For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
A) Find the energy in electron volts for a particle with this wavelength if the particle is a photon.
B) Find the energy in electron volts for a particle with this wavelength if the particle is an electron.
C) Find the energy in electron volts for a particle with this wavelength if the particle is an alpha particle (m=6.64×10−27kg).
Answers
Answer:
A) E = 4.96 x 10³ eV
B) E = 4.19 x 10⁴ eV
C) E = 3.73 x 10⁹ eV
Explanation:
A)
For photon energy is given as:
\(E = hv\)
\(E = \frac{hc}{\lambda}\)
where,
E = energy of photon = ?
h = 6.625 x 10⁻³⁴ J.s
λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Therefore,
\(E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}\)
\(E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})\)
E = 4.96 x 10³ eV
B)
The energy of a particle at rest is given as:
\(E = m_{0}c^2\)
where,
E = Energy of electron = ?
m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
\(E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\\)
\(E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\\)
E = 4.19 x 10⁴ eV
C)
The energy of a particle at rest is given as:
\(E = m_{0}c^2\)
where,
E = Energy of alpha particle = ?
m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
\(E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\\)
\(E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\\)
E = 3.73 x 10⁹ eV
A) The energy in electron volts for a particle with this wavelength if the particle is a photon is; .E = 4969.5 eV or 4.9695 keV
B) The energy in electron volts for a particle with this wavelength if the particle is an electron is; E = 23.58 eV
C) E = 0.003306 eV
A) The formula for the energy here is;
E = hc/λ
where;
h is planck's constant = 6.626 × 10⁻³⁴ J.s
c is speed of light = 3 × 10⁸ m/s
λ is wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Thus;
E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(0.25 x 10⁻⁹)
79.512 × 10⁻¹⁷ J
converting to eV gives;
E = (79.512 × 10⁻¹⁷)/(1.6 × 10⁻¹⁹)
E = 4969.5 eV or 4.9695 keV
B) Formula for the energy if the particle is an electron is;
E = h²/(2mλ²)
where m = 9.31 × 10⁻³¹ kg
E = (6.626 × 10⁻³⁴)²/(2 × 9.31 × 10⁻³¹ × (0.25 x 10⁻⁹)²)
E = 37.726 × 10⁻¹⁹ J
Converting to eV gives;
E = (37.726 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)
E = 23.58 eV
C) Mass of alpha particle is; m = 6.64 × 10⁻²⁷ kg
E = h²/(2mλ²)
where m = 6.64 × 10⁻²⁷ kg
E = (6.626 × 10⁻³⁴)²/(2 × 6.64 × 10⁻²⁷ × (0.25 x 10⁻⁹)²)
E = 52.896 × 10⁻²³ J
Converting to eV gives;
E = (52.896 × 10⁻²³)/(1.6 × 10⁻¹⁹)
E = 0.003306 eV
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Related Questions
At what temperature will water begin to boil and turn to steam?
212 degrees Celsius
100 degrees Fahrenheit
212 kelvins
100 degrees Celsius
Answers
Answer:
100 degrees Celsius
Explanation:
Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.
At 100 degrees Celsius water begin to boil and turns to steam.
What are the boiling point and melting point of water?The melting point for water is 0 degrees C (32 degrees F). The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower. At sea level, pure water boils at 212 °F (100°C).
Is boiling water always 212?If the temperature is much above 212°F, the water will boil. That means that it won't just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. If the water has very few dust flecks etc.
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When a falling body is at distance above the surface which is 4 times the Earth radius what is the acceleration due to Earth gravitation
Answers
Answer:
Explanation:
g = GM/R²
as gravity decreases by the inverse of the square of the distance, increasing the distance by 4 times will reduce gravity to 1/16 that at the surface
9.8 / 16 = 0.61 m/s²
The speed of sound in air is approximately 340 m/s. The speed of light in air is approximately 3 x 108 m/s. If 10 seconds elapses between seeing a lightning strike and hearing the thunder, how far away was the lightning strike?
Answers
Answer:
3400 m
Explanation:
Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;
speed = distance/time, hence, distance = speed x time.
For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as;
distance = 340 m/s x 10 s = 3400 m
A superconductor performs best at ______.
Extremely high temperatures
Room temperature
Very cold temperatures
Any temperature
Answers
Answer:
c
Explanation:
Superconductors perform best at very cold temperatures because resistivity of this kind of materials decays drastically with temperature. Chromium is most likely to be the best conductor of electricity because it belongs to the Transition Metal group of the periodic table
What force acts on a projectile in the horizontal direction?
Answers
The force that acts on a projectile in the horizontal direction is Gravitational force.
A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.
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Can someone explain how to do the algebra for this question? I know everything else, I just don’t know how to rearrange the question to solve for v.
Answers
Answer:
Refer to the step-by-step Explanation.
Step-by-step Explanation:
Simplify the equation with given substitutions,
Given Equation:
\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)
Given Substitutions:
\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)
Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)
Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the left-hand side of the equation:
\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
Simplifying the third term.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)
"R²'s" cancel, we are left with:
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)
We have like terms, combine them.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)
Each term has an "m" in common, factor it out.
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)
Now we have the following equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the right-hand side of the equation:
\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)
\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)
Now we have the equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
\(\hrulefill\)
Now solving the equation for the variable "v":
\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
Dividing each side by "m," this will cancel the "m" variable on each side.
\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)
Subtract the term "gh" from either side of the equation.
\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)
Multiply each side of the equation by "10/7."
\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)
Now squaring both sides.
\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)
Thus, the simplified equation above matches the simplified equation that was given.
A certain type of lens changes the direction of light rays from A to B. What property does
this illustrate?
Answers
The figure shows the polarization of light in the image.
What is light?We have to note that light is the form of energy that we can be able to see by the use of the optical eyes that we have. There are several propoerties of light that are also the properties of waves.
We should know that light is an electromagnetic wave and the polarization of waves can only occur in an electromagnetic wave. From A to B in the figure that have been shown, we can see the polarization of waves.
Thus the light waves are polarized.
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A plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.
Answers
Answer:
-22.1
Explanation:
1 / 4
Step 1. List horizontal (xxx) and vertical (yyy) variables
xxx-direction yyy-direction
t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text
a_x=0a
x
=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a
y
=−9.8
s
2
m
a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction
\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text
v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v
y
=?v, start subscript, y, end subscript, equals, question mark
v_{0x}=120\,\dfrac{\text m}{\text s}v
0x
=120
s
m
v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v
0y
=0v, start subscript, 0, y, end subscript, equals, 0
Note that there is no horizontal acceleration, so v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.
Also, the package has no initial vertical velocity.
Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:
\Delta x=v_xtΔx=v
x
tdelta, x, equals, v, start subscript, x, end subscript, t
Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv
y
v, start subscript, y, end subscript:
\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v
0y
t+
2
1
a
y
t
2
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared
Hint #22 / 4
Step 2. Find ttt from horizontal variables
\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}
Δx
t
t
=v
x
t
=
v
0x
Δx
=
120
s
m
255m
=2.125s
Hint #33 / 4
Step 3. Find \Delta yΔydelta, y using ttt
Using ttt to solve for \Delta yΔydelta, y gives:
\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}
Δy
=v
0y
t+
2
1
a
y
t
2
=
(0)t
+
2
1
(−9.8
s
2
m
)(2.125s)
2
=−22.1m
Hint #44 / 4
The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.
Moving current has electrical energy.
Answers
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time? Group of answer choices
Answers
Answer:
Dear Kaleb
Answer to your query is provided below
Acceleration of the vehicle is 12m/s^2
Explanation:
Explanation for the same is attached in image
Is it acceleration speed or velocity
Answers
Answer:
1st speed 2.velocity 3.speed4.velocity5.velocity6.speed
3. The property of sound called intensity is proportional
to the rate at which energy flows through
A.an area perpendicular to the direction of
propagation.
B.an area parallel to the direction of propagation.
C.a cylindrical tube.
D.a sound wave of a certain frequency.
Answers
Answer:
A. an area perpendicular to the direction of propagation.
Explanation:
The property of sound called intensity is proportional
to the rate at which energy flows through an area perpendicular to the direction of propagation.
A2.00 kg block is pulled across a
flat, frictionless floor with a 4.17 N
force directed 40.0° above
horizontal. What is the normal force
acting on the block?
(Hint: It is NOT = mg)
normal force (ND)
Answers
Answer:
Explanation:
R + 4.17sin(40) = mg
R = mg - 4.17sin(40)
= 2(9.8) - 4.17sin(40)
= 16.91957567
= 16.92 N
A 2.00 kg block is pulled across a flat, friction less floor with a 4.17 N force directed 40.0° above horizontal. The normal force acting on the block 16.92 N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Given in the question,
R + 4.17 sin(40) = mg
R = mg - 4.17 sin(40)
= 2(9.8) - 4.17 sin(40)
= 16.91957567
= 16.92 N
A 2.00 kg block is pulled across a flat, friction less floor with a 4.17 N force directed 40.0° above horizontal. The normal force acting on the block 16.92 N.
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The nearest neighboring star to the Sun is about 4 light-years away. If a planet happened to be orbiting this star at an orbital radius equal to that of the Earth-Sun distance, what minimum diameter would an Earth-based telescope's aperture have to be in order to obtain an image that resolved this star-planet system? Assume the light emitted by the star and planet has a wavelength of 550 nm
. The Earth-Sun distance is 149.6×106km
, and 1ly=9.461×1015m
.
Answers
To resolve the star-planet system at a distance of 4 light-years, a telescope on Earth would need an aperture with a minimum diameter of 55.88 mm.
What does microscopy's Rayleigh criterion mean?In optical microscopy, the Rayleigh criterion is frequently used to estimate the resolution of the microscope. The resolution limit imposed by this criterion has long been regarded as a roadblock to using an optical microscope to study biological phenomena at the nanoscale.
We can use the Rayleigh criterion,
θ = 1.22 λ / D
θ = angular resolution
λ = wavelength of light
D = diameter of the telescope's aperture
θ = arctan (r / d)
r = radius of the planet's orbit
d = distance to the star
Now, we use the given values,
r = 149.6×106 km = 149.6×109 m
d = 4 × 9.461×1015 m = 3.7844×1016 m
λ = 550 nm = 550×10-9 m
θ = arctan (r / d)
=arctan (149.6×109 / 3.7844×1016) = 0.000012 radians
we can use the Rayleigh criterion,
θ = 1.22 λ / D
D = 1.22 λ / θ
D = 1.22 × 550×10-9 / 0.000012
D = 55.88 mm
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what is meant by reticlinear propatigation of light?
Answers
Answer:
The property of light going in a straight lines in a homogenous straightforward medium is known as rectilinear engendering of light.
Answer:
the property of light travelling in a straight lines in a homogenous transparent medium
Explanation:
According to Howard Gardner's theories, who among the following would need to have good linguistic intelligence to be successful?
Answers
A psychologist will need to have good linguistic intelligence in other to be successful.
Who is a Psychologist?This is referred to as a professional who specializes in the handling of mental health challenges in individuals.
It is best for such professional to have a good linguistic intelligence as the right words being said to the patient will solve the problem thereby bringing in more success.
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the conductor offers some amount of ohmic resistance to the flow of electrons, depending on the type of material from which the conductor is made, its length, and its .
Answers
Depending on the type of material used, the length, and other factors, the conductor provides some ohmic resistance to the flow of electrons.
Because electrons in a conducting wire reject one another, a conducting wire provides resistance to the flow of electrons. The electrolyte resistance, the current collector resistance, the active mass, and the transition resistance between the current collector and active mass are added to create the ohmic resistance, or RB.
Theoretically, in accordance with Ohm's rule, the voltage at the ohmic resistance instantly follows the battery current. A substance or material that permits the flow of electricity is known as an electrical conductor. Electrical charge carriers, often electrons or ions, flow freely in a conductor.
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Contingent upon the sort of material utilized, the length, and different variables, the guide gives ohmic protection from the progression of electrons.
Since electrons in a directing wire reject each other, a leading wire gives protection from the progression of electrons. The electrolyte obstruction, the ongoing authority opposition, the dynamic mass, and the progress obstruction between the ongoing gatherer and dynamic mass are added to make the ohmic obstruction or RB.
Hypothetically, as per Ohm's standard, the voltage at the ohmic obstruction in a split second follows the battery current. A substance or material that allows the progression of power is known as an electrical transmitter. Electrical charge transporters, frequently electrons or particles, stream uninhibitedly in a transmitter.
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The greatest speed recorded by a baseball thrown by a pitcher was 162.3 km / h, obtained by Nolan Ryan in 1974. If the ball leaves the pitcher's hand with a horizontal speed of this magnitude, how much will the ball have fallen? by the time you have traveled 20 m horizontally?
Answers
Answer:
0.96 m
Explanation:
First, convert km/h to m/s.
162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s
Now find the time it takes to move 20 m horizontally.
Δx = v₀ t + ½ at²
20 m = (45.08 m/s) t + ½ (0 m/s²) t²
t = 0.4436 s
Finally, find how far the ball falls in that time.
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²
Δy = -0.96 m
The ball will have fallen 0.96 meters.
Revolution: One orit of an object in space around another_ days?
Answers
Answer:
365 days..
Explanation:
Hope it may help you シ︎♡︎
\( \huge\mathfrak\green{365 \: or \: 366 \: Days.}
\)
\(Hope \: it \: helps\)
A scientist makes a model of Earth's water by drawing 100 drops of water, all the same size. How many of the 100 drops represent ocean water?
A.3
B.50
C.75
D.97
Answers
Answer:
D
Explanation:
If the water represents the oceans water then you'd would need to calculate how much of earth is water (96.5)
Find the Magnitude of the resultant vector (the actual
path of the boat).
The picture is a little blurry, so here are the stats:
Velocity of the boat is 0.75 m/s
Velocity of the river is 1.2 m/s
Answers
The magnitude of the resultant vector, representing the actual path of the boat, is approximately 1.42 m/s.
To find the magnitude of the resultant vector, we need to consider the boat's velocity and the velocity of the river. The boat's velocity is given as 0.75 m/s, and the river's velocity is given as 1.2 m/s.
Since the boat is moving in a river, we can think of the boat's velocity as a combination of two velocities: its own velocity and the velocity of the river. The resultant vector represents the actual path of the boat, considering both velocities.
To calculate the resultant vector, we can use vector addition. The magnitude of the resultant vector can be found by taking the square root of the sum of the squares of the boat's velocity and the river's velocity. Mathematically, we have:
Resultant magnitude = √(boat velocity^2 + river velocity^2)
Plugging in the given values, we have:
Resultant magnitude = √(0.75^2 + 1.2^2)
= √(0.5625 + 1.44)
= √2.0025
≈ 1.42 m/s
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A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 31.0 m/s. Then the vehicle moves for 35.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.(a) How long is the ride-sharing car in motion (in s)? (b) What is the average velocity of the ride-sharing car for the motion described?
Answers
Answer:
A) total time = 55.5 seconds
B) average velocity = 25.27 m/s
Explanation:
It starts from rest, so initial velocity, u = 0 m/s
We are given;
acceleration; a = 2 m/s²
Final velocity; v = 31 m/s
From Newton's first law of motion,
v = u + at
So, 31 = 0 + 2t
t = 31/2
t = 15.5 sec
We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;
(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds
b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;
S = ut + ½at²
S1 = 0 + ½(2 * 15.5²)
S1 = 240.25 m
Distance traveled in 35 sec with with velocity of 31 m/sec is;
S2 = velocity x time
S2 = 35 × 31 = 1085 m
Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.
From the first equation of motion,
a = (v - u)/t
a = (0 - 31)/5
a = -6.2 m/s²
So, distance travelled is;
S3 = ut + ½at²
S3 = (31 × 5) + ½(-6.2 × 5²)
S3 = 155 - 77.5
S3 = 77.5 m
So overall total distance = S1 + S2 + S3
Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m
Average velocity = total distance/total time
Average velocity = 1402.75/55.5 = 25.27 m/s
Which is a form of energy that cannot be stored?
a. electric energy
b. mechanical energy
c. sound energy
d. thermal energy
Answers
Each of the following figures shows a person (not to scale) located on Earth at either 40°N or 40°S latitude. Rank the figures based on how much time the person spends in daylight during each 24-hour period, from most to least. To rank items as equivalent, overlap them.
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The ranking is based on the tilt of the Earth's axis and its orbit around the Sun. The figure at 40°N in June receives the most daylight because it is located at a high latitude during the summer solstice in the Northern Hemisphere. The Earth's axis tilts towards the Sun, resulting in longer days and shorter nights. The figure at 40°S in December receives a moderate amount of daylight as it is located at a lower latitude during the summer solstice in the Southern Hemisphere.
The figure at 40°N in December experiences less daylight because it is located at a high latitude during the winter solstice in the Northern Hemisphere, with shorter days and longer nights. Lastly, the figure at 40°S in June receives the least amount of daylight as it is located at a lower latitude during the winter solstice in the Southern Hemisphere, where the days are shortest and the nights are longest. Based on the information given, the ranking of figures based on the amount of daylight they experience in a 24-hour period, from most to least.
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the ratio of force between two charges in vacuum to that the force between two same charges when a medium is placed between them
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The ratio of the force between two charges in a vacuum to the force between two charges when a medium is placed between them is called relative permittivity
What should you know about relative permittivity?Relative permittivity has another term dielectric constant. The dielectric constant measures how well a material can store electrical energy in an electric field. Its equation is ε = ε₀ / εᵣ
ε₀ is the vacuum permittivity, and εᵣ is the relative permittivity or dielectric constant of the medium.
The dielectric constant is different depending on the material.
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a. When throwing a ball vertically upward, my hand moves through a distance of about 1.0 m before the ball leaves my hand. The 0.80 kg ball reaches a maximum height of about 20 m above my hand. while the ball is in my hand after the ball leaves my hand
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The required, it experiences a downward force due to gravity and a force due to air resistance.
What is the projectile motion?Projectile motion is the movement of an entity projected into space. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory.
Here,
When throwing a ball vertically upward, there is a displacement of about 1.0 m from the initial position of the hand to the position where the ball leaves the hand. The mass of the ball is 0.80 kg and it reaches a maximum height of about 20 m above the initial position of the hand. While the ball is in the hand after it leaves, it experiences a downward force due to gravity and a force due to air resistance.
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help pls!!
Fig above shows a wave traveling through a medium. Use the fig to answer the questions below.
A) What is the amplitude of the wave ? Include correct units.
B) Use the graph to determine the time of one wave. Use it to find the frequency.
C) If the speed of the wave is 25 cm/s, what is the wavelength of the wave ? Show data listing, equation , substitution leading to the answer for full credit.
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(a) The amplitude of the wave is determined as 8 cm.
(b) The period of the wave motion is 20 s and the frequency of the wave is 0.05 Hz
(c) The wavelength of the wave is 500 cm.
What is the amplitude of the wave ?(a) The amplitude of the wave is the maximum displacement of the wave.
from the graph, amplitude of the wave = 8 cm
(b) The period of the wave motion is calculated as;
T = 20 s
The frequency of the wave = 1/T = 1/20 s = 0.05 Hz
(c) The wavelength of the wave is calculated by applying the following wave formula.
λ = v / f
λ = 25 cm/s / 0.05 Hz
λ = 500 cm
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A 35kg box is pushed 14m by a force of 800n. The coefficient of kinetic friction is 0.4. How much work into accelerating?
Answers
Given data:
* The force applied on the box is,
\(F_a=800\text{ N}\)* The mass of the box is m = 35 kg.
* The distance traveled by the box is d = 14 m.
* The coefficient of kinetic friction is,
\(\mu_k=0.4\)Solution:
The normal force acting on the box is,
\(F_N=mg\)where g is the acceleration due to gravity,
Substituting the known values,
\(\begin{gathered} F_N=35\times9.8 \\ F_N=343\text{ N} \end{gathered}\)The kinetic frictional force acting on the box is,
\(F_k=\mu_kF_N\)Substituting the known values,
\(\begin{gathered} F_k=0.4\times343 \\ F_k=137.2\text{ N} \end{gathered}\)The net force acting on the box in terms of applied force and frictional force is,
\(\begin{gathered} F_{\text{net}}=F_a-F_k \\ F_{\text{net}}=800-137.2 \\ F_{\text{net}}=662.8\text{ N} \end{gathered}\)Thus, the work done in accelerating the box is,
\(\begin{gathered} W=F_{\text{net}}\times d \\ W=662.8\times14 \\ W=9279.2\text{ J} \end{gathered}\)Thus, the work done in accelerating the box is 9279.2 J or approximately 9.3 kJ.
which type of energy is the energy of a moving tennis ball
Answers
Answer:
Kinetic energy.
The explanation never existed
Answer: Kinetic Energy
The work done is a vector quantity and SI base unit is J
Answers
Answer:
Is this your question? Also I think work done is a scalar quantity.
Explanation: