Answer:
3.05×10⁵ Nm²C⁻¹
Explanation:
According to Gauss' law,
∅' = q/e₀............... Equation 1
Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space
From the question,
Given: q = 2.7 μC = 2.7×10⁻⁶ C,
Constant: e₀ = 8.85×10⁻¹² C²/N.m²
Substituting these values into equation 1
∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)
∅' = 3.05×10⁵ Nm²C⁻¹
190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?
Answer:
W = 4.75 KW
Explanation:
First, we will calculate the heat to be removed:
Q = (No. of students)(Metabolic Power of Each Student)
Q = (190)(125 W)
Q = 23750 W = 23.75 KW
Now the formula of COP is:
[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]
W = 4.75 KW
MCQ
................
Answer:
I think it would be (-7 C )..
In a
DC source, which has more cuwent?
(i)R L Circuit
(ii)RC Circuit (series)
(iii)LC Cirenit (series)
(iv)RLC Circuit (series)
Answer:
Answer is LC Cirenit (seres)
How do you find the product of gamma decay?
Answer:
The mass and atomic numbers don't change
Explanation:
An excited atom relaxes to the ground state emitting a photon...called a gamma ray.
The answer is that the mass and atomic numbers don't change.
In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.
To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.
During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.
The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).
For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.
It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).
Thus, the product nucleus remains unchanged in terms of atomic number and mass number.
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Why are objects measured?
In order to find out how long/wide/heavy/high/dense/deep/ massive/voluminous/reflective/opaque/ tansparent/warm/cold/hard/soft/ malleable/flexible/rigid/radioactive/old/ valuable/symmetrical/flat/regular/ irregular they are.
In a way that you can easily and conveniently describe to other people.
A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s
Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s
Answer:
1.43 s
Explanation:
Applying,
a = (v-u)/t........... Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (v-u)/a........... Equation 2
From the question,
Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²
Substitute into equation 2
t = (2-0)/1.4
t = 1.43 s
Investigators measure the size of fog droplets using the diffraction of light. A camera records the diffraction pattern on a screen as the droplets pass in front of a laser, and a measurement of the size of the central maximum gives the droplet size. In one test, a 690 nm laser creates a pattern on a screen 30 cm from the droplets. If the central maximum of the pattern is 0.24 cm in diameter, how large is the droplet?
Answer:
the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm
Explanation:
Given the data in the question;
Diameter of bright central maxima;
⇒ 2 × ( 1.22 × (λD/d) ) ⇒ 2.44( λD/d )
where D is the distance from the the droplet to the screen ( 30 cm )
d is the diameter of the droplet
λ is the wavelength of light ( 690 nm = 690 × 10⁻⁷ cm )
since the central maximum of the pattern is 0.24 cm in diameter,
we substitute
0.24 cm = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / d )
solve for d
d = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / 0.24 cm
d = 0.0050508 cm² / 0.24 cm
d = 0.021045 cm or 2.1 × 10⁻² cm
Therefore, the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm
why clinical thermometer cannot be used to measure the boiling point of water
Answer:
: No, a clinical thermometer cannot be used to measure the temperature of boiling water because it has a small range and might break due to extreme heat. ... The temperature is around 100 degrees Celsius.
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Explanation:
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10 A turning pork creates sound cares
with
Frequency of 170Hz: To the
speed of sound in is in 340mls
calculate the wave
wave length
of
in air is
the sound wales.
Answer:
2m
Explanation:
wavelength=speed/frequency
=340/170
=2m
Put the following energy sublevels in order from least to greatest energy
A None of these
BIS. 25. 20, 35, 38, 34, 45, 46, 4d. 48
Cisas is4s, 20, 30, 40, 30, 40, 4f
D. is 25, 20, 35, 3p. 45, 3d, 4p, 40, 48
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Answer:
sup qwertyasdfghjk
Explanation:
A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.
During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a tube 4 mm in diameter and 20 cm long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.a. What is the flow speed in the urethra?b. If we assume that the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder (a reasonable approximation), what bladder pressure would be necessary to produce this flow? (In fact, there are additional factors that require additional pressure; the actual pressure is higher than this.)
