Answer:
48.9 is the answer I think !
Answer:
28.4
Explanation:
planet smaller than earth but larger than mercury
Answer:
venus......................
8 points)Antireflection coating can be used on the eyeglasses to reduce the reflection of light: a) A 100nm thick coating is applied to the lens. What must be the coating’s index of refraction to be most effective at 500nm? (Assume the coating index of refraction is less than that of the lens). b) If the index of refraction of the coating is 1.20, find the necessary thickness of the coating at 500nm.
Answer:
- the coating’s index of refraction is 1.25
- the required thickness is 104.1667 nm
Explanation:
Given the data in the question;
Thickness of coating t = 100 nm
wavelength λ = 500nm
we know that refractive index is;
t = λ/4n
make n, the subject of formula
t4n = λ
n = λ / 4t
we substitute
n = 500 / ( 4 × 100 )
n = 500 / 400
n = 1.25
Therefore, the coating’s index of refraction is 1.25
2)
given that;
Index of refraction of the coating; n = 1.20
λ = 500 nm
thickness of coating t = ?
t = λ / 4n
we substitute
t = 500 / ( 4 × 1.2 )
t = 500 / 4.8
t = 104.1667 nm
Therefore, the required thickness is 104.1667 nm
A skateboarder is inside of a half pipe, shown here. Explain her energy transformations as she jumps off at point A, slides to point B, and finally reaches point C.
A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.
Answer:
V = 240V
Explanation:
V = I*R
V = 15A*16ohms
V = 240V
If ATM is 102 kPa, what force does the atmosphere exert on the palm of your hand which has an area of 0.016 meters?
Answer:
Force = 1.632 Newton
Explanation:
Given the following data;
Pressure = 102 kPa
Area = 0.016 m²
To find what force the atmosphere exert on the palm of your hand;
Mathematically, pressure is given by the formula;
[tex] Pressure = \frac {Force}{area} [/tex]
Force = 102 * 0.016
Force = 1.632 Newton
Consider a 92.0 kg ice skater who is spinning on the ice. What is the moment of inertia of the skater, if the skater is approximated to be a solid cylinder that has a 0.140 m radius and is rotating about the center axis of the cylinder.
Answer:
[tex]I=0.902kg*m^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=92.0kg[/tex]
Radius [tex]r=0.140m[/tex]
Generally the equation for moment of Inertia is mathematically given by
[tex]I = 0.5*m*r^2[/tex]
[tex]I=0.5*(92)*0.14^2[/tex]
[tex]I=0.902kg*m^2[/tex]
A person rolls a 7 kg bowling ball down a lane in a bowling alley. The lane is
18 m long. The ball is traveling at 7 m/s when it leaves the person's hand.
What is the ball's kinetic energy at this point?
Answer:
171.5J
Explanation:
K=1/2 *m *U²
K=1/2 *7 *7²
K=171.5 J
. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content of the system.
Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
What is this sport ⚽⚾
Answer:
sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.
hope it is helpful to you
What is the current in the 30 resistor?
A. 0.0833 A
B. 12 A
C. 2 A
D. 10 A
Answer:
Explanation:
Step 1) Combine all resistors into an equivalent overall resistor. These are all in series so you just add them up. Req = 10Ω + 20Ω + 30Ω = 60Ω:
Step 2) Using Ohm's Law, I = V/R = 120/60 = 2 A
Now you know how much current is flowing, and that current flows through each resistor the same. So the current in the 30 Ω resistor is 2.00 amps.
Nick and Chloe left their campsite by canoe and paddle downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average speed of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream
The time spent by the campers when they turn around downstream is 15 minutes.
Total distance traveled by Nick and Chloe
The concept of total distance traveled by Nick and Chloe can be used to determine the time they turn around downstream.
Let time for downstream = t1
Let time for upstream = t2
distance covered in upstream = distance covered in downstream = d
12(t1) = d
4(t2) = d
12t1 = 4t2
t1 + t2 = 1
t2 = 1 - t1
12t1 = 4(1 - t1)
12t1 = 4 - 4t1
16t1 = 4
t1 = 4/16
t1 = 0.25 hours
t1 = 0.25(60 min) = 15 mins
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What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal
Answer:
the electric field is 3.91 x 10⁶ N/C
Explanation:
Given the data in the question;
Electric field at a point due to point charge is;
E = kq/r²
where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator
Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C
so we substitute into the formula
E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²
E = 59400000 / 15.21
E = 3.91 x 10⁶ N/C
Therefore, the electric field is 3.91 x 10⁶ N/C
A 620 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 870 kg. As the elevator starts moving, the scale reads 450 N.
Required:
a. Find the acceleration of the elevator (magnitude and direction).
b. What is the acceleration if the scale reads 670 N?
c. If the scale reads zero, should the student worry? Explain.
d. What is the tension in the cable in parts (a) and (c)?
Answer:
(a) 9.28 m/s2
(b) 9.03 m/s2
(c) 9.8 m/s2
(d) 450 N, 670 N
Explanation:
mass of elevator + student, m = 870 kg
Reading of scale, R = 450 N
(a) When the elevator goes down, the weight decreases.
Let the acceleration is a.
By the Newton's second law
m g - R = m a
870 x 9.8 - 450 = 870 a
a = 9.28 m/s2
(b) R = 670 N
Let the acceleration is a.
870 x 9.8 - 670 = 870 a
a = 9.03 m/s2
(c) If the scale reads zero, it mean the elevator is falling freely. The acceleration is downwards and its value is 9.8 m/s2.
(d) Tension in cable is 450 N and 670 N.
A boxer punches a sheet of paper in midair from rest to a speed of 20 m/s in 0.05 s. If the mass of the paper is 0.01 kg, the force of the punch on the paper is
A) 0.08 N.
B) 4.0 N.
C) 8.0 N.
D) 40 N.
What must be true if energy is to be transferred as heat between two bodies in physical contact?
1-The two bodies must have different volumes.
2-The two bodies must be at different temperatures.
3-The two bodies must have different masses.
4-The two bodies must be in thermal equilibrium.
Answer:
answer is d
Explanation:
i hope this helps you
PLEASE HELP How does an object move when it is in linear motion?
in a straight line
up and down
in a circle
to the left
Answer:
In linear motion, the directions of all the vectors describing the system are equal and constant which means the objects move along the same axis and do not change direction. Correct Answer: In a straight line
Explanation:
Answer:
In a straight line. we can also have translational motion which is also a kind of linear motion .
They create a heat engine where the hot reservoir is filled with water and steam at equilibrium, and the cold reservoir is filled with ice and water at equilibrium. What is the Carnot efficiency for their heat engine if the pressure is constant at 1.0 atmospheres?
Answer:
The efficiency of Carnot's heat engine is 26.8 %.
Explanation:
Temperature of hot reservoir, TH = 100 degree C = 373 K
temperature of cold reservoir, Tc = 0 degree C = 273 K
The efficiency of Carnot's heat engine is
[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]
The efficiency of Carnot's heat engine is 26.8 %.
can someone please help me
F = 153 N
[tex]\theta = 11.3°[/tex]
Explanation:
Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:
[tex]F_{x}[/tex] = 350 N - 200 N = 150 N
[tex]F_{y}[/tex] = 180 N - 150 N = 30 N
The magnitude of the resultant force F is given by
[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]
[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]
[tex]\:\:\:\:\:\:=153\:N[/tex]
To find the direction [tex]\theta[/tex], we use
[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]
or
[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]
A crucible (container) of molten metal has an open top with an area of 5.000 m^2. The molten metal acts as a blackbody radiator. The intensity spectrum of its radiation peaks at a wavelength of 320 nm. What is the temperature of that blackbody?
Answer:
T = 9056 K
Explanation:
In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation
λ T = 2,898 10⁻³
where lam is the wavelength of the maximum emission
T = 2,898 10⁻³ /λ
let's calculate
T = 2,898 10⁻³ / 320 10⁻⁹
T = 9.056 10³ K
T = 9056 K
A bucket of mass m in a well is held up by a rope. the rope is wound around a drum of radius r there is also a handle of length R attached to the drum. the tension in the rope is equal to T. If the buket is allowed to fall into the well, which point will have the greatest angular acceleration, a point on the rim of the drum (at radius r) or a point on the end of the handle (at radius R)?
a. The point on the rim of the drum.
b. The point at the end of the handle.
c. They will both have the same angular acceleration.
Answer:
the correct answer is C
Explanation:
This is a system with circular motion, there is a relationship between the linear and angular variables
a = α r
with the cube going down the well, the tension of the leather is maintained therefore the acceleration of the cube is
W = m a
-mg = ma
a = -g
this acceleration a is the same as that at the edge of the drum.
α = a / r
where we can see that the angular acceleration is constant
consequently the correct answer is C
Part C
When only one color of light reflects from a piece of paper, what happens to the other colors of light?
Remember that light is energy, and energy cannot be created or destroyed.
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Answer:
the other colours get absorbed by the paper
Answer:
The other colors are absorbed by the paper and not reflected.
a car runs of a road and collides with a tree. glass pieces from the windscreen are projected forward and are found an average distance of 12m from the car. the average height of the windscreen is 1.2m.
Establish the speed of the car at the time of impact. assume g=10 m/s²
Answer:
v₀ₓ = 24.24 m / s
Explanation:
This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.
Y axis
initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
when it reaches the ground its height is zero and the initial height is y₀=1.2m
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2 y_o/g}[/tex]
t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]
t = 0.495 s
X axis
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 12 / 0.495
v₀ₓ = 24.24 m / s
1. A skydiver carries a buzzer which emits a steady 1800 Hz tone. A friend on the ground (directly below the skydiver) listens to the doppler shifted frequency. Assume the air is calm and that the speed of sound is independent of altitude. While the skydiver is falling at terminal velocity (the constant speed reached due to the balance of the downward gravitational force and the upward drag force due to air resistance), her friend on the ground hears waves of frequency 2150 Hz. a. How fast is the skydiver falling
Answer:
vs = 55.84 m/s
Explanation:
In this case, the source (skydiver with buzzer) is moving towards the observer (friend). The formula for this case will be:
[tex]f' = \frac{v}{v-v_s} f[/tex]
where,
f' = shifted frequency = 2150 Hz
f = actual frequency = 1800 Hz
v = speed of sound = 343 m/s
vs = speed of skydiver = ?
Therefore,
[tex]2150\ Hz = \frac{343\ m/s}{343\ m/s - v_s}(1800\ Hz)\\\\343\ m/s - v_s = \frac{343\ m/s}{2150\ Hz} (1800\ Hz)\\\\v_s = 343\ m/s - 287.16\ m/s\\[/tex]
vs = 55.84 m/s
Question 2:
Inclined Plane
A block (M) weighs 25-N, rests on an inclined plane when it is joined by a sting to a support
(S) as shown in the figure' below. Use g=10 N/Kg.
(S)
B
M
List and classify the forces acting on (M).
Représent, without scaling, the forces acting on (M).
Find the mass of (M).
74. If the string were cut, (M) does not slide. Explain this phenomenon.
15. Determine the mass and weight of (M) on moon.
06
Answer:
we need the block
Explanation:
1×2 =4 lest 74 =345
Find the X and Y components of the following:
A. 35 m/s at 57q from the x-axis.
Explanation:
Given that,
35 m/s at 57° from the x-axis.
Speed, v = 35 m/s
Angle, θ = 57°
Horizontal component,
[tex]v_x=v\cos\theta\\\\=35\times \cos(57)\\\\=19.06 m/s[/tex]
Vertical component,
[tex]v_y=v\sin\theta\\\\v_y=35\times \sin(57)\\\\=29.35\ m/s[/tex]
Hence, this is the required solution.
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.
A conducting sphere of radius R carries an excess positive charge and is very far from any other charges. Draw the graphs that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere?
Answer:
See annex
Explanation:
By convention potential at ∞ V(∞ ) = 0
As the distance from the sphere decreases the potential increases up to the point d = R ( R is the radius of the sphere. That potential remains constant while d = R and becomes 0 inside the sphere where there is not free charges and therefore the electric field is 0 and so is the potential.
I am sorry I could not make a better graph
The graph that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere is attached as an image below
[tex]V = \frac{KQ}{R}[/tex]
for r <= R
[tex]V = \frac{KQ}{r}[/tex]
for r > R
Therefore the graph will be
For more information on potentials as function of distance
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A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 830 kg car to rest from an initial speed of 1.4 m/s.
Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W =[tex]F^>[/tex] × [tex]x^>[/tex]
W = [tex]Fxcos\theta[/tex] ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = [tex]Fxcos([/tex] 0° [tex])[/tex]
W = [tex]Fx[/tex] ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
[tex]W_{net[/tex] = ΔKE
= [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu²
we make F, the subject of formula
F = [tex]\frac{m}{2x}[/tex]( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m
and slope angle 30 degree, suppose the coefficient of friction of the slope surface is
0.25, find (neglect air resistance,take g=10 m/s^2)
The friction force experienced by the box.
00) The acceleration of the box along the slope.
(1) The time T required for the object to reach the bottom of the slope from the slope top.
Answer:
Explanation:
The first thing we are asked to find is the Force experienced by the box. That is found in the formula:
F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:
[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:
First we need the weight of the box. Having the mass, we find the weight:
w = mg so
w = 25.14(10) so
w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:
251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so
F = 125.7 And in order to answer what a is equal to, we find f:
f = μ[tex]F_n[/tex] where Fn is the weight of the object.
f = .25(251.4) so
f = 62.85. Filling everything in now altogether to solve for a, the only missing value:
125.7 - 62.85 = 25.14a and
62.85 = 25.14a so
a = 2.5 m/s/s
Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that
a = -10 m/s/s
v₀ = 0 (it starts from rest)
Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:
Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:
[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and
[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so
t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.
