6⅔ m/s²
how far from the lens is the image of the house if the house is 16 ft from the thin convex lens with a focal length of 8 ft
This question involves the concepts of the thin lens formula, focal length, and image distance.
The image of the house is "16 ft" away from the lens.
According to the thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where,
f = focal length = 8 ft
p = object distance = 16 ft
q = image distance = ?
Therefore,
[tex]\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\[/tex]
q = 16 ft
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A car slams on its brakes creating an acceleration of -4.7 m/s^2. It comes to rest after traveling a distance of 235 m. What was its velocity before it began to accelerate?
Answer:
Explanation:
v² = u² + 2as
0² = u² + 2(-4.7)(235)
u² = 2209
u = 47 m/s
A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m
Answer:
P = 2 pi (L / g)^1/2
P2 / P1 = (8 / 2)^1/2 = 2
The period would be twice as long or 5.6 sec.
A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.
Answer:
a = 0.01m/s²
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
a = (V_f-V_0)/t
a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))
a = 0.01m/s²
A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal
acceleration if the circle has a radius of 27 m?
Explanation:
We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:
Vt = ωr with ω being the angular velocity.
We now calculate ω:
ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s
Now that we have ω we can calculate the centripetal aceleration:
a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2
Identify the type of chemical reaction:
CaCO3 -->CaO + CO2
Explanation:
decomposition reaction.....
Answer:
caco₃--cao+co₂
calcium oxide + carbon die oxide gives us calcium carbonate
this is the reaction of Acidic oxide(cao) and Basic oxide (co₂) to form salt.
22 Newton force is working on a 1,901 gram object. What is the acceleration in
meter/s^2 unit
Answer:
11.573
Explanation:
f = m*a
where f is the force in Newtons, m is the mass of the object (in kg) and a is the acceleration
so, we solve for a
a = f/m
a = 22/1.901
a = 11.573
K
Mission CG9: Weightlessness
Consider the several locations along a roller coaster
track. In which location(s) would the riders feel less
than their normal weight? Select all that apply.
Location A
Location B
Location C
a
=-10 m/s/s, dn
--2 m/s/s, up
a--6 m/s/s, dn
Location D
Location E
x=-12 m/s/s, dn
---6 m/s/s, up
The locations where the riders feel less than their normal weight are Location A, Location C and Location D.
The given parameters;
Location A, a = 10 m/s² downLocation B, a = 2 m/s² upLocation C, a = 6 m/s² downLocation D, a = 12 m/s² downLocation E, a = 6 m/s² upThe normal weight of the riders is calculated by applying Newton's second law of motion as follows;
W = mg
W = 9.8m
The apparent weight of the riders for the upward acceleration is calculated as follows;
[tex]R = m(g + a)[/tex]
The apparent weight of the riders for the downward acceleration is calculated as follows;
[tex]R = m(g - a)[/tex]
The apparent weight of the riders at location A is calculated as follows;
[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]
The apparent weight of the riders at location B is calculated as follows;
[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]
The apparent weight of the riders at location C is calculated as follows;
[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]
The apparent weight of the riders at location D is calculated as follows;
[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]
The apparent weight of the riders at location E is calculated as follows;
[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]
Thus, the locations where the riders feel less than their normal weight are;
Location ALocation CLocation D.Learn more about Newton's second law and reading of a scale here: https://brainly.com/question/11603452
For northern hemisphere observers, which celestial object would be above the horizon for the greatest
amount of time: one that is on the celestial equator, one that is 30° above the celestial equator, one that is
70° above the celestial equator, or one that is 40" below the celestial equator? Which one would be above
the horizon the greatest amount of time for southern hemisphere observers? Explain your answer.
Answer:
Explanation:
For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.
For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
What is the equator?The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.
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48.36
g.
MgSO4 to motes
Answer:120.3676
Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!
1. Wha' is the relationship between potential and kinetic energy?
As potential energy increases, kinetic energy increases.
b. As potential energy increases, kinetic energy decreases.
C. As potential energy decreases, kinetic energy decreases.
d. Potential and kinetic energy are two separate things and have no
relationship.
Answer:
B
Explanation:
The kinetic energy in an object is converted into potential energy. This makes the kinetic decrease, while the potential increases.
Why does the earth stay in orbit around the sun instead of drifting away from it into space?
A Electric force between the Sun and Earth
B Magnetic force between the Sun and Earth
Gravitational force between the Sun and Earth
Answer:
b it is b part answer i think so
A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. The momentum of the chair
a. was zero while stationary and increased when the person stood.
b. was greatest while the person sat in the chair.
c. remained the same.
d. was zero when the person got out of the chair and increased while the person sat.
Answer:
a. was zero while stationary and increased when the person stood.
Explanation:
momentum is mass times velocity.
initial velocity was zero
final velocity was NOT zero.
How much power does it take to lift 30.0 N 10.0 m high in 10.00 s?
Answer:
60w
Explanation:
The power required is 30 Watt.
