Answer:
[tex]I=2.67\times 10^{-6}\ W/m[/tex]
Explanation:
Given that,
Frequency of a radio antenna is 1 MHz
Power, P = 21 kW
We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :
[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m[/tex]
So, the wave intensity 25 km from the antenna is [tex]2.67\times 10^{-6}\ W/m[/tex].
A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
b
[tex]k =722.2 \ N/m[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]
The period is [tex]T = 17 \ s[/tex]
The test weight is [tex]W = 13 \ N[/tex]
Generally the radial acceleration is mathematically represented as
[tex]a = w^2 r[/tex]
at maximum angular acceleration
[tex]r = A[/tex]
So
[tex]a_{max} = w^2 A[/tex]
Now [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2 * \pi }{T}[/tex]
Therefore
[tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]
substituting values
[tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
Generally this test weight is mathematically represented as
[tex]W = k * A[/tex]
Where k is the spring constant
Therefore
[tex]k = \frac{W}{A}[/tex]
substituting values
[tex]k = \frac{13}{0.018}[/tex]
[tex]k =722.2 \ N/m[/tex]
When a ray of light traveling in air hits a tilted plane parallel slab (of glass, say), it emerges parallel to the original ray but shifted transversely. Carefully draw out the situation and use Snell’s law to derive the amount of the transverse shift, x, as a function of the tilt angle of the slab, θ, its thickness, d, and its index of refraction, n. Find the exact expression with no approximations. We recommend you do this out all in variables because it's a useful formula to have. Also, you will want this for the following questions. However, since the auto-grader has difficulty with these formulas, use n=1.5, d=1.0 cm, and θ = 45° and enter a numerical answer. Give your answer in cm to two significant figures.
Answer:
x = 0.4654 cm
Explanation:
In this exercise we use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
apply this formula to the first surface, where n₁ is the index of refraction of air (n₁ = 1) and n₂ is the index of refraction of glass (n₂ = n)
θ₂ = sin⁻¹ (sin θ₁ / n) (1)
having this angle we use trigonometry to find the value of the point where it comes out when we reach the other side
refracted ray
tan θ₂ = x₂ / d
x₂ = d tan θ₂
this value is the distance displaced by the refracted ray
now let's find the distance at which the incident beam should exit
tan θ₁ = x₁ / d
x₁ = d tan θ₁
the displacement of the ray is the difference between these two distances, we will call it x
x = x₁ - x₂
x = d tan θ₁ - d tan θ₂
x = d (tan θ₁ - tan θ₂) (2)
the easiest way to do the calculations is to find tea2 from the binding 1 and then perform the calculation with equation 2
calculate
θ₂ = sin⁻¹ (sin 45 /1.5)
θ₂ = 28.13º
x = 1.0 (tan 45 - tan 28.13)
x = 0.4654 cm
What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
c. The incident light must have at least as much energy as the electron work function
A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic
friction 0.50 and a coefficient of static friction 0.65. How much force is required to
get the block moving?
Answer:
The force is [tex]F = 172 \ N[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m_b = 27.0 \ kg[/tex]
The coefficient of static friction is [tex]\mu_s = 0.65[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.50[/tex]
The normal force acting on the block is
[tex]N = m * g[/tex]
substituting values
[tex]N = 27 * 9.8[/tex]
[tex]N = 294.6 \ N[/tex]
Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as
[tex]F_f = \mu_s * N[/tex]
substituting values
[tex]F_f = 0.65 * 264.6[/tex]
[tex]F_f = 172 \ N[/tex]
Now for this block to move the force require is equal to [tex]F_f[/tex] i.e
[tex]F= F_f[/tex]
=> [tex]F = 172 \ N[/tex]
Which one of the conditions can cause a particle to move with uniform circular motion in a uniform magnetic field
Given that,
A particle to move with uniform circular motion in a uniform magnetic field.
Suppose, The conditions are,
(I). The charged particle has to be positive and it should be moving in a direction opposite to a uniform magnetic field.
(II). The charged particle should be moving parallel to the magnetic force and perpendicular to the magnetic field.
(III). The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.
We know that,
An particle to move with uniform circular motion.
Here, electric force is perpendicular to velocity of particle.
The electric field is defined as,
[tex]F_{c}=\dfrac{mv^2}{r}[/tex].....(I)
Suppose, there is magnetic field, if a charge moving with velocity and the magnetic field exerts a field.
The magnetic force is defined as,
[tex]F_{m}=qvB[/tex].....(II)
We need to find the magnetic field
Using equation (I) and (II)
[tex]F_{c}=F_{m}[/tex]
[tex]\dfrac{mv^2}{r}=qvB[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
Hence, The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.
(III) is correct option.
The time constant of an RC circuit is 2.7 s. How much time t is required for the capacitor (uncharged initially) to gain 0.63 of its full equilibrium charge
Answer:
2.7s
Explanation:
The solution of time required is shown below:-
In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time
Now, the
63% that is equal to 0.63 which is full equilibrium charge
Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s
So, the correct answer is 2.7s
A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in metric tonnes per cubic metre? Give your answer to 1 significant figure.
Answer:
4 tonne/m³
Explanation:
ρ = m / V
ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))
ρ = 0.0041 g/mm³
Converting to tonnes/m³:
ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³
ρ = 4.1 tonne/m³
Rounding to one significant figure, the density is 4 tonne/m³.
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer:
A. It will be 0 meters per second per second.
Explanation:
The force and acceleration is in a proportional relationship, that means the line goes through the origin.
On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑ [tex]q_{i} / r_{i}[/tex]
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m
A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
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A rectangular loop of wire carries current I in the clockwise direction. The loop is in a uniform magnetic field B that is parallel to the plane of the loop, in the direction toward the bottom of the page. The length of the rectangle is b and the width is a. What is the net force on the loop by the magnetic field
Answer:
Explanation:
Area of the loop = a b
current = I
magnetic moment of the loop M = area x current
= ab I
Torque on the loop = MB sinθ
here θ = 90
Torque = MB
= abIB
In this case net force on the loop will be zero because here torque is created by two equal and opposite force acting on two opposite sides of the loop so net force will be zero .
Science activity
Imagine that some settlers have left Earth and gone to the Moon, taking
their recipe books with them. The first cake they baked was a disaster. It had
far too little moisture and was about six times the size they had expected.
the cake recipe was:
1.25 N butter
1.50 N sugar
4 eggs
1.50 N flour
20 ml milk
ANALYTICAL THINKING
Q. Why was the cake so big? Why was it se
dry?
Answer:
Answer in explanation
Explanation:
The reason for the big size and less moisture of the cake is due to difference in weight of the ingredients on the surface of moon. So, the same has the lesser weight on the surface of the moon than it has on the surface of earth. Or in other words, The same weight of the ingredients will have greater mass and thus the greater quantity on the surface of earth than the surface of earth. For example, on earth 1.25 N butter will have a mass:
m = W/g = 1.25 N/(9.8 m/s²) = 0.13 kg
But, on moon:
m = W/g = 1.25 N/(1.625 m/s²) = 0.77 kg
Hence, it is clear that the mass of the same weight of the substance becomes 6 times greater on the surface of moon. This explains why the cake was so big.
Now, coming to the second part about the dryness of the cake. The main and only source of moisture in recipe is the eggs bu the eggs are taken in a quantity of numbers. So they are exactly the same on moon as well. While all the other ingredients are increased, the same amount of eggs are not sufficient to provide them with enough moisture. Hence the cake was dry.
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m
Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm
Answer:
d = 1.0128×10⁻⁵m
Explanation:
given:
length L = 4.0m
maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m
wavelength λ = 633nm = 633×10⁻⁹m
note:
dsinθ = mλ (constructive interference)
where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength
for small angle
sinθ ≈ tanθ
[tex]d (\frac{y}{L}) =[/tex] mλ
[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]
[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]
d = 1.0128×10⁻⁵m
A 1-kilogram mass is attached to a spring whose constant is 18 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 11 times the instantaneous velocity. Determine the equations of motion if the following is true?
a. the mass is initially released from rest from a point 1 meter below the equilibrium position
b. the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 11 m/s
Answer:
Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.
a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].
b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].
Explanation:
Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)
[tex]x^\prime[/tex] would thus represent the velocity of this mass.[tex]x^{\prime\prime}[/tex] would represent the acceleration of this mass.Constructing the ODEConstruct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.
The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:
[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].
The question described another equation for the net force on this mass. This equation is the sum of two parts:
The restoring force of the spring: [tex]F(\text{spring}) = -k\, x[/tex], where [tex]k[/tex] denotes the constant of this spring.The damping force: [tex]F(\text{damping}) = - 11\,x^\prime[/tex] according to the question. Note the negative sign in this expression- the damping force should always oppose the direction of motion.Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:
[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].
Combine the two equations for the net force on this mass to obtain:
[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].
From the question:
Size of this mass: [tex]m = 1\; \rm kg[/tex].Spring constant: [tex]k = 18\; \rm N \cdot m^{-1}[/tex].Hence, the equation will become:
[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].
Rearrange to obtain:
[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].
Finding the general solution to this ODE[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:
[tex]m^2 + 11\, m + 18 = 0[/tex].
The two roots are:
[tex]m_1 = -2[/tex], and[tex]m_2 = -9[/tex].Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:
[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].
For this particular ODE, that general solution would be:
[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].
Finding the particular solutions to this ODENote, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time
The position at time [tex]t = 0[/tex] would be [tex]x(0) = c_1 + c_2[/tex].The velocity at time [tex]t = 0[/tex] would be [tex]x^\prime(0) = -2\, c_1 - 9\, c_2[/tex].For section [tex]\rm a.[/tex]:
[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].
Hence, the particular solution for section [tex]\rm a.[/tex] will be:
[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].
Similarly, for section [tex]\rm b.[/tex]:
[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].
Hence, the particular solution for section [tex]\rm b.[/tex] will be:
[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].
The first step to merging is entering the ramp and _____.
A. honking to indicate your location
B. matching your speed
C. signaling your intent
D. telling your passengers where you're going
Answer:
B. matching your speed
Explanation:
To merge safely, you must identify a gap in traffic and match your speed to the speed of the gap. Before you make your move to fill the gap, you should signal your intent.*
_____
* At least one resource says "The first step ... is to make sure you're traveling at the same speed ..." Then it goes on to say "Use your indicator. Do it early ...." The accompanying animation shows blinkers being activated on the ramp before the merge lane is entered. Apparently, "the first step" is not necessarily the first thing you do.
Answer:
It's C "signaling your intent"
Explanation:
The key thing to look at is they are asking the rest of the first step and that;s C
Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose screen is 65 cm away. What refractive power will these glasses require?
Answer:
Explanation:
Far point = 17 cm . That means he can not see beyond this distance .
He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula
1 / v - 1 / u = 1 / f
1 / - 17 - 1 / - 65 = 1 / f
= 1 / 65 - 1 / 17
= - .0434 = 1 / f
power = - 100 / f
= - 100 x .0434
= - 4.34 D .
Refractive power is the measure of degree of convergence by a lens. The required refractive power for the given glasses is -4. 34 D.
Using lens formula
[tex]\bold { \dfrac 1 v - \dfrac1 u = \dfrac {1}f}[/tex]
Where,
f- focal point
v - distance of the image
u - distance of the object
So,
[tex]\bold { \dfrac 1 {-17} - \dfrac1 {-65} = \dfrac {1}f}\\\\\bold { 0.434 = \dfrac {1}f}\\[/tex]
Since, [tex]\bold {power = \dfrac {- 100 }f}[/tex]
So,
[tex]\bold { power = - 100 \times 0.0434}}\\\\\bold { power = - 4.34\ D}[/tex]
Therefore, the required refractive power for the given glasses is -4. 34 D.
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A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2
Answer:
650W/m²Explanation:
Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²
Given parameters
Power rating = 6.50Watts
Cross sectional area = 100cm²
Before we calculate the intensity, we need to convert the area to m² first.
100cm² = 10cm * 10cm
SInce 100cm = 1m
10cm = (10/100)m
10cm = 0.1m
100cm² = 0.1m * 0.1m = 0.01m²
Area (in m²) = 0.01m²
Required
Intensity of the sunlight I
I = P/A
I = 6.5/0.01
I = 650W/m²
Hence, the intensity of the sunlight in W/m² is 650W/m²
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
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You stand near the edge of a swimming pooland observe through the water an object lying on the bottom of thepool. Which of the following statements correctly describes whatyou see?
a. The apparent depth of the object is less than thereal depth.
b. The apparent depth of the object is greater thanthe real depth.
c. There is no difference between the apparent depth and the actual depth of the object.
Answer:
a
Explanation:
The correct answer would be that the apparent depth of the object is less than the real depth.
The refractive property of light as it passes from air to water would make the depth of the pool appear less shallow than the actual depth to an observed. Hence, an object placed at the bottom of the pool will have an apparent depth that is shallower than its actual depth.
Due to the difference in the density of air and that of water, as the ray of light from an observer standing at the edge of a swimming pool travels from air into the water, it becomes refracted by bending away from the original traveling angle.
The same refraction occurs when light rays from an object inside the pool travel from water into the air. Hence, due to the refraction of the ray of light coming from the object at the bottom of the pool, the depth appears shallower than the actual depth.
Correct option: a
There are fiber optic telephone cables connecting North America and Europe, lying on the bottom of the Atlantic ocean. The wire is 4,500 km long how long and has an index of refraction of 1.5. How long will it take for the signal to cross the ocean? Give your answer in milliseconds.
Answer:
The time taken is [tex]t = 0.0225 \ s[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]l = 4500 \ km = 4500000 \ m[/tex]
The refractive index is [tex]n_f = 1.5[/tex]
The velocity of the signal is mathematically represented as
[tex]v = \frac{c}{n_f }[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
substituting values
[tex]v = \frac{3.0 *10^{8}}{1.5}[/tex]
[tex]v = 2.0*10^{8} \ m/s[/tex]
The time taken is mathematically evaluated as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{4500000}{2.0 *10^{8}}[/tex]
[tex]t = 0.0225 \ s[/tex]
Which of the following statements about stages of nuclear burning (i.e., first-stage hydrogen burning, second-stage helium burning, etc.) in a massive star is not true?
A) As each stage ends, the core shrinks further.
B) Each successive stage of fusion requires higher temperatures than the previous stages.
C) Each successive stage lasts for approximately the same amount of time.
D) Each successive stage creates an element with a higher atomic weight.
Answer:
C) Each successive stage lasts for approximately the same amount of time.
Explanation:
Nuclear burning is a series of nuclear processes through which a star gets its energy. The energy within a star is due to nuclear fusion of lighter elements (hydrogen) into more massive element (helium), with a release of a large amount of energy due to the conversion of some of the mass into energy. Each stage leads to a loss of some of the mass which is converted into energy (option A is valid).
The fusion of four hydrogen atoms into one helium atom means that there is a creation of element with a higher atomic weight (option D is valid), and the energy output of each stage exceeds its energy input, meaning that each stage will require a higher temperature than its previous stages (option B is valid).
A charged particle moves through a region of space at constant speed, without deflecting. From this, one can conclude that in this region: Select the correct answer
a) There must be no magnetic field in the region There could be electric and magnetic fields, oriented perpendicular to each other.
b) There could be electric and magnetic fields, oriented in opposite directions There must be no magnetic field and no electric field in the region.
c) There could be electric and magnetic fields, oriented in the same direction Your Ans
Answer:
There could be electric and magnetic fields, oriented in opposite directions
Explanation:
Lorentz force, is the force that may be exerted on a body of a specified magnitude of charge q, moving with a velocity v, in a magnetic field B and in an electric field of intensity E. This Lorentz force is given by; F= qE+qvBsin ϕ
However, if the motion of the particle is opposite to the magnetic field such a that ϕ = 0, then there is no net magnetic force on the charge and it moves freely, with a constant velocity and in a straight line. Hence, there is no magnetic field in the region.
The charge moves with constant speed due to same direction of magnetic and electric field.
There could be electric and magnetic fields that is oriented in the same direction or the other reason is that there is no magnetic field and electric field in that region where the charge moves. If the electric and magnetic field are present at the same direction then it means that it applies no force on the charge.
This is due to more distance from the charge as well as the charge travels away from the field occupies by the magnetic and electric field so we can conclude that the charge moves with constant speed due to same direction of magnetic and electric field.
Learn more: https://brainly.com/question/17132472
The voltage and power ratings of a particular light bulb, which are its normal operating values, are 110 V and 60 W. Assume the resistance of the filament of the bulb is constant and is independent of operating conditions. If the light bulb is operated at a reduced voltage and the power drawn by the bulb is 36 W. What is the operating voltage of the bulb?
a. 78 V
b. 72 V
c. 66 V
d. 90 V
e. 85 V
Answer:
c. 66 V
Explanation:
p =IV
I =P/V
P1/V1=P2/V2
60/110=36/V2
0.55 = 36/V2
V2 =36/0.55 = 65.5V
V2 = 66V
A particle with mass m = 700 g is found to be moving with velocity v vector (-3.50i cap + 2.90j cap) m/s. From the definition of the scalar product, v^2 = v vector. v vector.
a. What is the particle's kinetic energy at this time? J If the particle's velocity changes to v vector = (6.00i cap - 5.00j cap) m/s,
b. What is the net work done on the particle? J
Answer:
Explanation:
v₁² = v₁ . v₁
= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )
= 12.25 + 8.41
= 20.66 m /s
a ) kinetic energy = 1/2 m v₁²
= 1/2 x .7 x 20.66
= 7.23 J
b )
changed velocity v₂ = v₂.v₂
= (6i - 5 j ) . (6i - 5 j )
= 36 + 25
= 61 m /s
kinetic energy = 1/2 m v₂²
= 1/2 x .7 x 61
= 21.35 J
Work done = change in energy
= 21.35 - 7.23
= 14.12 J .
explain why cups of soup at a take away kiosk are often sold in white polystrene cups with a lid to stop spillage
Answer:
polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.
the cup helps it become more insulated
Suppose a proton moves to the right and enters a uniform magnetic field into the page. It follows trajectory B with radius rp. An alpha particle (twice the charge and 4 times the mass) enters the same magnetic field in the same way and with the same velocity as the proton. Which path best represents the alpha particle’s trajectory?
Answer:
R = r_protón / 2
Explanation:
The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement
F = m a
Since the magnetic force is perpendicular, the acceleration is centripetal.
a = v² / R
the magnetic force is
F = q v x B = q v B sin θ
the field and the speed are perpendicular so the sin 90 = 1
we substitute
qv B = m v² / R
R = q v B / m v²
in the exercise they indicate
the charge q = 2 e
the mass m = 4 m_protón
R = 2e v B / 4m_protón v²
we refer the result to the movement of the proton
R = (e v B / m_proton) 1/2
the data in parentheses correspond to the radius of the proton's orbit
R = r_protón / 2
calculate the upthrust aciting on a body if its
true weight is 550 N and apparent weight
lis 490 N
Answer:
As a body moving upward
T=real weight + apparent weight
T=550+490
T=1040
hope u will get the answer:)
Explanation:
which of the following is a physical change?
A. a newspaper burns when placed in a fire.
B.an iron chair rusts when left outside
C.a sample of water boils and releases gas.
D.a plant changes carbon dioxide and water into sugar
A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through
Answer:
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Explanation:
The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:
[tex]m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}[/tex]
Where:
[tex]m_{B}[/tex], [tex]m_{W}[/tex]- Masses of the bullet and the block of wood, measured in kilograms.
[tex]v_{B,o}[/tex], [tex]v_{W,o}[/tex] - Initial speeds of the bullet and the block of wood, measured in meters per second.
[tex]v_{B,f}[/tex], [tex]v_{W,f}[/tex]- Final speeds of the bullet and the block of wood, measured in meters per second.
The final speed of the block is cleared:
[tex]v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}[/tex]
[tex]v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})[/tex]
If [tex]v_{W,o} = 0\,\frac{m}{s}[/tex], [tex]m_{B} = 0.022\,kg[/tex], [tex]m_{W} = 2\,kg[/tex], [tex]v_{B,o} = 210\,\frac{m}{s}[/tex] and [tex]v_{B,f} = 150\,\frac{m}{s}[/tex], then the final velocity of the block after the bullet passes through is:
[tex]v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)[/tex]
[tex]v_{W,f} = 0.66\,\frac{m}{s}[/tex]
The final velocity of the block after the bullet passes through is 0.66 meters per second.
If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart? The density of blood plasma is about 1025 kg/m3 and a typical maximum (systolic) pressure of the blood at the heart is 120 mg of Hg (= 16 kP = 1.6 × 104 N/m2). Give your answer in mg of Hg.
Answer:
The correct answer is 88.84 mmHg.
Explanation:
The pressure differential between the brain and the heart while standing up will be 120 - rho × g (gravity) × h, here h is the distance from the brain to the heart. The h is 40 cm or 0.4 m.
rho×g×h = 1060 kg/m³×9.8 m/s²×0.4m
= 4155 Pa
Now converting Pa to mmHg we get:
4155 Pa × 760 mmHg / 1.01325 × 10⁵ Pa
= 31.16 mmHg
Thus, the pressure in the brain now is 120 - 31.16
= 88.84 mmHg (hypotension)