Answer:
Q = 0.83 μC
Explanation:
Assuming that plates are much larger than the distance between them, we can think the electric field as constant and perpendicular to the outer surface of the plates.Applying Gauss' law to a rectangular surface half inside one of the plates and half outside it, since E and A are parallel each other, and E is zero on the other three faces, we can find the following expression for the electric field created by the charge on the plate:[tex]E = \frac{Q*A}{\epsilon_{0} } (1)[/tex]
Solving for Q, replacing E by the maximum electric field that doesn't cause the dielectric strength to break (3*10⁶ N/C), we get:[tex]Q_{max} = E_{max} * \pi *\frac{d^{2}}{4} *\epsilon_{0} = 3e6N/C*\pi *\frac{(0.2m)^{2} }{4} * 8.85e-12C2/Nm2 = 0.83\mu C (2)[/tex]
Gamma rays are dangerous because they
a.can breakdown molecules and cells
b. fluoresce
c. cause food to spoil
d. cause sunburn
Answer:
A.
Explanation:
A large fraction of astronomical gamma rays are screened by Earth's atmosphere. Gamma rays are ionizing radiation and are thus biologically hazardous. Due to their high penetration power, they can damage bone marrow and internal organs.
The extremely high energy of gamma rays allows them to penetrate just about anything. They can even pass through bones and teeth. This makes gamma rays very dangerous. They can destroy living cells, produce gene mutations, and cause cancer.
1. Calculate the electric field due to a single +1nC point charge at a distance of lm, 2m, and 3m
Answer:
Approximately [tex]9.0\; \rm N \cdot C^{-1}[/tex] at [tex]1\; \rm m[/tex] from this charge, pointing away from the point charge.Approximately [tex]2.2\; \rm N \cdot C^{-1}[/tex] at [tex]\rm 2\; \rm m[/tex] from this charge, pointing away from the point charge.Approximately [tex]1.0\; \rm N \cdot C^{-1}[/tex] at [tex]3\; \rm m[/tex] from this charge, pointing away from the point charge.Assumption: there is no object between this point charge and the observer.
Explanation:
The electric field of a point charge is inversely proportional to the square of the distance from that point charge.
Let [tex]k[/tex] denote Coulomb's constant ([tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-1}[/tex].) Let the magnitude of that point charge be [tex]q[/tex]. At a distance of [tex]r[/tex] from this charge, the electric field due to this charge would be:
[tex]\displaystyle E = \frac{k \cdot q}{r^{2}}[/tex].
Convert the magnitude of the point charge in this question to standard units:
[tex]q = 1\; \rm nC = 10^{-9}\; \rm C[/tex].
Apply that equation to find the magnitude of the electric field due to this point charge:
[tex]r = 1\; \rm m[/tex]:
[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(1\; \rm m)^{2}} \\ &\approx 9.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].
[tex]r = 2\; \rm m[/tex]:
[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(2\; \rm m)^{2}} \\ &\approx 2.2\; \rm N \cdot C^{-1}\end{aligned}[/tex].
[tex]r = 3\; \rm m[/tex]:
[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(3\; \rm m)^{2}} \\ &\approx 1.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].
The direction of the electric field at a point is the same as the direction of a force from this field onto a positive point charge at this point.
Because the [tex](+1\; \rm nC)[/tex] point charge here is positive, the electric field of this charge would repel other positive point charges. Hence, the electric field around this [tex](+1\; \rm nC)\![/tex] point charge at any point in the field would point away from this charge.
A copper wire of length L and cross-sectional area A has resistance R. A second copper wire at the same temperature has a length of 2L and a cross-sectional area of A. What is the resistance of the second copper wire?
A. 2R
B. 4R
C. R
D. 1/2 R
Answer: 4R
Explanation:R=pL/A
R second wire=p2L/1/2A
=4R
A 20.0 newton force is used to push a 2.00 kilogram cart a distance of 5.00 meters how much work is done on the cart
Answer:
100
w=f*s
20*5=100....
100 Joules of work is done on the cart.
What is Work done?Work done by a force is defined as the product of the displacement and the component of the applied force on the object in the direction of displacement. When we push a block with some force, the body moves with some acceleration, so it is called work done.
Work done is expressed as W=Fd and its unit is joules which can be defined as the amount of work done by a force in Newton is applied to an object, as a result of which it is displaced in meter.
For above given information,
Force= 20 N
Distance= 5 m
So, work done= 20*5= 100 Joules
Thus, 100 Joules of work is done on the cart.
Learn more about Work done, here:
https://brainly.com/question/13662169
#SPJ2
often occurs at subduction zones. Check all that apply:
O convergence
trench
upward convection currents in the mantle below it
Orifts
most volcanoes...Pacific Ring of Fire
continental plate subducts because it's less dense
Answer:
convergence and trench
Explanation:
LOL
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance
Speed (m/s)
Glass
5,200
Aluminum
5,100
Iron
4,500
Copper
3,500
Salt water
1,530
Fresh water
1,500
Mercury
1,400
Hydrogen at 0°C
1,284
Ethyl Alcohol
1,125
Helium at 0°C
965
Air at 100°C
387
Air at 0°C
331
Oxygen at 0°C
316
Answer:
iron/copper
iron faster because it in the core so iron is the answer with would be 15C because its the hottest in the core and is very quick there to being moved by the core so iron being 15C is the answer.
Explanation:
A volleyball player spiking is it kinetic or potential
Answer:
It is kinetic energy.
Explanation:
The kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate the volleyball from rest to its stated velocity. Having gained this energy from the player spiking it during its acceleration, the body maintains this kinetic energy unless its speed changes.
Can heating cooking oil you’ll be classified as producing a chemical change?
yes it is boiling creating bubbles that aren't their originally
Explanation:
When cooking oils are subjected to heat in the presence of air and water (from food), such as in deep-fat frying and sautéing (pan frying), they can undergo at least three chemical changes: 1) oxidation of the fatty acids, 2) polymerization of the fatty acids, and 3) breaking apart of the triglyceride molecules into free fatty acids and glycerol by hydrolysis (reaction with water from the food being cooked) (Choe and Min 2007)
Suppose a meteor of mass 2.50 x 1013 kg is moving at 33.0 km/s relative to the center of the Earth and strikes the Earth. Suppose the meteor creates the maximum possible decrease in the angular speed of the Earth by moving toward the west and striking a point on the equator tangentially. What is the change in the angular speed of the Earth due to this collision
Answer:
change is imperceptible
w_f = 7.272 10⁻⁵ rad / s
Explanation:
For this exercise we can use the conservation of angular momentum.
Initial. Before the crash
L₀ = I w₀
final. After the crash
L_f = I w_f + p r
where the moment is
p = mv
As the system is formed by the two bodies, the forces during the impact are internal, therefore the angular momentum is conserved
L₀ = L_f
I w₀ = I w_f + m v r
w_f = w₀ - [tex]m \frac{ v \ r }{I}[/tex]
We can approximate the Earth to a sphere, so its angular momentum is
I = 2/5 M r²
we substitute
w_f = w₀ - [tex]\frac{5 \ m \ v}{2 \ M \ r}[/tex]
We can find the angular velocity of the Earth with the duration of a spin which is the period of one day
w₀ = 2π / T
T = 24 h (3600 s / 1h) = 86 400 s
w₀ = 2π / 86400
w₀ = 7.272 10⁻⁵ rad / s
let's calculate
w_f = 7.27 10⁻⁵ - [tex]\frac{5 \ 250 \ 10^{13} \ 33.0 \ 10^{3} }{ 2 \ 5.98 \ 10^{24} \ 6.37 10^{6} }[/tex]
w_f = 7.272 10⁻⁵ - 1.0829 10⁻¹³
w_f = 7.27199999 10⁻⁵
this change is imperceptible
w_f = 7.272 10⁻⁵ rad / s
Friction produces heat because it causes molecules to move faster and
have more energy.
True or false.
Answer:
True
Explanation:
Friction occurs because no surface is perfectly smooth. Even surfaces that look smooth to the unaided eye make look rough or bumpy when viewed under a microscope. You know that friction produces heat. That’s why rubbing your hands together makes them warmer. But do you know why? Friction causes the molecules on rubbing surfaces to move faster, so they have more energy. This gives them a higher temperature, and they feel warmer. Heat from friction can be useful. It not only warms your hands.
Help me guys please
Answer:
The answer is B
Explanation:
keep it up
Observer A, who is at rest in the laboratory, is studing a particle that is moving through the laboratory at a speed of 0.624c and determines its lifetime to be 159ns a) Obsever A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory? b) Obsever B, who observes the particle to be at rest and measures its lifetime to be 124ns. According to B, how far apart are the two markers in the laboratory?
Answer:
a) the markers are 29.76 m far apart in the laboratory
b) According to B, the markers are 23.21 m far apart in the laboratory
Explanation:
Given that;
speed of particle = 0.624c = (0.624 × 3×10⁸m/s)
{ where c is the speed of light}
lifetime of the particle t = 159ns = 1.59 × 10⁻⁷
a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory
lets apply the formula distance = velocity × time
lets substitute
distance = (0.624×3×10⁸ m/s) × (1.59 × 10⁻⁷ s)
distance = 29.76 m
Therefore, the markers are 29.76 m far apart in the laboratory
b) Observer B, who observes the particle to be at rest and measures its lifetime to be 124ns. According to B, how far apart are the two markers in the laboratory?
lifetime of the particle t = 124ns = 1.24 × 10⁻⁷
distance = (0.624×3×10⁸ m/s) × (1.24 × 10⁻⁷ s)
distance = 23.21 m
Therefore, According to B, the markers are 23.21 m far apart in the laboratory
During heavy rain, a section of a mountainside measuring 1.9 km horizontally (perpendicular to the slope), 0.55 km up along the slope, and 2.4 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 1.4 km x 1.4 km and that the mass of a cubic meter of mud is 1900 kg. What is the mass of the mud sitting above a 3.5 m2 area of the valley floor
Answer:
the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg
Explanation:
Given the data in the question;
first we compute the total volume of mud
V = ( 1.9×1000)m × (0.55×1000)m × (2.4×1000)m
V = 2.508 × 10⁹ m³
same volume of mud spread uniformly over 1.4 × 1.4 km² area
Hence,
V = (1.4×1000)m × (1.4×1000)m × depth
2.508 × 10⁹ = 1400m × 1400m × depth
2.508 × 10⁹ m³ = 1960000m² × depth
depth = 2.508 × 10⁹ m³ / 1960000 m²
depth = 1279.59 m
so volume of the mud sitting above 3.5 m² area of the valley floor will be;
⇒ 3.5 m² × 1279.59 m
⇒ 4478.565 m³
so mass will be
m = 4478.565 × 1900 kg
m = 8509273.5 kg
Therefore, the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg
A long, thin metal tube of cross-sectional area 1.60 cm2 is sealed into the top of a hollow cube of edge length 0.150 m. Water fills the cube and extends upward a distance h into the tube. Find h if the water exerts a force of 1.00 103 N on the base of the cube. What is the weight of the water?
Answer:
Explanation:
Given that:
The area of the metal tube = 1.60 cm² = 1.60 × 10⁻⁴ m²
Length of cube L = 0.150 m
The exerted force by water = 1.00 × 10³ N
At the top of the cube;
The force on water is F = pgL
F = 9.8 m/s² × (1000 kg/m³) (0.150 m)(0.150 m)²
F = 33.075 N
∴
[tex]pgh = \dfrac{F}{A}[/tex]
[tex]h = \dfrac{F}{Apg}[/tex]
[tex]h = \dfrac{33.075}{(1.60 \times 10^{-4} \ m^2) \times (1000 \ kg/m^3)(9.8)}[/tex]
h = 21.09 m
The volume of the water V = L³
V = (0.150 m)³
V = 0.003375 m³
weight of the water w = p(V + Ah)
w = 1000 (0.003375 + (1.60 × 10⁻⁴ × 21.09))
w = 6.75 kg
Calculate the amount of potential difference in a capacitor of 0.9 MF , If the amount of charge is 1.4x10^-4 C .(Show all the work)
Answer:
Potential difference [tex]V=1.6\times 10^{-10}\ V[/tex]
Explanation:
Capacitance of capacitor, C = 0.9 MF
Charge, [tex]Q=1.4\times 10^{-4}\ C[/tex]
We need to find the amount of potential difference. We know that,
[tex]Q=CV[/tex]
V is potential difference
So,
[tex]V=\dfrac{Q}{C}\\\\V=\dfrac{1.4\times 10^{-4}}{0.9\times 10^6}\\\\V=1.6\times 10^{-10}\ V[/tex]
So, the required potential difference is [tex]1.6\times 10^{-10}\ V[/tex].
Review. As it passes over Grand Bahama Island, the eye of a hurricane is moving in a direction 60.08 north of west with a speed of 41.0 km/h. (a) What is the unit-vector expression for the velocity of the hurricane
Answer:
Velocity = -20.45 i + 35.54 j km/h
Explanation:
given data
speed v = 41.0 km/h
direction = 60.08 north of west
solution
first we get here Speed in west direction that is express as
Speed in west direction = v × cos(60.08) ..........................1
Speed in west direction = 41 × cos(60.08)
Speed in west direction = 20.45 km/h
now we get Speed in north direction that is
Speed in north direction = v × sin(60.08) ............................2
Speed in north direction = 41 × sin(60.08)
Speed in north direction = 35.54 km/h
so Velocity will be here
Velocity = -20.45 i + 35.54 j km/h
A Drive to the BeachMary drove from her home to the beach that is 30 mi from her house. The first 15 mi she drove at 60 mph, and the next 15 mi she drove at 30 mph. Would the trip take more, less, or the same time if she traveled the entire 30 mi at a steady 45 mph
Answer:
Explanation:
Let us calculate the average velocity .
Average velocity = total distance / total time
Time taken to drive first 15 mi
= 15 / 60 = .25 h
Time taken to drive next 15 mi
= 15 / 30 = .5 h
Total time = .25 + .5 = .75 h
Total distance = 30 mi
average speed = 30 / .75
= 40 mi / h
So average speed is 40 mph which is less than given velocity of 45 mph .
Hence if she travels at 45 mph , she will take less time .
James makes a treasure map to locate a secret item on McEachern Campus. He uses the
following displacement vectors to locate the treasure
A student wants to simplify this information
5 m East
7 m North
5 m South
7 m East
3 m South
2 m West
A. Determine the Displacement Vector Components. (x componenty component)
Show all work. 2 pts
B. Use the Pythagorean Theorem to calculate the resultant vector. Show all work. 2 pts
Answer format:
a. work & answer...
b. work and answer...
(4 Points)
Answer:
south
Explanation:
How much time will it take for a bug to travel 2 meters across the floor if it is traveling at 0.5 m/s?
Answer:
it will take 4 seconds to travel 2 meters.
Could a sea breeze occur at night? Explain why or why not.
The specific heat of water is 4,186 J/kg.'C. Approximately how much heat must
be removed from 0.500 kg of water to change its temperature from 24.0°C to
5.00°C?
Explanation:
Q= mc∆T
∆T= 5-24=- 19
Q= 0.5*4186*-19
Q= -39767 J
negative sign show heat releases
Two carts collide and have a perfectly inelastic collision. Cart1 has a mass of 500 grams and
Cart2 has a mass of 600 grams. Cart1 has an initial westward velocity of 1.6 m/s and collides
with Cart2 which was initially at rest. What is was the final velocity of the two cart system (SI
units)?
Answer: look at the file
Which activities demonstrate speed the most?
Answer:I think sprinting
Explanation:
Using a simple model of an electromagnet, describe the factors that increase the magnetic field strength of an electromagnet.(2 points)
Answer:
Factors Affecting the Strength of the Magnetic Field of an Electromagnet: Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core
Explanation:
Follow me for more
The following factors affect the strength of an electromagnet; number of turns, wire size, current and the presence of iron core.
What is electromagnet?This is a type of magnet in which the magnetic field is produced by an electric current.
Factor affecting electromagnetThe following factors affect the strength of an electromagnet;
Number of turnsWire sizeCurrentPresence of iron core.Learn more about electromagnet here: https://brainly.com/question/12555869
(b) The efficiency of the pump which operates the fountain is 70%.
Calculate the power supplied to the pump.
Answer:
Power supplied to the pump is 980000 W or 980 KW
Note: The question is incomplete. The complete question is given below:
On a day with no wind, a fountain in Switzerland propels 30000 kg of water per minute to a height of 140 metres.
a)Calculate the power used in raising the water?
b)The efficiency of the pump which operates the fountain is 70%
calculate the power supplied to the pump?
Explanation:
Efficiency = Power output/Power input * 100%
Power output = work done/ time
where work done = mass * acceleration due to gravity (g) * height
time = 1 minute = 60 seconds
mass of water = 30000 kg, height = 140 meters, g = 9.8 m/s²
Then, work done = 30000 * 140 * 9.8 = 41160000 J
Power output then = 41160000 J / 60 s
Power output = 686000 W
The power input can then be derived from the formula of efficiency
Power input = (Power output * 100)/Efficiency
Power input = (100 * 686000) / 70
Power input = 980000 W
Therefore, power supplied to the pump is 980000 W or 980 KW
A stone is dropped off of a 123 meter cliff and falls to the ground below. How long does it take for the stone to hit the ground
25.10 or 24.6m/s both answers are correct
Explanation:
firstly , divide the multiply with 2 and divide the result by acceleration due to gravity which is 9.8m/s or 10m/s
123*2=246
246/9.8 or 10
= 25.10 or 24.6m/s
Convert the arc length of 3.05 degrees into meters.
That depends on the circle's radius, which you neglected to mention.
If the radius is 1 meter, then an arc of 3.05 degrees is 5.32 centimeters long.
If the radius is 1 kilometer, then an arc of 3.05 degrees is 53.2 meters long.
Convert the arc length of 6.86 radians into cycles.
Plz convert this for me people have been putting random comments
Answer:
1.092 cycles
Explanation:
We need to convert the arc length of 6.86 radians into cycles.
We know that,
1 radian = 0.1592 cycles
To convert 6.86 radians to cycles, we use unitary method.
6.86 radian = (0.1592 × 6.86) cycles
= 1.092 cycles
Hence, there are 1.092 cycles in 6.86 radians. Hence, this is the required solution.
A pendulum Bob released from some initial height such that the speed of the bob at the bottom of the swing is 1.9m/s. What is the initial height of the bob?
Answer:
h = 18.4 cm
Explanation:
Given that,
The speed of the bob at the bottom of the swing is 1.9m/s.
We need to find the initial height of the bob. Let it is h.
We can find it using the conservation of energy i.e.
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
Where
v is speed of the bob
So,
[tex]h=\dfrac{v^2}{2g}\\\\h=\dfrac{(1.9)^2}{2\times 9.8}\\\\h= 0.184\ m[/tex]
or
h = 18.4 cm
So, the initial height of the bob is 18.4 cm.
The initial height of bob will be "18.4 cm".
Speed and height:Speed would be the scalar quantity that describes "how quickly an attribute moves." This could sometimes be defined as this same rate whereby an object travels a given distance.
According to the question,
The speed of the bob = 1.9 m/s
Let,
The initial height be "h".
By using the conservation of energy, we get
→ mgh = [tex]\frac{1}{2}[/tex]mv²
or,
→ h = [tex]\frac{v^2}{2g}[/tex]
By substituting the values, we get
= [tex]\frac{(1.9)^2}{2\times 9.8}[/tex]
= 0.184 m or,
= 18.4 cm
Thus the above answer is correct.
Find out more information about speed here:
https://brainly.com/question/13665920
Question 6
2 pts
Jenny loves puffed rice cereal. As she is pouring the cereal, two little pieces of cereal each with
equal charges of 1.85 x10-18 coulombs fall into the bowl 0.025 meters apart. Use Coulomb's Law
to find the electrostatic force between them. (Remember, Coulomb's constant is 8.99 x10
N*m²/C2)
4.9 x 10-23 N
6.2 x 10-15 N
9.0 x 10-3N
2.5 x 10-23 N
Answer:
4.9 x 10-23 N
Explanation:
I took the test