Answer:
(a) 1.14 s
(b) 0.87 s
Explanation:
A person moves by the help of frictional force, as a result of gtound reaction. So, the formula for frictional force is:
F = μR
where,
F = frictional force
μ = coefficient of friction
R = Normal Reaction = Weight of Body = W = mg
Therefore,
F = μmg
but, from Newton's 2nd Law of Motion:
F = ma
Comparing both equations, we get:
μmg = ma
a = μg ---------- equation (1)
Now, to calculate the distance moved by a body, we use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
using equation (1), we get:
s = (Vi)(t) + (0.5)μgt²
where,
s = distance moved by body
Vi = initial velocity of body
t = time taken to cover the distance
g = acceleration due to gravity
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.5
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)
t = √1.30612 s²
t = 1.14 s
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.5
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)
t = √1.30612 s²
t = 1.14 s
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.87
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.87)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.87)(9.8 m/s²)
t = √0.75 s²
t = 0.87 s
Write the second law of motion’s formula and its unit.
please answer this question, for 13 points!!!!!!!!!
Answer:
The Formula Is F = m * a And It's Units Is (kg)(m/s2)
Explanation:
As you are waiting at the Willard airport for a friend coming from Chicago, suddenly something goes wrong with the loudspeakers, and they start making a low pitch hum. You realized that you are standing exactly in the middle between the two loudspeakers on the opposite walls. Since the sound from both loudspeakers has the same intensity, you are standing in the interference maximum As you move toward one of the loudspeakers, the intensity decreases. When you have moved by 0.5 m , you hit the first point of zero intensity. Assuming that the speed of sound is v = 330 m/s , what is the frequency f of the sound in Hz ?
Answer:
The frequency is [tex]f = 165 Hz[/tex]
Explanation:
From the question we are told that
The position of zero intensity is [tex]L = 0.5 \ m[/tex] from the center
Now the wavelength of the sound is mathematical represented as
[tex]\lambda = 4 L[/tex]
[tex]\lambda = 4 * 0.5[/tex]
[tex]\lambda = 2 \ m[/tex]
Now the frequency of the sound is mathematically represented as
[tex]f = \frac{v}{\lambda}[/tex]
substituting values
[tex]f = \frac{330}{ 2}[/tex]
[tex]f = 165 Hz[/tex]
On the moon, what would be the force of gravity acting on an object that has a mass of 7kg?
Answer:
Force of gravity, F = 70 N
Explanation:
It is required to find the force of gravity acting on an object that has a mass of 7 kg. Force of gravity always acts in downward direction.
The force of gravity is equal to the weight of an object. It is given by :
[tex]F=mg[/tex]
g = acceleration due to gravity, for Earth, g = 10 m/s²
So,
[tex]F=7\times 10\\\\F=70\ N[/tex]
So, 70 N of force of acting on an object.
The inhabitants of a planet in another galaxy have their eyes at the exact center of their 4.0-m-long bodies. How long must a plane mirror be for such a creature to be able to see all of its body in the mirror
Answer:
The correct answer will be "2 m".
Explanation:
As we know,
⇒ [tex]\frac{Mirror's \ size}{Body's \ height} =\frac{1}{2}[/tex]
Now,
⇒ [tex]Size \ of \ a \ mirror = \frac{n}{2}[/tex]
Then,
⇒ [tex]Size \ of \ mirror = \frac{4}{2}[/tex]
⇒ [tex]=2 \ meter[/tex]
The speed of sound
Medium Air (0C) Air (20°C) Helium (0°C) Ethyl alcohol Water Human tissue (ultrasound Lead Aluminum Granite Diamond
Speed (m/s) 331 343 970 1170 1480 1540 1200 5100 6000 12.000
Oil explorers set off explosives to make loud sounds, then listen for the echoes from underground oil deposits. Geologists suspect that there is oil under 480-m-deep Lake Physics. It's known that Lake Physics is carved out of a granite basin. Explorers detect a weak echo 0.930 ss after exploding dynamite at the lake surface. Part A If it's really oil, how deep will they have to drill into the granite to reach it
Answer:
Explanation:
Let the required depth be d .
