Answer:
the answer is option A = photoelectric effect
Explanation:
If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.
Complete Question
The diagram for this question is showed on the first uploaded image (reference homework solutions )
Answer:
The velocity at the bottom is [tex]v = 11.76 \ m/ s[/tex]
Explanation:
From the question we are told that
The total distance traveled is [tex]d = 1.2 \ m[/tex]
The mass of the block is [tex]m_b = 0.3 \ kg[/tex]
The height of the block from the ground is h = 0.60 m
According the law of energy
[tex]PE = KE[/tex]
Where PE is the potential energy which is mathematically represented as
[tex]PE = m * g * h[/tex]
substituting values
[tex]PE = 3 * 9.8 * 0.60[/tex]
[tex]PE = 17.64 \ J[/tex]
So
KE is the kinetic energy at the bottom which is mathematically represented as
[tex]KE = \frac{1}{2} * m v^2[/tex]
So
[tex]\frac{1}{2} * m* v ^2 = PE[/tex]
substituting values
=> [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]
=> [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]
=> [tex]v = 11.76 \ m/ s[/tex]
Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble
Answer:
Johannes KeplerExplanation:
He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.Hope this helped,
Kavitha
A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/s. What is the average power done by the torque
Answer:
128.61 WattsExplanation:
Average power done by the torque is expressed as the ratio of the workdone by the toque to time.
Power = Workdone by torque/time
Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]
I is the rotational inertia = 16kgm²
[tex]\theta = angular\ displacement[/tex]
[tex]\theta = 2 rev = 12.56 rad[/tex]
[tex]\alpha \ is \ the\ angular\ acceleration[/tex]
To get the angular acceleration, we will use the formula;
[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]
[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]
Workdone by the torque = 16 * 1.28 * 12.56
Workdone by the torque = 257.23 Joules
Average power done by the torque = Workdone by torque/time
= 257.23/2.0
= 128.61 Watts
On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?
Answer:
T = 2.06h
Explanation:
In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex] (1)
T: time for a complete orbit = ?
r: radius of the orbit
G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2
Mm: mass of the moon = 7.34*10^22 kg
The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:
[tex]r=R_m+160km\\\\[/tex]
Rm: radius of the moon = 1737.1 km
[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]
Then, you replace all values of the parameters in the equation (1):
[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]
In hours you obtain:
[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]
The time that the Apollo takes to complete an orbit around the moon is 2.06h
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that is 0.60 m from the center is closest to
Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is [tex]E = 7500 \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner circle is [tex]r_i = 0.80 \ m[/tex]
The radius of the outer circle is [tex]r_o = 1.20 \ m[/tex]
The charge on the spherical shell [tex]q_n = -500nC = -500*10^{-9} \ C[/tex]
The magnitude of the point charge at the center is [tex]q_c = + 300 nC = + 300 * 10^{-9} \ C[/tex]
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as
[tex]E = \frac{k * q_c }{x^2}[/tex]
substituting values
[tex]E = \frac{k * q_c }{x^2}[/tex]
where k is the coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
substituting values
[tex]E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}[/tex]
[tex]E = 7500 \ N/C[/tex]
You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your normal weight is w= 635 N. If the elevator has an initial acceleration of magnitude 2.45 m/s2, what does the scale read? Express your answer in newtons.
Answer: 479. 425 N
Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.
It is given by N= m(g+a)
When it is accelerating downward, the scale reading is less than the true weight.
It so given by N = m(g-a)
The answer to the above questions is in the attached photo
Answer:
the scale will read 476.414 N
Explanation:
Weight = 635 N
mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)
mass m = 635 ÷ 9.81 = 64.729 kg
initial acceleration of the elevator a = 2.45 m/s^2
the force produced by the acceleration of the elevator downwards = ma
your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss
apparent weight = weight - ma
apparent weight = 635 - (64.729 x 2.45)
apparent weight = 635 - 158.586 = 476.414 N
A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
I2 = 3.076 A
Explanation:
In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:
[tex]B=\frac{\mu_oI}{2R}[/tex] (1)
I: current in the wire
R: radius of the wire
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:
[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex] (2)
I1: current of the first ring = 8A
R1: radius of the first ring = 0.078m
I2: current of the second ring = ?
R2: radius of the first second = 0.03m
To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:
[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]
The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.
Passengers in a carnival ride move at constant speed in a circle of radius 5.0 m, making a complete revolution in 4.0 s. As they spin, they feel their backs pressing against the wall holding them in the ride. A. What is the direction of the passengers' acceleration? a. No direction (zero acceleration) b. Directed towards center c. Directed away from center d. Directed tangentially B. What is the passengers' linear speed in m/s? C. What is the magnitude of their acceleration in m/s^2? D. What is their angular speed in rad/s?
Answer:
A. b) Directed towards center
B. [tex]v = 7.854\ m/s[/tex]
C. [tex]a_c = 12.337\ m/s^2[/tex]
D. [tex]w = 1.57\ rad/s[/tex]
Explanation:
The "force" that they feel pressing their backs against the wall is because the reaction to the centripetal acceleration .
A.
This acceleration has its direction towards the center of the circle. (option b)
B.
Their linear speed can be calculated with the equation:
[tex]v = (\theta/t)*r[/tex]
Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:
[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]
C.
The centripetal acceleration is given by the equation:
[tex]a_c = v^2/r[/tex]
[tex]a_c = 7.854^2/5[/tex]
[tex]a_c = 12.337\ m/s^2[/tex]
D.
Their angular speed is given by the equation:
[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion, after which a horizontal force of 60 N is required to keep the block in moving with constant speed. Find the coefficient of static and kinetic friction between the block and the surface.
Answer:
μs = 0.30
μk = 0.24
Explanation:
In order to calculate the kinetic friction and static friction between the block and the surface, you take into account that the kinetic friction is important when the block is moving and the static friction when the block is at rest.
You use the following formula to find the coefficient of static friction:
[tex]F_1=\mu_s Mg[/tex] (1)
F1 = 75N
μs: coefficient of static friction = ?
M: mass of the block = 25.0kg
g: gravitational acceleration = 9.8m/s^2
You solve for μs in the equation (1):
[tex]\mu_s=\frac{F_1}{Mg}=\frac{75N}{(25.0kg)(9.8m/s^2)}=0.30[/tex]
For the coefficient of kinetic friction you have:
[tex]F_2=\mu_k Mg[/tex] (2)
F2 = 60N
μk: coefficient of kinetic friction = ?
You solve for μk in the equation (2):
[tex]\mu_k=\frac{F_2}{Mg}=\frac{60N}{(25.0kg)(9.8m/s^2)}=0.24[/tex]
Then, you have:
coefficient of static friction = 0.30
coefficient of kinetic friction = 0.24
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
v = 1.7 m/s
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0215 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.15 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.
Answer:
h = 8.48*10^-3m
Explanation:
In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.
The total momentum of the system before the collision is equal to the total momentum after:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex] (1)
Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.
m1: mass of the marble = 0.0215kg
m2: mass of the pendulum concentrated at its end = 0.250kg
v1: horizontal speed of the arble before the collision = 5.15m/s
v2: horizontal speed of the pendulum before the collision = 0m/s
v: horizontal speed of both marble and pendulum after the collision = ?
You solve the equation (1) for v, and replace the values of the other parameters:
[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]
Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:
[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]
v = 0.40m/s
g: gravitational acceleration = 9,8m/s^2
[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]
Then, the height reached by marble and pendulum is 8.48*10^-3m
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?
Answer:
[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]
Explanation:
Hello,
In this case, since pressure is defined as the force applied over a surface:
[tex]P=\frac{F}{A}[/tex]
We can associate the force with the weight of the needle computed by using the acceleration of the gravity:
[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]
And the area of the the tip (circle) in meters:
[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]
Thus, the pressure exerted on the record turns out:
[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]
Which is truly a large value due to the tiny area on which the pressure is exerted.
Best regards.
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Answer:
Explanation:
F ×1 = 0.5×0.145×47×47
F = 160.15 N
Stress is a factor that contributes to heart disease risk.true or false
Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals
Answer:
producers make its own energy frominorganic substances.
A d'Arsonal meter with an internal resistance of 1 kohm requires 10 mA to produce full-scale deflection. Calculate thew value of a series
A d’Arsonval meter with an internal resistance of 1 kΩ requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.
Answer:4kΩ
Explanation:Given;
internal resistance, r = 1kΩ
current, I = 10mA = 0.01A
Voltage of full scale, V = 50V
Since there is full scale voltage of 50V, then the combined or total resistance (R) of the circuit is given as follows;
From Ohm's law
V = IR
R = [tex]\frac{V}{I}[/tex] [substitute the values of V and I]
R = [tex]\frac{50}{0.01}[/tex]
R = 5000Ω = 5kΩ
The combined resistance (R) is actually the total resistance of the series arrangement of the series resistance([tex]R_{S}[/tex]) and the internal resistance (r) in the circuit. i.e
R = [tex]R_{S}[/tex] + r
[tex]R_{S}[/tex] = R - r [Substitute the values of R and r]
[tex]R_{S}[/tex] = 5kΩ - 1kΩ
[tex]R_{S}[/tex] = 4kΩ
Therefore the series resistance is 4kΩ
what is the largest star in our night sky
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. What is the direction of the electric field vector?
Answer:
the electric field is pointing horizontal direction and in south direction
Explanation:
In an electromagnetic wave, the magnetic field and electrical field are perpendicular to each other and these are perpendicular to the direction of the waves.
Based on what you know about electricity, hypothesize about how series resistors would affect current flow. What would you expect the effective resistance of two equal resistors in series to be, compared to the resistance of a single resistor?
Answer:
Effective resistance of two equal resistors in series is twice that of a single resistor and in essence will reduce the amount of current flowing in the circuit.
Explanation:
When two resistors are connected in series, their effective resistance is the sum of their individual resistances. For example, given two resistors of resistance values R₁ and R₂, their effective resistance, Rₓ is given by;
Rₓ = R₁ + R₂ --------------(1)
If these resistors have equal resistance values, say R, then equation 1 becomes;
Rₓ = R + R
Rₓ = 2R
This means that their effective resistance is twice of their individual resistances. In other words, when two equal resistors are in series, their effective resistance is twice the resistance of each single one of those resistors.
Now, according to Ohm's law, voltage(V) is the product of current (I) and resistance (R). i.e
V = IR
I = [tex]\frac{V}{R}[/tex]
We can deduce that current increases as resistance decreases and vice-versa.
So, if the two equal resistors described above are connected in series, the amount of current flowing will be reduced compared to having just a single resistor.
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum
Explanation:
We have
A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.
The frequency of a pendulum is equal to the no of oscillation per unit time. so,
[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]
Tim period is reciprocal of frequency. So,
[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]
So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.
To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch
Answer:
By a factor of about 0.23
Explanation:
Pressure is force over an area: P=F/A
Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.
By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.
PressureFriction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.
Pressure exists as force over an area: P=F/A
Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
let's divide P₁ by P₂
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that, we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,
To learn more about Pressure refer to:
https://brainly.com/question/912155
#SPJ2
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field with a solenoid of the same size, carrying a current of 1.8 A , how many turns of wire would you need
Answer:
The number of turns of the solenoid is 3536 turns
Explanation:
Given;
magnetic field of the solenoid, B = 0.1 T
current in the solenoid, I = 1.8 A
length of the solenoid, L = 8cm = 0.08m
The magnetic field near the center of the solenoid is given by;
B = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
n is number of turns per length
I is the current in the coil
The number of turns per length is calculated as;
n = B / μ₀I
n = (0.1 ) / (4π x 10⁻⁷ x 1.8)
n = 44203.95 turns/m
The number of turns is calculated as;
N = nL
N = (44203.95)(0.08)
N = 3536 turns
Therefore, the number of turns of the solenoid is 3536 turns
An accelerating voltage of 2.25 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 36.4 cm away. (a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field
Answer:
s= 8.28×10⁻¹⁶m
Explanation:
given
V= 2.25×10³V
from conservation of energy
mv²/2=qΔV
v=√(2qΔV/m)
v= √(2×1.6×10⁻¹⁹×2.25×10³/9.1×10⁻³¹)
=√7.9×10¹⁴m/s
=2.8×10⁷m/s
the deflection of electron beam is
S= gt²/2
recall t= d/v
s=g([tex]\frac{d}{v}[/tex])²/2
s= [tex]\frac{1}{2}[/tex]×9.8×(0.364/2.8×10⁷)²
s= 8.28×10⁻¹⁶m
What is meant civilized?
Answer:
at an advanced stage of social and cultural development. "a civilized society"
Explanation:
polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)
Unpolarized light enters a polarizer with vertical polarization axis. The light that passes through passes another polarizer with transmission axis at 40 degrees to the horizontal. What is the intensity of the light after the second polarizer expressed as a fraction of the original intensity
Answer:
I = 0.2934 I₀
Explanation:
The expression that governs the transmission of polarization is
I = I₀ cos² θ
Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of
I₁ = I₀ / 2
this is the light that enters the second polarizer
I = I₁ cos² θ
we substitute
I = I₀ / 2 cos² 40
I = I₀ 0.2934
I = 0.2934 I₀
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N . Rolling friction can be neglected. You may want to review (Pages 165 - 168) . Part A What is the magnitude of the force of the car on the truck
Answer:
a) 3344 N
b) 3344 N
Explanation:
This is the complete question
1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected. A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. B. What is the magnitude of the force of the truck on the car?
Mass of the car = 1100 kg
Mass of the truck = 2200 kg
Force exerted on the ground by the car = 5000 N
The total mass in the system = 1100 + 2200 = 3300 Kg
Total force in the system = 5000 N
Recall that the force in the system = mass x acceleration
therefore,
5000 = 3300 x a
Total acceleration in the system = 5000/3300 = 1.52 m/s^2
The force on the truck individually fro the car, will be the product of this acceleration and its mass
Force on the truck = 2200 x 1.52 = 3344 N
b) Force on the car From the truck will be equal to this force but will act in the opposite direction.
Force on the car from the truck is 3344 N
A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 105 Pa for the atmospheric pressure and 1200 kg/m3 for the density of the sauce, find the absolute pressure in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.
Answer:
(a) P = 103064 Pa = 103.064 KPa
(b) P = 102476 Pa = 102.476 KPa
Explanation:
(a)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)
Pg = 1764 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1764 Pa
P = 103064 Pa = 103.064 KPa
(b)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)
Pg = 1176 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1176 Pa
P = 102476 Pa = 102.476 KPa
The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.
Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.
In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.
The absolute pressure in the bulb can be calculated using the following formula:
P = P₀ + ρgh
where:
P is the absolute pressure in the bulb,
P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),
ρ is the density of the sauce (1200 kg/m³),
g is the acceleration due to gravity (9.8 m/s²), and
h is the height of the sauce in the tube.
(a) When h = 0.15 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)
P ≈ 1.013 x 10⁵ Pa + 1764 Pa
P ≈ 1.031 x 10⁵ Pa
(b) When h = 0.10 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)
P ≈ 1.013 x 10⁵ Pa + 1176 Pa
P ≈ 1.025 x 10⁵ Pa
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