Attempting to impress the skeptical patrol officer with your physics knowledge, you claim that you were traveling so fast that the red light (693 nm) appeared yellow (582 nm) to you. How fast would you have been traveling (in mi/hr) if that had been the case?
Answer:
v_r = 1.268 × 10⁸ mi/hr
Explanation:
We are given;
wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m
wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m
Frequency is given by the formula;
f = v/λ
Where v is speed of light = 3 x 10^(8) m
Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz
Similarly, Frequency of yellow light;
f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz
To find the speed of the car, we will use the formula;
f = f_o[(c + v_r)/c)]
Where c is speed of light and v_r is speed of car.
Making v_r the subject;
cf/f_o = c + v_r
v_r = c(f/f_o - 1)
So, plugging in the relevant values, we have;
v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]
v_r = 3 × 10⁸(0.189)
v_r = 5.67 x 10⁷ m/s
Converting to mi/hr, 1 m/s = 2.23694 mile/hr
So, v_r = 5.67 × 10⁷ × 2.23694
v_r = 1.268 × 10⁸ mi/hr
What is unique about visible
light in the electromagnetic
spectrum?
Answer:
The visible light spectrum is the segment of the electromagnetic spectrum that the human eye can view. More simply, this range of wavelengths is called visible light. Typically, the human eye can detect wavelengths from 380 to 700 nanometers.
Explanation:
Answer:
Visible light, which travels at a dizzying 186,282 miles per second through space, is just one part of light's broad spectrum, which encompasses all electromagnetic radiation. We can detect visible light because of cone-shaped cells in our eyes that are sensitive to the wavelengths of some forms of light
Explanation:
During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electromagnetic wave. During normal beating of the heart, the maximum value of the oscillating potential difference across 0.305 m of a person’s chest is 3.95 mV. This oscillating potential difference produces a 0.75-Hz electromagnetic wave.
What is the maximum electric field created?
Answer:
E = 0.0130 V/m.
Explanation:
The electric field is related to the potential difference as follows:
[tex] E = \frac{\Delta V}{d} [/tex]
Where:
E: is electric field
ΔV: is the potential difference = 3.95 mV
d: is the distance of a person's chest = 0.305 m
Then, the electric field is:
[tex]E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m[/tex]
Therefore, the maximum electric field created is 0.0130 V/m.
I hope it helps you!
If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?
What are found in nucleus and atoms?
Answer:
The nucleus of an atom contains the majority of the atom’s mass, and is composed of protons and neutrons, which are collectively referred to as nucleons. The much-lighter electrons orbit their atom’s nucleus. The Protons. Protons are positively charged particles found in an atom’s nucleus.
I hope u liked my answer. please mark me as branliest x
A child at the top of a slide has a gravatational store of 1800j what is the childs maximum kinetic store as he slides down explain your answer
Answer:
1800 J
Explanation:
Energy is conserved, so the maximum kinetic energy equals the change in gravitational energy.
Exercise 2: A 0.6 kg particle has a speed of 2 m / s at point A and kinetic energy of 7.5 J at point B. What is
(a) its kinetic energy at A?
(b) its speed at B?
(c) the total work done on the particle as it moves from A to B?
Explanation:
We have,
Mass of a particle is 0.6 kg
Speed at A is 2 m/s
Kinetic energy at B is 7.5 J
(a) The kinetic energy at A is given by :
[tex]K_A=\dfrac{1}{2}mv^2\\\\K_A=\dfrac{1}{2}\times 0.6\times 2^2\\\\K_A=1.2\ J[/tex]
(b) Kinetic energy at B is given by
[tex]K_B=\dfrac{1}{2}mV^2\\\\V=\sqrt{\dfrac{2K_B}{m}} \\\\V=\sqrt{\dfrac{2\times 7.5}{0.6}} \\\\V=5\ m/s[/tex]
(c) The work done on the particle as it moves form A to B is given by work energy theorem as :
[tex]W=\dfrac{1}{2}m(V^2-v^2)\\\\W=\dfrac{1}{2}\times 0.6\times (5^2-2^2)\\\\W=6.3\ J[/tex]
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement.
Φ = −10.3 ✕ 103 N · m2/C for r < r1Φ = 0 for r1 < r < r2Φ = 36.8 ✕ 103 N · m2/C for r2 < r < R1Φ = 0 for R1 < r < R2Φ = −36.8 ✕ 103 N · m2/C for r > R2
Complete Question
The complete question is shown on the first uploaded image
Answer:
The point charge is [tex]Q_z = -0.0912 \ \mu C[/tex]
The inner shell is [tex]Q_t = 0.4168 \ \mu C[/tex]
The outer shell is [tex]Q_w = -0.6514 \ \mu C[/tex]
Explanation:
From the question we are told that
The inner radius of thin first spherical conducting shell is [tex]r_1[/tex]
The outer radius of thin first spherical conducting shell is [tex]r_2[/tex]
The inner radius of second thin spherical conducting shell is [tex]R_1[/tex]
The outer radius of second thin spherical conducting shell is [tex]R_2[/tex]
The magnetic flux for different region is [tex]\phi = -10.3 *10^3 N\cdot m^2 /C \ for \ r < r_1[/tex]
The magnetic flux for first shell is [tex]\phi = 36 * 10^3 N \cdot m^2 /C \ for \ r_2 < r <R_1[/tex]
The magnetic flux for second shell is [tex]\phi = -36 * 10^3 N \cdot m^2 /C \ for \ r <R_1[/tex]
The magnitude of the point charge is mathematically represented as
[tex]Q_z = \ \phi_z * \epsilon _o[/tex]
[tex]Q_z = -10.3*10^{3} * 8.85 *10^{-12}[/tex]
[tex]Q_z = -9.115*10^{-8} \ C[/tex]
[tex]Q_z = -0.0912 \ \mu C[/tex]
Considering the inner shell
[tex]Q_a = \phi_a * \epsilon _o[/tex]
=> [tex]Q_a = 36 .8 * 10^3 * 8.85*10^{-12}[/tex]
[tex]Q_a = 32.56*10^{-8} \ C[/tex]
[tex]Q_a =0.326} \ \mu C[/tex]
Charge on the inner shell is
[tex]Q_t = Q_a - Q_z[/tex]
[tex]Q_t = 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]
[tex]Q_t = 0.4168 \ \mu C[/tex]
Considering the outer shell
[tex]Q_y = \phi_y * \epsilon_o[/tex]
=> [tex]Q_y = -36.8 *10^{3} * 8.85*10^{-12}[/tex]
[tex]Q_y = -32.56*10^{-8} \ C[/tex]
[tex]Q_y = - 0.326} \ \mu C[/tex]
Charge on the outer shell is
[tex]Q_w = Q_y - Q_z[/tex]
[tex]Q_w =- 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]
[tex]Q_w = -0.6514 \ \mu C[/tex]
A submarine dives from rest a 100-m distance beneath the surface of an ocean. Initially, the submarine moves at a constant rate of 0.3 m/s2 until reaches a speed of 4 m/s and then lowers at a constant speed. The density of salt water is 1030 kg/m3. The submarine has a hatch with an area of 2 m2 located on the top of the submarine’s body.
a. How much time does it take for the submarine to move down 100 m?
b. Calculate the gauge pressure applied on the submarine at the depth of 100
m.
c. Calculate the absolute pressure applied on the submarine at the depth of
100.
d. How much force is required in order to open the hatch from the inside of
the submarine?
answer this step-by-step, please.
Answer:
a. Time = 16.11 s
b. Gauge Pressure = 1009400 Pa = 1 MPa
c. Absolute Pressure = 1110725 Pa + 1.11 MPa
d. Force = 2.22 MN
Explanation:
a.
For the accelerated part of motion of submarine we can use equations of motion.
Using 1st equation of motion:
Vf = Vi + at₁
t₁ = (Vf - Vi)/a
where,
t₁ = time taken during accelerated motion = ?
Vf = final velocity = 4 m/s
Vi = Initial Velocity = 0 m/s (Since, it starts from rest)
a = acceleration = 0.3 m/s²
Therefore,
t₁ = (4 m/s - 0 m/s)/(0.3 m/s²)
t₁ = 13.33 s
Now, using 2nd equation of motion:
d₁ = (Vi)(t₁) + (0.5)(a)(t₁)²
where,
d₁ = the depth covered during accelerated motion
Therefore,
d₁ = (0 m/s)(13.33 s) + (0.5)(0.3 m/s²)(13.33 s)²
d₁ = 88.89 m
Hence,
d₂ = d - d₁
where,
d₂ = depth covered during constant speed motion
d = total depth = 100 m
Therefoe,
d₂ = 100 m - 88.89 m
d₂ = 11.11 m
So, for uniform motion:
s₂ = vt₂
where,
v = constant speed = 4 m/s
t₂ = time taken during constant speed motion
11.11 m = (4 m/s)t₂
t₂ = 2.78 s
Therefore, total time taken by submarine to move down 100 m is:
t = t₁ + t₂
t = 13.33 s + 2.78 s
t = 16.11 s
b.
The gauge pressure on submarine can be calculated by the formula:
Pg = ρgh
where,
Pg = Gauge Pressure = ?
ρ = density of salt water = 1030 kg/m³
g = 9.8 m/s²
h = depth = 100 m
Therefore,
Pg = (1030 kg/m³)(9.8 m/s²)(100 m)
Pg = 1009400 Pa = 1 MPa
c.
The absolute pressure is given as:
P = Pg + Atmospheric Pressure
where,
P = Absolute Pressure = ?
Atmospheric Pressure = 101325 Pa
Therefore,
P = 1009400 Pa + 101325 Pa
P = 1110725 Pa + 1.11 MPa
d.
Since, the force to open the door must be equal to the force applied to the door by pressure externally.
Therefore, the force required to open the door can be found out by the formula of pressure:
P = F/A
F = PA
where,
P = Absolute Pressure on Door = 1110725 Pa
A = Area of door = 2 m²
F = Force Required to Open the Door = ?
Therefore,
F = (1.11 MPa)(2 m²)
F = 2.22 MN
Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as object A to reach the ground, what is the ratio of the heights from which A and B fell
Answer:
1:4
Explanation:
We have, two identical objects A and B fall from rest from different heights to the ground.
Object B takes twice as long as object A to reach the ground. It is required to find the ratio of the heights from which A and B fell. Let [tex]h_A\ \text{and}\ h_B[/tex] are the height for A and B respectively. So,
[tex]\dfrac{h_A}{h_B}=\dfrac{(1/2)gt_A^2}{(1/2)gt_B^2}\\\\\dfrac{h_A}{h_B}=\dfrac{t_A^2}{t_B^2}[/tex]
We have,
[tex]t_B=2t_A[/tex]
So,
[tex]\dfrac{h_A}{h_B}=\dfrac{t_A^2}{(2t_B)^2}\\\\\dfrac{h_A}{h_B}=\dfrac{1}{4}[/tex]
So, the ratio of the heights from which A and B fell is 1:4.
A wire runs from the top of a pole that is h feet tall to the ground. The wire touches the ground a distance of d feet from the base of the pole. The wire makes an angle of theta with the top of the pole. Express h in terms of theta and d.
Answer:
A = Tan{-1} d/h
Explanation
Let the angle Theta be A
From identity the angle A has an opposite side d and an adjacent h.
From trigonometry ratio
Tan A = d/h
A = Tan{-1} d/h
A body of mass m=1kg is moving straight-line and its path as a function of time is given by the following
function: s= A - Bt+Ct - Dr, here C=10m/s2 and D=1m/s3 are some constants. What is the
magnitude of the force acting on the body at instant t=1s?
Answer: The force at t = 1s is 14 N
Explanation:
I guess that the position equation actually is: (by looking at the units of C and D)
S = A - B*t + C*t^2 - D*t^3
As you may know by Newton's third law, if we want to find the force, we first need to find the acceleration, and before that, the velocity.
the velocity can be found by integrating over time, the velocity is:
V = 0 - B + 2*C*T - 3*D*t^2
For the acceleration we integrate again:
A = 0 + 2*C - 6D*T
and we know that F = m*A
then:
Force(t) = 1kg*(2*10m/s^2 - 6*1m/s^3*t)
We want the force at t = 1s, so we replace t by 1s.
F(1s) = 1kg*(20m/s^2 - 6m/s^2) = 14 N
A candle is placed 14 cm in front of a concave mirror. The image of the candle is focused on a sheet of paper that is exactly 21 cm in front of the mirror. what is the magnification of the image
Answer:
The magnification of the image is [tex]M = -1.5[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 14 cm[/tex]
The image distance is [tex]v = 21 \ cm[/tex]
The magnification of the image is mathematically represented as
[tex]M = - \frac{v}{u}[/tex]
substituting values
[tex]M = - \frac{21}{14}[/tex]
[tex]M = -1.5[/tex]
The negative value means that the image is real , inverted and enlarged
⦁ Match the following terms:
⦁ Mass number
⦁ Isotopes
⦁ Nitrogen
⦁ Atomic number
⦁ The number of protons in the nucleus of an atom.
⦁ The number of protons and neutrons in the nucleus of an atom.
⦁ The name of the element with atomic number 7.
⦁ Atoms with the same number of protons, but different number of neutrons.
Answer:
Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.
Isotopes - ⦁ Atoms with the same number of protons, but different number of neutrons.
Nitrogen - ⦁ The name of the element with atomic number 7.
Atomic number - ⦁ The number of protons in the nucleus of an atom.
Think of something from everyday life that follows a two-dimensional path. It could be a kicked football, a bus that's turning a corner, or a person jogging around a track, etc. Describe your scenario in detail, and then identify the acceleration at each point. When is the acceleration vector not aligned with the direction of travel
Answer:
Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve. Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner.
The acceleration is not aligned with the direction of travel because the change in velocity is at a tangent (directed away) to the direction of travel of the bus.
Consider the boundary-value problem introduced in the construction of the mathematical model for the shape of a rotating string: T d2y dx2 + rhoω2y = 0, y(0) = 0, y(L) = 0. For constants T and rho, define the critical speeds of angular rotation ωn as the values of ω for which the boundary-value problem has nontrivial solutions. Find the critical speeds ωn and the corresponding deflections yn(x). (Give your answers in terms of n, making sure that each value of n corresponds to a unique critical speed.)
Answer:
[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]
[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]
Explanation:
The given differential equation is
[tex]T\frac{d^2y}{dx^2} + \rho w ^2y=0[/tex] and y(0) = 0, y(L) =0
where T and ρ are constants
The given rewrite as
[tex]\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0[/tex]
auxiliary equation is
[tex]m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi[/tex]
Solution of this de is
[tex]y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]
y(0)=0 ⇒ C₁ = 0
[tex]y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]
y(L) = 0 ⇒
[tex]C_2 \sin \sqrt{\frac{\rho}{T} } wL=0[/tex]
we need non zero solution
⇒ C₂ ≠ 0 and
[tex]\sin \sqrt{\frac{\rho}{T} } wL=0[/tex]
[tex]\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi[/tex]
[tex]w_n = \sqrt{\frac{T}{\rho} } \frac{n \pi}{L}[/tex]
solution corresponding these [tex]w_n[/tex] values
[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]
[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]
A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2Q. 1)How will the test charge move immediately after being released?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is the second option
Explanation:
Generally the electric force exerted by the charge Q on the charge (q) is mathematically represented as
[tex]F_Q = \frac{kqQ}{r^2}[/tex]
Generally the electric force exerted by the charge 2Q on the charge (q) is mathematically represented as