Sound will first travel trough 480 m deep lake . Then it will enter granite layer . Sound travelling through granite will reach oil level , get reflected and come back to the surface of lake as echo . Total time taken by sound to travel total distance is .93 s
Total distance = 2d + 2 x 480 m
= 2d + 960 m
speed of sound in granite is given as 6000 m / s and speed through water is 1480 m /s
total time taken
= 2d / 6000 + 960 / 1480 = .93
2d / 6000 + .6486 = .93
2d / 6000 = .2814
d= 844.2 m
A ball is kicked horizontally with a speed of 5.0 ms-1 from the roof of a house 3 m high. When will the ball hit the ground?
Answer:
the time taken for the ball to hit the ground is 0.424 s
Explanation:
Given;
velocity of the ball, u = 5 m/s
height of the house which the ball was kicked, h = 3m
Apply kinematic equation;
h = ut + ¹/₂gt²
where;
h is height above ground
u is velocity
g is acceleration due to gravity
t is the time taken for the ball to hit the ground
Substitute the given values and solve for t
3 = 5t + ¹/₂(9.8)t²
3 = 5t + 4.9t²
4.9t² + 5t -3 = 0
a = 4.9, b = 5, c = -3
Solve for t using formula method
[tex]t = \frac{-5 +/-\sqrt{5^2-4(4.9*-3)}}{2(4.9)} = \frac{-5+/-(9.154)}{9.8} \\\\t = \frac{-5+9.154}{9.8} \ or \ \frac{-5-9.154}{9.8} \\\\t = \frac{4.154}{9.8} \ or \ \frac{-14.154}{9.8} \\\\t = 0.424 \ sec \ or -1.444 \ sec\\\\Thus, t = 0.424 \ sec[/tex]
Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0.034 m. You wish the mass to have a kinetic energy of 13.0 Joules when released. How strong will the string need to be
Answer:
T = 764.41 N
Explanation:
In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:
[tex]F_c=m\frac{v^2}{r}[/tex] (1)
m: mass object = 2.3 kg
r: radius of the circular orbit = 0.034 m
v: tangential speed of the object
However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:
[tex]K=\frac{1}{2}mv^2[/tex]
You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:
[tex]v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}[/tex]
you replace this value of v in the equation (1). Also, you replace the values of r and m:
[tex]F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N[/tex]
hence, the tension in the string must be T = Fc = 764.41 N
2. If electrons are removed from an object, is the object positively or negatively
charged
positive
Because adding electrons makes an object to be negative and removing makes it to be positive
Suppose that a uniform electric field exists in a certain region of space. Now consider a mathematical plane surface of area A. To maximize the flux through this surface, the face of the plane (not its normal)
Answer: The normal of the plane must be parallel to the electric field vector.
Explanation:
the normal to the surface is defined as a versor that is perpendicular to the plane.
Now, if the angle between this normal and the line of the field is θ, we have that the flux can be written as:
Φ = E*A*cos(θ)
Where E is the field, A is the area and θ is the angle already defined.
Now, this maximizes when θ = 0.
This means that the normal of the surface must be parallel to the electric field
When time is measured in days, the decay constant for a particular radioactive isotope is 0.16. Determine the time required for a confined sample of the isotope to fall to 80% of its original value.
Answer:
The time take is [tex]t = 1.3964 \ days[/tex]
Explanation:
From the question we are told that
The decay constant is [tex]\lambda = 0.16[/tex]
The percentage fall is [tex]c = 0.80[/tex]
The equation for radioactive decay is mathematically represented as
[tex]N(t) = N_o * e^{- \lambda t }[/tex]
Where is [tex]N(t)[/tex] is the new amount of the new the isotope while [tex]N_o[/tex] is the original
At initial [tex]N_o = 100[/tex]% = 1
At [tex]N(t ) = 80[/tex]% = 0.80
[tex]0.80 = 1 * e^{- 0.16 t }[/tex]
=> [tex]-0.223 = -0.16 t[/tex]
=> [tex]t = 1.3964 \ days[/tex]
Answer:
t = 1.4 days
Explanation:
The law of radioactive decay gives the amount of radioactive substance, left after a certain amount of time has passed. The formula of law of radioactive decay is given as follows:
N = N₀ (e)^-λt
where,
λ = decay constant = 0.16
N₀ = Initial Amount of the Substance
N = The Amount of Substance Left after Decay = 80% of N₀ = 0.8 N₀
t = Time Required by the Substance to decay to final value = ?
Substituting these values in the law of radioactive decay formula, we get:
0.8 N₀ = N₀ (e)^-0.16 t
0.8 = (e)^-0.16 t
ln (0.8) = -0.16 t
t = - 0.2231/-0.16
t = 1.4 days
Mike ran ten laps around his school's 1/4 mile track in 15 minutes. He
finishes his run at the exact same point that he started. What was his average
velocity in miles per hour?
Answer:
v = 0
Explanation:
It is given that,
Mike ran ten laps around his school's 1/4 mile track in 15 minutes. He finishes his run at the exact same point that he started. We need to find his average velocity. Average velocity is equal to net displacement divided by total time taken.
As he returns exactly at the same point that he started, it means his displacement or the shortest distance covered is equal to 0. As a result, its average velocity is equal to 0.
a small sphere of
mass 0.25 g that carries a charge of 9.0 × 10−10 C.
Two parallel vertical infinite charged sheets of
charge densities σ1= -30 × 10−6 C/m2 and σ2= ab ×
10−6 C/m2 respectively. The sphere is attached to
one end of a very thin silk string 5.0 cm long. The
other end of the string is attached to the 2nd sheet as
shown in the figure. At equilibrium, the string will
make an angle (ϴ) with the vertical. Calculate the
angle that the string makes with the vertical?
Answer:
θ = 39.7º
Explanation:
In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left
Y Axis
[tex]T_{y}[/tex] -W = 0
[tex]T_{y}[/tex] = W
X axis
-[tex]F_{e1}[/tex] - F_{e2} + Tₓ = 0
let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical
sin θ = Tₓ / T
cos θ = T_{y} / t
Tₓ = T sin θ
T_{y} = T cos θ
let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.
F = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate
F = 2E A = q_{int} / ε₀
let's use the concept of surface charge density
σ = q_{int} / A
we substitute
2E A = σ A /ε₀
E = σ / 2ε₀
this is the field created by each plate. The electric force is
[tex]F_{e}[/tex] = q E
for plate 1 with σ₁ = -30 10⁻⁶ C / m²
F_{e1} = q σ₁ /2ε₀
for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²
F_{e2} = q σ₂ /2ε₀
we substitute and write the system of equations
T cos θ = mg
- q σ₁ / 2ε₀ - q σ₂ /2ε₀ + T sinθ = 0
we introduce t in the second equations
- q /2 ε₀ (σ₁ + σ₂) + (mg / cos θ) sin θ = 0
mg tan θ = q /2ε₀ (σ₁ + σ₂)
θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)
data indicates the mass of 0.25 g = 0.25 10⁻³ kg
give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²
let's calculate
θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2 8.85 10⁻¹² 0.25 10⁻³ 9.8))
θ = tan⁻¹ 8.3 10⁻¹)
θ = 39.7º
The carbon isotope C-14 is used for carbon dating. C-14 (mass 2.34 x 10-26 kg) decays by beta-decay, in which the nucleus emits an electron (the beta particle) and a subatomic particle called a neutrino. In one such decay, the electron and neutrino are emitted at right angles to each other. The electron (mass 9.11 x 10-31 kg) has a speed of 5.0 x 107 m/s and the neutrino has a momentum of 8.0 x 10-24 kg m/s. What is the recoil speed of the daughter nucleus?
Answer:
Explanation:
We shall apply the law of conservation of momentum to calculate the momentum and the speed of daughter nucleus .
Since the velocity of electron is very high we shall apply relativistic formula to calculate its momentum .
