Answer:
Explanation:
Ignoring friction and assuming one ramp end rests on level ground.
gravity acceleration acting parallel to the ramp is gsinθ.
F = mgsinθ
mg = 250 N, sinθ = 3/25
F = 250(3/25)
F = 30 N
A 25 meter long ramp strong enough to hold a person plus a 250 N refrigerator would weigh much more than 250 N.
A double-pane glass window is 60.0 cm x 90.0 cm and has 3.00-mm window panes. If the temperature difference between inside and outside is 24.0 K, how far apart should the panes be to have a heat loss of 4.09 W? Assume there is air in the gap.
The distance between the glass to have the given heat loss is 2.54 m.
The given parameters:
dimension of the window, = 60 cm by 90 cmtemperature, T = 24 Kheat lost, Q = 4.09 Wthermal conductivity of glass, k = 0.8 W/mKThe area of the glass window is calculated as follows;
[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]
The distance between the glass is calculated as follows;
[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]
Thus, the distance between the glass to have the given heat loss is 2.54 m.
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We have seen in earlier readings how to determine the speed of a wave on a string. What will happen to the wavelength of a sinusoidal wave on a string if the tension in the string is increased (assuming we keep the frequency of the wave the same)
This question involves the concepts of tension in a string and the wavelength of the wave in a string.
The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.
The speed of a wave on a string is given by the following formula:
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where,
v = speed of wave = fλ
f = frequency of the wave
λ = wavelength of the wave
T = tension force
μ = linear mass density of the string
Therefore,
[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]
It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,
[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]
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3. What can you infer about the position of the galaxies 100 million years before this telescope photo was taken
The galaxies are in constant motion so it can be inferred that their position will not be the same as a hundred million years ago.
A galaxy is a term to refer to the set of stars, gas clouds, planets, cosmic dust, dark matter, and energy gravitationally united in a more or less defined structure in the universe.
Galaxies are in constant motion because the elements that make them up are not static at any time; The galaxies rotate around the center of the galaxy. If they were still, the gravitational attraction would cause them to immediately fall towards the center of the galaxy: it is the same that would happen to the Earth and the other planets if they stopped rotating around the Sun, they would fall towards the Sun.
According to the above, it can be inferred that the galaxies after a while have moved from the place where they were last seen.
Note: This question is incomplete because there is some missing information. However, I can answer based on my general prior knowledge.
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can someone help me answer this question? A student connected an ammeter as shown in this picture. Did the student connect the ammeter correctly? Explain
Answer:
No - It is connected in parallel instead of series
Explanation:
The answer is No, the reason behind the answer that the ammeter is connected in parallel which is wrong it must be connected in series.
What is Ammeter?An ammeter is a device for detecting electric current in amperes, either directly (DC) or alternating (AC). Due to the fact that a shunt running parallel to the meter carries the majority of the electricity at high current values, the digital multimeter can measure a wide range of current values. A circle with the capital A in it serves as the icon for an ammeter for circuit diagrams.
The moving coil of the electrodynamics ammeter rotates in the field created by the fixed coil. It measures alternating and direct current with 0.1 to 0.25 percentage points (by converting Ac power To DC power using a rectifier).
The ammeter is connected in series and, on the other hand, the voltmeter is connected in parallel.
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Some amount of ideal gas with internal energy U was heated from 100^0C to 200^0C. We can predict that internal energy after heating in terms of U is:
The internal energy after heating in terms of U is 100U.
The given parameters;
initial temperature of the gas, T₁ = 100 ⁰Cfinal temperature of the gas, T₂ = 200 ⁰CAssuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.
[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]
Thus, we can conclude that the internal energy after heating in terms of U is 100U.
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Question 2 (4 points)
Listen
A 14,000kg freight train, coasting at 8.1m/s, strikes another car, and the two move
forward together at 3.6m/s.
What is the mass of the second car?
Paragraph
B
1
U
A
+
Answer:
Explanation:
I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was
Conservation of momentum
14000(8.1) + m(0) = (14000 + m)(3.6)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
The mass of the second car is 17500 kg.
