2. two dimensional motion
Because it has just 2 dimensions x and y
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
Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?
Answer:
T = 1/f = 1/50(s)
ω = 2πf = 100π (rad/s)
(vote 5 sao nhó :3 )
3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.
Answer:
The distance between mirror and you is 2 ft.
Explanation:
diameter, d = 8 ft
radius of curvature, R = 4 ft
magnification, m = 0.5
focal length, f = R/2 = 4/2 = 2 ft
let the distance of object is u and the distance of image is v.
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]
Use the formula of magnification
[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making
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
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.
which of the following cannot be increased by using a machine of some kind? work, force, speed, torque
Explanation:
Work cannot be increased by using a machine of some kind.
Work cannot be increased by using a machine of some kind.
A machine is any device in which the effort applied at one end overcomes a load at the other end.
Machines are generally used to perform different tasks faster.
However, a simple machine can not be used to increase the amount of work done at any time.
Force, speed and torque can all be increased using machines.
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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
a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection
Answer:
Answer:30 cm
Answer:30 cmExplanation:
Answer:30 cmExplanation:Given=ROC= 60cm
Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.
Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for longer wavelengths of light. The work function of the surface is
Explanation:
Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]
[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]
[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]
The work function [tex]\phi[/tex] is then
[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]
[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]
The work function of the surface is equal to 3.98 × 10⁻¹⁹J.
What are frequency and wavelength?The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.
The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.
The expression between wavelength (λ), frequency, and speed of light (c) is:
c = νλ
Given, the wavelength of the light, ν = 500 nm
The frequency of the light can determine from the above-mentioned relationship:
ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz
The work function = h ν = 6 × 10¹⁴ × 6.626 × 10⁻³⁴
φ = 3.98 × 10⁻¹⁹J
Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.
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Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?
Answer:
The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.
Explanation:
Consider newtonian mechanics here.
Dynamic equation is
The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.
we'll consider newtonian mechanics here.
so the dynamic equations is S = ut + 0.5at^2
we know u=0.8
S=1.2+0.9
t=1.2/1.5
find a.
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
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!
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
A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]
Answer:
Gap left = Change in length on heating
Gap=Initial length×Coefficient of linear expansion×change in temperature
Gap=10×0.000012×15m
⟹Gap=0.0018 m
this is an example u have to put your equation in it
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
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]
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.Question 5 of 10
What must be the same for two resistors that are connected in parallel?
Answer:
in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .
Explanation:
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Flag question
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
A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz
The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.
Given the following data:
Period = 645 μsNote: μs represents microseconds.
Conversion:
1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds
645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
[tex]Frequency = \frac{1}{Period}[/tex]
Substituting the value into the formula, we have;
[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
Conversion:
1 hertz = 0.001 kilohertz
1550.39 hertz = X kilohertz
Cross-multiplying, we have;
X = [tex]0.001[/tex] × [tex]1550.39[/tex]
X = 155039 kHz
To 3 significant figures;
Frequency = 155 kHz
Therefore, the corresponding frequency of this sinewave, in kHz is 155.
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Which of the following groups is the largest ?
population
community
ecosystem
biome
Answer:
B. Community
Took science classes for 6 years now
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.
Answer:
the distance from the cornea where a very distant object will be imaged is 23.35 mm
Explanation:
Given the data in the question;
For a spherical refracting surface;
[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium
[tex]d_0[/tex] is the object distance
[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium
[tex]d_i[/tex] is the image distance
R is the radius of curvature
Now, let [tex]d_0[/tex] = ∞, such that;
[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
we make [tex]d_i[/tex] subject of the formula
[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )
[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )
given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00
so we substitute
[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )
[tex]d_i[/tex] = 9.165 / 0.41
[tex]d_i[/tex] = 23.35 mm
Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Answer:
[tex]v=(6ti+6k)\ m/s[/tex]
Explanation:
Given that,
The position of a particle is given by :
[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]
Let us assume we need to find its velocity.
We know that,
[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]
So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].
A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event
Answer:
The rise in temperature is 0.06 K.
Explanation:
mass of bullet, m = 15 g
initial speed, u = 865 m/s
final speed, v = 534 m/s
mass of water, M = 13.5 kg
specific heat of water, c = 4200 J/kg K
The change in kinetic energy
[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]
According to the conservation of energy, the change in kinetic energy is used to heat the water.
K = m c T
where, T is the rise in temperature.
3473 = 13.5 x 4200 x T
T = 0.06 K