= [tex]\frac{mv}{\sqrt{1-(\frac{v}{c})^2 } }[/tex]
[tex]= \frac{9.11 \times 10^{-31}\times 5\times 10^7}{\sqrt{1-(\frac{5\times10^7}{3\times 10^8})^2 } }[/tex]
45.55 x 10⁻²⁴ x 1.176
= 53.58 x 10⁻²⁴ .
momentum of neutrino = 8 x 10⁻²⁴ . They are perpendicular to each other so total momentum
= √ [( 53.58 x 10⁻²⁴ )²+(8 x 10⁻²⁴)²]
= 54.17 x 10⁻²⁴
Hence the momentum of recoiled daughter nucleus will be same but in opposite direction
velocity of recoil = momentum / mass
= 54.17 x 10⁻²⁴ / 2.34 x 10⁻²⁶
= 23.15 x 10² m /s
A. Suppose the laser light has wavelength 400 nm, and that the two small slits, which act as point sources of light, are separated by 0.1 mm. (Recall that 1 nm = 10-9 meters.) Approximately how many nodal lines would be present in the pattern of overlapping light waves when the laser light emerges from the two slits in the mask?
Answer:
200 nodal lines
Explanation:
To find the number of lines you first use the following formula for the condition of constructive interference:
[tex]dsin\theta=m\lambda[/tex] (1)
d: distance between slits = 0.1mm = 0.1*10^-3 m
θ: angle between the axis of the slits and the m-th fringe of interference
λ: wavelength of light = 400 nm = 400*10^-9 m
You obtain the max number of lines when he angle is 90°. Then, you replace the angle by 90° and solve the equation (1) for m:
[tex]dsin90\°=m\lambda\\\\d=m\lambda\\\\m=\frac{d}{\lambda}=\frac{0.1*10^{-3}m}{500*10^{-9}m}=200[/tex]
hence, the number of lines in the interference pattern are 200
If the circuit in the motorcycle runs off a standard 12 V battery, and one of the headlights is switched on, current flowing through the headlight is measured at 3.75 A. What is the power usage of the headlamp? Report the answer to two significant digits
Answer:
45 W
Explanation:
Power: This can be defined as the rate at which electrical energy is consumed in a circuit. The unit of power is Watt (W).
From the question,
The expression for power is given as,
P = VI.................. Equation 1
Where V = Voltage, I = current.
Given: V = 12 V, I = 3.75 A.
Substitute into equation 1.
P = 12(3.75)
P = 45 W.
Answer:
P = 45 Watt
Explanation:
The electrical power used by an electrical device or the electrical circuit is given by the following formulae:
P = VI
P = I²R
P = V²/R
where,
P = Electrical Power Consumed by the Device
V = The Voltage applied to the circuit or device
R = Resistance of device or circuit
I = Current passing through the device or circuit
We have the following data for our circuit:
V = 12 volts
I = 3.75 A
Therefore, it is clear from the data that we will use the first formula:
P = VI
P = (12 volts)(3.75 A)
P = 45 Watt
A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before coming to rest. Assume that the acceleration of the bullet is constant. What force is exerted by the wood on the bullet
Answer:
The force exerted by the wood on the bullet is 399.01 N
Explanation:
Given;
mass of bullet, m = 0.0021 kg
initial velocity of the bullet, u = 497 m/s
final velocity of the bullet, v = 0
distance traveled by the bullet, S = 0.65 m
Determine the acceleration of the bullet which is the deceleration.
Apply kinematic equation;
V² = U² + 2aS
0 = 497² - (2 x 0.65)a
0 = 247009 - 1.3a
1.3a = 247009
a = 247009 / 1.3
a = 190006.92 m/s²
Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;
F = ma
F = 0.0021 x 190006.92
F = 399.01 N
Therefore, the force exerted by the wood on the bullet is 399.01 N
One of the advantages of skates is that they allow you to push against ice, which typically has low friction. Let’s consider an hockey player on frictionless ice who is going to shoot a slap shot. The player weighs Mp = 100kg. He is going to fire the puck, which weighs Mpuck = 0.5kg with a velocity of v = 50(m/s). How fast does the hockey player recoil in the direction opposite to the puck’s trajectory. If the force he exerts on the puck occurs over a time, ∆t = 0.1s what is the average force needed to create the impulse that accelerates the puck to that velocity?
