Answer:
Explanation:
In a Solid sphere; the moment of inertia around its geometrical axis can be expressed by using the formula:
[tex]\mathtt{I_s = \dfrac{2}{5} M_s R^2_s}[/tex]
For the solid disk; the moment of inertia around the central axis is:
[tex]\mathtt{I_D= \dfrac{1}{2}M_DR_D^2}[/tex]
Suppose [tex]M_D = M_S[/tex]; then we can say both to be equal to M
As well as [tex]R_D = R_S[/tex]; then that too can be equal to R
Now;
[tex]\mathtt{I_s = \dfrac{2}{5} M R^2} --- (1)[/tex]
[tex]\mathtt{I_D= \dfrac{1}{2}MR^2}---(2)[/tex]
Multiplying equation (1) by 2, followed by dividing it by 2; we have:
[tex]\mathtt{I_s= \dfrac{2}{5}MR^2} \times \dfrac{2}{2}[/tex]
[tex]I_s = \dfrac{4}{5} \times \dfrac{1}{2}MR^2 \\ \\ I_s = \dfrac{4}{5}\times I_D \\ \\ I_s > I_D[/tex]
how can you prove that acceleration is a derived unit
a = (dx / dt)²
Explanation: Unit of distance is m (metres) and unit of time is s (seconds) speed v is first derivative of distance x versus time:
v = dx / dt, unit is m/s. Acceleration is second derivative of
speed versus time a = (dx / dt)² = (dv/dt) , unit is m/s²
Answer:
Explanation:
Acceleration is derived unit because it has two fundamental units involved i.e. meter and second square.
As a new electrical engineer for the local power company, you are assigned the project of designing a generator of sinusoidal ac voltage with a maximum voltage of 120 V. Besides plenty of wire, you have two strong magnets that can produce a constant uniform magnetic field of 1.5 T over a square area with a length of 10.2 cm on a side when the magnets are separated by a distance of 12.8 cm . The basic design should consist of a square coil turning in the uniform magnetic field. To have an acceptable coil resistance, the coil can have at most 400 loops.
What is the minimum rotation rate of the coil so it will produce the required voltage? Express your answer using two significant figures.
Answer:
The rotation rate is 15.3 rad/s.
Explanation:
maximum voltage, V = 120 V
Magnetic field, B = 1.5 T
length, L = 10.2 cm
width, W = 12.8 cm
Number of loops, N = 400
Let the rate of rotation is w.
The maximum voltage is given by
V = N B A w
120 = 400 x 1.5 x 0.102 x 0.128 x w
w = 15.3 rad/s
what is the magnitude of an electric field (in 106 n/c) that balances the weight of a plastic sphere of mass 2.1 g that has been charged to 3.0 nc
Answer:
[tex]E=6.86\times 10^6\ N/C[/tex]
Explanation:
Given that,
Mass of the sphere, m = 2.1 g = 0.0021 kg
Charge, q = 3 nC
We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,
qE = mg
[tex]E=\dfrac{mg}{q}[/tex]
Put all the values,
[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]
So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].
5N
5 N
19 N
19 N
Pls help look at the pic
Answer:
b. is the correct answer ....
By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects
Answer:
[tex]Weight\ loss=1.6321N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=85.9kg[/tex]
Altitude [tex]h= 6.33 km[/tex]
Let
Radius of Earth [tex]r=6380km[/tex]
Gravity [tex]g=9.8m/s^2[/tex]
Generally the equation for Gravity at altitude is mathematically given by
[tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]
[tex]g_s=9.781m/s^2[/tex]
Therefore
Weight at sea level
[tex]W_s=9.8*85.9[/tex]
[tex]W_s=841.82N[/tex]
Weight at 6.33 altitude
[tex]W_a=9.781*85.9[/tex]
[tex]W_a=840.2N[/tex]
Therefore
[tex]Weight loss=W_s-W_b[/tex]
[tex]Weight loss=841.82-840.2[/tex]
[tex]Weight loss=1.6321N[/tex]
compare the time period of two pendulums of length 4m and 9m
area= length × length
area = 4m × 9m
ans 36
The index of refraction of n-propyl alcohol is 1.39. Find the angle of refraction of light in that medium if light comes from air with an angle of incidence of 55 degrees.
Answer:
36.11 degrees
Explanation:
index of refraction n = sin i/sinr
i is the angle of incidence
r is the angle of refraction
Substitute into the expression
1.39 = sin55/sin(r)
1.39 = 0.8191/sin(r)
sin(r) = 0.8191/1.39
sin(r) = 0.5893
r = arcsin(0.5893)
r = 36.11
hence the angle of refraction of light is 36.11 degrees
an object moves clockwise around a circle centered at the origin with radius m beginning at the point (0,). a. find a position function r that describes the motion of the object moves with a constant speed, completing 1 lap every s. b. find a position function r that describes the motion if it occurs with speed .
Answer:
Answer to An object moves clockwise around a circle centered at the origin with radius 6 m beginning at ... 6 M Beginning At The Point (0,6) B. Find A Position Function R That Describes The Motion If It Occurs With Speed E T A. R(t)= S The Motion Of The Object Moves With A Constant Speed, Completing 1 Lap Every 12 S.
Explanation:
If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N
The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.
What is Force?The force is the action of push or pull which makes an object to move or stop.
Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by
F = m(g-a)
F = 13.9 x (9.81 - (-0.898))
F = 148.84 N
Thus, the magnitude of the force is 148.84 N.
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A 3.50 kg basket of cookies sits on a 2.00 m high shelf. What is the gravitational potential energy of the basket?
pls help
Answer:
68.6 J
Explanation:
Applying,
P.E = mgh............... Equation 1
Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket
From the question,
Given: m = 3.5 kg, h = 2.00 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
P.E = 3.5×2×9.8
P.E = 68.6 J
Hence the potential energy of the basket is 68.6 J
One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A>s. (a) What is the mutual inductance of these solenoids
Answer: The mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].
Explanation:
Given: Length = 50.0 cm (1 cm = 0.01 m) = 0.50 m
[tex]N_{1}[/tex] = 6750
[tex]N_{2}[/tex] = 15
Radius = [tex]\frac{0.120 cm}{2} = 0.6 cm = 6 \times 10^{-4} m[/tex]
As inner of a solenoid resembles the shape of a circle. So, its area is calculated as follows.
[tex]Area = \pi \times r^{2} = \pi \times (6 \times 10^{-4})^{2}[/tex]
Formula used to calculate mutual conductance of two solenoids is as follows.
[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}[/tex]
where,
M = mutual conductance
A = area
[tex]\mu_{o}[/tex] = relative permeability = [tex]4 \pi \times 10^{-7} Tm/A[/tex]
[tex]N_{1}[/tex] = no. of coils in outer solenoid
[tex]N_{2}[/tex] = no. of coils in inner solenoid
l = length
Substitute the values into above formula as follows.
[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}\\= \frac{4 \pi \times 10^{-7} Tm/A \times \pi (6 \times 10^{-4})^{2} \times 6750 \times 15}{0.5 m}\\= 2.88 \times 10^{-7} H[/tex]
Thus, we can conclude that the mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].
Many adventures like to go rafting on the Colorado River through Grand Canyon National Park. There are many locations where the river becomes more narrow, both the distance between the canyon walls as well as the depth changes due to debris like boulders on the bottom of the river; this leads to changes in the water speed. In the park, the Colorado River has an average width of 100m and an average depth of 8m, and an average speed of 3 m/s. At the Lava Falls Rapids, the river has an average width of about 25m and an average depth of about 15m. What is the approximate speed of the water in this location
Answer:
6.4 m/s
Explanation:
Given that :
The average width of the Colorado river = 100 m
Average depth of the river is = 8 m
Therefore, area = [tex]$A_1= 100 \ m \times 8 \ m$[/tex]
Speed of the river, [tex]$v_1 = 3 \ m/s$[/tex]
After the lava falls on the river,
Width of the river becomes = 25 m
Depth of the river became = 15 m
Therefore, area = [tex]$A_2= 25 \ m \times 15 \ m$[/tex]
Now, since the volume flow rate of the Colorado river is same, then from the Continuity equation,
[tex]$Q_1=Q_2$[/tex]
[tex]$A_1v_1=A_2v_2$[/tex]
∴ [tex]$100 \times 8 \times3 = 25 \times 15 \times v_2$[/tex]
[tex]$v_2=\frac{100 \times 8 \times 3}{25 \times 15}$[/tex]
= 6.4 m/s
Therefore, the speed of the river in this location is 6.4 m/s
What is sieving? Give an example where this method is used. (2)
Answer:
sieving is when you separate particles of different sizes.
Explanation:
separating sand mixtures
separating chaffs from local garri
Answer this
a) which ink is likely to be pure? Why?
b) What does the chromatography tell us about ink Y
c) Why are the three different spots separated out from ink Y found at different heights?
Answer:
a) Ink X is likely to be pure because it only contain 1 spot.
b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.
c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.
The different spots from Y are found at various heights because they represent different compounds.
What is chromatography?The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.
We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.
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Give an example of a vertical motion with a positive velocity and a negative acceleration. Give an example of a vertical motion with a negative velocity and a negative acceleration.
Answer:
An example of positive velocity is throwing a ball upwards
An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height
Explanation:
In a vertical movement the acceleration is always downwards, therefore negative since it is created by the attraction of the Earth on the body.
An example of positive velocity is throwing a ball upwards
An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height
If two dogs are pulling a bone with force-20Newtons in opposite direction, then the resultant force is
Answer:
Newtons third law of motion: Balanced forces
Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs. Once more, tug-of-war is a prime illustration.
What force in opposite direction follow newton law?The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.
The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.
Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.
Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.
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One coulomb represents how many electrons?
a. 1 electron
b. 100 electrons
C. 6.25 quintillion electrons
d. 6.25 million-million electrons
e, none of the above
Answer:
6.24 x 1018 electrons.
Explanation:
So I think C
Isotopes of the same element always have the same
(2 points)
O atomic mass number
O A-number
O Z-number
O neutrinos
Answer:
Z-number
Explanation:
The Z number is the number of protons in an atom, and this does not change when an isotope is created. I got it right on the test.
You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?
Answer:
(a) I = 1650000 A
(b) 4.125 T
Explanation:
Magnetic field, B = 5.5 T
distance, r = 0.06 m
(a) Let the current is I.
The magnetic field due to a long wire is given by
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]
(b) Let the magnetic field is B' at distance r = 0.08 m.
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]
A parallel plate vacuum capacitor has 8.40 J of energy stored. The separation between plates is 2.30 mm. If the separation is decreased to 1.15 mm what is the energy stored if (a) the charge Q on the plates is held constant, and (b) the voltage V across the plates is held constant
(a) 4.20 J
(b) 16.74 J
Explanation:For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;
C = A∈₀ / d --------------------(i)
Where;
∈₀ = constant called permittivity of vacuum.
The energy U stored in such capacitor is given by;
U = [tex]\frac{1}{2}[/tex]CV² ----------------------(ii)
or
U = [tex]\frac{1}{2}[/tex](Q²/C) -------------------(**)
Where;
V = potential difference or voltage across the plates.
Q = charge on the plates.
(a) If the charge is held constant
Combine equations (i) and (**) to give;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d) -----------------------(iii)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iii)
8.40 = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)
8.40 = [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)
Multiply through by 2
2 x 8.40 = Q² x (0.023 / A∈₀)
16.80 = Q² x (0.023 / A∈₀)
Divide through by 0.023
16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023
730.4 = Q² / (A∈₀)
Make Q² subject of the formula
Q² = 730.4(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
Q = constant [this means that Q² still remains 730.4(A∈₀) ]
The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)
U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)
U = 4.20J
Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J
(b) If the voltage is held constant
Combine equations (i) and (ii) to give;
U = [tex]\frac{1}{2}[/tex](A∈₀ / d)V² -----------------------(iv)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iv)
8.40 = [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²
Multiply through by 2 x 0.023
2 x 0.023 x 8.40 = (A∈₀)V²
2 x 0.023 x 8.40 = (A∈₀)V²
0.385 = (A∈₀)V²
Make V² subject of the formula
V² = 0.385/(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
V = constant [this means that V² still remains 0.385/(A∈₀) ]
The energy stored is found by substituting these values of d and V² into equation (iv) as follows;
U = [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]
U = [tex]\frac{1}{2}[/tex](0.385/0.0115)
U = 16.74
Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J
magnetism/ magnetic field ana magnetic forces
Answer:
Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.
two 100 ohm resistors are connected inparallel and one identical resister in series. The maximum power that can be delivered to any one resistor is 25W. What is the maximum voltage that can be applied between the terminals A and B ?
A. 50V
B. 75V
C. 100V
D. 125V
SOLVED DOWN BELOW
Explanation:
In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law
I = 25/200
I= .125 amps or 125 ma
__________
R= r1 * r2 / r1 +r2
R= 100 * 100 / 100 + 100
R= 10000 / 200
R= 50 ohms
What is the first quantum number of a 252 electron in phosphorus,
1322s22p3s23p3?
A. n=0
B. n= 3
O
C. n = 1
O D. n = 2
Answer:
the correct answer is B
Explanation:
The quantum numbers are the constants obtained when solving the Schrodinger equation, the first quantum number or principal quantum number (n), can take values from zero to infinity.
This quantum number is placed as a coefficient in the quantum distribution.
In this case for phosphorus, the number is n = 3
the correct answer is B
Compared to its weight on Earth, a 5kg object on the moon will weigh
The same amount
Less
More
Answer:
Less
Explanation:
Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.
Dos cargas puntuales iguales y negativas, q1=q2=-24micro C se localizan en x=0 y y=38m y x=0 y y=-7m, respectivamente. Calcula la magnitud de la fuerza electrica total en N que ejercen estas dos cargas sobre una tercera, tambien puntual, Q=26micro C en y=0 y x=16m
Answer:
F_net = 9.87 10⁻⁴ N
Explanation:
Let's use that force is a vector magnitude
∑ F = F₁₃ + F₂₃
De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract
the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m
∑ F = F₁₃ - F₂₃
F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]
in this case q₁ = q₂ = q
F_net = k q q₃ ( )
let's look for the distance
r₂₃ = y₂ - y₃
r₂₃ = -7 -16
r₂₃ = - 23 m
r₁₃ = 38 - 16
r₁₃ = 22 m
let's calculate
F_net = 9 10⁹ 24 26 10⁻¹² ( )
F_net = 5.616 ( 1.758 10⁻⁴ )
F_net = 9.87 10⁻⁴ N
A 4.9 A current is set up in a circuit for 4.7 min by a rechargeable battery with a 12 V emf. By how much is the chemical energy of the battery reduced
Answer:
E = 16581.6 J
Explanation:
Given that,
Current, I = 4.9 A
Time for which the current is set up, I = 4.7 min = 282 s
The voltage of the battery, V = 12 V
We need to find how much chemical energy of the battery reduced. Let It is E. We know that,
E = P t
Where
P is power of battery, P = VI
So,
[tex]E=VIt[/tex]
Put all the values,
[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]
So, 16581.6 J of chemical energy of the battery is reduced.
Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.
A voltage of 75 V is placed across a 150 Ω resistor. What is the current through the resistor?
Answer:
0.5 A
Explanation:
Applying,
V = IR.................. Equation 1
Where V = Voltage, I = current, R = Resistance.
make I the subject of the equation
I = V/R............... Equation 2
From the question,
Given: V = 75 V, R = 150 Ω
Substitute these values into equation 2
I = 75/150
I = 0.5 A.
Hence the cuurent through the resistor is 0.5 A
Which statement is true?
a particle of violet light has less energy than a particle of red light
a particle of violet light has more energy than a particle of red light
a particle of violet light has exactly the same energy as a particle of red light
particles of light do not have any energy, regardless of what color the light is
a particle of violet light has exactly the same energy as a particle of red light
A 40-kg crate is being lowered with a downward acceleration is 2.0 m/s2 by means of a rope. (a) What is the magnitude of the force exerted by the rope on the crate
Answer:
F = 312 N
Explanation:
Given that,
The mass of a crate, m = 40 kg
Acceleration of the crate, a = 2 m/s²
As the carte is falling downward, the net force exerted by the rope on the carte is given by :
F = m(g-a)
Put all the values,
F = 40(9.8-2)
F = 312 N
Hence, the required force exerted by the rope on the crate is equal to 312 N.