Explanation:
how can I help you that?
1) What do (x) and (y) symbolize?
a) What equations belong to each one?
b) How does horizontal motion effect vertical motion and vis versa?
c) What do both horizontal and vertical motion use?
Answer:
x is vertical and y is horizontal
Explanation:
A 25 kg child is sitting at the top of a 4 m tall slide, what is his potential energy?
An engineer measures the velocity of a remote-controlled cart on a straight track at regular time intervals. The data are shown in the graph above. During which of the following time intervals did the cart return to its position at time t=0 s?
A. 7 s ≤ t < 10 s
B. 10 s ≤ t < 12 s
C. 5 s ≤ t < 6 s
D. 3 s ≤ t < 5 s
Answer:
A. 7 s ≤ t < 10 s
Explanation:
From Definite Integral Theory and Mechanical Physics we remember that change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:
[tex]A_{1}+A_{2} = 0[/tex] (1)
Where:
[tex]A_{1}[/tex] - Change in position from [tex]t = 0\,s[/tex] and [tex]t = 3\,s[/tex], measured in meters.
[tex]A_{2}[/tex] - Change in position from [tex]t = 5\,s[/tex] and [tex]t = T[/tex], measured in meters.
By definitions of triangle, we expand the equation above:
[tex]\frac{1}{2}\cdot (3\,s)\cdot \left(1.2\,\frac{m}{s} \right) -\frac{1}{2}\cdot (T-5\,s)\cdot v = 0\,m[/tex]
[tex]1.8\,m-0.5\cdot T \cdot v +2.5\cdot v = 0[/tex]
And the time is now cleared:
[tex]1.8\,m+2.5\cdot v = 0.5\cdot T\cdot v[/tex]
[tex]T = \frac{1.8+2.5\cdot v}{0.5\cdot v}[/tex]
Where:
[tex]v[/tex] - Final velocity of the cart, measured in meters per second.
[tex]T[/tex] - Time, measured in seconds.
If we know that [tex]v = 1\,\frac{m}{s}[/tex], then the intended instant is:
[tex]T = \frac{1.8+2.5\cdot (1)}{0.5\cdot (1)}[/tex]
[tex]T = 8.6\,s[/tex]
The value does not coincide with the time from the graph.
If we know that [tex]v = 1.5\,\frac{m}{s}[/tex], then the intended instant is:
[tex]T = \frac{1.8+2.5\cdot (1.5)}{0.5\cdot (1.5)}[/tex]
[tex]T = 7.4\,s[/tex]
This value does not coincide with the time from the graph.
If we know that [tex]v = 1.1\,\frac{m}{s}[/tex], then the intended instant is:
[tex]T = \frac{1.8+2.5\cdot (1.1)}{0.5\cdot (1.1)}[/tex]
[tex]T = 8.273\,s[/tex]
This results offer a reasonable approximation, which that the correct answer is A.
The car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.
The change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:
A + A' = 0 ..................................................(1)
A is the change in position from t = 0 and t = 3 s, measured in metres.
A' is the change in position from t = 5 and t = T s, measured in metres.
As per the triangle law,
1/2 ×(3 s)(1.2) - 1/2 ×(T - 5)v = 0
1.8 - 0.5T(v) +2.5(v) =0
T = (1.8 + 2.5v)/ (0.5v)
Here, v is the final velocity of cart.
For v = 1 m/s, we have,
T = (1.8 + 2.5(1))/ (0.5(1))
T = 8.6 s
Since, the values do not coincide with the time from the graph, so taking another value, v = 1.5 m/s to obtain the time as,
T = (1.8 + 2.5(1.5))/ (0.5(1.5))
T = 7.4 s
This value does not coincide with the time from the graph. If we know that , v = 1.1 m/s, then the intended instant is:
T = (1.8 + 2.5(1.1))/ (0.5(1.1))
T = 8.273 s
Thus, we can conclude that the car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.
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What is the mass, in kg, of a 136 pound gymnast on Earth?
Answer:
61.6886 kg
Explanation:
Answer:
61 kg
Explanation:
there ya go hope this was useful
1. An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s. How fast will he be moving backward just after releasing the ball?2. Which of the following best describes why you can analyze this event using conservation of momentum?A. The throwing action is quick enough that external forces may be ignored.B. External forces don't act on the system during the jump.C. Conservation of momentum is always the best way to analyze motion.
Answer:
1) 0.0806 m/s
2) External forces don't act on the system during the jump
Explanation:
velocity of ball ( Vbf )= 15 m/s
mass of quarter back ( Mq )= 80 kg
mass of football ( Mb ) = 0.43 kg
A) How fast will be be moving backward just after releasing the ball
we can determine this speed with the use of conservation of momentum equation
= Mb ( Vbf - Vbi ) = Mq ( Vq )
where Vq = Mb ( Vbf - Vbi ) / Mq
= 0.43 ( 15 m/s - 0 ) / 80 kg
= 0.0806 m/s
B) External forces don't act on the system during the jump
A ferris wheel with radius 12 m makes a revolution every 3 minutes. Find the linear (tangental) speed of a passenger. How far does a person move in a 5 minute ride?
Answer:
The linear (tangential) speed of a passenger is 0.4188 m/s
The distance traveled by the person in 5 minutes ride is 125.64 m
Explanation:
Given;
radius of the Ferris, r = 12 m
1 revolution per 3 minutes, [tex]\omega = \frac{2\pi (radian)}{3\ (minutes)} *\frac{1\ minute}{60 \ seconds} = 0.0349 \ rad/s[/tex]
The linear (tangential) speed of a passenger is given by;
v = ωr
v = (0.0349)(12)
v = 0.4188 m/s
The distance traveled by the person in 5 minutes ride is given by;
d = vt
d = (0.4188)(5 x 60)
d = 125.64 m
Which of the following represents a chemical change when using bread in a meal?
F. Removing the crust of the bread when making a sandwich.
G. Placing the bread in a toaster to make toast and applying butter on it afterwards
H. Cutting the bread in half to make two sandwiches
J. Placing mayonnaise, ketchup, and mustard on the bread before the meat
Answer:Well toasting it makes it a solid so that’s a chemical change that happens! Hope this helped!
What is the relationship between lightning and atoms?
A. Electrons colliding in the air create lightning
B. The nucleus of atoms colliding creates an atom
C. Lightning is made of atoms
D. Lightning is created by the collision of protons
Answer:
C. Lighting is made of atoms
Explanation:
Everything is made of atoms that are, in turn, made out of charged particles. All charged particles come in one of two types: positive and negative (or plus and minus). The minus particles are the electrons, and the plus particles are the much heavier protons which are buried deep in the nucleus.
Which symbol and unit of measurement are used for the electric current?
symbol: A; unit: I
symbol: C; unit: A
symbol: I; unit: C
symbol: I; unit: A
Answer:
Symbol: I unit: A
Explanation:
the formula for current is I = v/r
I: Current
V: voltage
R; Resistance
Symbol and unit of measurement are used for the electric current are symbol: I; unit: A
What is electric current ?"Electric Current is the rate of flow of electrons in a conductor. The SI Unit of electric current is the Ampere. Electrons are minute particles that exist within the molecular structure of a substance. Sometimes, these electrons are tightly held, and the other times they are loosely held."
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An object weighs 3 lb. at 10 earth radii from its center. What is the object's weight on the earth's surface?
______ lb.
300
30
3
3,000
Answer:
3
Explanation:
What did Rutherford conclude from the motion of the particles when shot through a thin gold foil
Answer:
He concluded the fact that most alpha particles went straight through the foil is evidence for the atom being mostly empty space. A small number of alpha particles being deflected at large angles suggested that there is a concentration of positive charge in the atom.
Explanation:
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).
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:
[tex]E = hv[/tex]
[tex]E = \frac{hc}{\lambda}[/tex]
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,
[tex]E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}[/tex]
[tex]E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})[/tex]
E = 4.96 x 10³ eV
B)
The energy of a particle at rest is given as:
[tex]E = m_{0}c^2[/tex]
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,
[tex]E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\[/tex]
[tex]E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]
E = 4.19 x 10⁴ eV
C)
The energy of a particle at rest is given as:
[tex]E = m_{0}c^2[/tex]
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,
[tex]E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\[/tex]
[tex]E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]
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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The difference between a law and a theory is the diffrence between what and why . Explain
Answer:
A law is defined as a description or a statement given after an observed phenomenon. A theory is a simplification of certain observational data as to how and why it happened.
Explanation:
I hope this helped solve your question.
A 5 kg block is released from rest at the top of a quarter- circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom o the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is 0.02.
a. What is the kinetic energy of the block at the bottom of the curved surface?
b. What is the speed of the block at the bottom of the curved surface?
c. Find the stopping distance of the block?
d. Find the elapsed time of the block while it is moving on the horizontal part of the track.
e. How much work is done by the friction force on the block on the horizontal part of the track?
Answer:
a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J
Explanation:
a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.
So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.
So, mgh + 0 = 0 + K'
K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m
So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J
b. Since the kinetic energy of the block K = 1/2mv² where m = mass of block = 5 kg, v = velocity of block at bottom of curve
So, v = √(2K/m)
= √(2 × 186.2 J/5 kg)
= √(372.4 J/5 kg)
= √(74.48 J/kg)
= 8.63 m/s
c. To find the stopping distance, from work-kinetic energy principles,
work done by friction = kinetic energy change of block.
So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance
ΔK = -fd
K" - K' = - μmgd
d = -(K" - K')/μmg
Substituting the values of the variables, we have
d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)
d = -(- 186.2 J)/(0.98 kg m/s²)
d = 190 m
d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m
So, a = (v² - u²)/2d
substituting the values of the variables, we have
a = (0² - (8.63 m/s)²)/(2 × 190 m)
a = -74.4769 m²/s²/380 m
a = -0.2 m/s²
Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.
t = (v - u)/a
t =(0 m/s - 8.63 m/s)/-0.2 m/s²
t = - 8.63 m/s/-0.2 m/s²
t = 43.2 s
e. The work done by friction W = fd where
= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m
W = 0.02 × 5 kg × 9.8 m/s² × 190 m
W = 186.2 J
The potential energy of the loss of the block will be equal to the kinetic energy gain. The kinetic energy of the block is 186.2 J at the bottom of the curved surface.
The potential energy of the loss of the block will be equal to the kinetic energy gain.
So,
[tex]U = mgh[/tex]
Where,
[tex]U[/tex] - potential energy
[tex]m[/tex] - mass of block = 5 kg
[tex]g[/tex] - gravitational acceleration = 9.8 m/s²
[tex]h[/tex] = height = radius of curve = 3.8 m
Put the values in the formula,
[tex]U = 5 \times 9.8 \times 3.8 \\\\ U = 186.2 \rm \ J[/tex]
Therefore, the kinetic energy of the block is 186.2 J at the bottom of the curved surface.
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A student trying to calculate the coefficient of friction between a box and a surface. She measures that the 80kg box will slide if the student pushes with a force greater than 400n.
Answer:
[tex]\mu_s=0.51[/tex]
Explanation:
Coefficient of friction
The static coefficient of friction is a measure of the force needed to start moving an object from rest along a rough surface.
It's calculated as:
[tex]\displaystyle \mu_s=\frac{F_r}{N}[/tex]
Where Fr is the friction force and N is the normal force exerted by the surface over the object.
In the absence of any other vertical force, the normal is equal to the weight of the object:
[tex]N = W = m.g[/tex]
The box has a mass of m=80 Kg, thus the normal force is:
[tex]N = 80\ Kg * 9.8\ m/s^2[/tex]
N = 784 N
The student needs to push with a force of 400 N to make the box move. That is the force that barely outcomes the friction force. Thus:
[tex]F_r=400\ N[/tex]
Calculating the coefficient:
[tex]\displaystyle \mu_s=\frac{400}{784}[/tex]
[tex]\mathbf{\mu_s=0.51}[/tex]
mention three features of the Constitution and a note about them
Answer:
Explanation:(1) a constitution is a supreme law of the land, (2) a constitution is a framework for government; (3) a constitution is a legitimate way to grant and limit powers of government officials. Constitutional law is distinguished from statutory law.
It the same child has a velocity of 2 m/s half-way down the slide, what is his kinetic energy?
Answer:
[tex] \frac{1}{2} \times m \times {v}^{2} \\ \frac{1}{2} \times m \times {2}^{2} \\ 2 \times m[/tex]
I dont the child mass...you should substitute that value to (m)
then you can get your answer
How does the moon cause tides?
Answer:
High and low tides are caused by the Moon. The Moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the Moon and the side farthest from the Moon. These bulges of water are high tides.
Explanation:
If Batman jumps off a 50 m tall building at 30m/s to save a civilian...
a) How long does it take him to reach the civilian?
b) How far from the bottom of the building does he go?
Answer:
Use
[tex]t = \: \sqrt{2h \div g } [/tex]
[tex]x = u \times \sqrt{2h \div g} [/tex]
Which physical activity is NOT aerobic exercise?
A) Hip-hop dancing
B) Jump roping
C) Yoga
D) Jogging
Answer:
C
Explanation:
Answer:
Hip-hop Dancing
Explanation:
Aerobic exercises are light (somewhat, a more accurate word would be bearable) workouts that you can endure over long periods of time. Anaerobic workouts describe workouts that require sudden burst of energy, like the ones you seen in most forms of dance, such as Hip-hop dance.
Hope this helps!
Two satellites are in orbit around a planet. One satellite has an orbital radius of 8.0×1 0 6 m. The period of revolution for this satellite is 1.0×1 0 6 s. The other satellite has an orbital radius of 2.0×1 0 7 m. What is this satellite’s period of revolution?
This question involves the concept of the orbital period.
The period of revolution of the second satellite is "3.95 x 10⁶ s".
ORBITAL PERIODFirst, we will consider the orbital period of the first satellite:
[tex]T_1=\sqrt{\frac{4\pi^2 r_1^3}{GM}}[/tex]
where,
T₁ = orbital period of frirst satellite = 1 x 10⁶ sr₁ = orbital radius for first satellite = 8 x 10⁶ mM = mass of planet = ?G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²Therefore,
[tex]1\ x\ 10^6\ s=\sqrt{\frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(M)}}\\\\M = \frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1\ x\ 10^6\ s)^2}\\\\M = 3.03\ x\ 10^{20}\ kg[/tex]
Now, we find out the orbital period of the second satellite:
[tex]T_2=\sqrt{\frac{4\pi^2 r_2^3}{GM}}[/tex]
where,
T₂ = orbital period of second satellite = ?r₁ = orbital radius for second satellite = 2 x 10⁷ mTherefore,
[tex]T_2=\sqrt{\frac{4\pi^2(2\ x\ 10^7\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(3.03\ x\ 10^{20}\ kg)}}[/tex]
T₂ = 3.95 x 10⁶ s
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90% of cancer originate from
Carcinoma refers to a malignant neoplasm of epithelial origin or cancer of the internal or external lining of the body. Carcinomas, malignancies of epithelial tissue, account for 80 to 90 percent of all cancer cases. Epithelial tissue is found throughout the body.
Three beakers are of identical shape and size , one beaker is pained matt black, one is dull white and one is gloss white . The beakers are filled with boiling waterin which beaker does the water cool more quickly? Give reason. ( i really need the answer quickly)
Answer:
The beaker that is matt black heats more quickly because black attracts more heat.
Explanation:
Three beakers are of identical shape and size, one beaker is pained matt black, one is dull white and one is gloss white . The beakers are filled with boiling water. The water in the matt black beaker cools more quickly.
The rate of cooling of an object is influenced by its surface properties and color. In this case, the matt black beaker will cool the water more quickly compared to the dull white and gloss white beakers.
The reason behind this lies in the concept of thermal radiation. Darker colors, such as matt black, absorb and emit thermal radiation more efficiently than lighter colors. When the boiling water is placed in the matt black beaker, the black surface absorbs a greater amount of thermal radiation from the water and its surroundings. This increased absorption accelerates the transfer of heat energy from the water to the beaker, leading to faster cooling.
On the other hand, the dull white and gloss white beakers reflect more thermal radiation, absorbing and emitting less heat. As a result, the water in these beakers cools at a slower rate compared to the matt black beaker.
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The acceleration due to gravity on Jupiter is 23.1 m/s2, which is about twice the acceleration due to gravity on Neptune.
Which statement accurately compares the weight of an object on these two planets?
An object weighs about one-fourth as much on Jupiter as on Neptune.
An object weighs about one-half as much on Jupiter as on Neptune.
An object weighs about two times as much on Jupiter as on Neptune.
An object weighs about four times as much on Jupiter as on Neptune.
Answer:
I believe its C
Explanation:
Answer:
C-An object weighs about two times as much on Jupiter as on Neptune.
Explanation:
on edg
Why do I feel so sad? I'm not sure if its depression, because i will be fine not happy, but fine all day than randomly in the day I will start to feel sad. I've tried to take my life twice before the past few monthes
You are most likely going through depression. I think you should talk to someone, it could help or try talking to one of us on brainly. I hope you feel better eventually. Your life is valuable try and realize that.
Answer:
You don't need a reason to be sad. It sound as though you are suffering from depression. I understand that. Last year I thought about taking my life, but my parents found out how I was feeling and have been so supportive. Something My mom said that really got to me was, if you killed yourself, who would find you. For me it would have been my little sister. I would have scared her for life. I have come a long way since then, I am doing family therapy, as well as persona, therapy. If you are feeling sad, there is a hotline you can text rather than call. Let them talk you out of it. They won't call anyone if you don't ask them to. Talking to a complete stranger helps, they don't know anything about you so they can't judge you. Also try listening to music, find an artist that you can relate to. For me I like listening to NF when I am feeling down. It helps.
A 5 kg rock is dropped down a vertical mine shaft. How long does it take to reach the bottom, 79 meters below?
Answer:
The time for the rock to reach the bottom is 4.02 seconds.
Explanation:
Given;
mass of the rock, m = 5 kg
height of the rock fall, h = 79 meters
The time to drop to the given height is given by;
[tex]t = \sqrt{\frac{2h}{g} }[/tex]
where;
t is the time to fall to the bottom
g is acceleration due to gravity = 9.8 m/s²
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*79}{9.8} }\\\\t = 4.02 \ s[/tex]
Therefore, the time for the rock to reach the bottom is 4.02 seconds.
The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it.
Part 1) What should be the focal length of this lens?
Answer in units of cm.
Part 2) What is the power of the needed corrective lens?Answer in units of diopters.
Answer:
a) 0.3 m
b) 3.3 diopters
Explanation:
Given that
Object distance is 26cm from the eye.
Image distance is 105 cm
From the above, we will use the formula
1/f = 1/v + 1/u
where
f is the focal length of the lens,
v is the image distance,
u is the object distance.
Since the image is a virtual image, we will give our v a negative sign. So, on calculating, we have
1/f = 1/25 - 1/151
1/f = 0.0333
f = 1/0.0333
f = 30.03 cm
f = 0.3 m
b) The power of the needed corrective lens is the reciprocal of the focal length in metres;
1/0.3 = 3.3 diopters
This is usually in diopters
A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?
Answer:
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
Explanation:
We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:
[tex]y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (1)
Where:
[tex]y(t)[/tex] - Position of the mass as a function of time, measured in meters.
[tex]A[/tex] - Amplitude, measured in meters.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]t[/tex] - Time, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
The spring is now calculated by Hooke's Law, that is:
[tex]k = \frac{m\cdot g}{\Delta y}[/tex] (2)
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]\Delta y[/tex] - Deformation of the spring due to gravity, measured in meters.
If we know that [tex]m=1.65\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta y = 0.260\,m[/tex], then the spring constant is:
[tex]k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}[/tex]
[tex]k = 62.237\,\frac{N}{m}[/tex]
If we know that [tex]A = 0.130\,m[/tex], [tex]k = 62.237\,\frac{N}{m}[/tex], [tex]m=1.65\,kg[/tex], [tex]x(t) = 0\,m[/tex] and [tex]\phi = 0\,rad[/tex], then (1) is reduced into this form:
[tex]0.130\cdot \cos (6.142\cdot t)=0[/tex] (1)
And now we solve for [tex]t[/tex]. Given that cosine is a periodic function, we are only interested in the least value of [tex]t[/tex] such that mass reaches equilibrium position. Then:
[tex]\cos (6.142\cdot t) = 0[/tex]
[tex]6.142\cdot t = \cos^{-1} 0[/tex]
[tex]t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s[/tex]
[tex]t \approx 0.255\,s[/tex]
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
The time taken for the spring to reach new equilibrium position is 0.26 s.
The given parameters;
mass, m = 1.65 kgextension of the string, x = 0.26 mdisplacement with time x(t) = 0.13The general wave equation is given as;
[tex]y(t) = A\ cos(\omega t \ + \phi)[/tex]
The angular frequency is given as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\[/tex]
The spring constant is calculated as;
[tex]F = mg\\\\kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{1.65 \times 9.8}{0.26} \\\\k = 62.2 \ N/m[/tex]
The angular frequency is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{62.2}{1.65} }\\\\\omega = 6.14 \ rad/s[/tex]
The time taken for the spring to reach new equilibrium position is calculated as follows;
[tex]y(t) = A \ cos(\omega t)\\\\0 = 0.13 \times cos(6.14t)\\\\6.14t = cos^{-1}(0)\\\\6.14t = 1.57 \ rad\\\\t = \frac{1.57 \ rad}{6.14 \ rad/s} \\\\t = 0.26 \ s[/tex]
Thus, the time taken for the spring to reach new equilibrium position is 0.26 s.
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Which type of force is the weakest?
The correct option is A i.e. Van der Waals forces.
what is Van der Waals forces?
Van der Waals forces are weak intermolecular forces that are depending on atom or molecule distance. Interactions between uncharged atoms/molecules produce these forces.
The variation in the polarizations of two particles close to each other can produce Van der Waals forces.Van der Waals forces are substantially weaker than covalent and ionic bonding.The Van der Waals forces are additive in nature, consisting of multiple distinct interactions.These forces are inexhaustible.These forces have no discernible directional characteristics.They aren't affected by temperature (except dipole-dipole interactions)Short-range forces are Van der Waals forces. When the atoms/molecules in question are near together, their magnitude is large.To know more about Van der Waals forces here
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A 150.0 kg cart rides down a set of tracks on four solid steel wheels, each with radius 20.0 cm and mass 45.0 kg. The tracks slope downward at an angle of 17 ∘ to the horizontal. If the cart is released from rest a distance of 27.0 m from the bottom of the track (measured along the slope), how fast will it be moving when it reaches the bottom
Answer:
13.4 m/sExplanation:
given
Mass of cart= 150kg
mass of each wheel=45kg
mass of 4 wheels= 180kg
angle of the track= 17 ∘
distance of track= 27m
The height of the tracl is calculated thus:
sin 17° = h / 27
h = sin 17*27
h=7.89m
"Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom "
1. Potential energy at top= (M+4m)gh
2. kinetic energy of cart and wheels at bottom= 1/2 (M+4m) v²
3. rotational energy of wheels at bottom= 4(1/2 Iω²)
The total is expressed as
(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²) -------------1
we know that I = mr² / 2
Put I= mr² / 2
(M+4m)gh = 1/2 (M+4m)v² + 4(1/2 (mr² / 2) ω²)
(M+4m)gh = 1/2 (M+4m)v² + m r² ω²
we know that v²= r² ω²
(M+4m)gh = 1/2 (M+4m)v² + m v²
(M+4m)gh = v² (M/2 + 2m + m)
(M+4m)gh = v² (M/2 + 3m)
v = √[(M+4m)gh / (3m + M/2)]
v = √[(150 + 4*45) * 9.8 m/s² * 7.89 m / (3*45 + 150/2)]
v=√25516.26/142.5
v=√179.06
v = 13.4 m/s