Answer:
[tex]Q_{net}=4.38\,\,10^{-8}\,C[/tex]
Explanation:
We use Gauss's Law for the flux over a closed surface equal the net charge inside the closed surface divided the permitivity of space [tex]\epsilon_0=8.85\,\,10^{-12}\,\,\frac{C^2}{N*m^2}[/tex]
Therefore, by knowing the flux, we can estimate the net charge inside the cubic box with the product:
[tex]\Phi=\frac{Q_{net}}{\epsilon_0} \\Q_{net}=4950\,*\,8.85\,\,10^{-12} \,\,C\\Q_{net}=4.38\,\,10^{-8}\,C[/tex]
Which do you suppose exerts more pressure on the ground—an elephant or a lady standing on spike heels? (Which will be more likely to make dents in a linoleum floor?) Can you approximate a rough calculation for each?
Answer:
A woman with a spike heel
Explanation:
Since we're assuming here, we proceed to assuming that the mass of the elephant is 6000 kg, and the mass of the lady is 60 kg. Then,
a woman with a spike heel will exert more pressure on the ground than an elephant weighing 6000 kg. This is because, the force exerted by the woman is a 588 N and she has a, say 1 cm² spike heel. This puts half her weight on each foot, and if evenly distributed to half on her heel and half on her sole, the pressure being exerted by each heel is 147 N/ 2cm² = 147 N/cm². On the other hand, if a 58,860 N elephant with an 1000 cm² feet is exerting 1/4 its weight on each foo. Then it has 14715 N/1000 cm² = 14.7 N/cm²; which is exactly 10 times lesser pressure than what the woman is exerting. So, the woman with the hell exerts more pressure, than the elephant will.
Calculate the momentum of a 1150 kg car traveling with a velocity of 24.5 m/s.
Answer:
P = 28175 Ns
Explanation:
Momentum is defined as the quantity of motion contained in a body. Mathematically, momentum is defined as the product of the mass and velocity of an object. So we will use the following formula to find the momentum of the car:
[tex]P = mv\\[/tex]
where,
P = momentum of the car = ?
m = mass of the car = 1150 kg
v = speed of the car = 24.5 m/s
Therefore,
[tex]P = (1150\ kg)(24.5\ m/s)[/tex]
P = 28175 Ns
Sound waves can be modeled by the equation of the form y=20sin(3t+theta). Determine what type of interference results when sound waves modeled by the equations y=20sin(3t+90) and y=20sin(3t+270) are combined.
Answer:
go to the link quizzlet it will give you tha answer
Explanation:
Match each word to its correct meaning.
1. humus
the thin upper layer of Earth's crust that supports plant life
2. inorganic
a vertical section of soil that shows the horizon and parent material
3. organic
the layer of loose rock on the surface of the earth; also called mantle rock
4. parent material
dark colored organic material in soil; it is left over from the decay of living things
5. regolith
a physical property of soil that describes how the soil feels, and the relative components of sand, silt, and clay sized particles
6. soil
a chemical property of soils that describes the measure of hydrogen ions in a soil sample; how acidic or basic a soil sample is
7. soil horizon
a substance that does not contain carbon and hydrogen atoms, such as salts, rocks, and minerals
8. soil ph
a distinct layer of soil that has characteristic properties
9. soil profile
the rock material that was weathered to form the sediments in a given soil
10. soil texture
a substance that contains carbon and hydrogen atoms, such as carbon dioxide, glucose, methane, and nucleic acid
Answer:
1. Soil.
2. Soil profile.
3. Regolith.
4. Humus.
5. Soil texture.
6. Soil pH.
7. Inorganic.
8. Soil horizon.
9. Parent material.
10. Organic.
Explanation:
1. Soil: the thin upper layer of Earth's crust that supports plant life. There are three (3) main types of soil; sandy, clay and loamy soil.
2. Soil profile: a vertical section of soil that shows the horizon and parent material.
3. Regolith: the layer of loose rock on the surface of the earth; also called mantle rock.
4. Humus: dark colored organic material in soil; it is left over from the decay of living things.
5. Soil texture: a physical property of soil that describes how the soil feels, and the relative components of sand, silt, and clay sized particles.
6. Soil pH: a chemical property of soils that describes the measure of hydrogen ions in a soil sample; how acidic or basic a soil sample is.
7. Inorganic: a substance that does not contain carbon and hydrogen atoms, such as salts, rocks, and minerals.
8. Soil horizon: a distinct layer of soil that has characteristic properties.
9. Parent material: the rock material that was weathered to form the sediments in a given soil.
10. Organic: a substance that contains carbon and hydrogen atoms, such as carbon dioxide, glucose, methane, and nucleic acid.
When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of the square. Suppose three identical charges of q are placed on the remaining three corners of the square (so that each corner has q). What is the magnitude of the net electric field at the center of the square
Answer:
0 N/C
Explanation:
Since the electric field due toeach charge q is at the same distance from the center of the square and has a magnitude E = 2 N/C, we resolve the components of each electric field to vertical and horizontal.
So, for each charge, the magnitude of the horizontal component is E' = Ecos45 and the magnitude of the vertical component is E" = Esin45.
For the charges on the bottom half of the square, the direction of the vertical component is upwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the bottom two charges is 2Esin45.
For the charges at the upper half of the square, the direction of the vertical component is downwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the top two charges is -2Esin45.
So, the resultant electric field at the center of the square is thus 2Esin45 + (-2Esin45) = 2Esin45 - 2Esin45 = 0 N/C
What is a Wave In your own words
An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat curve with radius of 65.0 m be rated
Answer:
The maximum speed is 21.39 m/s.
Explanation:
Given;
radius of the flat curve, r₁ = 150 m
maximum speed, [tex]v_{max}[/tex] = 32.5 m/s
The maximum acceleration on the unbanked curve is calculated as;
[tex]a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2[/tex]
the radius of the second flat curve, r₂ = 65.0 m
the maximum speed this unbanked curve should be rated is calculated as;
[tex]a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max} \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max} \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s[/tex]
Therefore, the maximum speed is 21.39 m/s.
An engine has an output energy of 2,400 J in 10 seconds. What is its average power in watts?
Answer:
P = 240 W
Explanation:
By definition, the average power is the rate of change of Energy (in Joules) regarding time (in seconds), as follows:[tex]P = \frac{\Delta E}{\Delta t} (1)[/tex]
Replacing in (1) by the givens of ΔE = 2, 400 J and Δt = 10 sec, we can find the average power in Watts as follows:[tex]P = \frac{\Delta E}{\Delta t} =\frac{2,400J}{10s} = 240 W (2)[/tex]
If the radius of curvature of the bump is 52 m, find the apparent weight of a 62-kg person in your car as you pass over the top of the bump. Express your answer using two significant figures.
Answer:
The answer is below
Explanation:
Driving in your car with a constant speed of 12m/s, you encounter a bump in the road that has a circular cross-section, as indicated in the figure . If the radius of curvature of the bump is 52 m, find the apparent weight of a 62-kg person in your car as you pass over the top of the bump.
Solution:
Centripetal force is the net force acting on a body which makes it move along a curved path. This force is always towards the center of curvature.
As the car passes over the bump, the centripetal acceleration acts downward towards the circle center.
The sum of all vertical forces is equal to zero, hence:
F - mg + ma = 0
where F is the apparent weight of the person, m is the mass of the person, ma = centripetal force = mv²/r
Given that:
m = 62 kg, v = velocity = 12 m/s, r = radius of curvature of bump = 52 m, g = acceleration due to gravity = 10 m/s. Therefore:
F - mg + ma = 0
F - mg + mv²/r = 0
F = mg - mv²/r
F = m(g - v²/r)
Substituting:
F = 62(10 - 12²/54)
F = 456.67 N
The apparent weight of a 62-kg person as the top of the bump is passed = 456.67 N
But the weight of the person = mg = 62* 10 = 620 N
6. Thomas sits on a small rug on a polished wooden floor. The coefficient to kinetic friction between the rug and
the slippery wooden floor is only 0.12. If Thomas weighs 650 N, what horizontal force is needed to pull the rug
and Thomas across the floor at a constant speed?[78N]
1
Answer:
F=78 N
Explanation:
Taking to Thomas and the rug as a single system, if they slide across the floor at constant speed, this means that their acceleration is just zero.According to Newton's 2nd Law, if the acceleration is zero, this means that the net force applied is zero too.In the horizontal direction, there are two forces acting on Thomas and the rug (as a single system), the applied force, and the kinetic friction force, which must be equal and opposite each other:[tex]F_{app} = F_{kfr} (1)[/tex]
By definition, as the friction force is the horizontal component of the contact force, it can be expressed as follows:[tex]F_{1kfr} = \mu_{k} * F_{n} (2)[/tex]
where μk = coefficient of kinetic friction = 0.12
Fn = normal force
In this case, as the system boy+rug is not accelerating in the vertical direction, and the surface is level, the normal force (which is always perpendicular to the surface), must be equal to the force of gravity.Assuming that the mass of the rug is neglectable, we can write:[tex]F_{n} = F_{g} = m*g = 650 N (3)[/tex]
Replacing (3) and μk in (2)[tex]F_{1kfr} = \mu_{k} * F_{n} = 0.12 * 650 N = 78 N (4)[/tex]
From (1), we finally get:[tex]F_{app} = F_{kfr} = 78 N (5)[/tex]
7. A girl pushes her little brother on his sled with a force of 300. N for 750. m. How much work is this if the force of friction acting on the sled is (a) 200. N. (b) 300. N?
Answer:
a) 75000Joules
b) 0Joules
Explanation:
Workdone = Force * Distance
Given
distance= 750m
Force = 300N
a) If the frictional force = 200N
The Total force = 300N - 200N = 100N
Work done = 100 * 750
Workdone = 75,000Joules
Hence the workdone if the force of friction is 200N is 75,000Joules
b) If the frictional force = 300N
The Total force = 300N - 300N = 0N
Work done = 0* 750
Workdone = 0Joules
Hence the workdone if the force of friction is 300N is 0Joules i.e no work will be done on the sled
5. Candance got $0.22 in change when she bought chips at the
store. What percent of a dollar does her change represent?
Answer:
22%
Explanation:
Given parameters:
The amount of change = $0.22
Unknown
What percentage of a dollar the change represents = ?
Solution:
A dollar is made up of 100cents;
So;
$0.22 represents 22cents
Percentage of dollar this represents = [tex]\frac{22}{100}[/tex] x 100 = 22%
What causes the movement of electrical current in the circuit within the device shown in the diagram?
NASA
Movement of protons
Movement of electrons
Heating of silicon
The anti-reflection coating
Answer:
B. Movement of electrons
Hope this helps!! may I have brainliest? :)
Answer:
Movenment of electrons
Explanation:
A 70 kg object strikes the ground with 2500 J of KE after falling freely from rest. How far above the ground was the object when it was released?
Answer:
3.64 m
Explanation:
m = Mass of object = 70 kg
Kinetic energy of the object = 2500 J
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
h = Height from which the object is dropped
Kinetic energy is given by
[tex]\dfrac{1}{2}mv^2=2500\\\Rightarrow v^2=\dfrac{2\times 2500}{70}[/tex]
From conservation of energy we get kinetic energy equal to potential energy.
[tex]\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{1}{2g}v^2\\\Rightarrow h=\dfrac{1}{2\times 9.81}\times \dfrac{2\times 2500}{70}\\\Rightarrow h=3.64\ \text{m}[/tex]
The object was released from a height of 3.64 m.
In October 1997, Andy Green broke the sound barrier on land in a jet-powered car. Green's car had accelerated from rest to 763 mi/h (341 m/s) over a 5.0-mi (8047-m) stretch. Assuming constant acceleration, over what time interval had he reached the top speed
Answer: the time interval Green had to reached the top speed is 47.2 secs
Explanation:
Given the data in the question;
first we determine Green's acceleration.
from third equation of motion, v² - u² = 2as
a = v² - u² / 2s
we substitute,
a = ( (341 m/s)² - (0)² ) / 2(8047)
a = (116281 - 0) / 16094
a = 7.23 m/s²
Now we determine the time interval need to reach top or maximum speed,
v = u + at
at = v - u
t = (v - u) / a
so we substitute
t = (341 - 0) / 7.23
t = 341 / 7.23
t = 47.2 secs
Therefore, the time interval Green had to reached the top speed is 47.2 secs
The acceleration due to gravity on or near the surface of Earth is 32 ft./s/s. Neglecting friction, from what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft./s
Given :
The acceleration due to gravity on or near the surface of Earth is 32 ft/s/s
To Find :
From what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft/s.
Solution :
Initial velocity of stone, u = 0 ft/s.
Now, by equation of motion :
[tex]2as = v^2 -u^2 \\\\2\times 32 \times s = 136^2 -0^2\\\\s = \dfrac{136^2}{2\times 32}\ ft\\\\s = 289 \ ft[/tex]
Therefore, height from which stone is thrown is 289 ft.
On a day when the wind is blowing toward the south at 4 m/s, a runner jogs east at 5 m/s. What is the velocity (speed and direction) of the air relative to the runner
Answer:
[tex]v=5m/s[/tex]
Direction:SW(south-west)
Explanation:
From the question we are told that
Velocity and direction of Runner [tex]V_r= 5m/s ,East[/tex]
Velocity and direction of Air[tex]V_a= 4m/s ,South[/tex]
Generally the the resultant velocity v is mathematically given as
[tex]v=\sqrt{V_a^2+V_r^2}[/tex]
[tex]v=\sqrt{4^2+5^2}[/tex]
[tex]v=5m/s[/tex]
The resultant velocity is towards the south-west
1. Which of the following will be surrounded by a magnetic field? *
1Two copper wires held close together.
2A pot being heated on an electric stove.
3A wire that is conducting an electrical current.
4An aluminum rod that has been touched by a magnet.
Answer:
4th answer
Explanation:
if we are making koolaid with sugar, koolaid powder and water whitch part is the solvent
Answer:The powder of Kool Aid crystals are the solute. The water is the solvent and the delicious Kool Aid is the solution.... .-.
Explanation:
Soccer fields vary in size. A large soccer field is 115 m long and 85.0 m wide. Assume that 1 m equals 3.281 ft. What are its dimensions in feet
Answer:
The dimensions are 377.3 feet by 278.9 feet.
Explanation:
Given that,
Length of a soccer field = 115 m
Width of a soccer field = 85 m
We need to assume that, 1 m equals 3.281 ft
We need to find the dimensions in feet.
As 1 m = 3.281 ft
⇒ 115 m = (115×3.281) feet
= 377.315 feet ≅ 377.3 feet
⇒ 85 m = (85×3.281) feet
= 278.885 feet ≅ 278.9 feet
Hence, the dimensions are 377.3 feet by 278.9 feet.
Determine the values of mm and nn when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,000,000,000 kgkg.
Answer:
"m" and "n" are 5.97 and 24 respectively.
Explanation:
Standard form which is a scientific notation [m × 10^n] can be regarded as way to reduce large figures to small one in a decimal firm and this is usually done for conviniency sake.
✓Let us find "m" , to do this we will shorten the long number so that it will be from 1-9. And this is 5.97, hence
m = 5.97
✓ " n" can be determined by counting the digits from our right hand then stop where the decimal point was put when we were finding our "m". Hence
n= 24
✓ if we input the values to the scientific expresion above, we have
m = 5.97 and n = 24
Hence, the Standard form = 5.97 × 10^24 kg which is the scientific notation.
Tripling the wavelength of the radiation from a monochromatic source will change the energy content of the individually radiated photons by what factor
Answer: By a factor of 1/3.
Explanation:
For a photon with wavelength λ, the energy is written as:
E = h*c/λ
where:
h is the Planck's constant:
h = 6.63*10^(-34) Js
c is the speed of light:
c = 3*10^8 m/s
Now, if we triple the wavelength of this photon, then the energy will be:
E' = (h*c)/(3*λ)
We rewrite this as:
E' = (1/3)*(h*c/λ)
And (h*c/λ) was the previous energy:
(h*c/λ) = E
Then we can replace that in the above equation to get:
E' = (1/3)*(h*c/λ) = (1/3)*E
Then if we triple the wavelength, it will change the energy content of the individually radiated photons by a factor of 1/3.
a substance has a pH of 9.what type of substance is it
Answer:
I think its Baking soda, antacids, Sorry if i'm wrong!
Explanation:
Answer:
it base
Explanation:
because i passed the test on that
A 235 kg object and a 1.37×1012 kg are located 2.59×104 m away from each other. What is the force due to gravity between the two objects?
Answer:
F = 3.2 x 10⁻⁵ N
Explanation:
The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:
[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]
where,
F = gravitational force = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²
m₁ = mass of object 1 = 235 kg
m₂ = mass of object 2 = 1.37 x 10¹² kg
r = distance between objects = 2.59 x 10⁴ m
Therefore,
[tex]F = \frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(235\ kg)(1.37\ x\ 10^{12}\ kg)}{(2.59\ x\ 10^{4}\ m)^{2}}[/tex]
F = 3.2 x 10⁻⁵ N
Answer:
the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.
Explanation:
Given;
mass of the first object, m₁ = 235 kg
mass of the second object, m₂ = 1.37 x 10¹² kg
distance between the two object, r = 2.59 x 10⁴ m
The gravitational force between the two object is calculated as;
[tex]F= \frac{Gm_1m_2}{r^2}[/tex]
where;
G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²
[tex]F= \frac{(6.67\times 10^{-11})(235)(1.37\times 10^{12})}{(2.59\times 10^4)^2} \\\\F = 3.2 \times 10^{-5} \ N[/tex]
Therefore, the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.
An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length. The distance between the lenses is 5.0 cm. Find the location and size of the final image.
Answer:
122.735 behind converging lens ; 2.16
Explanation:
Given tgat:
Object distance, u = 29 cm
Image distance, v =
Focal length, f = - 19 (diverging lens)
Mirror formula :
1/u + 1/v = 1/f
1/29 + 1/v = - 1/19
1/v = - 1/19 - 1/29
1/v = −0.087114
v = −11.47916
v = -11.48
Second lens
Object distance :
u = 11.48 + 11 = 22.48 cm
1/v = 1/19 - 1/22.48
1/v = 0.0081475
v = 1 / 0.0081475
v = 122.735 cm
122.735 behind second lens
Magnification, m
m = m1 * m2
m = - v / u
Lens1 :
m1 = -11.48 / 29 = - 0.3958620
m2 = - 122.735 / 22.48 = - 5.4597419
Hence,
- 0.3958620 * - 5.4597419 = 2.16
If an object is thrown downward at 4.73 m/s and it falls for 6.21 seconds before landing, how fast is it falling the instant before it lands?
Answer:
65.59 m /s
Explanation:
Initial velocity u = 4.73 m/s
Final velocity v = ?
time t = 6.21 s
acceleration = g = 9.8 m /s
v = u + gt
= 4.73 + 9.8 x 6.21
= 65.59 m /s .
A hunter points a rifle directly at a coconut that he intends to shoot off a tree. It so happens that the coconut falls from the tree at the exact instant the hunter pulls the trigger. What happens to the bullet?
Answer:
Explanation:
The bullet will hit the coconut .
Both coconut and bullet will fall under the gravitational pull of the earth . They will experience same acceleration because acceleration does not depend upon mass . So both bullet and coconut will fall by the same distance vertically .
So bullet will exactly hit the coconut .
A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.10 103 N/C, while the magnetic field is 0.360 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.20 cm. Find the charge-to-mass ratio of the particle.
Answer:
The charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg
Explanation:
From the formulae
F = qvB and F = mv²/r
Where F is Force
q is charge
v is speed
B is magnetic field strength
m is mass
and r is radius
Then,
qvB = mv²/r
qB = mv/r
We can write that
q/m = v/rB ---- (1)
Also
From Electric force formula
F = Eq
Where E is the electric field
and magnetic force formula
F = Bqv
Since, electric force = magnetic force
Then, Eq = Bqv
E = Bv
∴ v = E/B
Substitute v = E/B into equation (1)
q/m = (E/B)/rB
∴ q/m = E/rB²
(NOTE: q/m is the charge to mass ratio)
From the question,
E = 3.10 ×10³ N/C
r = 4.20 cm = 0.0420 m
B = 0.360 T
Hence,
q/m = 3.10 ×10³ / 0.0420 × (0.360)²
q/m = 569517.9306 C/kg
q/m = 5.7 × 10⁵ C/kg
Hence, the charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg.
A wooden block is let go from a height of 5.80 m. What is the velocity of the block just before it hits the ground?
Given :
A wooden block is let go from a height of 5.80 m.
To Find :
The velocity of the block just before it hits the ground.
Solution :
We know, by equation of motion :
[tex]v^2 - u^2 = 2as[/tex]
Here, a = g = 9.8 m/s²( Acceleration due to gravity )
Putting all given values in above equation, we get :
[tex]v^2 - u^2 = 2as\\\\v^2 -0 = 2\times 9.8 \times 5.8 \\\\v = \sqrt{2\times 9.8 \times 5.8 } \ m/s\\\\v = 10.66\ m/s[/tex]
Hence, this is the required solution.
A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field at distance r from the axis inside the cylinder in terms of r/R.
Answer:
[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]
Explanation:
Given that:
[tex]\rho_o = (10^{-3} ) \ c/m^3[/tex]
R = (0.1) m
To find the electric field for r < R by using Gauss Law
[tex]{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)[/tex]
For r < R
[tex]Q_{enclosed}=(\rho) ( \pi r^2 ) l[/tex]
[tex]E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}[/tex]
[tex]E= \dfrac{\rho ( r)}{2 \varepsilon_o}[/tex]
where;
[tex]\varepsilon_o = 8.85 \times 10^{-12}[/tex]
[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]
[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]
[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]
