Answer:
in the downward movement of the movement when the constant is lost
Explanation:
When the coin is on the piston it has a relationship given by
a = d²x / dt²
the piston position is
x = A cos wt
a = - A w² cos wt
the maximum acceleration is
a = - A w²
When the piston raises the acceleration of gravity and that of the piston go in the same direction, when the piston descends they relate it is contrary to gravity, therefore when the frequency increases, the point where the acceleration of the piston is greater than gravity arrives and the coin loses contact.
The point where you lose contact is
a = g
g = A w²
In short, in the downward movement of the movement when the constant is lost
In terms of running the Earth's atmospheric processes, the significant energy comes from the Sun and ________. a. the Moon b. the next closest star c. reflection from Venus d. no other source
Answer: no other source
Explanation:
The Earth gets energy from the sun which includes both heat and light and these are necessary for chemical reaction.
When the energy from the Sun gets to the Earth, they come as solar radiation and included in such solar radiation are infrared, invisible light, X-rays, ultraviolet light, radio waves, and gamma rays.
We should note that no other source is required for the Earth to get energy. The sun supplies it its energy.
Why do you think the process of photosynthesis and respiration are interrelated?
Answer:
The processes of photosynthesis and respiration are interrelated because the products of such processes are used as reactants for one process. One product of photosynthesis is oxygen. This oxygen is used in the process of respiration as a reactant producing carbon dioxide which is used as a reactant for photosynthesis.
Energy from a battery powers an electromagnet Which of the
following energy transformations takes place?
a
magnetic to heat
b magnetic to electrical
С
electrical to magnetic
d light to sound
Answer: it’s actually C, electrical to magnetic
Explanation:
To get an idea of how much thermal energy is contained in the world's oceans, estimate the heat liberated when a cube of ocean water, 3 km on each side, is cooled by 4 K. (Approximate the ocean water as pure water for this estimate.)
Answer:
Q = 4.52 10¹⁷ J
Explanation:
Thermal energy can be calculated with
Q = m c_{e} ΔT
in this case it indicates that we approximate seawater to pure water with
c_{e} = 4186 J/ kg K
with the density
ρ = m / V
m = ρ V
V = L³
we substitute
m = ρ L³
Q = ρ L3 c_{e} ΔT
calculate
Q = 1000 (3 103) 3 4186 4
Q = 4.52 10¹⁷ J
How is the volume flow rate of water out of the tank, dVdt, related to the flow speed v ? Express your answer in terms of some, all, or none of the variables v , d, the acceleration due to gravity g, and the constant π.
Answer:
[tex]\frac{dV}{dt}= \frac{\pi d^2}{4}v[/tex]
Explanation:
The rate of volume flow out of tank can be expressed as:
[tex]\frac{dV}{dt} = A\frac{dL}{dt}[/tex]
where,
dV/dt = Volume flow rate
A = Cross-sectional area of outlet = πd²/4
d = diameter of circular outlet
dL = Displacement covered by water
dt = time taken
but we know that:
Velocity = υ = displacement/time = dL/dt
Substituting the values of "dL/dt" and "A" in the equation, we get:
[tex]\frac{dV}{dt} = \frac{\pi d^2}{4}v[/tex]
This is the expression for volume flow rate dV/dt, on terms pf v, d.
When a battery-powered flashlight is turned on, it releases more energy than it had when it was off.
True or False?
Answer:
True
Explanation:
When it's turn off, chemical energy is stored (the battery) in the flashlight, when it's turn on the chemical energy transfers into light and thermal energy.
What Initial Velocity would you have to give the Marble to make it hit the ground in 5 s
Answer:
Traveling 381m in 5 seconds means the average velocity is: 381m/5s = 76.2 m/s.
Explanation:
Letting x be the initial velocity and v the final velocity, we have:
12(x+v)=76.2m/s
and (since v=x+50),
12(2x+50)=76.2m/s
x+25=76.2m/s
x=51.2m/s
Answer:
-49 m/s
Explanation:
You would use the formula vf = vi + at, where vf stands for final velocity, vi stands for initial velocity, a stands for acceleration, and t stands for time. Assuming you're dropping the marble on Earth, the acceleration would be -9.8m/s², and the final velocity would be 0 m/s, as it stops after hitting the ground. (0 = vi + (9.8)(5))
I would like to know why this is the correct answer
-A spring scale shows a net force of 0.8 N acting on a 1.5-kg mass. What happens to the acceleration of the object if the net force is decreased to 0.2 N?
*The acceleration decreases to a quarter of its original value, or about 0.13 m/s2.
The acceleration of the object if the net force is decreased = 0.13 m/s²
Further explanationGiven
A net force of 0.8 N acting on a 1.5-kg mass.
The net force is decreased to 0.2 N
Required
The acceleration of the object if the net force is decreased
Solution
Newton's 2nd law :
[tex]\tt \sum F=m.a[/tex]
The mass used in state 1 and 2 remains the same, at 1.5 kg
state 1ΣF=0.8 N
m=1.5 kg
The acceleration, a:
[tex]\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.8}{1.5}\\\\a=0.53`m/s^2[/tex]
state 2ΣF=0.2 N
m=1.5 kg
The acceleration, a:
[tex]\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.2}{1.5}\\\\a=0.13~m/s^2[/tex]
Answer:
A spring scale shows a net force of 0.8 N acting on a 1.5-kg mass. What happens to the acceleration of the object if the net force is decreased to 0.2 N? A, The acceleration decreases to a quarter of its original value, or about 0.13 m/s².
The highest surface temperature on any of the solar system’s planets is found on Venus. Partly because of its nearness to the sun and partly because of the extreme pressure of its atmosphere, the average daytime temperature on Venus is 453°C. What is this temperature in degrees Fahrenheit and in kelvins?
Answer:
Fahrenheit ⇒ 847.4 °FKelvin ⇒ 726.15KExplanation:
To convert Celsius to Fahrenheit, use the following formula;
= (°C * 9/5) + 32
= (453 * 9/5) + 32
= 847.4 °F
To convert Celsius to Kelvin, use the following;
= °C + 273.15
= 453 + 273.15
= 726.15K
4
2 points
Which of the following is a vector quantity?
O Velocity
O
Time
Speed
оо
Distance
Answer:
Velocity Has vector Quantity
I need help with this answer
Answer:
double replacement
Explanation:
sorry if im wrong
A person pushes a 10 kg box from rest and accelerates it to a speed of 4 m/s with a constant force. If the box is pushed for a time of 2.5 s, what is the force exerted by the person?
The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration a is such that
4 m/s = a (2.5 s) → a = (4 m/s) / (2.5 s) = 1.6 m/s²
Then the force applied to the box has a magnitude F such that
F = (10 kg) (1.6 m/s²) = 16 N
Which force represented by the arrow at B? a. force of friction b. force of gravity c. tension force d. normal force
Answer:
A. Force of friction
Explanation:
Find the diagram attached
From the diagram we can see that the force B is acting up on the body along the plane. This force is the force that prevents the body from sliding down the plane. Originally, the body on the line will tend to slide down the incline even without application of a force hence the reason of an opposing force that must act in the opposite direction is counter this motion to make the body be in a state of equilibrium. Hence the required force is force of friction since it is am opposing force acting in the direction opposite to the moving force acting on the body.
Pls help quick too.................
Answer:
Pulley = Flag Pole
Screw = Side latch
Wedge = Scissors
Wheel and axle = Car
a ball is thrown with a speed of 17.7 m/s at an angle of 49.8° above the horizontal. how much time does the ball need to reach a height of 4.7 m above the release point ?
Answer:
The ball needs approximately [tex]0.41\; \rm s[/tex] to reach that height for the first time.
Explanation:
The initial speed of the ball [tex]17.7\; \rm m \cdot s^{-1}[/tex]. However, what would be the initial vertical speed of this ball?
The angle of elevation is [tex]\theta = 49.8^\circ[/tex]. Consider the initial speed of this ball as the length of the hypotenuse of a right triangle. If the angle of elevation is one of the two acute angles of this triangle, the initial vertical speed of this ball would be the leg opposite to that angle.
[tex]\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}[/tex].
[tex]\displaystyle \sin \theta = \frac{v(\text{vertical, initial})}{v(\text{initial})}[/tex].
Therefore:
[tex]\begin{aligned}&v(\text{vertical, initial})\\ &= v(\text{initial}) \cdot \sin\theta \\ &= 17.7\; \rm m \cdot s^{-1} \times \sin \left(49.8^\circ\right) \approx 13.5\; \rm m \cdot s^{-1}\end{aligned}[/tex].
Let [tex]t[/tex] denote the time (in seconds) required for the ball to reach a height of [tex]4.7\; \rm m[/tex].
Let [tex]g[/tex] denote the acceleration because of gravity (typically [tex]g \approx 9.81\; \rm m \cdot s^{-2}[/tex] near the surface of the earth.) The height of the ball [tex]t[/tex] seconds after it was thrown would be: [tex]\displaystyle \frac{1}{2}\, g \cdot t^{2} + v(\text{vertical, initial}) \cdot t[/tex].
Assume that [tex]g \approx 9.81\; \rm m \cdot s^{-2}[/tex]. Set the value of this expression for height to [tex]4.7\; \rm m[/tex] and solve for [tex]t[/tex]:
[tex]\displaystyle \frac{1}{2} \times 9.81 \, t^{2} + 13.5 \, t = 4.7[/tex].
Either [tex]t \approx 0.41[/tex] or [tex]t \approx 2.3[/tex] will satisfy this equation. Both of these two values are reasonable. The first value for [tex]t[/tex] ([tex]0.41\; \rm s[/tex]) is the time required for the ball to reach a height of [tex]4.7\; \rm m[/tex] for the first time. The second value ([tex]2.3\; \rm s[/tex]) is the time required for the ball to come under that height on its way back to the ground. The question seems to be asking only for the first (the smaller one) of these two times.
What causes sound energy to be produced? *
Speed
instruments
Vibration
Fluency
The writing on the passenger-side mirror of your car says "Warning! Objects are closer than they appear." There is no such warning on the driver's mirror. Consider a typical convex passenger-side mirror with a focal length of -80 cm . A 1.5 m -tall cyclist on a bicycle is 30 m from the mirror. You are 1.4 m from the mirror, and suppose, for simplicity, that the mirror, you, and the cyclist all lie along a line.
- You are 2.2 m from the image of the cyclist
A) What is the image height?
B) What would the image height have been if the mirror were flat?
C) What is the angular size of the image of the cyclist?
D) What would the angular size of the cyclist's image have been if the mirror were flat?
E) Explain why is there a warning label on the passenger-side mirror?
Answer:
kdfjnfkasd
Explanation:
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. The proton must impact the nucleus with a kinetic energy of 2.30 MeV. Assume the nucleus remains at rest. With what speed must the proton be fired toward the target?
Answer:
The value is [tex]u = 3.23 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The diameter of the nucleus is [tex]d = 5.50 \ fm = 5.50 *10^{-15} \ c[/tex]
The charge of the proton that makes up the nucleus is [tex]Q_2 = \frac{12}{2} * 1.60 *10^{-19} =9.6*10^{-19} \ C[/tex]
The energy to be impacted is [tex]KE_f = 2.30 \ MeV = 2.30 *10^{6} \ eV = 2.30 *10^{6} * 1.60 *10^{-19} = 3.68*10^{-13} \ J[/tex]
Generally the radius of the nucleus is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{5.50 *10^{-15}}{2}[/tex]
=> [tex]r = 2.75 *10^{-15} \ m[/tex]
Generally from the law energy conservation we have that
[tex]Initial \ total \ energy \ of the \ proton = final \ total \ energy \ of the \ proton[/tex]
i.e
[tex]T_i = T_f[/tex]
Here
[tex]T_i = KE_i + PE_i[/tex]
Here [tex]KE_i[/tex] is the initial kinetic energy which is mathematically represented as
[tex]KE_I = \frac{1}{2} * m * u ^2[/tex]
Here [tex]PE_i[/tex] is the initial potential energy of the proton and the value is 0 J given that the proton is moving
Also [tex]T_f[/tex] is mathematically represented as
[tex]T_f = KE_f + PE_f[/tex]
Here
[tex]PE_f[/tex] is the final potential energy which is mathematically represented as
[tex]PE_f = \frac{k * Q_1 * Q_2}{r}[/tex]
Here [tex]Q_1[/tex] is the charge on the proton with a value of [tex]Q_1 = 1.60 *10^{-19} \ C[/tex]
So
[tex]PE_f = \frac{9*10^{9} *(1.60 *10^{-19} ) * ( 9.6 *10^{-19})}{ 2.75 *10^{-15}}[/tex]
=> [tex]PE_f = 5.027 *10^{-13 } \ J[/tex]
So
[tex]KE_i + PE_i = KE_f + PE_f[/tex]
=> [tex]\frac{1}{2} * m * u ^2 +0 = 3.68*10^{-13} + 5.027 *10^{-13 }[/tex]
Here m is the mass of the moving proton with value [tex]m = 1.67*10^{-27} \ kg[/tex]
So
[tex]\frac{1}{2} * 1.67*10^{-27} * u ^2 +0 = 3.68*10^{-13} + 5.027 *10^{-13 }[/tex]
=> [tex]u = \sqrt{\frac{3.68*10^{-13} + 5.027 *10^{-13 }}{0.5 * 1.67*10^{-27}} }[/tex]
=> [tex]u = 3.23 *10^{7} \ m/s[/tex]
Describe the role of the
nucleus in cell activities.
Answer:the nucleus controls and regulates the activities of the cell
Explanation:trust me bro
Answer:
The nucleus controls and regulates the activities of the cell and carries the genes, structures that contain the hereditary information.
A 25-g string is stretched with a tension of 43 N between two fixed points 12 m apart. What is the frequency of the second harmonic?
Answer:
The frequency of the second harmonic ([tex]2f_o[/tex]) is 11.97 Hz.
Explanation:
Given;
mass of the string, m = 25 g = 0.025kg
tension on the string, T = 43 N
length of the string, L = 12 m
The speed of wave on the string is given as;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length = 0.025 / 12 = 0.002083 kg/m
[tex]v = \sqrt{\frac{43}{0.002083} }\\\\v = 143.678 \ m/s[/tex]
The wavelength of the first harmonic wave is given as;
[tex]L = \frac{1}{2} \lambda _o\\\\\lambda _o = 2L \\\\\lambda _o = 2 \ \times \ 12\\\\\lambda _o = 24 \ m[/tex]
The frequency of the first harmonic is given as;
[tex]f_o = \frac{v}{\lambda _o} = \frac{v}{2L} = \frac{143.678}{24} = 5.99 \ Hz\\\\[/tex]
The wavelength of the second harmonic wave is given as;
[tex]L = \lambda_1 \\\\\lambda_1 = 12 \ m[/tex]
The frequency of the second harmonic is given as;
[tex]f_1 = \frac{v}{\lambda _1} = \frac{143.678}{12} = 11.97 \ Hz = 2(\frac{v}{\lambda _0}) = 2f_o[/tex]
Therefore, the frequency of the second harmonic ([tex]2f_o[/tex]) is 11.97 Hz.
An object is placed on the moveable piston of a cylinder filled with a gas. The object exerts a downward force F on the piston that accelerates the piston downward, compressing the gas. Which of the following is a correct description of the magnitude of the upward force exerted by the piston on the object?
a. The force exerted by the piston on the object is equal to FF.
b. The force exerted by the piston on the object is less than FF because the piston is accelerating downward.
c. The force exerted by the piston on the object is greater than F because the gas exerts an upward force on the piston.
d. The force exerted by the piston cannot be compared to FF without knowing the magnitude of the acceleration of the piston and pressure of the gas.
Answer:
b. The force exerted by the piston on the object is less than F because the piston is accelerating downward.
Explanation:
Given that object exerts a downward force F on the piston that accelerates the piston downward.
An object can accelerate if the net force acting on it is non-zero.
Here, the piston is acceleration in the downward direction, so the net for on the piston is in the downward direction.
Force due to object, F, is in the downward direction while the force exerted by the piston on the object is in the upward direction.
As the acceleration of the piston is in the downward direction, so the downward force is more.
Therefore, the force exerted by the piston on the object is less than F because the piston is accelerating downward.
Hence, option (b) is correct.
Which one of the the following occurs when sodium forms an ionic bond?
Answer:
An atom of sodium (Na) donates one of its electrons to an atom of chlorine (Cl) in a chemical reaction, and the resulting positive ion (Na+) and negative ion (Cl−) form a stable ionic compound (sodium chloride; common table salt) based on this ionic bond.
Explanation:
All the info i can give you
2. What is the kinetic energy of an 80 kg football player running at 8m/s? *
O A. 2560J
O B. 320J
O C. 2560 W
O D.320 W
HELPPPPP
Answer:
the answer is A!! 2560 J
A 1.2-kg object moving at 5.0 m/s collides with and sticks to a 4.8-kg object initially at rest. What is the amount of kinetic energy lost by the system during this collision?
Answer:
-15J
Explanation:
Step one:
given data
m1= 1.2kg
v1= 5 m/s
m2= 4.8kg
v2= 0 m/s
Required
The kinetic energy lost
Step two:
The kinetic energy before impact is KE1
[tex]KE1= 1/2mv^2\\\\KE1= 1/2*1.2*5^2\\\\KE1= 15J[/tex]
KE after impact is KE2
[tex]KE2= 1/2*(m1+m2)v2\\\\KE2= 1/2*(6)0\\\\KE2= 0 J[/tex]
The loss in Kinetic energy is
Loss in KE= KE2-KE1
Loss in KE= 0-15
Loss in KE= -15J
Two masses sit at the top of two frictionless inclined planes that have different angles,__deleted9917f34947e1359b5705bbac0d9227f8ec5df862078213a2db19c72f00189109deleted__ 0N86-C1-52-40-A837-22820 50% Part (a) What can be said about the speeds of the two masses at the bottom of their respective paths
Answer:
v = [tex]\sqrt{2gh}[/tex]
the speed in the two planes will be the same since it does not depend on the angle of the same
Explanation:
In this exercise we are told that the two inclined planes have no friction force, so we can apply the conservation of energy for each one, we will assume that the initial height in the two planes is the same
starting point. Highest part of each plane
Em₀ = U = m g h
final point. Lowest part of each plane
[tex]Em_{f}[/tex] = K = ½ m v²
as there is no friction, the mechanical energy is preserved
Em₀ = Em_{f}
mg h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
As we can see, the speed in the two planes will be the same since it does not depend on the angle of the same
How do insects aid in crime scene investigations
Answer:
They can figure out when the person died, possibly the nature of crime, it can also detect if the body was moved in certain area.
Explanation:
A satellite has a mass of 3.25 x 10^3 kg, while earth has a mass of 5.97 x 10^24 kg
Answer:
200N ;)
Explanation:
Answer:
200N
Explanation:
What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to 14 REarth before coming to rest momentarily
Answer:
v = 1.078 10⁴ m / s
Explanation:
To solve this exercise we can use conservation of mechanical energy
Starting point. Just clearing
Em₀ = K + U = ½ m v² - G m M / [tex]R_{e}[/tex]
Final point. At r = 14R_{e}
Em_{f} = U = -G m M / 14R_{e}
how energy is conserved
Em₀ = [tex]Em_{f}[/tex]
½ m v² - G m M /R_{e} = -G m M / 14R_{e}
½ v² = G M / R_{e} (-1/14 + 1)
v² = 2 G M / R_{e} 13/14
we calculate
v² = 2 6.67 10⁻¹¹ 5.98 10²⁴ / 6.37 10⁶ 13/14
v = √ (1,16 10⁸)
v = 1.078 10⁴ m / s
The launch speed of a projectile that rises above the earth will be
v = 1.078 10⁴ m / s
What is speed?
Speed is defned as the movement of any object with respect to the time the formula of speed is given as the ratio of the distance with time.
To solve this exercise we can use conservation of mechanical energy
Starting point. Just clearing
[tex]E_{mo} = K + U = \dfrac{1}{2}m v^2 - \dfrac{G m M} { r}[/tex]
Final point. At[tex]r = 14R_{e}[/tex]
[tex]Em_{f} = U = \dfrac{-G m M }{14R_{e}}[/tex]
how energy is conserved
[tex]E_{mo }= \dfrac{1}{2} m v^2 -\dfrac{-G m M }{R_{e}} = \dfrac{-G m M }{ 14R_{e}}[/tex]
[tex]\dfrac{1}{2} v^2 = \dfrac{G M }{ R_{e} (\dfrac{-1}{14} + 1)}[/tex]
[tex]v^2 = \dfrac{2 G M} { R_{e} \dfrac{13}{14}}[/tex]
we calculate
[tex]v^2 = \dfrac{2 6.67 \times10^{-11}\times 5.98 \times 10^{24}} { 6.37 \times 10^{6} \times \dfrac{13}{14}}[/tex]
[tex]v = \sqrt{ (1,16\times 10^8)}[/tex]
v = 1.078 10⁴ m / s
Hence the launch speed of a projectile that rises above the earth will be
v = 1.078 10⁴ m / s
To know more about speed follow
https://brainly.com/question/4931057
SUBJECT: ASTRONOMY -- TOPIC: Time Dilation
Find the time interval measured by your stationary watch if your friend’s watch measures 1.0 year while she is traveling at 0.87c aboard a futuristic interstellar rocket.
Answer:
You can determine if the ship is moving by lying down and measuring your height.
Explanation:
What type of energy transfer occurs
when two objects touch?
O conduction
O induction
O convection
Answer:
conduction
Explanation:
