Answer:
C
Explanation:
because of the kids are doing well and staying in the middle of the day early to get a new phone number for home now deep
50 POINTS!!!!!!!
Two charged objects are positioned 5 cm away from each other. Describe the change in the force between these two objects when:
(a) the charge on one of the objects is increased
(b) the distance between the objects is increased to 10 cm
(c) the charge on both of the objects is decreased
Please answer all 3
Answer:
A. The shorter the distance the greater the force, so now the force would have increased because they haven't moved away from each other.
B. The force would decrease because they are now further apart.
C. the force wouldn't be as great as before but the force would still be high because they are not far apart from each other.
I hope this helps you( ◜‿◝ )♡
1.) Rn-222 decays from 400 grams to 6.25 grams in 240 minutes. How long is one “half-life.
Answer:
40
Explanation:
!! How much voltage is needed to generate a current of 20 Amps if a line has a resistance of 10 ohms ? How much power does the appliance from question number one give off ? If the appliance runs for 3 minutes , how much energy is used ? Please help me
Answer:
Power = 4000watts
Energy = 22.22Joules
Explanation:
Power = I²R
I is the current
R is the resistance
t is the time
Given the following
I = 20Amps
R = 10ohms
t = 3miuntes = 180secs
Substitute
P = 20²*10
P = 400*10
P = 4000Watts
Hence the amount of power used is 4000Watts
Energy used = Power/time
Energy used= 4000/180
Energy used = 22.22Joules
A microscope has an objective lens with diameter 1.04 cm. You wish to resolve an object 9.09 micrometers in size. You are using visible light of wavelength 562 nm. Using the Rayleigh criterion, what is the distance that the objective lens must be from the object to resolve it
Answer:
L = 0.1379 m = 13.79 cm
Explanation:
The Rayleigh criterion establishes that two objects are separated when the maximum of diffraction for slits coincides with the minimum of the other point, therefore the expression for the diffraction
a sin θ = m λ
the first zero occurs when m = 1
let's use trigonometry to find the angle
tan θ = y / L
as in these experiments the angles are very small
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
we substitute
a y /L = λ
In the case of circular aperture the system must be solved in polar coordinates, for which a numerical constant is introduced
a y / L = 1.22 λ
L = a y / 1.22 λ
We search the magnitudes to the SI system
a = 1.04 cm = 1.04 10⁻² m
y = 9.09 10⁻⁶ m
λ = 562 10⁻⁹ m
let's calculate
L = [tex]\frac{1.04 \ 10^{-2} \ 9.09 \ 10^{-6} }{1.22 \ 562 \ 10^{-9} }[/tex]
L = 1.379 10⁻¹ m
L = 0.1379 m = 13.79 cm
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross-sectional area in 10s is
Answer:
n = 1.56 x 10¹⁷ electrons
Explanation:
First of all, we will calculate the current passing through wire:
[tex]I = \frac{q}{t}[/tex]
where,
I = current = ?
q = charge = 9 mC = 0.009 C
t = time = 3.6 s
Therefore,
[tex]I = \frac{0.009\ C}{3.6\ s}\\\\I = 0.0025\ A = 2.5\ mA[/tex]
Now, for the same current in 10 s time the charge will be:
q = It = (0.0025 A)(10 s)
q = 0.025 C
Now, the number of electrons can be given as:
[tex]q = ne\\\\n = \frac{q}{e}\\\\[/tex]
where,
n = no. of electrons = ?
q = charge = 0.025 C
e = charge on single electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]n = \frac{0.025\ C}{1.6\ x\ 10^{-19}\ C}[/tex]
n = 1.56 x 10¹⁷ electrons
Which of the following statements correctly describes the index of refraction of a material? Select all that apply.
The index of refraction is the ratio of the speed of light in a material to the speed of light in a vacuum.
The index of refraction of a material must be greater than 1.
The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the material.
The index of refraction of a material must be less than 1.
The index of refraction of water is less than the index of refraction of air.
