Answer:
The volume required is
Explanation:
From the question we are told that
The cars mileage is v = 28.0 mi/gal
The distance is d = 142 km
Converting the distance from km to miles
Generally the volume of gasoline needed is mathematically represented as
=>
=>
Converting to Liters
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Answer:
0.00124 V
Explanation:
Parameters given:
Initial circumference = 162 cm
Rate of decrease of circumference = 14 cm/s
Magnetic field, B = 0.5 T
Time, t = 8 secs
The magnitude of the EMF induced in the loop is given as:
V = (-NBA) / t
Where N = number of turns = 1
B = magnetic field
A = area of loop
t = time taken
First, we need to find the area of the loop.
To do this, we will find the radius after the loop circumference has decreased for 8 secs.
The rate of decrease of the circumference is 14 cm/s and 8 secs has passed, which means after 8 secs, it has decreased by:
14 * 8 = 112 cm
The new circumference is:
162 - 112 = 50 cm = 0.5 m
To get radius:
C = 2 * pi * r
r = C / (2 * pi)
r = 0.5 / (2 * 3.142)
r = 0.0796 m
The area is:
A = pi * r²
A = 3.142 * 0.0796²
A = 0.0199 m²
Therefore, the EMF induced is:
V = (-1 * 0.5 * 0.0199) / 8
V = -0.00124V
This is the EMF induced in the coil.
The magnitude is |-0.00124| V = 0.00124 V.
Answer:
force = 11.33
Explanation:
given data:
sled mass = 17.0 kg
inital velocity (U) = 4.10 m/s
elapsed time (T) 6.15 s
final velocity (V) = 0
final momentum P2 = 0
Initial momentum of sledge is
from newton second law of motion
Kgm/s^2
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Answer:
This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve
Explanation:
The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.
Let's apply this principle to our case.
One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.
qA = ½ q
qB = ½ q
qC = 0
The total charge is q
we make a second contact
If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half
qA = 1/2 (q / 2) = ¼ q
qC = ¼ q
qB = ½ q
At this point all spheres have a charge,
qA = ¼ q
qb = ½ q
qC = ¼ q
The total charge is q
Now let's contact spheres B and one of the other two
Q = ½ q + ¼ q = ¾ q
When splitting the charge
qB = ½ ¾ q = 3/8 q
qC = ½ ¾ q = 3/8 q
qA = ¼ q
The total charge is q
Note that the total load is always equal to q
Now let's analyze the given configuration
Let's look for the total load
Q = qA + QB + QC
Q = ½ q + 3/8 q + ¼ q
Q = 9/8 q
This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve
The principle of charge distribution among conductive materials is violated in qA = 1/2q, qB=3/8q, qC=1/4q, as the sum of charges on B and C doesn't equate to the charge on A and 3/8q isn’t a multiple of halving the original charge.
The answer lies in the fact that balls made of conducting materials when in contact, distribute charges evenly among them. This is due to the free movement of electrons within the conductive material that seeks to equalize potential difference, a property exploited in charge distribution problems of this sort.
Given the scenario, every time a charged ball, A, touches an uncharged ball (B or C) the net charge is evenly split between them. Hence, each subsequent distribution halves the charge of the originating ball (A) and gives the complementary half to the ball it's being touched to (B or C).
In the distribution, qA = 1/2q, qB=3/8q, qC=1/4q, the sum of charges on B and C does not equate to A, which is a violation of the charge conservation principle. Moreover, 3/8q isn’t a multiple of halving the original charge q, which negates the manner in which the charge is distributed (i.e., by halving).
Learn more about Charge Distribution here:
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Answer:
σ = 0.8 N/m
Explanation:
Given that
L = 12 cm
We know that 1 m = 100 cm
L = 0.12 m
The force ,F= 0.096 N
Lets take surface tension = σ
We know that surface tension is given as
Therefore the surface tension σ will be 0.8 N/m .
σ = 0.8 N/m
The surface tension of the liquid in air is 0.8 N/m.
To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.
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Explanation:
Given data:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Distance d between the plates = 1 mm = 1×10⁻³m
Voltage of the battery is emf = 100 V
Resistance = 1025 ohm
Solution:
In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by
Taking natural log on both sides,
(1)
Now we can calculate the capacitance by using the area of the plates.
C = ε₀A/d
=
= 18×10⁻¹²F
Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)
= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
Motion is detected when an object changes its position with respect to a reference point. Coordinate system is basically used to represent motion. A coordinate system uses numbers or coordinates which represent position of the reference points on a two-dimensional or three-dimensional space. The trajectory of a point or line can be studied on a coordinate system which describes various aspects of motion like velocity, acceleration, distance, displacement etc. Coordinate system is important because it helps to choose a starting point and the direction (which will be positive).