Answer:
Explanation:
Given:
volume of urine discharged, [tex]V=400~mL=0.4~L=4\times 10^{-4}~m^3[/tex]
time taken for the discharge, [tex]t=30~s[/tex]
diameter of cylindrical urethra, [tex]d=4\times10^{-3}~m[/tex]
length of cylindrical urethra, [tex]l=0.2~m[/tex]
density of urine, [tex]\rho=1000~kg/m^3[/tex]
a)
we have volume flow rate Q:
[tex]Q=A.v[/tex] & [tex]Q=\frac{V}{t}[/tex]
where:
[tex]A=[/tex] cross-sectional area of urethra
[tex]v=[/tex] velocity of flow
[tex]A.v=\frac{V}{t}[/tex]
[tex]\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}[/tex]
[tex]v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}[/tex]
[tex]v=1.06~m/s[/tex]
b)
The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:
[tex]P=\rho.g.l[/tex]
[tex]P=1000\times 9.8\times 0.2[/tex]
[tex]P=1960~Pa[/tex]
If an object of a constant mass experiences a constant net force, it will have a constant what?
Explanation:
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If an object of a constant mass experiences a constant net force, it will have a constant acceleration.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
According to Newton's second law of motion:
Applied force = mass × acceleration.
Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.
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why kg is a fundamental unit?
This above answer helps a lot.
The value of mass remains constant but weight changes place to place why
Explanation:
No matter where you are in the universe, your mass is always the same: mass is a measure of the amount of matter which makes up an object. Weight, however, changes because it is a measure of the force between an object and body on which an object resides (whether that body is the Earth, the Moon, Mars, et cetera).
Explanation:
Hence, weight of a body will change from one place to another place because the value of g is different in different places. For example, the value of g on moon is 1/6 times of the value of g on earth. As mass is independent of g , so it will not change from place to place.
A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how much work has she done
Answer:
W = 343.2 J
Explanation:
Given that,
Mass of bale of hay = 26 kg
Horizontal force exerted = 88 N
Distance moved, d = 3.9 m
Work done, W = Fd
Put all the values,
W = 88 N × 3.9 m
= 343.2 J
So, the work done is 343.2 J.
A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the container the pressure is 116 kPa.
(a) What is the depth of the fluid?
(b) Find the pressure at the bottom of the container after an additional 2.05 X 10^-3 m^3 of this fluid is added to the container. Assume that no fluid spills out of the container.
Two substances, M and N, have specific heats c and 2c. if heats Q and 4Q are supɔlied to Mand N, respectively, their changes in temperature become equal. If substance M has mass m, find the mass of substance N in terms of m
Answer:
If the mass of B is m and the temperature change is the same, the mass of B will be 2m.
Explanation:
Q = mcT
T = mc/Q
M = 4Q/2cT........... (1)
T = Q/mc
Plug this in equation 1.
M = 4Q/(2c × Q/mc) = 4Q ÷ 2Q/m = 4Q × m/2Q = 2m
Characteristics or properties of matter or energy that can be measured
Answer:
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Explanation:
If the length of the rod is 2.65 m, and the mass of the bob and the rod are both 1.4 kg, what is the period of this pendulum
Answer:
T = 5.66 s
Explanation:
The system formed by the bar plus ball forms a physical pendulum
w = [tex]\sqrt{mgd/I}[/tex]
the moment of inertia of a rod held at one end is
I = [tex]\frac{1}{3}[/tex] m L²
we substitute
w = [tex]\sqrt{\frac{d \ d}{ 3 L^2 } }[/tex]
in this case the turning distance and the length of the rod are equal
d = L
w = [tex]\sqrt{\frac{g}{3L} }[/tex]
angular velocity and period are related
w = 2π / T
2π / T = [tex]\sqrt{\frac{g}{3L} }[/tex]
T = 2π [tex]\sqrt{3L/g}[/tex]
let's calculate
T = 2π [tex]\sqrt{3 \ 2.65 / 9.8}[/tex]
T = 5.66 s
An object with mass m is located halfway between an object of mass M and an object of mass 3M that are separated by a distance d. What is the magnitude of the force on the object with mass m?A) 8GMm/d^2B) GMm/(4d^2)C) 4GMm/d^2D) GMm/(2d^2)E) 3GMm/2d^2
Answer:
A) 8GMm/d^2
Explanation:
We are given that
[tex]m_1=M[/tex]
[tex]m_2=3M[/tex]
[tex]m_3=m[/tex]
Distance between m1 and m2=d
Distance of object of mass m from m1 and m2=d/2
Gravitational force formula
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
Using the formula
Force acting between m and M is given by
[tex]F_1=\frac{GmM}{d^2/4}[/tex]
Force acting between m and 3M is given by
[tex]F_2=\frac{Gm(3M)}{d^2/4}[/tex]
Now, net force acting on object of mass is given by
[tex]F=F_2-F_1[/tex]
[tex]F=\frac{Gm(3M)}{d^2/4}-\frac{GmM}{d^2/4}[/tex]
[tex]F=\frac{12GmM}{d^2}-\frac{4GmM}{d^2}[/tex]
[tex]F=\frac{12GmM-4GmM}{d^2}[/tex]
[tex]F=\frac{8GmM}{d^2}[/tex]
Hence, the magnitude of the force on the object with mass m=[tex]\frac{8GmM}{d^2}[/tex]
Option A is correct.
As it pulls itself up to a branch, a chimpanzee accelerates upward at 2.4 m/s2 at the instant it exerts a 260-N force downward on the branch.Find the magnitude of the force the chimpanzee exerts on the Earth.
Answer:
[tex]F=208.83N[/tex]
Explanation:
From the question we are told that:
Acceleration [tex]a=2.4m/s^2[/tex]
Force of Branch [tex]F=260N[/tex]
Generally the Newton's equation second law for Force is mathematically given by
[tex]ma=F-mg[/tex]
[tex]m=\frac{260}{2.4+9.8}[/tex]
[tex]m=21.31kg[/tex]
Therefore
[tex]F=mg[/tex]
[tex]F=(21.31)(9.8)[/tex]
[tex]F=208.83N[/tex]
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What is the linear speed (in m/s) of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, [tex]\alpha=3.3 rad/s^2[/tex]
Diameter of the wheel, d=21 cm
Radius of wheel, [tex]r=\frac{d}{2}=\frac{21}{2}[/tex] cm
Radius of wheel, [tex]r=\frac{21\times 10^{-2}}{2} m[/tex]
1m=100 cm
Magnitude of total linear acceleration, a=[tex]1.7 m/s^2[/tex]
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,[tex]a_t=\alpha r[/tex]
[tex]a_t=3.3\times \frac{21\times 10^{-2}}{2}[/tex]
[tex]a_t=34.65\times 10^{-2}m/s^2[/tex]
Radial acceleration,[tex]a_r=\frac{v^2}{r}[/tex]
We know that
[tex]a=\sqrt{a^2_t+a^2_r}[/tex]
Using the formula
[tex]1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}[/tex]
Squaring on both sides
we get
[tex]2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}[/tex]
[tex]\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}[/tex]
[tex]v^4=r^2\times 2.7699[/tex]
[tex]v^4=(10.5\times 10^{-2})^2\times 2.7699[/tex]
[tex]v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}[/tex]
[tex]v=0.418 m/s[/tex]
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
1. Compare and contrast the SI and the English systems of measurement.
Answer:The SI system is based on the number 10 as well as multiples and products of 10. This makes it much easier to use, and so it has been the accepted system in scientific and technical applications. The English system is more complicated as relationships between units of the same quantity aren't uniform.
Explanation:
Answer:
The metric system is an internationally agreed decimal system of measurement while The International System of Units (SI) is the official system of measurement in almost every country in the world
In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 s. Assume that during this race he ran in a straight line with constant acceleration a. What would be the required constant acceleration a
In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 seconds, assuming that he ran this race with constant acceleration, then the required constant acceleration would have been
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
By using the second equation of motion given by Newton,
S = ut + 1/2at²
100= 0 + 0.5*a*9.58²
a = 2.17 meters / second²
Thus,the required constant acceleration of Usain Bolt would have been 2.17 meters / second².
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Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 309 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars
Answer:
The velocity is 2661.5 m/s.
Explanation:
Radius, horizontal distance, d = 309 km
height, h = 25 km
acceleration due to gravity on moon, g =3.71 m/s^2
Let the time taken is t and the horizontal velocity is u.
horizontal distance = horizontal velocity x time
309 x 1000 = u t .... (1)
Use second equation of motion in vertical direction.
[tex]h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s[/tex]
So, put in (1)
309 x 1000 = u x 116.1
u = 2661.5 m/s
Find the refractive index of a medium
having a velocity of 1.5 x 10^8*
Explanation:
refractive index ,is the ratio of velocity of light in vacuum to the velocity of light a medium
) Efficiency of a lever is always less than hundred percent.
Yes. Because it opposes the law of friction
I hope this helps.
Explanation:
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