Answer:
The force must be applied during 8 seconds to reach trhe same final speed.
Explanation:
By Impulse Theorem, a change in the magnitude of linear momentum of a system with constant mass can be done by applying a force during a given time. That is:
[tex]m\cdot (v_{f}-v_{o}) = F \cdot \Delta t[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speed, measured in meters per second.
[tex]F[/tex] - Net external foce, measured in newtons.
[tex]\Delta t[/tex] - Time, measured in seconds.
We can eliminate mass and speeds by constructing the following relationship:
[tex]F_{1}\cdot \Delta t_{1} = F_{2}\cdot \Delta t_{2}[/tex] (2)
If we know that [tex]F_{1} = F[/tex], [tex]\Delta t_{1} = 4\,s[/tex] and [tex]F_{2} = \frac{F}{2}[/tex], then the time is:
[tex]4\cdot F = \frac{F\cdot \Delta t_{2}}{2}[/tex]
[tex]\Delta t_{2} = 8\,s[/tex]
The force must be applied during 8 seconds to reach trhe same final speed.
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is [01] m. (a) What is the force (N) that the track must exert on the car? (positive is up) (b) What must be the force (N) that the car exerts on a 61 kg passenger?
This question is incomplete, the complete question is;
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.
(a) What is the force (N) that the track must exert on the car? (positive is up)
(b) What must be the force (N) that the car exerts on a 61 kg passenger?
Answer:
a) the force (N) that the track must exert on the car is -6139.14 N
b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
Explanation:
Given the data in the question;
Let N represent the force that the track must exerted on the car
Net force on the car Fnet = Mg + N
so
M × a = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
we substitute
N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )
N = 3136 - ( 320 × 28.9848 )
N = 3136 - 9275.136
N = -6139.14 N
Therefore, the force (N) that the track must exert on the car is -6139.14 N
b) What must be the force (N) that the car exerts on a 61 kg passenger?
Let N represent the force that the car exerts on 61kg passengers
so
Net force of passengers Fnet = mg + N
Ma = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )
N = 597.8 - ( 61 × 28.9848)
N = 597.8 - 1768.0728
N = -1170.27 N
Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].
The force (N) that the car exerts on a 61 kg passenger is 597.8 N.
Centripetal force of the track
The centripetal force of the track on the car moving in the circular path is calculated as follows;
[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]
Normal force of the passengerThe force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.
F = mg
F = 61 x 9.8
F = 597.8 N
Learn more about centripetal force here: https://brainly.com/question/20905151
PLEASE ANSWER. I WILL MAKE YOU THE BEST ANSWER BRAINLIEST. I PROMISE.
Anna weighs 132 lb. Determine her mass in kilograms using the conversion 1 kg equal 2.2 lb. Use this mass to answer this question.
calculate Anna's weight on the Moon. (G = 1.6 m / S2) must include a unit with your answer
1). If 2.2 lb actually = 1 kg, then
(132lb) x (1 kg/2.2lb) = 60 kg.
(Sadly, 2.2 lb doesn't " = " 1 kg. The explanation, unfortunately, is beyond the scope of this discussion.)
2). Weight = (mass) x (gravity)
Moon weight = (60 kg)x(1.6 m/s^2)
Moon weight = 96 Newtons
A cylindrical metal rod has a resistance . If both its length and its diameter are quadrupled, its new resistance will be:________.
A. 16R
B. R/4
C. R
D. 4R
Answer:
R' = R/4
Explanation:
The resistance of a metal rod is R. It is given by the relation as follows :
[tex]R=\rho\dfrac{l}{A}[/tex]
Where
l is the length and A is the area of cross-section
[tex]A=\pi r^2=\pi (\dfrac{d}{2})^2[/tex]
If both its length and its diameter are quadrupled, it means,
l' = 4l
and d'= 4d
It means,
[tex]A'=\pi (\dfrac{4d}{2})^2[/tex]
Let new resistance be R'. So,
[tex]R'=\rho\dfrac{l'}{A'}\\\\R'=\rho\dfrac{4l}{\pi (\dfrac{4d}{2})^2}\\\\=\rho \dfrac{4l}{\pi \dfrac{16d^2}{2}}\\\\=\dfrac{4}{16}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\=\dfrac{1}{4}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\R'=\dfrac{R}{4}[/tex]
So, the correct option is (B) "R/4".
walking dancing and even some household chores are?
Answer:
Actions
Explanation:
if u mean what are these called then it's actions
Answer: Regular Physical Activity
Explanation:
Why is the total solar eclipse important for scientists? explain
Answer:
hello
Explanation:
For scientists, they offer a unique opportunity to study aspects of the sun like its corona, the layer of plasma surrounding the star. ... By studying the inner regions of the corona that we can generally only see and photograph well during a total solar eclipse, scientists can learn more about how weather works in space.
have a nice day
hope it helps
byee
Answer:
For scientists,they offer a unique opportunity to study aspects of the sun like it's Corona, the layer of plasma surrounding the star......By studying the inner regions of the Corona that we can generally only see and photograph well during a total solar eclipse, scientists can learn more about how weather works in space
8+10÷5(5×4+2)=?
it ıs said that this question was very diffıcult
can you slove?
Answer:
220 bastanyan sagot ko yawa
A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective valley width of 5000 m. The height of the ceiling due to atmospheric conditions is 1000 m. During the fall and spring, a breeze that travels at a velocity of 4 m/s flows through the valley. Fortunately, the breeze coming in contains no pollutants. Based on the information given, the steady state concentration of pollutants in the valley, in micrograms per cubic meter is.
Answer:
[tex]1.25\ \mu\text{g/m}^3[/tex]
Explanation:
v = Velocity of the breeze = 4 m/s
w = Width of the valley = 5000 m
h = Height of the valley = 1000 m
Volumetric flow rate is given by
[tex]\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}[/tex]
[tex]\dot{m}[/tex] = Mass flow rate of pollutant = 25 g/s = [tex]25\times 10^6\ \mu\text{g/s}[/tex]
Concentration is given by
[tex]C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3[/tex]
The steady state concentration of pollutants in the valley, is [tex]1.25\ \mu\text{g/m}^3[/tex].
on both sides.
F
10cm
2cm
(2 marks)
(a)
a
State the type of the lens in the box and explain your answer.
Answer:
please put pic of the questions
Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be +(30±5)pC+(30±5)pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain). What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)?
(a) Because air is a good conductor, the positive charge on the bee’s surface flowed through the air from bee to plant.
(b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee.
(c) The plant became electrically polarized as the charged bee approached.
(d) Bees that had visited the plant earlier deposited a positive charge on the stem.
Answer:
a) True
Explanation:
There are several possible explanations for this positive charge
* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,
to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct
* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.
When reviewing the different statements we have
a) True, it agrees with the second explanation of the phenomenon
b) False. The earth is a deposit of negative charge
c) false. If this is the case the charge should be negative
d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.
Answer:
Explanation:
.
During sexual reproduction, each parent contributes
Answer:
each parents contributes the dad gives the seed and the mom gives the egg
Explanation:
and then you are born
Need help only in part C!
An electric space heater draws 15.0 A from a 120-V source. It is operated, on the average, for 5.0 h each day. a. How much power does the heater use ? b. How much energy in kWh does it consume in 30 days? C. AED 0.30 per kWh , how much does it cost to operate the heater for 30 days ?
Answer:
A. 1800 Watt
B. 10.8 KW/h
C. AED 3.24
Explanation:
A. Determination of the power.
Voltage (V) = 120 V
Current (I) = 15 A
Power (P) =?
P = IV
P = 15 × 120
P = 1800 Watt
B. Determination of the energy consumption in KW/h for 30 days
We'll begin by converting 1800 Watt to kilowatt. This can be obtained as follow:
1000 W = 1 KW
Therefore,
1800 W = 1800 W × 1 KW / 1000 W
1800 W = 1.8 KW
Next, we shall determine the energy consumption per day. This can be obtained as follow:
Power (P) = 1.8 KW
Time (t) = 5 hour per day
Energy (E) =?
E = P /t
E = 1.8 / 5
E = 0.36 KW/h per day.
Finally, we shall determine the energy consumption in 30 days. This can be obtained as follow:
1 day = 0.36 KW/h
Therefore,
30 days = 30 × 0.36 KW/h
30 days = 10.8 KW/h
Thus, the energy consumption in 30 days is 10.8 KW/h
C. Determination of the cost of operation in 30 days.
1 KW/h = AED 0.30
Therefore,
10.8 KW/h = 10.8 KW/h × AED 0.3 / 1 KW/h
10.8 KW/h = AED 3.24
Thus, the cost of operation in 30 days is AED 3.24
Answer:
1800 W
Explanation:
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
This question is incomplete, the complete question is;
A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Answer:
the required heat energy is 16 J
Explanation:
Given the data in the question;
Lets consider the ideal gas equation;
PV = nRT
from the image, we calculate initial pressure;
Pi = ( 2000N/M × 0.06m) / 0.0008 m²
Pi = 15 × 10⁴ Pa
next we find Initial velocity
Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²
now we find the number of moles
n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K
N = 6.6 × 10⁻³ mol
next we calculate the final temperature;
Pf = ( 2000N/m × 0.08) / 0.0008 m²
Pf = 2 × 10⁵ Pa
Calculate the final Volume
Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³
we also determine the final temperature
[tex]T_{f}[/tex] = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK
[tex]T_{f}[/tex] = 466.8 K
so change in temperature ΔT
ΔT = 466.8 K - 300K = 166.8 K
we then calculate the change in thermal energy
ΔU = nCΔT
ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K
ΔU = 13.761 J
C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic
now we calculate the work done;
W = 1/2 × K( x[tex]_{i\\}[/tex]² - x[tex]_{f\\}[/tex]² )
W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )
= - 2.8 J
and we then calculate the heat energy using the following expression;
Q = ΔU - W
we substitute
Q = 13.761 - (- 2.8 J)
Q = 13.761 + 2.8 J)
Q = 16 J
Therefore, the required heat energy is 16 J
What are two benefits of scientists using a diagram to model the water cycle?
A. It can be used to show how the parts of the cycle relate to one
another.
B. It can be used to show as much detail ahis present in the actual
water cycle.
c. Only a few factors in the water cycle can be shown on the
diagram
D. It can show changes that occur in many different parts of Earth at
the same time.
Answer:
Options A. and D. are correct.
Explanation:
The water cycle shows the continuous movement of water within the Earth and atmosphere.
The two benefits of scientists using a diagram to model the water cycle are as follows:
It can be used to show how the parts of the cycle relate to one another.
It can show changes that occur in many different parts of Earth at the same time.
Options A. and D. are correct.
Two forces are exerted on an object in the vertical direction: a 20 N force downward and a 10 N force upward. The mass of the object is 25 kg. (1) What are some possibilities about the motion of this object? (2) Represent the motion of the object with a force diagram and a motion diagram.
Answer:
They are equal.
Explanation:
a. If we increase the wind velocity, the maximum vertical dispersal height and rate of diffusion will decrease____.
b. If we increase the humidity, the maximum vertical dispersal height will ___________increase____ after 24 hours.
c. If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will ____________________.
Answer:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Explanation:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Answer:
a decrease
b increase
c increase
What is the density of a 36 g object with a volume of 15 cm3? (Density: D = )
0.42 g/cm3
0.54 g/cm3
2.4 g/cm3
5.4 g/cm3
Answer:
density = mass/volume
so . . .
density = (36 g)/(15 cm³) = 2.4 g/cm³
Explanation:
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
A) Determine how long the clock run on the battery. use the relation below
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
B) Determine how many electrons per second flowed
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
Stored energy due to vertical position is known as
Elastic Potential energy
Vibrational energy
Kinetic energy
O Gravitational Potential energy
1
2
3
4
5
Answer: gravitational potential energy
Explanation:
what is the distance of truck and travel if it moves 7MS for 20 seconds
A meteorologist who has been analyzing weather data from the last fifty years found evidence of climate change. An ecologist has been studying deposits of pollen from the last fifty years to track the types of plants that grew in certain regions. She has also found evidence of climate change. What does this imply about science?
A. Meteorology and ecology are the same type of science.
B. When scientific disciplines overlap, researchers should choose new topics.
C. Different scientific disciplines can be used to investigate similar questions.
D. There should be fewer sciences, as they seem to answer the same questions.
Explain
Answer:
Probably (C), but don't forget the argument that:
Roses are red, and roses are flowers; therefore all flowers are red.
Which pair of labels is correct? A: Maximum kinetic energy C: Maximum gravitational potential energy B: Maximum kinetic energy D: Maximum gravitational potential energy A: Maximum gravitational potential energy C: Maximum kinetic energy B: Maximum gravitational potential energy D: Maximum kinetic energy
Answer: C. A:Maximum gravitational potential energy C: Maximum kinetic energy
Explanation:
Consider a uniformly charged sphere of total charge Q and radius R centered at the origin. We want to find the electric field inside the sphere (r
Answer:
Hello your question is incomplete attached below is the complete question
answer :
Total charge enclosed within the sphere : [tex]\frac{q_{r1} }{4\pi e_{0}R^3 } . r[/tex]
Total charge enclosed outside the sphere : [tex]\frac{q}{4\pi e_{0}r^2 } .r[/tex]
Explanation:
Given data:
Total charge of a uniformly charged sphere = Q
radius = R
first step : find the electric field inside and outside the uniformly charged sphere
2nd step : determine the total charge enclosed within and outside the sphere
make a sketch of the uniformly charged sphere
Attached below is a detailed solution
A 2[kg] rock and a 4[kg] rock are lifted to a height of 10[m].
How much PE does each rock have?
Answer:
See below
Explanation:
Potential Energy = mgh
for 2 rock = 2 * 9.81 * 10 = 196.2 j
for 4 rock = 4 * 9.81 * 10 = 392.4 j
Name the nutrients required for the body
Answer:
Explanation:
Water.
Carbohydrates.
Protein-Amino acids.
Fat.
Vitamins.
Minerals.
Omega-3 fatty acids.
A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the pendulum, g is gravity, andT is the period of pendulum.Twenty measurements of T give a mean of 1.823 seconds and a standard deviation of 0.0671 s. The device used to measure time has a resolution of 0.02 s. The pendulum length is measured once to be 0.823 m (with a scale having a resolution of 0.001 m). Determine the value of g and its uncertainty (assume 90% confidence where necessary). You may use any method of uncertainty propagation that we covered in class.
Answer:
g ±Δg = (9.8 ± 0.2) m / s²
Explanation:
For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use
T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]
T² = [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g
g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]
They indicate the average time of 20 measurements 1,823 s, each with an oscillation
let's calculate the magnitude
g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2
g = 9.7766 m / s²
now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation
for the period
T = t / n
ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn
In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently
ΔT = Δt / n
ΔT = Δt
now let's look for the uncertainty of g
Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex] ΔT
Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex] ΔL + 4π²L (-2 T⁻³) ΔT
a more manageable way is with the relative error
[tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]
we substitute
Δg = g ( \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}DL / L + ½ Dt / T)
the error in time give us the stanndard deviation
let's calculate
Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])
Δg = 9.7766 (0.001215 + 0.0184)
Δg = 0.19 m / s²
the absolute uncertainty must be true to a significant figure
Δg = 0.2 m / s2
therefore the correct result is
g ±Δg = (9.8 ± 0.2) m / s²
A beam of light, incident on a flat water surface, reflects from the mirror-like surface so that the angle of incidence equals the angle of reflection. The water has waves. Would individual light beams obey the law of reflection in this case?
Answer:
a protractor
Explanation:
because protractors measure angles
In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of a highway. Knowing that the speed of the bus was 80 mph as it begins to go up the hill and that the driver does not change the setting on his throttle or shift gears, determine the distance traveled (in miles) by the bus up the hill when its speed decreased to 50 mph.
Answer:
The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles
Explanation:
The parameters of the motion of the driver are;
The upgrade of the road, θ = 10°
The rate of constant acceleration of the bus driver = 5 ft./s²
The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s
The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s
The acceleration due to gravity, g ≈ 32.1740 ft./s²
Therefore, we have;
The acceleration due to gravity down the incline plane, gₓ = g·sinθ
∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²
The net acceleration of the bus, on the incline plane, [tex]a_{Net}[/tex] = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²
The vertical component of the velocity, [tex]v_y[/tex] = v × sin(θ)
∴ [tex]v_y[/tex] = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s
vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s
The velocity of the car, v₂, on the inclined plane is given as follows;
v₂ = v₁ - [tex]a_{Net}[/tex] × t
∴ t = (v₁ - v₂)/[tex]a_{Net}[/tex] = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s
The distance covered, 's', is given as follows;
s = v₁·t - 1/2·[tex]a_{Net}[/tex]·t²
∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.
The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles
Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the between the two wires where the magnetic field is zero?
a. yes, midway between the wires
b. yes, 12 cm from the 5-A wire
c. yes, 7.7 cm from the 5-A wire
d. no
Answer:
c. yes, 7.7 cm from the 5-A wire
Explanation:
Given;
distance between the two wires, r = 20 cm = 0.2 m
first current, I₁ = 5.0 A
second current, I₂ = 8.0 A
The magnetic field due to the two wires occurs in different directions and it can be zero at this region.
Let x be the distance from 5 A wire where the magnetic field is zero.
[tex]B = \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{5}{x} -\frac{8}{0.2-x} ] = 0\\\\\frac{5}{x} -\frac{8}{0.2-x} = 0\\\\\frac{5}{x} = \frac{8}{0.2-x}\\\\5(0.2-x) = 8x\\\\1 -5x = 8x\\\\1 = 5x \ + \ 8x\\\\1 = 13x\\\\x = \frac{1}{13} \\\\x = 0.077 \ m\\\\x = 7.7 \ cm[/tex]
Therefore, the correct option is c. yes, 7.7 cm from the 5-A wire
FIND THE DISTANCE BETWEEN TWO GIVEN POINTS.
1. S (5, -1) and T (5, 7)
Answer:
8 units
Explanation:
Which of the following would MOST likely slow Earth's tectonic activity?
O A. Earth's crust becomes cooler.
O B. Earth's mantle becomes warmer.
O C. Earth's mantle becomes cooler.
O D. Earth's outer core becomes warmer.
The Answer to your question is:
A.
The Earth's tectonic activity occurs by the movement of the fourteen main tectonic plates of the planet, which move over the mantle continuously, and result in the formation of mountains, earthquakes, tsunamis, volcanic activities, etc.
The decrease in the Earth's tectonic activity would therefore occur if the Earth's mantle became cooler, as the tectonic plates move over the magma, which is a paste formed by silicate of iron and magnesium and whose temperature reaches 600° and 1,200° Celsius , driven by forces from inside the planet.
Therefore the letter C is correct, as with the Earth's mantle cooler, the magma would become more solid which would decrease the Earth's tectonic activity.
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