Answer:
Option C and D only
Explanation:
Option A is incorrect because refractive index of a material is the ratio of speed of light in vacuum to the speed of light in a any given medium
Option B is correct as the speed of light in vacuum is always greater than the speed of light in any given medium.
Option C is correct
Option D is incorrect
Option E is incorrect because the denser the medium the more is the refractive index. Water is denser than air, hence it should have more refractive index as compared to that of air.
These steps are followed when using the half-life of carbon-14 to determine
the age of an object that contains carbon. What is the correct order of these
steps?
A. Use the half-life of carbon-14 to determine the number of half-lives
that have passed.
B. Measure the ratio of parent nuclei to daughter nuclei.
C. Use the number of half-lives that have passed to determine the age
of the object.
A. A,B,C
B. A,C,B
0 0
C. B, A,C
D. C, A, B
Answer: a different one is a.b.c
Explanation: still for ape.x
The correct order to determine the age of the an object using carbon-14 is C, A, B. Thus, option D is correct.
What is half life?
The half-life time is defined as the time taken by the radioactive element to reduce one half of its initial value. It is denoted by t(1/2).
To measure the age of an object, a radioactive isotope called carbon-14 is used. The half-life of carbon-14 is 5,730 years. All the objects in the universe consumes carbon in their lifetime and hence, carbon-14 is used to measure the age of the objects.
The process of determining the age of objects using carbon-14 is called Radiocarbon dating. All living organisms consume carbon in means of food and from atmosphere and when the plant and animals dies, the radioactive carbon atoms start decaying.
When it starts decaying, by using Carbon-14 the age of an object is calculated. The age is estimated by measuring the amount of carbon-14 present in the sample and comparing this carbon with the reference Carbon-14 isotope.
The amount of carbon in preserved plants is identified by:
f(t) = 10e {₋ct}
t = time in years when the plant dies( t= 0)
c = the amount of carbon-14 remaining in preserved plants.
The steps include to find the age of an object is :
1. Use the number of half-lives that have passed to determine the age of the object.
2. Use the half-life of carbon-14 to determine the number of half-lives that have passed.
3.Measure the ratio of parent nuclei to daughter nuclei.
Hence, from these steps the age of an object is determined. Therefore the correct solution is D) C, A, B.
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Which of the following describes an electric insulator?
O A. A material that has a low resistance and prevents charges from
moving freely
B. A material that has a low resistance and allows charges to move
freely
C. A material that has a high resistance and prevents charges from
moving freely
D. A material that has a high resistance and allows charges to move
freely
The following that describes an electric insulator are :
C. A material that has a high resistance and prevents charges from moving freely.
The following that describes an electric insulator is a material that has a high resistance and prevents charges from moving freely. The conductors exhibit low resistance, which allows for the flow of current through it freely.Thus, the correct answer is C.
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A bicycle possesses 1000 units of momentum. what would be the bicycle's momentum if,
A.its velocity is doubled
B. its mass is tripled
use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5
The school is 3, 000m meters from the mall. What is the distance in kilometers?
Answer:
3,000 meters = 3 kilometers
Explanation:
NO LINKS; The graph shows the motion of a train first moving, then stopping, then traveling again at a slower speed. Calculate the average speed for the entire trip.
20 m/s
8.3 m/s
10 m/s
0 m/s
Answer: 0 m/s. that is your answer i hope this help sorry if i am wrong
Explanation:
What does the area under the curve on a velocity-versus-time graph represent? ... then slows down to travel the last 40 miles in three hours. ... 20. A bicyclist travels the first 700 m of a trip at an average speed of 8 m/s, travels the ... to complete the trip at an average speed, for the entire trip of 440 km/h. ... 125) v1= 0 m/s.
A rock is thrown by applying a force of 2.50N while doing 6.00J of work. Over what distance was the force applied?
O 0.417m
8.50m
2.40m
15.0m
Answer:
[tex]\boxed {\boxed {\sf 2.40 \ meters}}[/tex]
Explanation:
Work is the product of force and distance, so the formula is:
[tex]W= F \times d[/tex]
The work is 6.00 Joules, but
1 Joule (J) is equal to 1 Newton meter (N*m). Therefore, the work of 6.00 J is equal to 6.00 N*mThe force is 2.50 Newtons.
