Answer:
A denser object will usually have a high index of
refraction.
A denser object will usually have a high index of refraction.
What is index of refraction?The refractive index is the ratio of the speed of light in vacuum and speed of light in any medium.
n = c/v
The density is greater for the denser medium (water, oil, mercury, etc) then the rarer medium (any gas or air).
When a light ray travels through denser medium, its velocity is reduced and the refracted ray bends towards normal.
As, the index of refraction is inversely proportional to the velocity of light in the medium, index of refraction will be high for denser object.
Thus, denser object have high index of refraction.
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A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
A light bulb, attached (by itself) to an ideal 10 V battery, becomes hotter as time goes on. The bulb's filament is made of tungsten, a metal, which becomes more resistive as its temperature increases. Which statement below is true?
a. As time goes on and the bulb grows hotter, the voltage across the bulb increases so the bulb would grow brighter.
b. As time goes on and the bulb grows hotter, the current through the bulb increases so the bulb would grow brighter.
c. As time goes on and the bulb grows hotter, the voltage across the bulb decreases so the bulb would grow dimmer.
d. As time goes on and the bulb grows hotter, the current through the bulb decreases so the bulb would grow dimmer.
Answer:
the correct one is d
Explanation:
The filament of the bulb fulfills the law of ohm
V = i R
indicate that the filament resistance increases with temperature, as the voltage remains constant if the resistance of the filament increases the current should decrease,
When examining the different statements, the correct one is d
Where is the center of mass of homogeneous body which has a regular
Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."
(see explanation below)
A bicyclist moves along a straight line with an initial velocity vo and slows downs. Which of the following the best describes the signs set for the initial position, initial velocity and the acceleration ?
The sign set after the slowdown of the bicycle will be positive for the position, negative for velocity, and negative for acceleration.
What is velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.
According to Que, when a bicyclist moves in a straight line and slows down, then the velocity decrease as displacement is decreasing, and the acceleration also decreases only displacement increases.
Therefore, the sign set for the position is +ve, for velocity it is -ve, and for acceleration also -ve
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A biker slows down after traveling in a long, straight line at initial velocity v0. Which of the following the best \sdescribes the signs set for the initial position, initial velocity and the acceleration? Initial position Initial velocity Acceleration
A. Positive Negative Negative
B. Positive Positive Negative
C. Negative Positive Negative
D. Negative Negative Positive
E. Negative Negative Negative
what do we mean by thrust?
Answer:
the answer is push example: she thrust her hand into her pocket
A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]
Let d is the distance moved in 2.25 s. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]
So, it will move 6.32 m from rest in 2.25 seconds.
a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.
Explanation:
It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is
[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]
[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]
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Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?
Answer:
P = density (p) * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2
since kg m / s^2 = Newtons
The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density
Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
Learn more about object kinetic energy of objects in free fall here;
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what is meant by specific latent heat of vaporization of water is -2.26mjkg^-1 or -2.26mj/kg?
Answer:
The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
[tex]V=1.4*10^5m/s[/tex]
Explanation:
From the question we are told that:
Electric field [tex]B=1.5*10N/C[/tex]
Distance [tex]d=2 x 10^{-3}[/tex]
At negative plate
Generally the equation for Velocity is mathematically given by
[tex]V^2=2as[/tex]
Therefore
[tex]V^2=\frac{2*e_0E*d}{m}[/tex]
[tex]V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}[/tex]
[tex]V=\sqrt{19.2*10^9}[/tex]
[tex]V=1.4*10^5m/s[/tex]
what is the value of x if x-36=5?
Answer:
Therefore, the value of x is 41
Explanation:
x=5+36
x=41
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88°C I am not 100% sure this is right but I am 98% sure this IS right
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720
An object moving with a constant
acceleration changes its velocity from
10ms' to 20 ms' in five seconds. What is the
distance travelled in five seconds
Answer:
Acceleration:
[tex]{ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}[/tex]
From third equation:
[tex]{ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}[/tex]
Answer:
Formula = m/s
Explanation:
The answer is 10 m / 5 seconds = 2 meters distance
The answer is 20 m / 5 seconds = 4 meters distance
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]
Speed of rocket 1 with respect to rocket 2 :
[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]
[tex]$u' = \frac{0.7 c}{1.12}$[/tex]
[tex]u'=0.625 c[/tex]
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
A magnifying glass produces a maximum angular magnification of 5.4 when used by a young person with a near point of 20 cm. What is the maximum angular magnification obtained by an older person with a near point of 65 cm
Physics question plz help ASAP
If the mass of an object is 15 kg and the velocity is -4 m/s, what is the momentum?
momentum p= m x v = 15 x -4 = -60 N.s
The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.
Answer:
The speed of proton is 2.1 x 10^5 m/s .
Explanation:
potential difference, V = 234 V
let the initial speed of the proton is v.
The kinetic energy of proton is
KE = q V
[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!
Answer:
y = 1/2at^2
we could also write it as-
y = (at^2)/2
2y = at^2
2y/a = t^2
√2y/a = t
hope it helps
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force
Explanation:
Hey there!
According to the question;
A person can lift mass of 60 kg on earth.
mass(m1) = 60kg
acceleration due to gravity on earth (a) = 9.8m/s²
Now;
force (f) = m.a
= 60*9.8
= 588 N
Since, there is application of same magnitude of force on moon,
mass(m) =?
acceleration due to gravity on moon (a) = 1.67m/s²
Now;
force (f) = m.a
588 = m*1.67
m = 352.09 kg
Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.
Hope it helps!
Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at a given distance:___________.
A) must be directed inward.
B) acts like it originated in a point charge Q at the center of the distribution.
C) must be directed outward.
D) must be greater than zero.
E) must be zero.
Answer:
Q at the center of the distribution.
Explanation:
The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2
Answer:
148.5 N
Explanation:
Given that,
The mass of a bungee jumper, m = 55 kg
The downward acceleration, a = 7.1 m/s²
We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :
T = m(g-a)
Put all the values,
T = 55(9.8 - 7.1)
= 148.5 N
So, the force exerted on the bungee cord is 148.5 N.
Answer:
The downward force is 148.5 N.
Explanation:
mass, m = 55 kg
downwards acceleration, a = 7.1 m/s^2
Let the force is F.
According to the newton's second law
m g - F = m a
F = m (g - a)
F = 55 (9.8 - 7.1)
F = 148.5 N
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.
Answer:
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex] (1)
Where:
I is the moment of inertiaM is the mass of the pendulumd is the distance from the center of mass to the pivotg is the gravityLet's solve the equation (1) for I
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex]
[tex]I=Mgd(\frac{T}{2\pi})^{2}[/tex]
Before find I, we need to remember that
[tex]T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s[/tex]
Now, the moment of inertia will be:
[tex]I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}[/tex]
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
I hope it helps you!
A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast
Answer:
x = 9.32 cm
Explanation:
For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation
Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar
- W l / 2 - W_{child} x + N₂ l = 0
x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex] 1)
now let's use the expression for translational equilibrium
N₁ - W - W_(child) + N₂ = 0
indicate that N₂ = 4 N₁
we substitute
N₁ - W - W_child + 4 N₁ = 0
5 N₁ -W - W_{child} = 0
N₁ = ( W + W_{child}) / 5
we calculate
N₁ = (450 + 250) / 5
N₁ = 140 N
we calculate with equation 1
x = -250 1.50 + 4 140 3) / 140
x = 9.32 cm
An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.
Answer:
the body is subjected to a continuous rotation and the body is not in rotational equilibrium
Explanation:
For an object to have a static equilibrium, it must meet two relationships
∑ F = 0
∑ τ =0
force acting on a body fulfills the relation of
sum F = F - F = 0
when two forces do not move from position.
To find the torque we assume that the counterclockwise rotations are positive
Σ τ = - F r - F r
Στ = -2 Fr <> 0
consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium
a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