Answer:
a) v1f = 0.25 m/s
b) F = 50000N
Explanation:
a) In order to calculate the speed of the player after he fires the puck, you use the conservation of momentum law. Before the puck is in motion and after the total momentum of both player and puck must conserve:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex] (1)
m1: mass of the player = 100kg
v1i: initial velocity of the player = 0m/s
v2f: final velocity of the puck = ?
m2: mass of the puck = 0.5kg
v2i: initial velocity of the puck = 0 m/s
v2f: final velocity of the puck = 50 m/s
You replace these values into equation (1) and you solve for v1f (final velocity of player):
[tex]0kgm/s+0kgm/s=(100kg)v_{1f}+(0.5kg)(50m/s)\\\\v_{1f}=-\frac{(0.5kg)(50m/s)}{100kg}\\\\v_{1f}=-0.25\frac{m}{s}[/tex]
The minus sign means that player moves in a opposite direction to the motion of the puck
The velocity of the player after he fires the puck is 0.25 m/s
b) The force needed is given by the change in time , of the momentum of the player, which is given by:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex]
The change on the velocity of the puck is 50m/s, and the time interval is 0.1s.
[tex]F=(100kg)\frac{50m/s}{0.1s}=50000N[/tex]
The force needed to create the needed impulse, to accelerate the puck is 50000N
A tank holds a 2.38-m thick layer of oil that floats on a 1.24-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.27 and 1.81, respectively. A ray originating at O reaches the brine-oil interface at the critical angle. What is the distance of this point from the axis?
Answer:
1.22m
Explanation:
Since
sinθ = refraction-of-the-oil/refraction-of-the-brine =1.27/1.81 = 0.702
θ = [tex]sin^-^1[/tex](0.702)
Hence
Critical angle = θ = 44.58°
tan(θ) = d/1.24
tan(44.58°) = d/1.24
Hence, 0.98 = d/1.24
The distance d = 0.98 x 1.24 = 1.22m
You are throwing a stone straight-up in the absence of air friction. If the stone is caught at the same height it was thrown from, which of the following is true of the stones motion?
1. the acceleration is zero at the top point of the motion.
2. the acceleration is minimum at the top point of the motion.
3. acceleration and velocity always point in opposite directions.
4. the time going up is equal to the time coming down the acceleration is directly proportional to the velocity.
Answer:
4. the time going up is equal to the time coming down the acceleration is directly proportional to the velocity.
Explanation:
1. FALSE
The acceleration at the top point is equal to the acceleration due to gravity (9.8 m/s²). It is the velocity of the stone that becomes zero for a moment, at the top point.
2. FALSE
The acceleration of the body in a free fall motion always remains constant, and its value is equal to the acceleration due to gravity (9.8 m/s²)
3. FALSE
The velocity and acceleration point in opposite direction, during upward motion only (Velocity points up and acceleration points down). But they point in same direction during the downward motion (Both velocity and acceleration point down).
4. TRUE
Since, there is no air friction. Therefore, the acceleration for both upward an downward motion will be constant, without any opposing force. Hence, the time taken by the stone to go up will be equal to time taken by the stone to come down.
Give 2 examples for Newton’s first law of motion.
please answer this question!!!!!
Consider two sinusoidal waves traveling along a string, modeled as y1(x, t) = (0.4 m)sin[(3 m−1)x + (2 s−1)t] and y2(x, t) = (0.8 m)sin[(6 m−1)x − (5 s−1)t]. What is the vertical displacement (in m) of the resultant wave formed by the interference of the two waves at the position x = 0.9 m at time t = 0.4 s? (Indicate the direction with the sign of your answer.)
Answer:
y(x,t)=-0.395m
Explanation:
The wave function modeling waves are:
[tex]y_1(x,t)[/tex]= (0.4 m)sin[(3 [tex]m^-^1[/tex])x + (2 [tex]s^-^1[/tex])t]
[tex]y_2(x,t)[/tex]=(0.8 m)sin[(6 [tex]m^-^1[/tex])x − (5 [tex]s^-^1[/tex])t]
The principle of superposition can be defined as the resultant of [tex]y_1[/tex] and [tex]y_2[/tex] is equal to their algebraic sum
[tex]y(x,t)=y_1(x,t)+y_2(x,t)[/tex]
y(x,t)= (0.4 m)sin[(3 [tex]m^-^1[/tex])x + (2 [tex]s^-^1[/tex])t] + (0.8 m)sin[(6 [tex]m^-^1[/tex])x − (5 [tex]s^-^1[/tex])t]
Substitute 0.9m for x and 0.4s for t as required
y(x,t)= (0.4 )sin[(3 [tex]m^-^1[/tex])(0.9) + (2 [tex]s^-^1[/tex])(0.4)] + (0.8 m)sin[(6 [tex]m^-^1[/tex])(0.9)− (5 [tex]s^-^1[/tex])(0.4)]
y(x,t)= -0.14 - 0.255
y(x,t)=-0.395m
The vertical displacement of the resultant wave formed by the interference of the two waves is -0.395m
Interference of waves:Two sinusoidal waves are given such that:
y₁(x, t) = (0.4 m)sin[(3 m⁻¹)x + (2s⁻¹)t] and
y₂(x, t) = (0.8 m)sin[(6 m⁻¹)x − (5s⁻¹)t]
The superposition of the two waves gives the outcome of the interference at x = 0.9 m and t = 0.4 s.