The known values are:
W= 6.00 N*m F= 2.50 NSubstitute the values into the formula.
[tex]6.00 \ N*m= 2.50 \ N *d[/tex]
We are solving for distance, so we must isolate the variable, d. It is being multiplied by 2.50 Newtons and the inverse of multiplication is division. Divide both sides by 2.50 N.
[tex]\frac {6.00 \ N*m}{2.50 \ N}=\frac{2.50 \ N * d}{2.50 \ N}[/tex]
[tex]\frac {6.00 \ N*m}{2.50 \ N}=d[/tex]
The units of Newtons cancel out.
[tex]\frac {6.00 m}{2.50 }=d[/tex]
[tex]2.40 \ m =d[/tex]
The force was applied to the rock over a distance of 2.40 meters and choice C is correct.
Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five seconds.
Answer:
E = 75 J
Explanation:
First, we will calculate the total power consumed by the five lamps:
[tex]Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W[/tex]
Now, the energy supply can be calculated as follows:
[tex]E = Pt[/tex]
where,
E = Energy = ?
t = time = 5 s
Therefore,
E = (15 W)(5 s)
E = 75 J
How does the intensity of a sound wave change if the distance from the
source is increased by a factor of 3?
O A. The intensity decreases by a factor of 3.
O B. The intensity decreases by a factor of 9.
O c. The intensity increases by a factor of 9.
O D. The intensity increases by a factor of 3.
Answer: C- The intensity increases by a factor 9
Explanation: The intensity of a sound wave follows an inverse square law, that means that it is inversely proportional to the square of the distance: so the new distance is the intensity will increase by a factor 9.
My sentence- I hope that helped!
Intensity of a sound wave decreases by a factor of 9, if the distance from the source is increased by a factor of 3. Hence option B is correct.
Intensity of Sound is inversely proportional to the square of the distance from the sound source. Since sound waves carry its energy though a two-dimensional or three-dimensional medium, the intensity of the sound wave decreases with increasing as second power of distance form the source.
Mathematically,
Intensity I ∝ 1/D²
If the distance from the source is increased by a factor of 3, Then
I ∝ 1/3² ∝ 1/9
Intensity ∝ 1/9
Hence option B is correct.
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A force of 15N is action on an area with pressure of 10pa. What is the area covered?
Answer:
150
Explanation:
P=F/A
10=15/A
10*15=15/A*15
150=A
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Three liquids that do not mix are poured into a cylindrical container with a diameter of 10.0 cm. The densities and volumes of the liquids are as follows.
Liquid 1: ????1 = 2.80 ✕ 103 kg/m3 and V1 = 2.00 ✕ 10−3 m3
Liquid 2: ????2 = 1.00 ✕ 103 kg/m3 and V2 = 1.50 ✕ 10−3 m3
Liquid 3: ????3 = 0.600 ✕ 103 kg/m3 and V3 = 1.00 ✕ 10−3 m3
Determine the pressure on the bottom of the container.
Answer:
P = 9622.9 Pa = 9.62 KPa
Explanation:
First, we will calculate the mass of all three liquids:
m = ρV
where,
m = mass of liquid
ρ = density of liquid
V = Volume of liquid
FOR LIQUID 1:
m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg
m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg
m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg
The total mass will be:
m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg
m = 7.7 kg
Hence, the weight of the liquids will be:
W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N
Now, we calculate the base area:
A = πr² = π(0.05 m)²
A = 7.85 x 10⁻³ m²
Now the pressure will be given as:
[tex]P = \frac{F}{A}\\\\P = \frac{75.54\ N}{7.85\ x\ 10^{-3}\ m^2}[/tex]
P = 9622.9 Pa = 9.62 KPa
18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
heno desde una altitud de 60.0 m. Si la paca de heno pesa 175 N, ¿cuál es el momentum
de la paca antes de que golpee el suelo?
Answer:
Definimos momento como el producto entre la masa y la velocidad
P = m*v
(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)
Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.
Peso = m*9.8m/s^2 = 175N
m = (175N)/(9.8m/s^2) = 17.9 kg
Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.
Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:
Vx = 36m/s
Mientras que para la velocidad vertical, usamos la conservación de la energía:
E = U + K
Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)
Entonces al principio solo hay energía potencial:
U = m*g*h
donde:
m = masa
g = aceleración gravitatoria
h = altura
Sabemos que la altura inicial es 60m, entonces la energía potencial es:
U = 175N*60m = 10,500 N
Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:
10,500N = (m/2)*v^2
De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.
√(10,500N*(2/ 17.9 kg)) = 34.25 m/s
La velocidad vertical es 34.25 m/s
Entonces el vector velocidad se podrá escribir como:
V = (36 m/s, -34.25 m/s)
Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.
Reemplazando esto en la ecuación del momento obtenemos:
P = 17.9kg*(36 m/s, -34.25 m/s)
P = (644.4 N, -613.075 N)
wegut.
c) A body weighs 1.2N on the moon and 120N on
the earth. Calculate the density of the moon,
taking acceleration of free fall as 10ms on the
earth surface and gravitational constant as
6.67 x 10Nm’kg. The radius of the moon is
2740 km.
We/Wm = ge/gm = 120N/1.2N
or
gm = ge/100 = 0.1 m/s^2
density = mass/volume = 3M/(4pir^3)
Re-arranging this equation, we get
M/r^2 = (4/3)×pi×(density)×r
From Newton's universal law of gravitation, the acceleration due to gravity on the moon gm is
gm = G(M/r^2) = G×(4/3)×pi×(density)×r
Solving for density, we get the expression
density = 3gm/(4×pi×G×r)
= 3(0.1)/(4×3.14×6.67×10^-11×2.74×10^6)
= 130.6 kg/m^3
two 0.5 kg carts, one red and one green, sit about half a meter apart on a low friction track, you push on the red one with the constant force of 4N for 0.17m and then remove your hand. the cart moves 0.33 m on the track and then strikes the green cart. what is the work done by you on the two cart system?
Answer:
The work done by you on the two cart system is 2 N-m
Explanation:
Work done is the product of force and displacement.
W = F * D
Substituting the given values we get -
W =
[tex]4 * (0.17+0.33)\\= 2[/tex]
The work done by you on the two cart system is 2 N-m
What is the answer to this problem?
Answer:
The answer is A
Explanation:
An insulator doesn't have to be a metal
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Answer:
a substance which does not readily allow the passage of heat or sound.
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11. A candle is placed in front of a plane mirror. The calculated value of m,
the lateral magnification, is positive. What does the positive sign indicate
about the image?
O The image is enlarged.
The image is on the same side of the mirror as the object.
The image distance is greater than the object distance.
The image is upright.
Answer: If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.
(I think, I was also stuck on this question for a bit)
The positive sign indicate about the image that
b) The image is on the same side of the mirror as the object.
d) The image is upright.
What is magnification ?
Magnification is a quantification of comparing the size of the image with respect to the size of the object. It gives us information about the image in terms of how large or small is the image formed.
magnification = height of image / height of object
since , height of object is always positive as it is always kept upright hence , in order to make m positive , height of image need to be positive
also magnification = -(v/u )
v = image distance
u = object distance
u is always positive hence , in order to make m positive , v needs to be negative , which implies on the same side of the mirror as the object
correct option
b) The image is on the same side of the mirror as the object.
d) The image is upright.
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PLEASE HELP WITH THIS ONE QUESTION
Which of the following must be true for a current to be induced in a wire passing through a magnetic field?
A) the magnetic field and the direction of motion must be perpendicular
B) the magnetic field and direction of motion must be parallel
Answer:
A) the magnetic field and the direction of motion must be perpendicular
For a current to be induced in a wire passing through a magnetic field, the magnetic field and the direction of motion must be perpendicular. The correct option is A.
What is magnetic field?The magnetic field is the region of space where another object experiences magnetic force and the current is induced in it.
According to the Fleming's right hand rule, the direction of motion, magnetic force and magnetic field are mutually perpendicular.
