For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.
When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.
For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.
Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.
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what is the largest star in our night sky
A passenger jet flies from one airport to another 1,233 miles away in 2.4 h. Find its average speed. = ____ m/s
Speed = (distance) / (time)
Speed = (1,233 mile) / (2.4 hour)
Speed = 513.75 mile/hour
Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)
Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)
Speed = 229.7 meter/second
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
Unpolarized light enters a polarizer with vertical polarization axis. The light that passes through passes another polarizer with transmission axis at 40 degrees to the horizontal. What is the intensity of the light after the second polarizer expressed as a fraction of the original intensity
Answer:
I = 0.2934 I₀
Explanation:
The expression that governs the transmission of polarization is
I = I₀ cos² θ
Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of
I₁ = I₀ / 2
this is the light that enters the second polarizer
I = I₁ cos² θ
we substitute
I = I₀ / 2 cos² 40
I = I₀ 0.2934
I = 0.2934 I₀
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?
Answer:
[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]
Explanation:
Hello,
In this case, since pressure is defined as the force applied over a surface:
[tex]P=\frac{F}{A}[/tex]
We can associate the force with the weight of the needle computed by using the acceleration of the gravity:
[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]
And the area of the the tip (circle) in meters:
[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]
Thus, the pressure exerted on the record turns out:
[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]
Which is truly a large value due to the tiny area on which the pressure is exerted.
Best regards.
Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
v = 1.7 m/s
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your normal weight is w= 635 N. If the elevator has an initial acceleration of magnitude 2.45 m/s2, what does the scale read? Express your answer in newtons.
Answer: 479. 425 N
Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.
It is given by N= m(g+a)
When it is accelerating downward, the scale reading is less than the true weight.
It so given by N = m(g-a)
The answer to the above questions is in the attached photo
Answer:
the scale will read 476.414 N
Explanation:
Weight = 635 N
mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)
mass m = 635 ÷ 9.81 = 64.729 kg
initial acceleration of the elevator a = 2.45 m/s^2
the force produced by the acceleration of the elevator downwards = ma
your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss
apparent weight = weight - ma
apparent weight = 635 - (64.729 x 2.45)
apparent weight = 635 - 158.586 = 476.414 N
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element
Answer:
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.Explanation:
The formula for calculating the power expended in a circuit is P = I²R... 1
i is the current (in amperes)
R is the resistance (in ohms)
If a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I
New power dissipated P₂ = (I₂)²R
P₂ = (2I)²R
P₂ = 4I²R ... 2
Dividing equation 2 by 1 will give;
P₂/P = 4I²R/I²R
P₂/P = 4
P₂ = 4P
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.
Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals
Answer:
producers make its own energy frominorganic substances.
Which describes any compound that has at least one element from group 17? Halide;noble gas; metalliod; transition metal
Answer:
Halide
Explanation:
It has at least one element from the halogen group (17)
Halide describes any compound that has at least one element from group 17, therefore the correct option is option A.
What are halides?When the elements belonging to group 17 of the periodic table form ionic compounds with other electropositive elements, then these compounds are known as halides.
These elements from group 17 are also known as halogens. Generally, these halides have very high electronegativity as they reside on the right side of the periodic table.
Generally, the valency of the halogens element involved in the halide compound is one and they form ionic compounds with the alkali and alkaline earth metals.
Thus, halides are compounds that have at least one element from group 17.
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What is the equivalent resistance between the points A and B of the network?
Explanation:
First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
what tools use cut wood
Answer:
hand saws
power saws
Circular Saw
Explanation:
that is all that i know
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 105 Pa for the atmospheric pressure and 1200 kg/m3 for the density of the sauce, find the absolute pressure in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.
Answer:
(a) P = 103064 Pa = 103.064 KPa
(b) P = 102476 Pa = 102.476 KPa
Explanation:
(a)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)
Pg = 1764 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1764 Pa
P = 103064 Pa = 103.064 KPa
(b)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)
Pg = 1176 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1176 Pa
P = 102476 Pa = 102.476 KPa
The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.
Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.
In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.
The absolute pressure in the bulb can be calculated using the following formula:
P = P₀ + ρgh
where:
P is the absolute pressure in the bulb,
P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),
ρ is the density of the sauce (1200 kg/m³),
g is the acceleration due to gravity (9.8 m/s²), and
h is the height of the sauce in the tube.
(a) When h = 0.15 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)
P ≈ 1.013 x 10⁵ Pa + 1764 Pa
P ≈ 1.031 x 10⁵ Pa
(b) When h = 0.10 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)
P ≈ 1.013 x 10⁵ Pa + 1176 Pa
P ≈ 1.025 x 10⁵ Pa
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A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
nvbnncbmkghbbbvvvvvvbvbhgggghhhhb
HELP ILL MARK BRAINLIEST PLS!!!!
A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
What is the equivalent temperature in kelvin if you have a metal at 50°F?
Answer:
The required temperature is 283 K.
Explanation:
[tex]T\:=\:\left(50-32\right)\times \frac{5}{9}+273\\\\T=283\:K[/tex]
Best Regards!
Stress is a factor that contributes to heart disease risk.true or false
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Answer:
Explanation:
F ×1 = 0.5×0.145×47×47
F = 160.15 N
An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?
Answer:
The power used is 0.96 watts.
Explanation:
Recall the formula for electric power (P) as the product of the voltage applied times the circulating current:
[tex]P=V\,\,I[/tex]
and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:
[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]
Then we can re-write the power expression as:
[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]
which in our case becomes:
[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]
When moving to a new apartment, you rent a truck and create a ramp with a 244 cm long piece of plywood. The top of the moving ramp lies on the edge of the truck bed at a height of 115 cm. You load your textbooks into a wooden box at the bottom of the ramp (the coefficient of kinetic friction between the box and ramp is = 0.2). Then you and a few friends give the box a quick push and it starts to slide up the ramp. A) What angle is made by the ramp and the ground?B) Unfortunately, after letting go, the box only tables 80cm up the ramp before it starts coming back down! What speed was the box initially traveling with just after you stopped pushing it?
Answer:
A) θ = 28.1º , B) v = 2.47 m / s
Explanation:
A) The angle of the ramp can be found using trigonometry
sin θ = y / L
Φ = sin⁻¹ y / L
θ = sin⁻¹ (115/244)
θ = 28.1º
B) For this pate we can use the relationship between work and kinetic energy
W =ΔK
where the work is
W = -fr x
the negative sign is due to the fact that the friction force closes against the movement
Lavariacion of energy cineta is
ΔEm = ½ m v² - mgh
-fr x = ½ m v² - m gh
the friction force has the equation
fr = very N
at the highest part there is no speed and we take the origin from the lowest part of the ramp
To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular
Axis y . perpendicular
N- Wy = 0
cos tea = Wy / W
Wy = W cos treaa
N = mg cos tea
we substitute
- (very mg cos tea) x = ½ m v²2 - mgh
v2 = m (gh- very g cos tea x)
let's calculate
v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)
v = RA (7.84 -1.729)
v = 2.47 m / s
Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble
Answer:
Johannes KeplerExplanation:
He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.Hope this helped,
Kavitha
On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?
Answer:
T = 2.06h
Explanation:
In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex] (1)
T: time for a complete orbit = ?
r: radius of the orbit
G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2
Mm: mass of the moon = 7.34*10^22 kg
The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:
[tex]r=R_m+160km\\\\[/tex]
Rm: radius of the moon = 1737.1 km
[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]
Then, you replace all values of the parameters in the equation (1):
[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]
In hours you obtain:
[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]
The time that the Apollo takes to complete an orbit around the moon is 2.06h
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]
In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum
Explanation:
We have
A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.
The frequency of a pendulum is equal to the no of oscillation per unit time. so,
[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]
Tim period is reciprocal of frequency. So,
[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]
So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.