[tex]F_{2Q} = \frac{kq2Q}{2r^2}[/tex]
Now the net force exerted on q is
[tex]F_{net} = \frac{kqQ}{r^2} - \frac{2k q Q}{4r^2}[/tex]
[tex]F_{net} = \frac{4kqQ- 2kqQ}{4r^2}[/tex]
[tex]F_{net} = \frac{kqQ}{2r^2}[/tex]
Looking at the resulting equation we see that [tex]F_{net} > 0[/tex]
This implies that the charge q would move to the right
An aluminium bar 600mm long with diameter 40mm has a hole drilled in the centre of the bar.The hole is 3omm in diameter and is 100mm long.If the modulus of elasticity for the aluminium is 85GN/m^2. Calculate the total contraction on the bar due to a compresive load of 180kn
Given that,
Length of bar = 600 mm
Diameter of bar = 40 mm
Diameter of hole = 30 mm
Length of hole = 100 mm
Modulus of elasticity = 85 GN/m²
Load = 180 kN
We need to calculate the area of cross section without hole
Using formula of area
[tex]A=\dfrac{\pi\times d^2}{4}[/tex]
Put the value into the formula
[tex]A=\dfrac{\pi\times40^2}{4}[/tex]
[tex]A=1256.6\ mm^2[/tex]
We need to calculate the area of cross section with hole
Using formula of area
[tex]A=\pi\times\dfrac{(d_{b}^2-d_{h}^{2})}{4}[/tex]
Put the value into the formula
[tex]A=\pi\times\dfrac{(40^2-30^2)}{4}[/tex]
[tex]A=549.77\ mm^2[/tex]
We need to calculate the total contraction on the bar
Using formula of total contraction
Total contraction = contraction in bar without hole part + contraction in bar with hole part
[tex]Total\ contraction = \dfrac{F\times L_{1}}{A_{1}\times E}+\dfrac{F\times L_{2}}{A_{2}\times E}[/tex]
Where, F = load
L = length
A = area of cross section
E = modulus of elasticity
Put the value into the formula
[tex]Total\ contraction=\dfrac{180\times10^3}{85\times10^{3}}(\dfrac{500}{1256.6}+\dfrac{100}{549.77})[/tex]
[tex]Total\ contraction = 1.227\ mm^2[/tex]
Hence, The total contraction on the bar is 1.227 mm²
g A ceiling fan is rotating counterclockwise with a constant angular acceleration of 0.35π rad/s2 about a fixed axis perpendicular to its plane and through its center. Assume the fan starts from rest. (a) What is the angular velocity of the fan after 2.0 s? (Enter the magnitude.) rad/s (b) What is the angular displacement of the fan after 2.0 s? (Enter the magnitude.) rad
Answer:
The correct answer will be:
(a) 2.2 rad/s
(b) 2.2 rad
Explanation:
The given values are:
Angular acceleration, [tex]\alpha = 0.35 \pi \ rad/s^2[/tex]
(a)...
At time t = 2.0 s,
The angular velocity will be:
⇒ [tex]\omega = \alpha t[/tex]
On putting the estimated values, we get
⇒ [tex]=0.35\pi \times 2.0[/tex]
∴ [[tex]\pi = 3.14[/tex]]
⇒ [tex]=0.35\times 3.14\times 2.0[/tex]
⇒ [tex]=2.2 \ rad/s[/tex]
(b)...
The angular displacement will be:
⇒ [tex]\theta = \frac{1}{2}\alpha t^2[/tex]
On putting the estimated values, we get
⇒ [tex]=\frac{1}{2}\times 0.35\pi\times (2.0)^2[/tex]
⇒ [tex]=\frac{1}{2}\times 0.35\times 3.14\times 4[/tex]
⇒ [tex]=1.099\times 2[/tex]
⇒ [tex]=2.2 \ rad[/tex]
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 26.0 m/s. The first one is thrown at an angle of 67.5° with respect to the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first?
Answer:
22.5°
Explanation:
Short answer: The trajectories will hit the same target when the projectile is launched at complementary angles. The second angle is 90° -67.5° = 22.5°.
__
Longer answer: The horizontal speed of the snowball launched at angle α with speed s is ...
vh = s·cos(α)
Then the horizontal distance at time t is ...
x = vh·t
and the time taken to get to some distance x is ...
t = x/vh = x/(s·cos(α))
The equation for the vertical motion of the projectile is ...
y = -4.9t² +s·sin(α)·t
Substituting the above expression for t, we have y in terms of x:
y = -4.9x²/(s·cos(α))² +(s·sin(α)·x)/(s·cos(α))
Factoring gives ...
y = (x/(cos(α))(-4.9x/(s²·cos(α)) +sin(α))
The height y will be zero at x=0 and at ...
0 = -4.9x/(s²·cos(α)) +sin(α)
x = s²·sin(α)·cos(α)/4.9 = (s²/9.8)sin(2α)
So, for some alternate angle β, we want ...
sin(2α) = sin(2β)
We know this will be the case for ...
2β = 180° -2α
β = 90° -α
The second snowball should be thrown at an angle of 90°-67.5° = 22.5° to make it hit the same point as the first.
A flat plate of polished copper of surface emissivity 0.1 is 0.1 m long and 0.1 m wide. The plate is placed vertically, with one side heated to a surface temperature of 500 K, and the other side remaining insulated. The heated side is exposed to quiescent air at 300 K and the surroundings are also at 300 K. Assume that air can be taken as an ideal gas. Estimate the heat rate from the flat plate.
Answer:
The heat rate is
Explanation:
From the question we are told that
The surface emissivity is [tex]e=0.1[/tex]
The length is [tex]L = 0.1 \ m[/tex]
The width is [tex]W = 0.1 \ m[/tex]
The surface temperature of one side is [tex]T_1 = 500 \ K[/tex]
The temperature of the quiescent air [tex]T_c = 300 \ K[/tex]
The temperature of the surrounding is [tex]T_s = 300 \ K[/tex]
The heat rate from the flat plate is mathematically represented as
[tex]Q = \sigma A e (T_1^4 - T_a^4)[/tex]
Where [tex]\sigma[/tex] is the quiescent air Stefan-Boltzmann constant and it value is
[tex]\sigma = 5.67*10^{-8} m^{-2} \cdot K^{-4}[/tex]
A is the area which is mathematically evaluated as
[tex]A = W * L[/tex]
substituting values
[tex]A = 0.1 * 0.1[/tex]
[tex]A = 0.01 \ m^2[/tex]
substituting values
[tex]Q = 5.67 *10^{-8} * (0.01) *(500^4 -300^4)[/tex]
[tex]Q =3.045 \ Watt[/tex]
How can you identify a moveable pulley?
A. It has a fixed axle.
B. It moves up and down with the load.
C. It is anchored.
D. It has been relocated from one location to another.
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km/hr (69.4 m/s) how far in advance of the recipients (horizontal distance) must the goods be dropped
Answer:
481 m
Explanation:
To fall 235 m, the time required is
t = √(2H/g)
t= √(2[tex]\times[/tex]235/9.8)
t=6.92 seconds.
The supplies will travel forward
6.92 [tex]\times[/tex] 69.4 ≈ 481 m
Therefore, the goods must be dropped 481 m in advance of the recipients.
consider an electric dipole lying along x-axis with mid-point O as the origin of coordinate system.find the electric potential V at point P due to dipole.in addition to this,find electric potentials for special cases. when P lies on the axial line and when P lies on the equatorial line.
Answer:
axial V = 0
equatorial V = k q 2a / (x² -a²), V = k q 2x / (a² -x²)
Explanation:
A dipole is a system formed by two charges of equal magnitude, but different sign, separated by a distance 2a; let's look for the electrical potential in an axial line
V = k (q / √(a² + y²) - q /√ (a² + y²))
V = 0
the potential on the equator
we place the positive charge to the left and perform the calculation for a point outside the dipole
V = k (q / (x-a) - q / (x + a))
V = k q 2a / (x² -a²)
we perform the calculation for a point between the dipo charges
V = k (q / (a-x) - q / (a + x))
V = k q 2x / (a² -x²)
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. The hoist weighs 350 N. The ropes, fastened at different heights, make angles of 50° and 38° with the horizontal. Find the tension in each rope and the magnitude of each tension.
Answer:
276 N and 225 N
Explanation:
Draw a free body diagram. There are three forces on the hoist:
Weight force 350 N pulling down,
Tension force T₁ pulling up and left 50° from the horizontal,
Tension force T₂ pulling up and right 38° from the horizontal.
Sum of forces in the x direction:
∑F = ma
T₂ cos 38° − T₁ cos 50° = 0
T₂ cos 38° = T₁ cos 50°
T₂ = T₁ cos 50° / cos 38°
Sum of forces in the y direction:
∑F = ma
T₂ sin 38° + T₁ sin 50° − 350 = 0
T₂ sin 38° + T₁ sin 50° = 350
Substitute:
(T₁ cos 50° / cos 38°) sin 38° + T₁ sin 50° = 350
T₁ cos 50° tan 38° + T₁ sin 50° = 350
T₁ (cos 50° tan 38° + sin 50°) = 350
T₁ = 350 / (cos 50° tan 38° + sin 50°)
T₁ = 276 N
T₂ = T₁ cos 50° / cos 38°
T₂ = 225 N
momentum is closely related to
Answer:
Momentum is mainly related to forces
Enterprising students set an enormous slip-n-slide (a plastic sheetcovered in water to reduce friction) on flat ground. If the slip-n-slideis 250 m long, how small does the average acceleration have to be fora student starting at 5 m/s to slide to the end
Answer:
a = -0.05 m/s² (negative sign shows deceleration)
Explanation:
In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:
2as = Vf² - Vi²
where,
a = minimum acceleration required = ?
s = minimum distance covered = 250 m
Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)
Vi = Initial Velocity = 5 m/s
Therefore,
2a(250 m) = (0 m/s)² - (5 m/s)²
a = - (25 m²/s²)/(500 m)
a = -0.05 m/s² (negative sign shows deceleration)
linear momentum is the product of mass and acceleration
Answer:
This statement is false , linear momentum is the product of mass and velocity
Answer:
Linear momentum is defined as the product of a system's mass multiplied by its velocity. In symbols, linear momentum is expressed as p = mv. ... Thus the greater an object's mass or the greater its velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v.
Explanation:
hope this helps u stay safe
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.5 s for the boat to travel from its highest point to its lowest, a total distance of 0.69 m. The fisherman sees that the wave crests are spaced 6.6 m apart. Part A How fast are the waves traveling
Answer:
v= 0.94 m/s
Explanation:
In order to calculate the speed of the waves, you use the following formula for the speed of the waves:
[tex]v=\lambda f[/tex] (1)
v: speed of the wave
λ: wavelength of the wave
f: frequency of the wave
The frequency is calculated by using the information about the time that boat takes to travel from its highest point to its lowest point. This time is a half of a period:
[tex]\frac{T}{2}=3.5s\\\\T=7.0s[/tex]
Then, the frequency is:
[tex]f=\frac{1}{T}=\frac{1}{7s}=0.142Hz[/tex]
The wavelength of the wave is the distance between crests of the wave
[tex]\lambda=6.6m[/tex]
With the values of the frequency and the wavelength, you can find the speed of the wave by using the equation (1):
[tex]v=(6.6m)(0.142Hz)=0.94\frac{m}{s}[/tex]
The speed of the wave is 0.94m/s
A student is asked to design an experiment to determine the change in angular momentum of a disk that rotates about its center and the product of the average torque applied to the disk and the time interval in which the torque is exerted. A net force is applied tangentially to the surface of the disk. The rotational inertia of the disk about its center is I = 1/2 MR^2
Which two of the following quantities should the student measure to determine the change in angular momentum of the disk after 10 s? Select two answers
A. The magnitude of the net force exerted on the disk
B. The distance between the center of the disk and where the net force is applied to the disk
C. The radius of the disk
D. The mass of the disk
Answer:
A. The magnitude of the net force exerted on the disk
B. The distance between the center of the disk and where the net force is applied to the disk
Explanation:
To determine the change in angular momentum of the disk after a stipulated time, one must measure the above options.
The radius of the disk is fixed and does not vary with the experiment, and the mass of the disk is also constant and known.
One must first measure the magnitude of the net force exerted on the disk, and determine the torque as a result of this torque from the distance between the center of the disk and the point where the net force is applied. The above statement also points out the necessity of measuring the distance between the center of the disk and the point where the net force is applied on the disk, as both the torque, and the moment of inertia is calculated from this point.
torque T = Force time distance of point of action of force from mid point of the disk
T = F X r
T x t = Δ(Iω)
Where t is the time,
and Δ(Iω) is change in angular momentum.