Thus, the correct option is A.
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The dwarf planet Ceres contains over 50% of the mass of the main asteroid belt.
True
False (if is why)
False
Explanation:
Called an asteroid for many years, Ceres is so much bigger and so different from its rocky neighbors that scientists classified it as a dwarf planet in 2006. Even though Ceres comprises 25 percent of the asteroid belt's total mass, tiny Pluto is still 14 times more massive.
Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground? Justify your answer.
Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = [tex]\sqrt{ \frac{k}{m} }[/tex] x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
Based on the law of conservation of energy, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
What is the energy in a compressed spring?The energy in a compressed spring is elastic potential energy given by the formula:
Ek = 1/2 Kx^2where
K is spring constant x is displacement of the springWhat is the kinetic energy of a body?The kinetic energy of a body is the energy the body the has due to it's motion.
Kinetic energy, KE, is givenby the formula below:
KE = 1/2mv^2How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground?From the law of conservation of energy, the total energy in a closed system is conserved.
Based on this law, all the energy in the compressed spring is converted to the kinetic energy of the box just before it reaches the ground.
Therefore, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
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A wave is a disturbance that carries
A. water from one place to another.
B. sound from one place to another.
C. matter from one place to another.
D. energy from one place to another.
Answer:
(D) energy from one place to another
Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (568 nm) on a double slit with a separation of 0.112 mm. The diffraction pattern shines on the classroom wall 3.5 m away. Calculate the fringe separation between the second order and central fringe.
Answer:
Y = 17.75 x 10⁻³ m = 17.75 mm
Explanation:
Using Young's Double Slit formula below:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = fringe separation = ?
λ = wavelength = 568 nm = 5.68 x 10⁻⁷ m
L = distance between slits and screen = 3.5 m
d = slit separation = 0.112 mm = 1.12 x 10⁻⁴ m
Therefore,
[tex]Y = \frac{(5.68\ x\ 10^{-7}\ m)(3.5\ m)}{1.12\ x\ 10^{-4}\ m}[/tex]
Y = 17.75 x 10⁻³ m = 17.75 mm
A car travels 8km in 7 minutes. Find the speed of the car.
Answer:
42.6083 mi/h
Explanation:
Given: A car travels 8km in 7 minutes.
To find: Find the speed of the car.
Formula: [tex]Speed = \frac{Distance}{Time}[/tex]
Solution: Since the formula for the speed of an object (which is the car) is speed = distance ÷ time, divide the distance (8km) by the time (7min)
Speed = 42.6083 miles per hour
Intensity: Radiation of a single frequency reaches the upper atmosphere of the earth with an intensity of 1350 W/m2. What is the maximum value of the electric field associated with this radiation? (c = 3.00 × 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
The right answer is "1010 V/m".
Explanation:
The given values are:
Intensity,
[tex]I=1350 \ W/m^2[/tex]
[tex]c = 3.00\times 108 \ m/s[/tex]
[tex]\mu_0= 4\pi\times 10^{-7} \ T.m/A[/tex]
Now,
The electric field's maximum value will be:
= [tex]\sqrt{2\times u\times c\times I}[/tex]
On substituting the values in the above formula, we get
= [tex]\sqrt{2\times 4\times \pi\times 10^{-7}\times 3\times 10^8\times 1350}[/tex]
= [tex]\sqrt{32400\times 3.14\times 10^{-7}\times 10^8}[/tex]
= [tex]1010 \ V/m[/tex]
The maximum value of the electric field associated with this radiation is :
1010 V/mGiven data :
Intensity ( I ) = 1350 W/m²
C = 3.00 * 10⁸ m/s
The maximum value of the electric field can be calculated using the equation below
[tex]E_{max} = \sqrt{2*u*c*I}[/tex] ----- ( 1 )
where : I = 1350 W/m², μ = 4π * 10⁻⁷, c = 3.00 * 10⁸
Insert values into equation ( 1 )
∴ [tex]E_{max}[/tex] = 1010 V/m.
Hence we can conclude that the maximum value of the electric field is 1010 V/m .
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