Answer:
ΔT = 25°C
Explanation:
Given that.
The mass of a bullet, m₁ = 30 g = 0.03 kg
The speed of the bullet, v = 900 m/s
Mass of soft iron, m₂ = 1 k
The specific heat of iron, c=490J/kg°C
We need to find the increase in temperature of iron. using the conservation of energy,
Kinetic energy = heat absorbed
[tex]\dfrac{1}{2}m_1v^2=m_2c\Delta T\\\\\Delta T=\dfrac{\dfrac{1}{2}m_1v^2}{m_2c}\\\\\Delta T=\dfrac{\dfrac{1}{2}\times 0.03\times 900^2}{1\times 490}\\\\=24.79^{\circ} C\\\\or\\\\\Delta T=25^{\circ} C[/tex]
So, the correct option is (A).
which unit would be most suitable for its scale?
A mm
B
с
crn?
D
cm
[0625_504_9p_1].
8
A piece of cotton is measured between two points on a ruler.
1
coton
BAS
2
4
5
6
7
8
9
10
11
12
13
14
15 16
when the lenge of coton is wound closely around a pen, goes round six times.
pen
six turns of coton
दे-
What is the distance onde round the pen?
4 2.2 m
B 26 cm
с
13.2 cm
D 15.6 cm
Answer:
Mm, thats the answer trust me men
Three forces are pulling on the same object such that the system is in equilibrium. Their magnitudes are F1 = 2.83 N.F= 3.35 N. and F3 = 3.64 N, and they make angles of 0, = 45.0°, 02 = -63.43 and 03 =164.05° with respect to the x-axis, respectively.
Required:
a. What is the x-component of the force vector F1?
b. What is the y-component of the force vector F1?
(a) 2.001N
(b) 2.001N
Explanation:A sketch of the scenario has been attached to this response.
Since only the force vector F₁ is required, the only force shown in the sketch is F₁.
As shown in the sketch;
The x-component of the force vector F₁ = [tex]F_{x}[/tex]
The y-component of the force vector F₁ = [tex]F_{y}[/tex]
The magnitude of F₁ as given in the question = 2.83N
The angle that the force makes with respect to the x-axis = 45.0°
Using the trigonometric ratio, we see that;
(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]
=> [tex]F_{x}[/tex] = F₁ cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 x 0.7071
=> [tex]F_{x}[/tex] = 2.001N
(b) Also;
sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]
=> [tex]F_{y}[/tex] = F₁ sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 x 0.7071
=> [tex]F_{y}[/tex] = 2.001N
Therefore, the x-component and y-component of the force vector F₁ is 2.001N
The x and y component of vector F1 is mathematically given as
F_x = 2.001N
F_y= 2.001N
What is the x and y component of vector F1?Question Parameters:
Generally, the equation for the x-component is mathematically given as
x=Fsin\theta
Therefore
F_x = F₁ cos 45.0°
F_x = 2.83 x 0.7071
F_x = 2.001N
For y component
x=Fcos\theta
F_y = F₁ sin 45.0
F_y = 2.83 x 0.7071
F_y= 2.001N
Read more about Cartesian
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A 5.0-kg solid cylinder of radius 0.25 mis free to rotate about an axle that runs along the cylinders length and passes through its center. A thread wrapped around the cylinder is weighed down by a mass of 2.0 kg so as to unwrap and make the cylinder rotate as this mass falls. Ignore any friction in the axle. If there is no slippage between the thread and the cylinder, and the cylinder starts from rest (a) Calculate the velocity of the block after it has fallen a distance of 2.0m. Give your answer in m.s (b) Calculate the total work done by the rope on the cylinder after the block has fallen a distance of 2.0 m. Give your answer in Joule.
Answer:
157n is the correct answer
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
To solve more questions on Work, Energy and Power, visit the link below -
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Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
An exoplanet has three times the mass and one-fourth the radius of the Earth. Find the acceleration due to gravity on its surface, in terms of g, the acceleration of gravity at Earth's surface. A planet's gravitational acceleration is given by gp = G Mp/r^2p
a. 12.0 g.
b. 48.0 g.
c. 6.00 g.
d. 96.0 g.
e. 24.0 g.
Answer:
b. 48.0 g.
Explanation:
Given;
mass of the exoplanet, [tex]M_p = 3M_e[/tex]
radius of the exoplanet, [tex]r_p = \frac{1}{4} r_e[/tex]
The acceleration due to gravity of the planet is calculated as;
[tex]g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48 \ g[/tex]
Therefore, the correct option is b. 48.0 g
What is the change in internal energy if 70 J of heat is added to a system and
the system does 30 J of work on the surroundings. Uze al-Q-W.
O A. 40 J
O B. -40.3
O C. 100.