What is law of conservation of momentum?The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.
Newton's third law of motion has a direct impact on the idea of momentum conservation.
mass of the first car = 14000 kg.
Initial speed of first car = 8.1 m/s.
Final speed of the two cars = 3.6 m/s.
From law of conservation of momentum; we can write:
Total initial momentum = total final momentum
14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
Hence, the mass of the second car is 17500 kg.
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when a torque is acting on a fly wheel the angular velocity of the fly wheel changes from 10rad/sec to 25rad/sec in 5sec.what will be the magnitudes of the angular acceleration of the fly wheel?
Hello!
We can use the following angular kinematic equation to solve:
α = Δω/Δt, or (ωf-ωi)/t
Plug in the given values:
α = (25 - 10)/5 = 15/5 = 3 rad/sec²
Which light is most sensitive to the eyes?
Answer:
Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.
A camera takes a properly exposed photo at f/5.6 and 1/250 s. What shutter speed should be used if the lens is changed to f/4.0?
a.1/65 s
b.1/125 s
c.1/250 s
d.1/500 s
e.1/1000 s
The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.
The relationship between the shutter speed and the focal ratio is an inverse relationship.
Ratio of Areas = 5.6²/4²
A1/A2 = 1.96 = 2Since; T1A1 = T2A2
T2 = T1(A1/A2)T2 = 250 × 2 = 500 secondsIn essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.
Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.
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 A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.
Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?
The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
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49. A particle starts from rest at time t=0 and movies along the x axis. if the net force on is proportional to t its kinetic energy is proportional to?
Answer:
F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]
Explanation:
In pic
_________________
(hopet his helps can I pls have brainlist (crown)☺️)
How does climate change?
Answer:
Con el cielo del agua
Explanation:
espero que te ayude
A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the gas is the difference between these two masses. The gas is squeezed out of the bag to determine its volume by the displacement of water. What is the actual weight of the gas
The actual weight of the gas = apparent weight + weight.
The actual weight = [tex]W_{A}[/tex] + W
Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.
If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]
The gas is squeezed out of the bag to determine its volume by the displacement of water. Since
density = mass / volume
The density of water is 1000 kg/[tex]m^{2}[/tex]
we can get the mass of the gas by making m the subject of the formula.
W = mg
The actual weight of the gas = apparent weight + weight
That is,
The actual weight = [tex]W_{A}[/tex] + W
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Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball is thrown and travels 30 inches before bouncing, It bounces and travels 50% of the distance traveled. It continues to do this until coming to rest. What is the total distance traveled by ball? Also it's cold and dark and you have no phone nor anyone to ask. You have no ball
60 inches
because if 30=50% then 60=100% so its 60 inches
The sum of the series allows to find the result for the total distance that the ball bounces is:
total distance = 59.52 in
A series is a set of things or numbers related by a specific operation.
They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.
Let's build a table to observe the sequence.
drop height rebound
1 30 15
2 15 7.5
3 7.5 3.75
If we call the first term y₀
The first bounce can be found.
y₁ = [tex]\frac{y_o}{2}[/tex]
The second bounces.
[tex]y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}[/tex]
The third bounce.
[tex]y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}[/tex]
By observing this table we can construct a series of the form
Total distance = [tex]y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )[/tex]
The sum of the serie has a result of
sum = 127/64 = 1,984
Let's calculate
distance total = 30 1,984
Distance total = 59.52 cm
In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:
total distance = 59.52 in
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I need the answer to this question
Answer:
I don't see anything so you should repost
tính các mô men quán tính chính trung tâm của tiết diện, cho biết a = 10cm
Answer:
i dont know need points
Explanation:
1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out in front of you, so you slam on the brakes and go from from 30.0 m/s to 18.0 m/s. Luckily your date brought a stop watch and told you the whole thing took place in 10.5s. What is your acceleration and how far did you go?
Acceleration = (change in velocity ( final speed - starting speed))/ (time)
Acceleration = (18-30)/10.5
Acceleration = -12/10.5
Acceleration = -1.14 m/s^2
Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2
Distance = 252.2 meters
If you were told an atom was an ion, you would know the atom must have a?
A neutral charge
B charge
C negative charge
D positive charge
Answer:
its b it must be charge
Explanation:
A car of mass 5000 kg was initially moving at 100 km/h and stops at a distance of 55 m. Find the
magnitude of the net force (in N) acting to stop the car.
Answer:
|F| = 35 kN
Explanation:
a = F/m
100 km/hr(1000 m/km / 3600 s/hr) = 27.8 m/s
v² = u² + 2as
a = (v² - u²) / 2s
F/m = (v² - u²) / 2s
F = m(v² - u²) / 2s
F = 5000(0² - 27.8²) / 2(55)
F = - 35,072.9517...
Heat is transferred from molecules with more kinetic energy to molecules with _________kinetic energy.
Answer:
less
Explanation:
heat travels from hotter to colder object. hotter object has more kinetic energy.
An astronaut is doing thee simple pendulum experiment on a different planet to measure the acceleration due to gravity. If the length of the pendulum is 45 cm and the period of oscillations is equal to 1.428 s, what is the acceleration due to the gravity of the planet
The acceleration due to gravity on the given planet is 8.71 m/s².
The given parameters;
Length of the pendulum, L = 45 cm Period of the oscillation, T = 1.428 sThe acceleration due to gravity on the planet is calculated by applying following formula as follows;
[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2 \pi } = \sqrt{\frac{l}{g}} \\\\\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\g = \frac{4\pi^2 l}{T^2} \\\\g = \frac{4 \times (\pi)^2 \times 0.45}{1.428^2} \\\\g = 8.71 \ m/s^2[/tex]
Thus, the acceleration due to gravity on the given planet is 8.71 m/s².
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A country is deciding what to do about pollution glven off by power plants.
Which of the following is an example of how science can be used to make
this decision?
O A. Scientists own most of the power plants in the United States.
O B. Scientists can measure how much pollution is given off
O C. Scientists know what kind of power plants people want.
OD. Scientists are allowed to pass laws about pollution
Answer:
option B is the correct answer
Explanation:
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5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
Explain different layers of atmosphere and the pressure in each layer. Draw diagram
Answer:
Our atmosphere has five different layers. They are:
1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.
2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.
3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.
4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.
5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.
HOW DO U FEEL WHEN U PLAY OR WATCH BADMINTON?
Answer:
I feel exited and happy I enjoy it with my friend
Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples
The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
An applied force accelerates a 4.00 kg block to an initial velocity of 11 m/s across a rough horizontal surface, in the positive x direction. As the block reaches 11 m/s, the applied force is removed. The block then slows to 1.5 m/s at a distance of 4.00 m beyond where the applied force was removed. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.
The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is
∑ F = -f = ma = (4.00 kg) a
Assuming constant acceleration a , the acceleration applied by friction is such that
(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)
Solve for the acceleration :
a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²
Then the frictional force exerted a magnitude of
-f = (4.00 kg) (-14.8 m/s²)
f ≈ 59.4 N
and was directed opposite the block's motion.
calculate the change in gravitational potential energy when a 10kg object falls from 6m on earth
Answer:
600
Explanation:
GPE=mgh (mass*gravitational force*height)
m=10 kg
g=10
h=6m
10*10*6= 600 Joules
Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer
Answer:
thx for the points
Explanation:
no need brainliest
In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead to be 5 N/m. Since in our model each atom is connected to two springs, each half the length of the interatomic bond, the effective "interatomic spring stiffness" for an oscillator is 4*5 N/m = 20 N/m. The mass of one mole of lead is 207 grams (0.207 kilograms).
What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead?
Answer:
8.01e-22
Explanation:
The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.
Extension produced by one mole of leadThe extension produced by one mole of lead atom is calculated by applying Hooke's law;
F = kx
mg = kx
x = mg/k
x = (0.207 x 9.8) / (20)
x = 0.101 m
Energy stored in the lead blockThe Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;
E = ¹/₂kx²
E = ¹/₂ (20)(0.101)²
E = 0.102 J
Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.
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