Answer:
a) [tex]W=9810\: N[/tex]
b) [tex]F_{1}=251.14\: N[/tex]
c) [tex]V_{g}=0.012\: m^{3}[/tex]
Explanation:
a)
El peso del cuerpo es:
[tex]W=mg[/tex]
g es la gravedad (9.81 m/s²)
[tex]W=1000*9.81[/tex]
[tex]W=9810\: N[/tex]
b)
Usando el principio de Pascal tenemos:
[tex]P_{1}=P_{2}[/tex]
y la presion es la fuerza sobre el area.
[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}[/tex]
F(1) es la fuerza aplicada en el embolo pequeñoA(1) es el area del disco pequeñoF(2) es la fuerza aplicada en el embolo grandeA(2) es el area del disco grandeDespejando F(1):
[tex]F_{1}=F_{2}\frac{A_{1}}{A_{2}}[/tex]
el area del plato es: [tex]A=\pi R^{2}[/tex]
[tex]F_{1}=F_{2}\frac{\pi R_{1}^{2}}{\pi R_{2}^{2}}[/tex]
[tex]F_{1}=F_{2}\frac{R_{1}^{2}}{R_{2}^{2}}[/tex]
F(2) es el peso del cuerpo de 1000 kg (W)
[tex]F_{1}=9810\frac{8^{2}}{50^{2}}[/tex]
Por lo tanto, la fuerza que se debe hacer es:
[tex]F_{1}=251.14\: N[/tex]
c)
Como tenemos un sistema cerrado el volumen de agua que desciende por el embolo pequeño debe ser igual al que sube por el grande, por lo tanto:
[tex]V_{p}=V_{g}[/tex]
Vp es el volumen de agua en el émbolo pequeño
Vg es el volumen de agua en el émbolo grande
Como sabemos que son cilindros (V=πR²h)
[tex]\pi R^{2}h=V_{g}[/tex]
Entonces el volumen del émbolo mayor será:
[tex]V_{g}=\pi 0.08^{2}0.6[/tex]
[tex]V_{g}=0.012\: m^{3}[/tex]
Espero te haya sido de ayuda!
A wet towel spread out and hung outside on a day without wind dries faster than an identical wet towel left rolled up in a plastic bag.
Explain why.
Even without wind the sun help evaporate the water so it would dry faster. The other towel is in a bag so the moist and water has nowhere to go, therefor staying in the towel.
. An object 8.5 cm high is placed 28 cm from a converging lens. The focal length of the lens is 12 cm. Calculate the image distance, di. Calculate the image height, hi.
The converging lens is also called a concave lens. The height of the image formed by the lens is 2.55 cm.
Using the lens formula;
1/f = 1/u + 1/v
f = focal length of the lens
u = object distance
v = image distance
Note that the focal length of a converging lens is positive
Substituting values;
1/12 = 1/28 + 1/v
1/v = 1/12 - 1/28
v = 8.4 cm
Magnification= image height/object height = image distance/object distance
image height = ?
object height = 8.5 cm
image distance = 8.4 cm
object distance = 28 cm
So
image height/8.5 = 8.4/28
image height = 8.5 × 8.4/28
image height = 2.55 cm
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Which molecules are in put in photosynthesis
Answer:
During the process of photosynthesis, cells use carbon dioxide and energy from the Sun to make sugar molecules and oxygen. These sugar molecules are the basis for more complex molecules made by the photosynthetic cell, such as glucose.
Explanation:
yes.
Answer:
Sunlight, Carbon Dioxide, and Water
Explanation:
Technically minerals are in there too but when I learned this it was just Sunlight, Carbon Dioxide, and Water
5
Select the correct answer.
What is the current in a parallel circuit which has two resistors (17.2 ohms and
22.4 ohms) and a power source of 6.0 volts?
ОА.
0.30 amps
OB.
9.8 amps
OC.
0.61 amps
D.
