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.
Where do animals such as snakes and frogs, which do not make their own body heat, usually get their heat?
fuel
the sun
hot lava
friction
plssssssss answer correctly
The answer to this question is the Sun.
As im writing this, 12,826 people vote the Sun.
animals such as snakes and frogs, which do not make their own body heat. They usually get their heat from the sun. Hence option B is correct.
What is Cold-blooded animals ?A body temperature that is only slightly higher than the ambient temperature. This condition differentiates cold-blooded, or homoiothermic, animals from fishes, amphibians, reptiles, and invertebrates (birds and mammals). Due of their reliance on external warmth for metabolic activity, terrestrial cold-blooded species are restricted to locations with temperatures ranging from 5-10° to 35-40° C (41-50° to 95-104° F).
Cold-blooded creatures cannot create their own body heat, but they may control it by modifying their surroundings. Alligators and other reptiles frequently lay in the sun to warm up. They, on the other hand, cool off by swimming, going into a burrow in the earth, or moving inside the sade of a rock.
Hence option B is correct.
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How much Work is done when a 5 kg object is picked up a distance of 8 m.
Answer:
392J?
Explanation:
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.
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
PLESE HELP !!!!!!!!!
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.
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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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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d. Two point charges, q1 = +25 nC and q2 = -75 nC, are separated by a distance of 3.0 cm. Find the magnitude and direction of; i. the electric force q1 exerts on q2 [5] ii. the force that q2 exerts on q1 [4] (take k = 9.0 x 109 N.m2 /C2 )
Answer:
a) F₂₁ = 0.02 N, attracting.
b) F₁₂ = 0.02 N, attracting.
Explanation:
a)
The magnitude of the force that q₁ exerts on q₂ (F₂₁) is given by Coulomb's Law, as follows:[tex]F_{21} = k * \frac{q_{1} *q_{2}}{r_{12}^{2} } = 9e9 N.m2/C2 * \frac{(25e-9C)*(75e-9C)}{(0.03m)^{2}} = 0.02 N (1)[/tex]
Since q₁ and q₂ have opposite signs, the force between them will be always attractive, i.e., from q₂ towards q₁, along the line that joins both charges.b)
The magnitude of the force on q₁ due to q₂ can be obtained applying Newton's 3rd Law, or using (1), because all parameters are the same, so F₁₂ (in magnitude) = F₂₁ = 0.02 NAs we have already said, it must be opposite to the one found in a) so it must go from q₁ towards q₂, it is an attracting force also.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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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
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
Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? It would point up. It would point east. It would point down. It would point west.
Answer:
It would point up.
Explanation:
Since I am at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward, the north pole of the compass would also point towards the earth's geographic north magnetic pole, since all other compasses point toward there.
Since the compass is free to swivel in any direction, the compass would point up, since it is at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward.
So, the compass would point up.
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
This is one popular brand of exercise machine for a crossword puzzle
Answer:
Aerobics I think.
Explanation:
A wave has a wavelength of 1.5 meters and frequency of 125 Hz. What is the wave speed?
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 .
. 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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when an object slides over a smooth horizontal surface, how does the force of friction depend on the surface area of blocks that's are in contact with the table?
Answer: with a greater surface area, there will be a greater force of friction
Explanation:
3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
(10 Points)
0.05 kg.m/s
8000 kg.m/s
80,000 kg.m/s
20 kg.m/s
Answer:
momentum=mass×velocity
momentum =400kg×20m/s=8000kg.m/s
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
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!^^
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 :)
Which type of biological molecule would contain fats?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.
B
Explanation:
lipids contains fat
hope it helps
Score
A 4400 Kg Track Travelling at Intial speed 52m\s can be stopeed in 42 sec. By Gently Break, also the track can be stoped
in 7.6 m\s if the driver hit the wall.
a. What is the Impulse Excerted on the Vehicle?
b. Whta is the Averge force is exeted on the Track in each stops ?
Answer:
(a) J = 10560 kg-m/s (b) 251.42 N
Explanation:
Given that,
Mass, m = 4400 kg
Initial speed, u = 5.2 m/s
Final speed, v = 7.6 m/s
Time, t = 42 s
(a) Let J be the impulse exerted on the vehicle. Impuse is equal to the change in momentum such as :
J = m(v-u)
J = 4400 (7.6-5.2)
J = 10560 kg-m/s
(b) Impulse = Force × t
[tex]F=\dfrac{J}{t}\\\\F=\dfrac{10560}{42}\\F=251.42\ N[/tex]
Hence, this is the required solution.
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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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
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]
Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader
Answer:
7.55 g
Explanation:
Given that:
Heat of fusion = 333kj/kg
Heat capacity, c = 4190 j/kg /k
The Number of grams of ice that will melt can be represented as y:
Number of grams of ice that will melt * heat of fusion = specific heat capacity * temperature change
y * 333 * 10^3 J = (4190) * (6 - 0)
333000y = 25140
y = 25140 / 333000
y = 0.0754954 kg
y = 0.0754954 * 100
y = 7.549 g
Hence, Number of grams of ice that will melt = 7.55 g
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
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