D. -1003
Answer:
A. 40 J
Explanation:
Given;
heat added to the system, Q = 70 J
work done by the system, W = 30 J
The change in the internal energy of the system is calculated using the first law of thermodynamic as shown below;
ΔU = Q - W
ΔU = 70 J - 30 J
ΔU = 40 J
Therefore, the change in the internal energy of the system is 40J
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
A 2 kg stone is dropped from a height of 100 m. How far does it travel in the third second? take g = 9.8 m/s2
Answer:
S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m
Explanation:
Convert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
Answer:
1) 0.3675
2) 0.367
3) 72.6186
4) 0.5
5) 0.000671
Answer:
1) 367.5 mg = 0.3675 g
2) 367 mL = 0.367 L
3) 28.59 in = 72.61 cm
4) 8 0z = 0.5 lb
5) 0.671 mm = 0.0000671 m
Question: A car of mass 500kg travelling at 12m/s enters a stretch of road where there's a constant resistive force of 8000N. The car comes to a stop due to this resistive force. Calculate the distance travelled by the car before stopping.
Answer:
ans: 2.25 meter
explanation
use following equations
F = ma
V = U + aT
S = UT + 1/2 aT^2
Baseball runner with a mass of 70kg, moving at 2.7m/s and collides head-on into a shortstop with a mass of 85kg and a velocity of 1.6m/s. What will be the resultant velocity of the system when they make contact with each other
Answer:
The speed of the combined mass after the collision is 2.1 m/s.
Explanation:
mass of runner, m = 70 kg
speed of runner, u = 2.7 m/s
mass of shortstop, m' = 85 kg
speed of shortstop, u' = 1.6 m/s
Let the velocity of combined system is v.
Use conservation of momentum
Momentum before collision = momentum after collision
m u + m' u' = (m + m') v
70 x 2.7 + 85 x 1.6 = (70 + 85) v
189 + 136 = 155 v
v = 2.1 m/s
Two small silver spheres, each of mass m=6.2 g, are separated by distance d=1.2 m. As a result of transfer of some fraction of electrons from one sphere to the other, there is an attractive force F=900 KN between the spheres. Calculate the fraction of electrons transferred from one of the spheres: __________
To evaluate the total number of electrons in a silver sphere, you will need to invoke Avogadro's number, the molar mass of silver equal to 107.87 g/mol and the fact that silver has 47 electrons per atom.
Answer:
4.60 × 10⁻⁸
Explanation:
From the given information;
Assuming that q charges are transferred, then:
[tex]F = \dfrac{kq^2}{d^2}[/tex]
where;
k = 9 ×10⁹
[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]
[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]
q = 0.012 C
No of the electrons transferred is:
[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]
[tex]= 7.5 \times 10^{16} \ C[/tex]
Initial number of electrons = N × 47 × no of moles
here;
[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]
no of moles = 0.0575 mol
∴
Initial number of electrons = [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]
= 1.63 × 10²⁴
The fraction of electrons transferred [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]
= 4.60 × 10⁻⁸
The slope at point A of the graph given below is:
WILL MARK BRAINLIEST TO CORRECT ANSWER
RQ/PQ I think
rise/run
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
A car has a mass of 900 kg is accelerated from rest at a rate of 1.2 m/s calculate the time taken to reach 30/s
Answer:
12+2=24+30+2=66
Explanation:
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 kg, up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00x102. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s
Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer numerically, in meters per second.
1. How to approach the problem
2. Find the total work done on the box
3. Initial kinetic energy
4. What is the final kinetic energy?
Answer:
v₀ = 2.67 m / s
Explanation:
This problem can be solved using the Kinetic Enemy Work Theorem
W = ΔK
Work is defined by the relation
W = fr. d
The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.
the friction force opposes the displacement therefore its angle is 180º
W = - fr d
Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane
Y axis
N- Wy = 0
N - W cos tea = 0
the friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
W = - μ W cos θ d
let's look for kinetic energy
the minimum velocity at the highest point is zero
K_f = 0
the initial kinetic energy is
K₀ = ½ m v₀²
we substitute energy in the work relationship
- μ W cos θ d = 0 - ½ m v₀²
v₀² = - μ W cos θ 2d / m
Let's use trigonometry to find distance d
sin θ= y / d
d = y /sin θ
d = 3.50 / sin 30
d = 7 m
let's calculate
v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50
v₀ = √7.129
v₀ = 2.67 m / s
Its Acceleration during the upward Journey ?
Mary applies a force of 73 N to push a box with an acceleration of 0.48 m/s^2. When she increases the pushing force to 84 N, the box's acceleration changes to 0.64 m/s^2. There is a constant friction force present between the floor and the box.
Required:
a. What is the mass of the box?
b. What is the coefficient of kinetic friction between the floor and the box?
Answer: [tex]68.75\ kg, 0.06[/tex]
Explanation:
Mary applies a force of 73 N to create an acceleration of [tex]0.48\ m/s^2[/tex]
When She increases force to 84 N, it creates an acceleration of [tex]0.64\ m/s^2[/tex]
Friction opposes the motion of box
[tex]\Rightarrow 73-f=m\times 0.48\quad \ldots(i)\\\Rightarrow 84-f=m\times 0.64\quad \ldots(ii)[/tex]
Subtract (i) from (ii)
[tex]\Rightarrow 11=m(0.64-0.48)\\\Rightarrow m=68.75\ kg[/tex]
Therefore friction is
[tex]\Rightarrow f=73-68.75\times 0.48\\\Rightarrow f=73-33\\\Rightarrow f=40\ N[/tex]
Here, friction is kinetic friction which is given by
[tex]\Rightarrow f=\mu_kmg\\\Rightarrow 40=\mu_k 68.75\times 9.8\\\Rightarrow \mu_k=0.061[/tex]
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
Answer:
Explanation:
Here's what we know and in which dimension:
y dimension:
[tex]v_0=30[/tex] m/s
v = 0 (I'll get to that injust a second)
a = -9.8 m/s/s
The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.
x dimension:
Δx = 70 m
v = ??
Velocity is our unknown.
Solving for the time in the y dimension:
[tex]v=v_0+at[/tex] and filling in:
0 = 30 + (-9.8)t and
-30 = -9.8t so
t = 3.1 seconds
We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.
In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.
Δx = vt and
70 = v(6.2) so
v = 11.3 m/s
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
How is fitness walking beneficial?
It can relieve stress and improve mood.
It can decrease energy levels.
It can decrease perspiration.
It can relieve allergy symptoms.
Answer:
It can relieve stress and improve mood.
What is the y component of a vector that is 673 m at -38o?
Answer:
D_y = 414.38m
Explanation:
D_y = D*sin(x)
D_y = 673m*sin(38°)
D_y = 414.38m
Diwn unscramble the word
Answer:
WIND Is what you're looking for
Explanation:
The word is WIND
A 92-kg man climbs into a car with worn out shock absorbers, and this causes the car to drop down 4.5 cm. As he drives along he hits a bump, which starts the car oscillating at an angular frequency of 4.52 rad/s. What is the mass of the car ?A) 890 kg
B) 1900 kg
C) 920 kg
D) 990 kg
E) 760 kg
Answer:
the mass of the car is 890 kg
Explanation:
Given;
mass of the man, m = 92 kg
displacement of the car's spring, x = 4.5 cm = 0.045 m
acceleration due to gravity, g = 9.8 m/s²
The spring constant of the car,
f = kx
where;
f is the weight of the man on the car = mg
mg = kx
k = mg/x
k = (92 x 9.8) / 0.045
k = 20,035.56 N/m
The angular speed of car, ω, when the is inside is given as 4.52 rad/s
The total mass of the car and the man is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\m = \frac{k}{\omega^2} = \frac{20,035.56}{(4.52)^2} = 980.7 \ kg[/tex]
The mass of the car alone = 980.7 kg - 92 kg
= 888.7 kg
≅ 890 kg
Therefore, the mass of the car is 890 kg
What must be the same for any two resistors that are connected in a series
Which of the following is acceleration toward the center of a circular motion? O A. Centripetal acceleration O B. Uniform circular motion O C. Centrifugal force D. Centripetal force
PLEASE HELP ASAP!!
We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...
There are only two charged particles in a particular region. Particle 1 carries a charge of 3q and is located on the negative x-axis a distance d from the origin. Particle 2 carries a charge of -2q and is located on the positive x-axis a distance d from the origin.
Required:
Where is it possible to have the net field caused by these two charges equal to zero?
Answer:
The net field will be the sum of the fields created by each charge.
where the charge Q in a position r' is given by:
E(r) = k*Q/(r - r')^2
Where k is a constant, and r is the point where we are calculating the electric field.
Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:
E₁(r) = k*3q/(r - r₁)^2
While for the other charge of -2q in the position r₂ = (d, 0, 0)
The electric field is:
E₂(r) = -k*2*q/(r - r₂)^2
Then the net field at the point r is:
E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2
E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)
Then if the we want to find the points r = (x, y, z) such that:
E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)
Then we must have:
0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)
Also remember that the distance between two points:
(x, y, z) and (x', y', z') is given by:
D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)
Then we can rewrite:
r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)
= √( (x + d))^2 + y^2 + z^2)
and
r - r₂ = √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)
= √( (x - d))^2 + y^2 + z^2)
Replacing that in our equation we get:
0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)
0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)
We want to find the values of x, y, z such that the above equation is true.
2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)
2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]
2*(x + d)^2 + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2
2*(x + d)^2 - 3*(x - d)^2 = 3*y^2 + 3*z^2 - 2*y^2 - 2*z^2
2*(x + d)^2 - 3*(x - d)^2 = y^2 + z^2
2*x^2 + 2*2*x*d + 2*d^2 - 3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2
-x^2 + 10*x*d - d^2 = y^2 + z^2
we can rewrite this as:
- ( x^2 - 10*x*d + d^2) = y^2 + z^2
now we can add and subtract 24*d^2 inside the parenthesis to get
- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) = y^2 + z^2
-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2
-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2
The thing inside the parenthesis is a perfect square:
-(x - 5d)^2 + 24d^2 = y^2 + z^2
we can rewrite this as:
24d^2 = y^2 + z^2 + (x - 5d)^2
This equation gives us the points (x, y, z) such that the electric field is zero.
Where we need to replace two of these values to find the other, for example, if y = z = 0
24d^2 = (x - 5d)^2
√(24d^2) = x - 5d
√24*d = x - 5d
√24*d + 5d = x
so in the point (√24*d + 5d, 0, 0) the net field is zero.
Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