Let us recall that power is defined as the rate of doing work. Hence, we can write as follows;
Power = Work done/ time taken
Now;
work done = Force × distance
Force = 30.0 N
Distance = 10.0 m
work done = 30.0 N × 10.0 m = 300 J
The power expended = 300 J/10.00 s = 30 Watt
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Shorter the vibrating part more will be the pitch. How?
Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.
This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Earth without flying off or hitting the ground.
Answer:
One force will be gravity & inertia.
Explanation:
Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth
Help me outtttt jejjejejeje
Answer:
do it got a picture
Explanation:
What is the approximate value of k when 30 = e^5k?
Answer:
Explanation:
30 = e^5k
ln30 = lne^5k
ln30 = 5k
k = ln30/5
k = 0.68023947...
round to your heart's content.
The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 69.5 Hz. At what speed are you driving
Answer:
Explanation:
ASSUMING your speed is constant
f₀ = f(v + vo)/(v + vs)
Δf = f approach - f depart
69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))
69.5 = 769(2vo/343)
vo = 15.5 m/s
The speed of car driving is 15.5 m/s as the car is parked and drive away.
What is speed?Speed is defined as a measurement of the length of time it takes for an object to travel a certain distance. You can determine an object's speed if you know how far it moves in a given amount of time. Time does not move, hence there is no concept of a speed of time. Time refers to how we move through the temporal realm. Speed is a unit of measurement for how quickly something is moving. A change in velocity results in a change in speed.
To calculate the speed we use the formula
f₀ = f (v + vo) / (v + vs)
Δf = f approach - f depart
69.5 = (769(343 + vo) / (343 + 0)) - (769(343 - vo)/(343 + 0))
69.5 = 769(2vo/343)
vo = 15.5 m/s
Thus, the speed of car driving is 15.5 m/s as the car is parked and drive away.
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If the velocity and frequency of a wave are both doubled, how does the wavelength change?
The wavelength will remain unchanged.
Explanation:
The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is
[tex]v = \lambda\nu[/tex] (1)
so if we double both the velocity and the frequency, the equation above becomes
[tex]2v = \lambda(2\nu)[/tex] (2)
Solving for the wavelength from Eqn(2), we get
[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]
We would have gotten the same result had we used Eqn(1) instead.
Answer:
the wavelength increases
Explanation:
please help 9.2.1 project in science just ned an example
Answer:
Give me what kind of example you need please so I can help you. Put it in the comments.
Explanation:
The qualitative equivalent of external validity is:
A- Credibility
B- Dependability
C- Transformability
D- Confirmability
the turns ratio for a transformer with 225 turns of wire in its primary winding and 675 turns in the secondary is: n
The ratio of the primary turns to the secondary turns is 1/3
The correct answer to the question is Option A. 1/3
From the question given above, the following data were obtained:
Primary turn (Nₚ) = 225 turnsSecondary turn (Nᵣ) = 675 turns Ratio of primary to secondary =?Ratio = Nₚ/Nᵣ
Nₚ/Nᵣ = 225 / 675
Nₚ/Nᵣ = 1/3
Therefore, the ratio of the primary turns to the secondary turns is 1/3
Complete question:
See attached photo
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What is work? A. What happens when energy is destroyed B. The number of molecules in a substance C. The total size of an object divided by its mass D. What happens when a force causes an object to move
Answer:
Which extended definition would be most helpful to add to this body paragraph?
Laughter Yoga, a new form of yoga, is becoming increasingly popular.
Laughter Yoga, which is becoming more and more popular, was created in the 1990s.
Laughter Yoga is modeled after traditional yoga but includes laughter exercises.
Laughter Yoga is a form of yoga that was created by Dr. Madan Kataria in the 1990s.
Explanation:
c
Would you please help me with this? I can't figure it out, please! I need to know what the E means!
Answer:
Without the full content of your question, I will have to GUESS at the context and assume
E = Energy
released when glucose is broken down.
which statement about metals is true?
All metals are solid at (our) living temperatures except for mercury which is a liquid. Metals are shiny for a while at least when properly finished. Metals are good conductors of both heat and electricity. This is due to the excess electrons in their valence clouds that facilitate energy transfer.What are 5 characteristics of metal?
Metals are lustrous, malleable, ductile, good conductors of heat and electricity
Silver conducts electricity better than any other metal.
Brass is an alloy made from zinc and copper.
Pure gold is too soft for many things so most gold is combined with other metals to make it stronger. ...
World time periods are often listed by the metal used.
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, how much time does it take the puck to come to rest?
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is 4.51 sec.
What is speed?The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.
Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,
force = ma = μmg, putting the value,
a = - 1.175 m/sec²
now using equation of motion
v = u + at
t = 4.51 sec
So, time taken by the puck to come to rest is 4.51 sec.
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I can learn new things but I cannot change how good I am at math. A strongly agree B. agree C disagree D. strongly disagree
Mathematics is a broad subject that anyone who constantly practices by learning from first principle and from example will grow in perfection of the subject as time goes by. Hence, I strongly disagree
During my schools days, I struggled learning mathematics, at the time it was difficult, I began practising and learning from text examples, with time I got a hang of it and my grade and confidence level in the subject increased.
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