y(x,t) = y₁(x, t) + y₂(x, t)
y(0.9,0.4) = y₁(0.9, 0.4) + y₂(0.9, 0.4)
y(0.9,0.4) = (0.4 m)sin[(3 m⁻¹)0.9 + (2s⁻¹)0.4] + (0.8 m)sin[(6 m⁻¹)0.9 − (5s⁻¹)0.4]
y(0.9,0.4) = -0.395m
Learn more about interference:
https://brainly.com/question/16098226?referrer=searchResults
A transmission line consisting of two concentric circular cylinders of metal, with radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0. A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0. The cylinders are symmetric about the z-axis. Calculate the time-averaged power flow, P, along the line.
Answer:
P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.
Explanation:
So, we will be making use of the data or parameters given in the question above;
=>" radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0."
=> " A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0."
So, we will be making use of the two equations below;
Ë = ( λ/ 2πEP) × P'. --------------------------(1)..
B' = √ μE × ( λ/ 2πEP) × P' --------------(2).
Where equation (1) and (2) represent Gauss' law and magnetic field equation respectively.
Jo = 1/√ μE × λ/2 × π × a.
When we solve for charge per unit length, we have;
λ = 2 × π × Jo × a × √ μE.
The energy flux,s = E' × J'= √μE× |Jo(Z) |^2 × a^2/b^2 { cos^2 kz - wt + Avg Jo} Z'.
Hence, the time. Average power flux = 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z'.
Therefore, P = ∫z' . <s> da
P = ∫ ∫ 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z' pd pd p Θ.
(Take limit on the first at second integration as : 2π,0 and b,a).
P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.
What is the momentum of a 2 kg ball traveling at 2m/s
Answer:
4
Explanation:
p=m×v
m=2kg
v=2m/s
2×2=4
Answer:
4kgms/1
Explanation:
p=m×v
=2kg×2m/s
=4kgms/1
The sound of a tuba is very low. Why?
0
O When you blow into a tuba the air vibrates very strongly.
) O When you blow into a tuba the air vibrates very quickly.
O When you blow into a tuba the air vibrates very softly.
O When you blow into a tuba the air vibrates very slowly.
Answer:
When you blow into a tuba the air vibrates very slowly.
Explanation:
Tuba is a buzz instrument ie sound is produced in it with the help of lip vibration . It is the lowest pitched musical instrument in the brass family .
Due to absence of resonance in it , it produces music of lowest pitch , So when one blows into it the air column of the instrument vibrates very slowly producing low pitched sound.
A ball is dropped from the top of a 91-m-high building. What speed does the ball have in falling 3.2 s?
Answer:
The speed of the ball is 28.4m/s.
Explanation:
Given that the formula of speed is Speed = Distance/Time. So you have to substitute the values into the formula :
[tex]speed = distance \div time[/tex]
Let distance = 91m,
Let time = 3.2s,
[tex]speed = 91 \div 3.2[/tex]
[tex]speed = 28.4 \: metrepersecond[/tex]
When solving problems involving forces and Newton's laws, the following summary of things to do will start your mind thinking about getting involved in the problem at hand.
When solving problems involving forces and Newton'
Apply these steps
Use the steps outlined above to find the magnitude of the acceleration a of a chair and the magnitude of the normal force FN acting on the chair: Yusef pushes a chair of mass m = 50.0 kg across a carpeted floor with a force Fp (the subscript 'p' here is lowercase and throughout the question) of magnitude Fp = 168 N directed at ? = 35.0 degrees below the horizontal. The magnitude of the kinetic frictional force between the carpet and the chair is Fk= 99.7 N .
