Answer:
Explanation:
Magnetic field due to current at a distance of 4.4 cm
B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻² [ B = 10⁻⁷ x 2i / r = ]
= 2.36 x 10⁻⁵ T.
Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .
Force = 2.36 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴
= 23.41 x 10⁻²⁰ N .
This force will be perpendicular to the direction of current .
A violin has a string of length
0.320 m, and transmits waves at
622 m/s. At what frequency does
it oscillate?
Answer:
1.9kHz
Explanation:
Given data
wavelength [tex]\lambda= 0.32m[/tex]
velocity [tex]v= 622 m/s[/tex]
We know that
[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]
substitute
[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]
Hence the frequency is 1.9kHz
Answer:
971.2
Explanation:
It was right on acellus :)
Two point charges are placed on the x-axis as follows: charge q1 = 3.95 nC is located at x= 0.198 m , and charge q2 = 4.96 nC is at x= -0.297 m. What are the magnitude and direction of the total force exerted by these two charges on a negative point charge q3=6.00nCq that is placed at the origin?
Answer:
F = 2.40 × [tex]10^{-6}[/tex] N
Explanation:
given data
charge q1 = 3.95 nC
x= 0.198 m
charge q2 = 4.96 nC
x= -0.297 m
solution
force on a point charge kept in electric field F = E × q ................1
here E is the magnitude of electric field and q is the magnitude of charge
and
first we will get here electric field at origin
So net field at origin is
E = (Kq2÷r2²) - (kq1÷r1²) ...............2
put here value
E = 9[(4.96÷0.297²)-(3.95÷0.198²)]
E = 400.72 N/C ( negative x direction )
so that force will be
F = 6 × [tex]10^{-9}[/tex] × 400.72
F = 2.40 × [tex]10^{-6}[/tex] N
The net force on the third charge is 2.404 x 10⁻⁶ N.
The given parameters:
Position of first point charge, x1 = 0.198 mPosition of second point charge, x2 = -0.297 mFirst point charge, q1 = 3.95 nCSecond point charge, q2 = 4.96 nCThird point charge, q3 = 6 nC Position of the third charge, = 0The force on the third charge due to first charge is calculated as follows;
[tex]F_{13} = \frac{kq_1 q_3}{r^2} \\\\F_{13} = \frac{9\times 10^9 \times 3.95 \times 10^{-9} \times 6 \times 10^{-9} }{(0.198)^2} (+i)= 5.44 \times 10^{-6} \ N \ (+i)[/tex]
The force on the third charge due to second charge is calculated as follows;
[tex]F_{23} = \frac{kq_2q_3}{r^2} \\\\F_{23} = \frac{9\times 10^9 \times 4.96 \times 10^{-9}\times 6 \times 10^{-9} }{(0.297)^2} (-i)\\\\F_{23} = (3.036 \times 10^{-6} ) \ N \ (-i)[/tex]
The net force on the third charge is calculated as follows;
[tex]F_{net} = 5.44 \times 10^{-6} - 3.036 \times 10^{-6} \\\\F_{net} = 2.404 \times 10^{-6} \ N[/tex]
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A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
1) What is the charge on the inner surface of the shell?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
2) What is the charge on the outer surface of the shell?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
3) The tack is now allowed to touch the interior surface of the shell. After this contact, what is the charge on the tack?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
4) What is the charge on the inner surface of the shell now?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
5) What is the charge on the outer surface of the shell now?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
The charge on the inner surface of the shell is -Q
The charge on the outer surface of the shell is Q
After this contact, the charge on the tack is 0
The charge on the inner surface of the shell now is 0
The charge on the outer surface of the shell now is Q
What is the charge on a shell ?The charge on a shell depends on the situation and the conditions of the shell. If the shell is an electrically neutral object, such as a metallic spherical shell, it has no net charge, meaning that the total positive charge is equal to the total negative charge. However, if the shell has an excess or deficit of electrons, it will have a net charge, either positive or negative, depending on whether it has an excess of electrons or a deficit of electrons.
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In the Bohr model of the atom, the electrons are considered to be particles moving in specific and discrete orbits.
a. True
b. False
Answer:
True
Explanation:
Bohr proposed an atomic model in which;
the electrons found in an atom can only occupy orbits for which the angular momentum is quantized, which results in discrete values of energy level. the electrons in such stationary state or ground state emits no light, but if an electron jumps to a lower state, it emits a photon whose energy equals the difference in energy between the two states.A ball is thrown straight upward and reaches the top of its path in 1.71 s (before it starts to come back down). A second ball is thrown at an angle of 34 degrees with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically
Answer:
The second ball must be thrown at 30.01 m/s.
Explanation:
First, we need to find the maximum height (H) reached by the ball 1:
[tex] v_{f_{1}}^{2} = v_{0_{1}}^{2} - 2gH [/tex]
Where:
[tex]v_{f_{1}}[/tex]: is the final speed of ball 1 = 0 (at the maximum height)
[tex]v_{0_{1}}[/tex]: is the initial speed of ball 1
g: is the gravity = 9.81 m/s²
We need to find the initial speed, by using the following equation:
[tex] v_{f_{1}} = v_{0_{1}} - gt [/tex]
Where t is the time = 1.71 s (when it reaches the maximum height)
[tex] v_{0_{1}} = gt = 9.81 m/s^{2}*1.71 s = 16.78 m/s [/tex]
So, the maximum height is:
[tex] H = \frac{v_{0_{1}}^{2}}{2g} = \frac{(16.78 m/s)^{2}}{2*9.81 m/s^{2}} = 14.35 m [/tex]
Finally, the speed at which ball 2 must be thrown is:
[tex]v_{f_{2y}}^{2} = (v_{0_{2y}}sin(\theta)})^{2} - 2gH[/tex]
[tex]v_{0_{2y}}= \frac{\sqrt{2gH}}{sin(\theta)} = \frac{\sqrt{2*9.81 m/s^{2}*14.35 m}}{sin(34)} = 30.01 m/s[/tex]
Therefore, the second ball must be thrown at 30.01 m/s.
I hope it helps you!
In the Faraday pail experiment a metal ball is lowered into a brass pail. In a variation of this experiment, suppose we charge the metal ball with Q and the brass pail with charge 2Q. The ball is slowly lowered into the pail. At no time does the sphere touch the pail. While the ball is inside, the charge on the outside of the pail is _____. The outer part of the pail is then touched/grounded. Then the ball is removed. After the ball is removed, the charge on the outer surface of the pail is ___. Briefly support your answers with reasoning from Gauss's Law.
Answer:
a)
*The charges of which and the cube are of the same sign..
Q_exterior = 3 Q
* the charge of the sphere has a different sign than the charge of the cube,
Q_exterior = Q
b) Q = 0
Explanation:
To correctly describe this experiment, you must remember that in metals charges are mobile and that charges of the same sign repel and of different signs attract.
Let's analyze each situation
a) Suppose that the charges of which and the cube are of the same sign.
When the ball is introduced without touching the walls, its charge Q attracts a charge of equal magnitude and different sign to the internal wall. If we create a Gaussian surface around the inner wall of the sphere the net charge between the ball and the inner wall is zero. Consequently, a face Q should have been generated in the outer wall, therefore in this wall it has a total load of
Q_exterior = 3 Q
Now suppose that the charge of the sphere has a different sign than the charge of the cube, for simplicity let's say that the charge of the sphere is -Q and the cube + 2Q,
Again we create a Gaussian surface outside the inner wall, now the charge on the ball attracts a charge of value + Q to neutralize the charge between the ball and the inner wall. Therefore a load remains on the outer wall
Q_exterior = + Q
b) The cube is connected to earth and it is touched with the ball, in this case the charge of the two bodies is neutralized by the Earth, therefore the bodies have zero charge
Q = 0
A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N
of friction. What distance does the box slide?
Answer:
the distance moved by the box is 70.03 m.
Explanation:
Given;
mass of the box, m = 35 kg
initial velocity of the box, u = 10 m/s
frictional force, F = 25 N
Apply Newton's second law of motion to determine the deceleration of the box;
-F = ma
a = -F / m
a = (-25 ) / 35
a = -0.714 m/s²
The distance moved by the box is calculated as follows;
v² = u² + 2ad
where;
v is the final velocity of the box when it comes to rest = 0
0 = 10² + (2 x - 0.714)d
0 = 100 - 1.428d
1.428d = 100
d = 100 / 1.428
d = 70.03 m
Therefore, the distance moved by the box is 70.03 m.
Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge 1.02 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C. What is the x component of the electric force?
Answer:
a) F_net = 30.47 N , θ = 10.6º
b) Fₓ = 29.95 N
Explanation:
For this exercise we use coulomb's law
F₁₂ = k [tex]k \frac{ q_{1} \ q_{2} }{ r^{2} }[/tex]
the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.
As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces
X axis
Fₓ = [tex]F_{bc x}[/tex]
Y axis
[tex]F_{y}[/tex]Fy = [tex]F_{ab} - F_{bc y}[/tex]
let's find the magnitude of each force
[tex]F_{ab}[/tex] = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²
F_{ab} = 2.82 10¹ N
F_{ab} = 28.2 N
[tex]F_{bc}[/tex] = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²
F_{bc} = 3.75 10¹ N
F_{bc} = 37.5 N
let's use trigonometry to decompose this force
tan θ = y / x
θ = tan⁻¹ and x
θ= tan⁻¹ ¾
θ = 37º
let's break down the force
sin 37 = F_{bcy} / F_{bc}
F_{bcy} = F_{bc} sin 37
F_{bcy} = 37.5 sin 37
F_{bcy} = 22.57 N
cos 37 = F_{bcx} /F_{bc}
F_{bcx} = F_{bc} cos 37
F_{bcx} = 37.5 cos 37
F_{bcx} = 29.95 N
let's do the sum to find the net force
X axis
Fₓ = 29.95 N
Axis y
Fy = 28.2 -22.57
Fy = 5.63 N
we can give the result in two ways
a) F_net = Fₓ i ^ + [tex]F_{y}[/tex] j ^
F_net = 29.95 i ^ + 5.63 j ^
b) in the form of module and angle
let's use the Pythagorean theorem
F_net = [tex]\sqrt{ F_{x}^2 + F_{y}^2 }[/tex]
F_net = √(29.95² + 5.63²)
F_net = 30.47 N
we use trigonometry for the direction
tan θ= [tex]\frac{ F_{y} }{ F_{x} }[/tex]
θ = tan⁻¹ \frac{ F_{y} }{ F_{x} }
θ = tan⁻¹ (5.63 / 29.95)
θ = 10.6º
The 49-g arrow is launched so that it hits and embeds in a 1.45 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.44 m higher than the block's starting point.
Required:
How fast was the arrow moving before it joined the block?
Answer:
the initial speed of the arrow before joining the block is 89.85 m/s
Explanation:
Given;
mass of the arrow, m₁ = 49 g = 0.049 kg
mass of block, m₂ = 1.45 kg
height reached by the arrow and the block, h = 0.44 m
The gravitational potential energy of the block and arrow system;
P.E = mgh
P.E = (1.45 + 0.049) x 9.8 x 0.44
P.E = 6.464 J
The final velocity of the system after collision is calculated as;
K.E = ¹/₂mv²
6.464 = ¹/₂(1.45 + 0.049)v²
6.464 = 0.7495v²
v² = 6.464 / 0.7495
v² = 8.6244
v = √8.6244
v = 2.937 m/s
Apply principle of conservation of linear momentum to determine the initial speed of the arrow;
[tex]P_{initial} = P_{final}\\\\mv_{arrow} + mv_{block} = (m_1 + m_2)V\\\\0.049(v) + 1.45(0) = (0.049 + 1.45)2.937\\\\0.049v = 4.4026\\\\v = \frac{4.4026}{0.049} \\\\v = 89.85 \ m/s[/tex]
Therefore, the initial speed of the arrow before joining the block is 89.85 m/s
The arrow moving as the speed of "76.36 m/s".
According to the question,
By using the conservation of energy, we have
→ [tex]K.E=P.E[/tex]
→ [tex]\frac{1}{2} (m_1+m_2)v_2^2= (m_1+m_2)gh[/tex]
or,
→ [tex]v_2 = \sqrt{2mgh}[/tex]
By substituting the values, we have
→ [tex]= \sqrt{2\times 9.8\times 0.44}[/tex]
→ [tex]=2.469 \ m/s[/tex]
Now,
By using the conservation of momentum, we get
→ [tex]m_1 v_1 = (m_1+m_2) v_2[/tex]
or,
→ [tex]v_1 = \frac{(m_1+m_2)v_2}{m_1}[/tex]
[tex]= \frac{1.45+0.049}{0.049}\times 2.469[/tex]
[tex]= 30.6\times 2.496[/tex]
[tex]= 76.36 \ m/s[/tex]
Thus the above approach is correct.
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A neutral metal bob is hanging on the bottom of a pendulum that is 15 cm long. A charged balloon is held near the metal bob and the pendulum is pulled up to a vertical angle of 20-deg. If the mass of the metal bob is 0.025kg, what is the charge on the balloon.
Answer:
Explanation:
See the figure attached
F is electrostatic force .
T cos20 = mg
T sin20 = F
Tan20 = F / mg
F = mg tan 20 = .025 x 9.8 tan20
= .09 N
Distance between bob and balloon
= 15 sin20 = 5.1 cm = .051 m
If q be the charge on balloon
F = 9 x 10⁹ x q² / .051²
= 3460 x 10⁹ q² = .09
q² = 26 x 10⁻⁶ x 10⁻⁹
q = 16.12 x 10⁻⁸ C .
4. A ball is dropped from height of 45 m.
Then distance covered in last 0.6 sec of
its motion will be
Answer:
1.76 m
Explanation:
Height from which the object is dropped = 45 m
Time (t) remaining for the ball to land = 0.6 s
Height (h) in the remaining =?
The height to which the object falls in the remaining time can be obtained as follow:
Time (t) remaining for the ball to land = 0.6 s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) in the remaining =?
h = ½gt²
h = ½ × 9.8 × 0.6²
h = 4.9 × 0.36
h = 1.76 m
Thus, the distance travelled in the last 0.6 s is 1.76 m
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm
Solution :
1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).
2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).
3. Non-selective scatter takes place when particle size in greater than the wavelength (λ).
We have the sizes of different particles :
[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]
Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]
Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]
Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]
Wavelength [tex]$ O_2 $[/tex] Smoke particles Cloud droplets Rain droplets
[tex]$10^{-10} \ m$[/tex] [tex]$ 3 \times 10^{-7} \ m$[/tex] [tex]$ 2 \times 10^{-5} \ m$[/tex] [tex]$ 3 \times 10^{-3} \ m$[/tex]
[tex]$5500 \times 10^{-4} \ m$[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$11 \times 10^{-6} \ m $[/tex] Rayleigh Rayleigh Non-selective Non-selective
[tex]$1600 \times 10^{-10} \ m $[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$10^{-2} \ m $[/tex] Rayleigh Rayleigh Rayleigh Mie
Determine the magnitude of the electric field at the point P. Express your answer in terms of Q, x, a, and k. Express your answer in terms of the variables Q, x, a, k, and appropriate constants.
Complete Question
The question image is in the first uploaded image
Answer:
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
Explanation:
From the question we are told that
Distance b/w Q mid point and P is given as x
Generally the equation for magnitude of the electric field at the point P is given as
[tex]E=\frac{kQ}{d^2}[/tex]
where
[tex]k=\frac{1}{4\pi e_0}[/tex]
[tex]d=x^2-a^2[/tex]
Therefore
[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]
[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]
Therefore equation for magnitude of the electric field at the point P is
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:
[tex]H_m=1.65m[/tex]
[tex]H_E=1.16307m[/tex]
Explanation:
From the question we are told that
Mass of ball [tex]M=2kg[/tex]
Length of string [tex]L= 2m[/tex]
Wind force [tex]F=13.2N[/tex]
Generally the equation for [tex]\angle \theta[/tex] is mathematically given as
[tex]tan\theta=\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{13.2}{2*2}[/tex]
[tex]\theta=73.14\textdegree[/tex]
Max angle =[tex]2*\theta= 2*73.14=>146.28\textdegree[/tex]
Generally the equation for max Height [tex]H_m[/tex] is mathematically given as
[tex]H_m=L(1-cos146.28)[/tex]
[tex]H_m=0.9(1+0.8318)[/tex]
[tex]H_m=1.65m[/tex]
Generally the equation for Equilibrium Height [tex]H_E[/tex] is mathematically given as
[tex]H_E=L(1-cos73.14)[/tex]
[tex]H_E=0.9(1+0.2923)[/tex]
[tex]H_E=1.16307m[/tex]
A 6.11-g bullet is moving horizontally with a velocity of 366 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1206 g, and its velocity is 0.662 m/s after the bullet passes through it. The mass of the second block is 1550 g. (a) What is the velocity of the second block after the bullet imbeds itself
Answer:
[tex]V=1.86m/s[/tex]
Explanation:
Mass of bullet [tex]M_B=6.11g[/tex]
Velocity of bullet [tex]V_B=366m/s[/tex]
Mass of first block [tex]M_b_1=1206g[/tex]
Velocity of block [tex]V_b=0.662m/s[/tex]
Mass of second block [tex]M_b_2=1550g[/tex]
Generally the total momentum before collision is mathematically given as
[tex]P_1=0.006kg*366+0+0[/tex]
[tex]P_1=2.196kg\ m/s[/tex]
Generally the total momentum after collision is mathematically given as
[tex]P_2=(1.206kg*0.633)+(1.550+0.00611)V[/tex]
[tex]P_2=0.763398+1.55611V[/tex]
Generally the total momentum is mathematically given as
[tex]P_1=P_2[/tex]
[tex]2.196=0.763398+1.55611V[/tex]
[tex]V=\frac{2.196+0.763398}{1.55611}[/tex]
[tex]V=1.86m/s[/tex]
An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours, (a) Find the charge in coulombs that passes through the range, (b) Find the power absorbed by the range, (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.
Answer:
A. 72000 C
B. 1100 W
C. 26.4 cents.
Explanation:
From the question given above, the following data were obtained:
Current (I) = 10 A
Voltage (V) = 110 V
Time (t) = 2 h
A. Determination of the charge.
We'll begin by converting 2 h to seconds. This can be obtained as follow:
1 h = 3600 s
Therefore,
2 h = 2 h × 3600 s / 1 h
2 h = 7200 s
Thus, 2 h is equivalent to 7200 s.
Finally, we shall determine the charge. This can be obtained as follow:
Current (I) = 10 A
Time (t) = 7200 s
Charge (Q) =?
Q = It
Q = 10 × 7200
Q = 72000 C
B. Determination of the power.
Current (I) = 10 A
Voltage (V) = 110 V
Power (P) =?
P = IV
P = 10 × 110
P = 1100 W
C. Determination of the cost of operation.
We'll begin by converting 1100 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
1100 W = 1100 W × 1 KW / 1000 W
1100 W = 1.1 KW
Thus, 1100 W is equivalent to 1.1 KW
Next, we shall determine the energy consumption of the range. This can be obtained as follow:
Power (P) = 1.1 KW
Time (t) = 2 h
Energy (E) =?
E = Pt
E = 1.1 × 2
E = 2.2 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost 12 cents.
Therefore, 2.2 KWh will cost = 2.2 × 12
= 26.4 cents.
Thus, the cost of operating the range for 2 h is 26.4 cents.
A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.
What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?
Answer:
V= -3.6*10⁻¹¹ V
Explanation:
Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:[tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]
At r= 0.3 m the spherical surface can be written as follows:[tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]
Replacing (2) in (1) and solving for E, we have:[tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]
Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:[tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]
V = -3.6*10¹¹ Volts.The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]
We can arrive at this answer as follows:
To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]
Solving these equations will have:[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]
As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]
More information about Gauss' law at the link:
https://brainly.com/question/14705081
PLEASE HELP PLEASEEEE
Answer:
How can I help you??? Plz insert some questions
HELP PLEASE!!!
Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s.If the collision between the players lasted for 0.04 s, determine the impact force on either during the collision
Answer:
10kN
Explanation:
Given data
m1= 50kg
u1= 3m/s
m2= 100kg
u2= 6m/s
v1= 2m/s
time= 0.04s
let us find the final velocity of Bruce v1
from the conservation of linear momentum
m1u1+m2u2=m1v1+m2v2
substitute
50*3+100*6= 50*v1+100*2
150+600=50v1+200
750-200=50v1
550= 50v1
divide both sides by 50
v1= 550/50
v1=11 m/s
From
F= mΔv/t
for Bruce
F=50*(11-3)/0.04
F=50*8/0.04
F=400/0.04
F=10000
F=10kN
for Max
F=100*(6-2)/0.04
F=100*4/0.04
F=400/0.04
F=10000
F=10kN
Work put into a machine can never be reclaimed.
True
False
Please answer?!
I am not sure but I think it is false
Fill in the blank with the correct word below (from the reading_):
helps you track your progress once you have made a lifestyle
change.
Self-monitoring
Healthy food
Regular xxercise
Goals
Answer:I think it’s self monitoring sorry if wrong
Explanation:
Answer:
It self monitoring
Explanation:
I took the test
two identical balls are rolling down a hill ball 2 is rolling faster than ball 1 which ball has more kinetic energy
A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.7 s the rocket is at a height of 93 m.
What are the magnitude and direction of the rocket's acceleration?
What is its speed at this elevation?
Answer:
The magnitude and direction of the rocket acceleration is 68.89 m/s² upward.
The speed of the rocket at the given elevation is 186 m/s.
Explanation:
Given;
time to reach the given height, t = 2.7 s
height reached, h = 93 m
initial velocity of the rocket, u = 0
The magnitude and direction of the rocket acceleration is calculated as;
h = ut + ¹/₂at²
h = 0 + ¹/₂at²
h = ¹/₂at²
a = 2h / t²
a = (2 x 93) / 2.7
a = 68.89 m/s²
the direction of the acceleration is upward.
The speed at this elevation, V = u + at
V = at
V = 68.89 x 2.7
V = 186 m/s
If there is "waste" energy, does the Law of Conservation of Energy still apply?
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.This table shows the mass and volume of four different objects.
A two-column table with 4 rows. The first column titled objects has entries W, X, Y, Z. The second column titled Measurements has entries Mass: 16 grams Volume: 84 centimeters cubed in the first cell, Mass: 12 grams Volume: 5 centimeters cubed in the second cell, Mass: 4 grams Volume: 6 centimeters cubed in the third cell, Mass: 408 grams Volume: 216 centimeters cubed in the fourth cell.
Which ranks the objects from most to least dense?
Answer:
Here its right but its also better than Barney's response
Explanation:
W, Y, Z, X or C
Answer:
W, Y, Z, X
Explanation:
is 0.8 kilograms bigger then 80 grams
Answer:
Yes
Explanation:
0.8 kilograms is equal to 800 grams
Answer:
Yes, 0.8 kilograms is greater than 80 grams
Explanation:
0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.
Sorry if I'm wrong, correct me.
In the measurement 365 cL the “c” stands for the___ And the “ L”stands for the___
Answers
A.base unit and prefix
B.suffix and prefix
C.prefix and suffix
D.prefix and base unit
Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with in a one-dimensional box 34.0 pm in length.
The question is incomplete. The complete question is :
Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 6 in a one-dimensional box 34.0 pm in length.
Solution :
In an one dimensional box, energy of a particle is given by :
[tex]$E=\frac{n^2h^2}{8ma^2}$[/tex]
Here, h = Planck's constant
n = level of energy
= 6
m = mass of particle
a = box length
For n = 6, the energy associated is :
[tex]$\Delta E = E_6 - E_1 $[/tex]
[tex]$\Delta E = \left( \frac{n_6^2h_2}{8ma^2}\right) - \left( \frac{n_1^2h_2}{8ma^2}\right) $[/tex]
[tex]$=\frac{h^2(n_6^2 - n_1^2)}{8ma^2}$[/tex]
We know that,
[tex]$E = \frac{hc}{\lambda} $[/tex]
Here, λ = wavelength
h = Plank's constant
c = velocity of light
So the wavelength,
[tex]$= \frac{hc}{E}$[/tex]
[tex]$=\frac{hc}{\frac{h^2(n_6^2 - n_1^2)}{8ma^2}}$[/tex]
[tex]$=\frac{8ma^2c}{h(n_6^2 - n_1^2)}$[/tex]
[tex]$=\frac{8 \times 9.109 \times 10^{-31}(0.34 \times 10^{-10})^2 (3 \times 10^8)}{6.626 \times 10^{-34} \times (36-1)}$[/tex]
[tex]$= \frac{ 8 \times 9.109 \times 0.34 \times 0.34 \times 3 \times 10^{-43}}{6.626 \times 35 \times 10^{-34}}$[/tex]
[tex]$=\frac{25.27 \times 10^{-43}}{231.91 \times 10^{-34}}$[/tex]
[tex]$= 0.108 \times 10^{-9}$[/tex] m
= 108 pm
For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball 3.2 s later at the same height from which it was thrown. What was the initial upward speed of the ball?
Answer:
15.68 m/s
Explanation:
Given that,
She catches the ball 3.2 s later at the same height from which it was thrown.
When it reaches the maximum height, its height is equal to 0.
It will move under the action of gravity.
[tex]t=\dfrac{2u}{g}[/tex]
2 here comes for the time of ascent and descent.
So,
[tex]u=\dfrac{tg}{2}\\\\u=\dfrac{3.2\times 9.8}{2}\\\\u=15.68\ m/s[/tex]
So, the initial upward speed of the ball is 15.68 m/s.
If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no longer any force on the ball due to the woman's hand), what is the relationship between aaa and ggg, the magnitude of the acceleration of gravity
Explanation:
At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.
g= acceleration due to gravity