1.2 amps
Reset
Next
Answer:
Current in a parallel circuit = 0.61 amps (Approx)
Explanation:
Given:
Voltage V = 6 volt
Two resistors = 17.2 , 22.4 in parallel circuit
Find:
Current in a parallel circuit
Computation:
1/R = 1/r1 + 1 / r2
1/R = 1/17.2 + 1 / 22.4
R = 9.73 ohms (Approx)
Current in a parallel circuit = V / R
Current in a parallel circuit = 6 / 9.73
Current in a parallel circuit = 0.61 amps (Approx)
This figure shows a sinusoidal wave that is traveling from left to right, in the +x-direction. Assume that it is described by a frequency of 57.1 cycles per second, or hertz (Hz).
7.60 cm4.80 cm
A sinusoidal wave lies on an unlabeled coordinate system. One of the wave's maxima lies on the vertical axis. The horizontal distance from the first maximum to the first minimum is labeled 4.80 cm and the vertical distance between a maximum and a minimum is labeled 7.60 cm.
(a)
What is the wave's amplitude (in cm)?
cm
(b)
What is the wavelength (in cm)?
cm
(c)
Calculate the wave's period (in s).
s
(d)
Compute the speed of this wave (in m/s).
m/s
Answer:
a) A = 3.80 cm, b) λ = 9.60 cm, c) T = 1.75 10⁻² s, d) v = 5.48 m / s
Explanation:
The wave is a way of transporting energy and moment without the need to transport the material. They are described by expressions of the type
x = A sin (kx - wt)
where the amplitude A is the distance from the point of zero intensity to the maximum.
Frequency is the number of times the wave oscillates per unit of time
the wavelength is the distance necessary for the wave to start repeating.
a) In the exercise it tells us that the vertical distance from a machismo to a minimum that is worth 7.60 cm
when checking the definition of amplitude is from zero to a maximum, therefore the value given is twice the amplitude
2A = 7.60
A = 3.80 cm
b) the distance between a minimum and the next maximum is 4.80 cm
Using the definition of wavelength the given value corresponds to half wavelength
λ/ 2 = 4.80
λ = 9.60 cm
c) frequency and period are related
f = 1 / T
T = 1 / f
we calculate
T = 1 / 57.1
T = 0.0175 s
T = 1.75 10⁻² s
d) the speed of the wave is related to the frequency and the wavelength
v = λ f
v = 0.0960 57.1
v = 5.48 m / s
what is the quantum and its types?
Answer:
It is the physics that explains how everything works, the nature of the particles that make up matter and the forces with which they interact.
Its types: Electromagnetism, the strong nuclear force, and the weak nuclear force.
Hope this help :)
a toy of mass 600 is whirled by a child in a horizontal circle using a string of length 2m with a linear speed of 5 m/s determine the angular velocity of the toy?
Explanation:
angular velocity = velocity/radius
= 5/2
= 2.5 rad/s
12. An organ pipe that is 1.75 m long and open at both ends produces sound of
frequency 303 Hz when resonating in its second overtone. What is the speed of
sound in the room?
295 m/s
328 m/s
354 m/s
389 m/s
401 m/s
Answer:
354 m/s
Explanation:
For the second overtune (Third harmonic) of an open pipe,
λ = 2L/3................................ Equation 1
Where L = Length of the open pipe, λ = Wave length.
Given: L = 1.75 m.
Substitute into equation 1
λ = 2(1.75)/3
λ = 1.17 m.
From the question,
V = λf.......................... Equation 2
V = speed of sound in the room, f = frequency
Given: f = 303 Hz.
Substitute into equation 2
V = 1.17(303)
V = 353.5
V ≈ 354 m/s
Hence the right answer is 354 m/s
You use 350 W of power to move a 7.0 N object 5 m.
How long did it take?
Answer:
0.1 second
Explanation:
We are given;
Power; P = 350 W
Force; F = 7 N
Distance; d = 5 m
Formula for power is;
P = workdone/time taken
Workdone = F × d
Thus;
350 = (7 × 5)/t
t = 35/350
t = 0.1 second
PLEASE HELP ME I WILL GIVE BRAINLY
Select five short rope exercises and describe how they are done.
Answer:
Jumping battle slams - just move the rope up and down
Alternating jump wave - jump and move the rope side to side
Alternating wide circles - move the rope in a circle position
Jumping jacks
Squat to sholder
Explanation:
The guy above me is correct give him Brainliest
What is a overly-simplified definition of Einstein's theory of general relativity?
Answer:
the laws of physics are the same for all non-accelerating observers
Explanation:
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light when it was hit from behind by car A, of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.30 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.65.
(a) What was the speed of car A just before the collision?
(b) If the speed limit was 35 mph, was car A speeding, and if so, by how many miles per hour was it exceeding the speed limit?
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
(a) The speed of car A just before the collision is 51.58 mph.
(b) With the given speed limit of 35 miles per hour, car A was crossing the speed limit by 16.58 mph.
What is collision?
The event when two objects strike each other from either direction, then such event is known as a collision. During the collision, the speed of colliding objects may vary according to the direction of the approach.
Given data -
The mass of car A is, mA = 1500 kg.
The mass of car B is, mB = 2100 kg.
The length of the skid mark is, d = 7.30 m.
The coefficient of kinetic friction between tires and road is, [tex]\mu = 0.65[/tex].
(a)
The combined kinetic energy of both cars is,
[tex]KE_{T}=\dfrac{1}{2} (mA+mB)v^{2}\\\\KE_{T}=\dfrac{1}{2} (1500+2100)v^{2}\\\\KE_{T}=1800v^{2}[/tex]
Applying the work-energy principle as,
Work done due to kinetic friction = Combined kinetic energy of cars
[tex]F \times d = KE_{T}\\\\(\mu \times (mA+mB)\times g) \times d = KE_{T}\\\\(0.65 \times (1500+2100)\times 9.8) \times 7.30 = 1800v^{2}\\\\v = 9.64 \;\rm m/s[/tex]
Converting into mph as,
[tex]v = 9.64 \times 2.23\\\\v = 21.49 \;\rm mph[/tex]
To know the speed of car A , we shall apply the law of conservation of momentum. Let the speed of car A before collision be v₁.
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 21.49
v₁ = 51.58 mph
Thus, we can conclude that the speed of car A just before the collision is 51.58 mph.
(b)
With the given speed limit of 35 mph, the obtained speed of car A before the collision is 51.58 mph. Clearly, car A is crossing the speed limit. And the difference is,
= 51.58 - 35 = 28.63 mph.
= 16.58 mph
Thus, we can conclude that car A was crossing the speed limit by 16.58 mph.
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Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect any change in its brightness as it moves toward or away from the earth. Instead we can use the Doppler effect to determine its relative speed. For this problem we are going to look at the spectral lines from hydrogen, specifically the one with a wavelength of 656.46 nm.
The hydrogen atoms in a star are also moving at high velocity because of the random motions caused by their high temperature. As a result, each atom is Doppler shifted a little bit differently, leading to a finite width of each spectral line, such as the 656.46-nm line we were just discussing. For a star like our sun, this leads to a finite width of the spectral lines of roughly Δλ=0.04nm.
If our instruments can only resolve to this accuracy, what is the lowest speed V, greater than 0, that we can measure a star to be moving?
Answer:
The answer is "[tex]\bold{18 \ \frac{km}{s}}[/tex]"
Explanation:
Its concern is not whether star speed is significantly lower than the light speed. Taking into consideration the relativistic tempo (small speed star)
[tex]\to \frac{\Delta \lambda}{\lambda} = \frac{v}{c}\\\\\to v = \frac{\Delta \lambda}{\lambda} \left (c \right ) \\\\[/tex]
[tex]= \left ( \frac{0.04}{656.46} \right ) (3 \times 10^8)\\\\ = 18280 \ \frac{m}{s} \approx 18 \ \frac{km}{s}[/tex]
What is the kinetic energy of a 10kg object that is moving with a speed of 60m/s.
The answer is 18000 J
I hope this helps!^^ , if you need the work to be shown please tell me, I hope you have a great day!^^
Your friend said that the star in this picture with the highest apparent magnitude must definitely have the highest absolute brightness as well.
Answer:
A white dwarf, also called a degenerate dwarf
Explanation:
sorry if im wrong im kind of du-m
Please help! Due in 5 min! I will pick brainiest! Thanks! YOU ROCK!
The resistance of an electric stove burner element is 11 ohms. What current flows through this
element when it runs off a 220 volt line?
Answer:
Current flow I = 20 ampere
Explanation:
Given:
Resistance R = 11 ohms
Voltage V = 220 volts
Find:
Current flow I
Computation:
Current flow I = V / R
Current flow I = 220 / 11
Current flow I = 20 ampere
The amount of current flow through the element is of 20 A.
Given data:
The magnitude of resistance of Electric stove is, R = 11 ohms.
The magnitude of potential difference in a line is, V' = 220 V.
Here we can simple go for Ohm's law. As per the Ohm's law, the potential difference across the element is proportional to the current flow and the resistance of the element.
The expression is,
V' = I × R
here, I is the amount of current flowing through the element.
Solving as,
220 = I × 11
I = 20 A
Thus, we can conclude that the amount of current flow through the element is of 20 A.
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What is the dependent variable of this testable question? How does the temperature of a tennis ball affect the height of its bounce?
Question 2 options:
brand of tennis balls
the age of the tennis ball
temperature of a tennis ball
height of its bounce
ILL GIVE BRANLIEST TO THE CORRECT ONE
Answer:
Height of its bounce
Explanation:
The dependent variable is always what is being measured or the data collected.
a toy of mass 600 is whirled by a child in a horizontal circle using a string of length 2m with a linear speed of 5 m/s determine the centripetal force experience by the toy?
there you go, 15N. I hope this helps
A wave has a wavelength of 1.5 meters and frequency of 125 Hz. What is the wave speed?
Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Clear button on the voltage tool. Place the first equipotential line 1 m\rm m away from the charge. It should have a value of roughly 9 V\rm V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V\rm V (e.g., one with 12 V\rm V, one with 15 V\rm V, and one with 6 V\rm V). Don�t worry about getting these exact values. You can be off by a few tenths of a volt.Which statement best describes the distribution of the equipotential lines?1-The equipotential lines are closer together in regions where the electric field is weaker.2-The equipotential lines are closer together in regions where the electric field is stronger.3-The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.
Answer:
the correct one is 2. the equipotential lines must be closer together where the field has more intensity
Explanation:
The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.
In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by
V = k q / r
also the electric field and the electuary potential are related
E = [tex]- \frac{dV}{dr}[/tex]
therefore the equipotential lines must be closer together where the field has more intensity
When checking the answers, the correct one is 2
2) How much work is required to pull a sled 15
meters if you use 30N of force?
2 people
Explanation:
This is one popular brand of exercise machine for a crossword puzzle
Answer:
Aerobics I think.
Explanation:
What happens to the energy of a rubber band when it is stretched?
The following statements are related to the force of a magnetic field on a current-carrying wire. Indicate whether each statement is true or false.
1) The magnetic force on the wire is independent of the direction of the current.
A) True
B) False
2) The force on the wire is directed perpendicular to both the wire and the magnetic field.
A) True
B) False
3) The force takes its largest value when the magnetic field is parallel to the wire.
A) True
B) False
Answer:
1) B: False
2) A: True
3) B: False
Explanation:
1) Statement is false because the force is not independent of the current but rather depends on the direction of the field and current.
2) Statement is true as per right hand thumb rule.
3) The statement is false because force takes its largest value when the magnetic field direction and electric current direction are perpendicular to each other.
a ford explorer traveled 100 miles the next day for 5 hours. What was the average speed of this vehicle?
Answer:
25 miles per hour
Explanation:
It was 20 miles per hour the next day. We don't have enough information to calculate the average speed for the whole trip.
An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at 5.0 Hz and 6.4 Hz , and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be 333 m/s .
Answer:
L = 116.6 m
Explanation:
For this exercise we approximate the tunnel as a tube with one end open and the other closed, at the open end there is a belly and at the closed end a node, therefore the resonances occur at
λ = 4L 1st harmonic
λ = 4L / 3 third harmonic
λ = 4L / 5 fifth harmonic
General term
λ = 4L / n n = 1, 3, 5,... odd
n = (2n + 1) n are all integers
They indicate that two consecutive resonant frequencies were found, the speed of the wave is related to the wavelength and its frequency
v = λ f
λ = v / f
we substitute
[tex]\frac{v}{f} = \frac{4L}{n}[/tex]
L = [tex]n \frac{ v}{4f}[/tex]
for the first resonance n = n
L = (2n + 1) [tex]\frac{v}{4f_1}[/tex]
for the second resonance n = n + 1
L = (2n + 3) [tex]\frac{v}{4f_2}[/tex]
we have two equations with two unknowns, let's solve by equating
(2n + 1) \frac{v}{4f_1}= (2n + 3) \frac{v}{4f_2}
(2n + 1) f₂ = (2n +3) f₁
2n + 1 = (2n + 3) [tex]\frac{f_1}{f_2}[/tex]
2n (1 - \frac{f_1}{f_2}) = 3 \frac{f_1}{f_2} -1
we substitute the values
2n (1- [tex]\frac{5}{6.4}[/tex]) = 3 [tex]\frac{5}{6.4}[/tex] -1
2n 0.21875 = 1.34375
n = 1.34375 / 2 0.21875
n = 3
remember that n must be an integer.
We use one of the equations to find the length of the Tunal
L = (2n + 1) \frac{v}{4f_1}
L = (2 3 + 1) [tex]\frac{333}{4 \ 5.0}[/tex]
L = 116.55 m
Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q1 and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.250 m. What is the net force on 92?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
-5.00 x 10-6
-5.00 x 10-6 C
91
92
93
0.500 m
0.250 m
q1 = -5.00 x 10-6 C
q2 = -5.00 x 10-6 C
q3 = -5.00 x 10-6 C
E1 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.5^2) = 180000 N/C to the left
E2 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.25^2) = 720000 N/C to the right
E net = 720000 - 180000 = 540000 N/C to the right
F = qE
F = (-5 x 10^6 C)(540000 N/C) = - 2.7 N
The force on q2 is 2.7 N to the left.
The net electrostatic force on the q2 is 2.7N owards left
The equation for electrostatic force is
[tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]
where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.
the force has to be calculated on a charge q2 = -5.0 ×[tex]10^{-6}[/tex] C by the charges q1= -5.0 ×[tex]10^{-6}[/tex] C and q3= -5.0 ×[tex]10^{-6}[/tex] C
distance between q1 and q2 is 0.5 m = 5×[tex]10^{-1}[/tex]m
distance between q2 and q3 is 0.25 m = 25×[tex]10^{-2}[/tex]m
force due to charge q1
[tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(-5)×(-5)×[tex]10^{-12}[/tex]/25×[tex]10^{-2}[/tex] N = +0.9N = 0.9N towards right
[tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/625×[tex]10^{-4}[/tex] N = -3.6N = 3.6N towards left
hence net force F = [tex]F_{1}+F_{2}[/tex]
= 0.9N - 3.6N = -2.7N
F = 2.7 N towards left
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The only way that heat can travel through outer space is ______
convection
radiation
conduction
none of the above
plssssssssssss answer correctly
tank contains 335 kg of water at a uniform temperature of 60oC. The tank is insulated and not heated; it neither loses nor gains heat through the walls of the tank. A valve is opened and water exits the tank at a rate of 0.5 kg/sec and a temperature of 60oC. After 10 seconds the valve is closed again . Using the assumption that water at zero degrees centigrade contains zero energy and considering only internal, how much energy left the tank through the valve during this 10 second period; report as kJ.
Answer:
Explanation:
Thermal energy or internal energy gain or loss = mass x specific heat x temperature
specific heat of water = 4.2 kJ / kg degree Celsius
mass of water lost in 10 second = rate of loss x time = .5 x 10 = 5 kg .
heat energy associated with lost water = 5 x 4.2 x ( 60 - 0 ) = 1260 kJ .
Heat energy lost = 1260 kJ .
A plane takes off at St. Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during the entire flight, and there are no head winds or tail winds. Since the earth revolves around its axis once a day, you might expect that the times for the outbound trip and the return trip differ, depending on whether the plane flies against the earth's rotation or with it. Is this expectation true or false
Answer:
In the Both time
Explanation:
A plane takes off at St.Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during ...
Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.
What is Plane?Physical quantities such as work, temperature, and distance can all be completely represented in daily life by their magnitude. The laws of arithmetic can, however, be used to explain how these physical values relate to one another.
Motion in two dimensions is another name for motion in a plane. For instance, a projectile moving in a circle. The origin, along with the two coordinate axes X and Y, will serve as the reference point for the investigation of this kind of motion.
Therefore, Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.
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