Part A
Identify and sketch all the external forces acting on the chair. Because the chair can be represented as a point particle of mass m, draw the forces with their tails centered on the black dot in the middle of the chair. Be certain to draw your forces so that they have the correct orientation.
Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded.
Answer:
Explanation:
There is no acceleration in vertical direction
Fn = mg + Fp sin 35
Frictional force = 99.7 N
Net force in forward direction = Fp cos 35 - 99.7
= 168 cos 35 - 99.7
= 137.61-99.7
= 37.91 N
net acceleration of chair = net force / mass
= 37.91 / 50
= .7582 m / s² .
Why intermediate elements has negative packing fraction?
Explanation:
The negative value of the packing fraction indicates that the actual isotopic mass is less the mass number.
Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI), a measure of the likelihood of injury, is defined as SI = a5/2t, where a is the acceleration in multiples of g and t is the time the acceleration lasts (in seconds). In one set of studies of rear end collisions, a person's velocity increases by 12 km/h with an acceleration of 35 m/s2.(a) What is the severity index for the collision? (s)(b) How far does the person travel during the collision if the car was initially moving forward at 7.0 km/h? (m)
Answer:
a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.
Explanation:
a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:
[tex]a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]a' = 3.569[/tex]
The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:
[tex]v_{f} = v_{o} + a \cdot t[/tex]
Where:
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v_{f}[/tex] - Final speed, measured in meter per second.
[tex]a[/tex] - Acceleration, measured in [tex]\frac{m}{s^{2}}[/tex].
[tex]t[/tex] - Time, measured in seconds.
[tex]t = \frac{v_{f}-v_{o}}{a}[/tex]
[tex]t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }[/tex]
[tex]t = 0.095\,s[/tex]
Lastly, the severity index is now determined:
[tex]SI = \frac{a'^{5}}{2\cdot t}[/tex]
[tex]SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}[/tex]
[tex]SI = 3047.749[/tex]
b) The initial and final speed of the injured are [tex]1.944\,\frac{m}{s}[/tex] and [tex]5.278\,\frac{m}{s}[/tex], respectively. The travelled distance can be determined from this equation of motion:
[tex]v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s[/tex]
Where [tex]\Delta s[/tex] is the travelled distance, measured in meters.
[tex]\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s = 0.345\,m[/tex].
Which angle is the angle of reflection? 1, 2, 0R 3?
Answer:
The answer is angle 2.
Explanation:
Angle 2 is the Reflected Ray of angle 1.
Angle 3 is the Refracted angle.
Angle 2 is the angle of reflection. the light incident will bounce to the opposite end after striking is called the reflection.
What is the law of reflection?The law of reflection specifies that upon reflection from a downy surface, the slope of the reflected ray is similar to the slope of the incident ray.
The reflected ray is consistently in the plane determined by the incident ray and perpendicular to the surface at the point of reference of the incident ray.
When the light rays descend on the smooth surface, the angle of reflection is similar to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in a similar plane.
In the reflection, the light incident will bounce to the opposite end after striking. Angle 2 satisfies the properties of the reflection.
Hence angle 2 is the angle of reflection.
To learn more about the law of reflection refer to the link;
brainly.com/question/12029226
Sometimes earthquakes occur at a boundary between the Indian and Eurasian Plates. The collisions of these two continental plates formed the Himalayan mountain range.
Based on this information, which type of boundary exists between the Indian and Eurasian Plates?
continental-continental convergent
continental-continental divergent
oceanic-continental divergent
oceanic-continental convergent
Answer:
Continental Continental convergent.
Explanation:
Continental Continental convergent is a type of earthquakes that occur between boundaries of two continents in which the tectonic plates move closer to each other or converge.
Eurasian plates refers to tectonic plates that is found in the Continent of Eurasia which include Asian and Europe excluding India subcontinent.
From the question, the earthquake occur between boundaries of two continents India and Eurasia and this converging and collision of the two continental plates formed the Himalayan mountain range.
Therefore, it is continental continental convergent.
Answer:
A
Explanation: