Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
In the diagram, the block is sliding at a constant velocity down a rough incline, which makes a 45 degree angle to the horizontal. Which of the following statements is NOT true?
A) The magnitude of the Net Force is equal to the magnitude of Force C.
B) The magnitude of Force A is equal to the magnitude of Force B
C) The magnitude of Force B is equal to the magnitude of Force D
D) The coefficient of friction between the block and the ramp is equal to 1.
Answer:
I haven't done physics for a while, but with the information given I don't understand how D could be correct, so I think D is your anwser
Explanation:
What is the KE of a 4 kg rock rolling at 10 m/s? KE = 1/2 mv2
O 160 J
O 2003
O 40 J
O 400 J
Answer:
200J
Explanation:
0.5 *4 * 10 * 10
=200J
According to Newton's Universal Law of Gravitation, when the distance between two interacting objects doubles, the gravitational force is
Answer:
If the distance doubles, the gravitational force is divided by 4
Explanation:
Newton’s Universal Law of Gravitation
Objects attract each other with a force that is proportional to their masses and inversely proportional to the square of the distance.
[tex]\displaystyle F=G{\frac {m_{1}m_{2}}{r^{2}}}[/tex]
Where:
m1 = mass of object 1
m2 = mass of object 2
r = distance between the objects' center of masses
G = gravitational constant: 6.67\cdot 10^{-11}~Nw*m^2/Kg^2
If the distance between the interacting objects doubles to 2r, the new force F' is:
[tex]\displaystyle F'=G{\frac {m_{1}m_{2}}{(2r)^{2}}}[/tex]
Operating:
[tex]\displaystyle F'=\frac{1}{4}G{\frac {m_{1}m_{2}}{r^{2}}}[/tex]
Substituting the original value of F:
[tex]\displaystyle F'=\frac{1}{4}F[/tex]
If the distance doubles, the gravitational force is divided by 4
The pendulum on a grandfather clock is 0.993 m long and swings to a maximum 4.57° angle. If the bob of the pendulum has mass = 0.415 kg, how much PE does it have at the top of its swing?
Answer:
0.0128
Explanation:
calculate the force required to take away a flat corcular plate of radius 0.01m from the surface of water. the surface tention of water is 0.075N/m.
Answer:
[tex]Force = 0.0047175\ N[/tex]
Explanation:
Given
[tex]T = 0.075N/m[/tex] --- Surface Tension
[tex]r = 0.01m[/tex] --- Radius
Required
Determine the required force
First, we calculate the circumference (C) of the circular plate
[tex]C= 2\pi r[/tex]
[tex]C= 2 * \frac{22}{7} * 0.01m[/tex]
[tex]C= \frac{2 * 22 * 0.01}{7}m[/tex]
[tex]C= \frac{0.44}{7}m[/tex]
[tex]C= 0.0629 m[/tex]
The applied force is then calculated using;
[tex]Force = C * T[/tex]
[tex]Force = 0.0629m * 0.075N/m[/tex]
[tex]Force = 0.0047175\ N[/tex]
Calculate the magnitude of the gravitational force exerted by Venus on a 60 kg human standing on the surface of Venus. (The mass of Venus is 4.9multiply1024 kg and its radius is 6.1multiply106 m.)
Answer:
The gravitational force exerted by Venus on the human is 527 N.
Explanation:
Given;
mass of the human, m₁ = 60 kg
mass of Venus, m₂ = 4.9 x 10²⁴ kg
gravitational constant, G = 6.67 x 10⁻¹¹ Nm²/kg²
radius of the Venus, r = 6.1 x 10⁶ m
The gravitational force exerted by Venus on the human is calculated as;
[tex]F = \frac{Gm_1m_2}{r^2} \\\\F = \frac{(6.67\times 10^{-11})(60)(4.9\times 10^{24})}{(6.1\times 10^6)^2} \\\\F = 527 \ N[/tex]
Therefore, the gravitational force exerted by Venus on the human is 527 N.
Two spherical objects are separated by a distance that is 1.94 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 7.61 x 10-21 N. How many electrons did it take to produce the charge on one of the objects
Answer:
n = 1 10¹⁰ electron
Explanation:
For this exercise we will use Coulomb's law
F = k q₁ q₂ / r²
where it indicates that the value of the force is 7.61 10⁻²¹ N, the distance between the objects is r = 1.94 10⁻³ m and the charge on the objects has the same charge q₁ = q₂ = q,
F = k q² / r²
q = [tex]\sqrt{ \frac{ k \ q^2}{k} }[/tex]
let's calculate
q = [tex]\sqrt{ \frac{ 7.61 \ 10^{-21} \ (1.93 \ 10^{-3})^2 }{9 \ 10^{9} } }[/tex]
q = √(3.182 10⁻¹⁸)
q = 1.783 10⁻⁹ C
the charge of an electron is q₀ = 1.6 10⁻¹⁹ C, therefore
q = n q₀
n = q / q₀
n = 1,78 10⁻⁹ / 1.6 10⁻¹⁰
n = 1.06 10¹⁰ electron
the number of electrons must be integer
n = 1 10¹⁰ electron
Coulomb's Law is a fundamental principle in physics that describes the electrostatic interaction between charged particles. It took approximately 3.62 electrons to produce the charge on one of the objects.
According to Coulomb's Law, the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
To solve this problem, we can use Coulomb's law, which relates the electrostatic force between two charged objects to the charges and the distance between them. Coulomb's law is given by:
[tex]F = k * (|q1 * q2|) / r^2[/tex]
Where:
F is the electrostatic force,
k is the electrostatic constant [tex](k = 8.99 x 10^9 Nm^2/C^2),[/tex]
q1 and q2 are the charges on the objects, and
r is the distance between the objects.
Given that the force F is [tex]7.61 x 10^{-21} N[/tex], and the distance r is [tex]7.61 x 10^{-21} N[/tex]m, and assuming the charges on both objects are equal (let's call it q), we can rewrite Coulomb's law as:
[tex]F = k * (q^2) / r^2[/tex]
Rearranging the equation, we get:
[tex]q^2 = (F * r^2) / k[/tex]
Substituting the given values:
[tex]q^2 = (7.61 x 10^{-21} N) * (1.94 x 10^{-3} m)^2 / (8.99 x 10^9 Nm^2/C^2)[/tex]
Calculating this expression:
[tex]q^2 = 3.352 x 10^{-37} C^2[/tex]
Taking the square root of both sides:
[tex]q = \sqrt {3.352 x 10^{-37} C^2}\\q = 5.79 x 10^{(-19)}C[/tex]
Now, we know that the elementary charge of an electron is approximately[tex]1.6 x 10^{(-19)} C[/tex]. Since each electron carries a charge of[tex]-1.6 x 10^{-19} C[/tex], we can calculate the number of electrons:
Number of electrons = q / (elementary charge)
Number of electrons =[tex](5.79 x 10^{-19} C) / (1.6 x 10^{-19} C)[/tex]
Number of electrons = 3.62
Therefore, it took approximately 3.62 electrons to produce the charge on one of the objects.
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what is the spring potential energy of a spring that is stretched 15 cm if its spring constant is 350 n/m?
how much force is required for this stretch?
Answer:
Spring potential energy = 7.875Nm
Explanation:
Given the following data;
Extension, e = 15cm to meters = 15/100 = 0.15m
Spring constant, k = 350n/m
To find the force;
Force = spring constant * extension
Force = 350 * 0.15
Force = 52.5 Newton.
Now to find the spring potential energy we would use the formula below;
Spring potential energy = force * extension
Substituting into the equation, we have;
Spring potential energy = 52.5 * 0.15
Spring potential energy = 7.875Nm
How much energy is needed to heat 1kg of copper by 20 degrees Celsius
Step by step explanation
Answer:
Q = 0.0077 J
Explanation:
Given that,
Mass of copper, m = 1 kg = 0.001 g
The change in temperature = 20°C
We need to find the energy needed to heat copper. The formula is given by :
[tex]Q=mc\Delta T[/tex]
c is the specific heat of copper, c = 0.385 J/g°C
Put values in the above formula
[tex]Q=0.001\ g\times 0.385\ J/g^{\circ} C\times 20^{\circ} C\\\\Q=0.0077\ J[/tex]
So, the energy needed to heat 1 kg of copper is 0.0077 J.
Which object has the most momentum?
A toy car rolling along the floor
A child peddling their bike as fast as possible
&o
A large moving truck stopped at a red light
PLZ Help ASAP
Peter wants to find out which boils faster: salt water or tap water. What procedure should he follow to conduct this experiment?
Answer:
C. Fill two identical pots with equal volumes of salt water and tap water and use a stopwatch to determine the time it takes each pot to boil.
Explanation:
A) is incorrect because Peter should have the same testing environment for both of his experiments.
He should choose the same method of boiling the salt water and tap water because the stovetop and the microwave could also affect the results and make them unreliable.
B) is incorrect because Peter should not estimate the time it takes the salt water and tap water to boil.
Peter should measure and record the amount of time that it takes these substances to boil in order to have an accurate, valid experimental thesis.
C) is correct because Peter uses the same volume of salt water and tap water, fills them into two identical pots, and uses a stopwatch to determine the amount of time it takes each pot to boil.
The stopwatch makes the experiment more valid and accurate compared to the previous methods, and the identical pots and amounts of water help this experiment become even more precise.
D) is incorrect because the variables in the experiment are not controlled amounts and will therefore produce an inaccurate and invalid experiment.
Pulling on a spring with a force of 1.2 N causes a stretch of 6.4 cm. What is the spring constant for this spring?
Answer:
k=19 N/m
Explanation:
Use the equation F=kx for this problem and isolate k since that is what you're solving for (k=F/x). Plug in your values (F=1.2 N and x=0.064 m) and solve for k which is 19 N/m.
By using Hooke's law formula, the spring constant K for the spring is 18.75 N/m
The parameters given are :
Force F = 1.2N
Extension e = 6.4 cm
convert cm to m
e = 6.4 / 100
e = 0.064 m
Spring constant K = ?
From Hooke's law, It states that the extension is proportional to the force applied provided that elastic limit is not exceeded. That is,
F = Ke
Substitute all the parameters into the formula.
1.2 = 0.064K
Make K the subject of the formula
K = 1.2/0.064
K = 18.75 N/m
Therefore, the spring constant K for this spring is 18.75 N/m
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Why is temperature scalar?
Help ASAP!
Everything on screenshot
Answer:
For the first one, its B) cities B and C
I'm not so sure, but I hope this helps.
Based on the information presented in the introduction of this problem, what is a sound wave? Based on the information presented in the introduction of this problem, what is a sound wave? Propagation of sound particles that are different from the particles that comprise the medium Propagation of energy that does not require a medium Propagation of pressure fluctuations in a medium
Answer:
sound is a wave that carries energy and the particles are oscillating around their equilibrium positions.
Explanation:
When the air particles are subjected to a periodic force they oscillate around their equilibrium possessions, this oscillation can propagate in the air creating a sound wave.
Therefore the sound wave is a wave that propagates energy, but the particles have only one oscillation around their equilibrium position.
These waves can be mathematically analyzed as displacement waves and since the oscillation is in the same line of the movement of the wave we can also analyze it as a pressure wave
In summary, sound is a wave that carries energy and the particles are oscillating around their equilibrium positions.
If you double the mass of both Mass 1 and the distance, how does the gravitational force change? A) The force doubles B) The force triples C) the force quadruples D) the force is half E) the force is 1/4
Answer:
Choice C)
Explanation:
Newton’s Universal Law of Gravitation
It states objects attract each other with a force that is proportional to their masses and inversely proportional to the square of the distance.
[tex]\displaystyle F=G{\frac {m_{1}m_{2}}{r^{2}}}[/tex]
Where:
m1 = mass of object 1
m2 = mass of object 2
r = distance between the objects' center of masses
G = gravitational constant: 6.67\cdot 10^{-11}~Nw*m^2/Kg^2
If m1 and r are doubled, then the new force F' is:
[tex]\displaystyle F'=G{\frac {2m_{1}m_{2}}{(2r)^{2}}}[/tex]
Operating:
[tex]\displaystyle F'=G{\frac {2m_{1}m_{2}}{4r^{2}}}[/tex]
[tex]\displaystyle F'=\frac{2}{4}G{\frac {m_{1}m_{2}}{r^{2}}}[/tex]
[tex]\displaystyle F'=\frac{1}{2}G{\frac {m_{1}m_{2}}{r^{2}}}[/tex]
Substituting the value of the original force:
[tex]\displaystyle F'=\frac{1}{2}F[/tex]
This means the force is halved
Choice C)
These two waves travel along the same string: y1 = (4.17 mm) sin(2.24?x - 300?t), y2 = (5.96 mm) sin(2.24?x - 300?t + 0.727?rad). What are (a) the amplitude and (b) the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 5.20 mm is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?
Answer:
a) Amplitude is 9.49 mm
b) ∅ = 0.43056 rad = 24.7°
c) phase angle = 2.24x - 300t + 0.43056 rad
Explanation:
Given that;
A1 = y1 = (4.17 mm) sin(2.24x - 300t)
A2 = y2 = (5.96 mm) sin(2.24x - 300t + 0.727rad)
now phase difference between y1 and y2 is;
Δ∅ = (2.24x - 300t + 0.727rad) - (2.24x - 300t) = 0.727 rad = 41.65°
a) the amplitude
Amplitude A = √( A1² + A2² + 2A1A2cosΔ∅)
we substitute
A = √( (4.17)² + (5.96)² + (2 × 4.17 × 5.96 × cos(41.65) )
A = √( 17.3889 + 35.5216 + ( 49.7064 × 0.7472 )
A = √(52.9105 + 37.1406 )
A = √90.0511
A = 9.49 mm
Therefore, Amplitude is 9.49 mm
b) the phase angle (relative to wave 1) of the resultant wave;
tan∅ = A2sinΔ∅ / ( A1 + A2cosΔ∅)
we substitute
tan∅ = 5.96sin41.65 / ( 4.17 + 5.96cos41.65)
tan∅ = 3.96088 / ( 4.17 + 4.4534)
tan∅ = 3.96088 / 8.6234
tan∅ = 0.4593
∅ = tan⁻¹ ( 0.4593 )
∅ = 0.43056 rad = 24.7°
c)
For the third wave, maximum amplitude. it should be in the direction of resultant of A1 and A2
so, phase angle in order to maximize the amplitude of the new resultant wave will be;
phase angle = 2.24x - 300t + 0.43056 rad
Which of the following is NOT a type of acceleration?
O A.
Positive acceleration
B.
Constant velocity
O c.
Change in direction
D.
Negative acceleration
Is cordyceps fungi a predator? If not, what is it?
Answer:
It is a predator.
Explanation:
They eat other insects so therefore I think they're a predator.
A burning candle provides :
a.radiant energy
b.solar energy
c.chemical potential energy
d.thermal energy
A 20 kg cart with frictionless bearings is initially at rest on a horizontal tabletop. The cart is connected by a light string to a hanging 5 kg block. The string passes over an ideal pulley. The system is released from rest and both objects have an acceleration of magnitude a 5 kg
(a) Draw the forces (NOT components) that act on each object.
(b) Calculate the acceleration ay of the system of masses once the system is released.
Answer:
[tex]a=1.96m/s^2[/tex]
Explanation:
From the question we are told that
Mass of cart =20kg
Mass of block =5kg
Magnitude of acceleration M_a= 5 kg
Generally Force on the 20kg mass in the [tex]\bar X[/tex] is mathematically given as
[tex]\sum F=ma\\t=(20kg)a[/tex]
Generally Force on the 2kg in the[tex]\bar Y[/tex] mass is mathematically given as
[tex]\sum F=ma\\(5kg)(9.8m/s^2)-(5kg)a[/tex]
Therefore
[tex](20kg)a=(5kg)(9.8m/s^2)-(5kg)a[/tex]
[tex]a=\frac{5kg)(9.8m/s^2)-(5kg)}{25}[/tex]
[tex]a=1.96m/s^2[/tex]
A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that is stationary. The carts stick together after the collision. What is the final kinetic energy of the system?
Answer:
0.006075Joules
Explanation:
The final kinetic energy of the system is expressed as;
KE = 1/2(m1+m2)v²
m1 and m2 are the masses of the two bodies
v is the final velocity of the bodies after collision
get the final velocity using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
0.12(0.45) + 0/12(0) = (0.12+0.12)v
0.054 = 0.24v
v = 0.054/0.24
v = 0.225m/s
Get the final kinetic energy;
KE = 1/2(m1+m2)v
KE = 1/2(0.12+0.12)(0.225)²
KE = 1/2(0.24)(0.050625)
KE = 0.12*0.050625
KE = 0.006075Joules
Hence the final kinetic energy of the system is 0.006075Joules
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass?
Answer:
The quarterback applies a force of 8.08 N to the ball during the pass.
Explanation:
To find the force we need to calculate the acceleration first so we can use the following equation:
[tex] F = ma [/tex]
Now, we can calculate the acceleration as follows:
[tex] x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2} [/tex]
Since the ball starts from rest [tex]v_{0} = 0[/tex] and [tex]x_{0} = 0[/tex]:
[tex] a = \frac{2*x_{f}}{t^{2}} = \frac{2*1.909 m}{(0.439 s)^{2}} = 19.81 m/s^{2} [/tex]
Now we can find the force:
[tex] F = 0.408 kg*19.81 m/s^{2} = 8.08 N [/tex]
Therefore, the quarterback applies a force of 8.08 N to the ball during the pass.
I hope it helps you!
A man walks 8 m east in 12 seconds . What is the man's velocity ?
Answer:
0.67m/s due east
Explanation:
Given parameters:
Displacement of the man = 8m east
Time taken = 12s
Unknown:
Velocity of the man = ?
Solution:
The velocity of a body is the rate of displacement per time;
Velocity = [tex]\frac{Displacement}{time}[/tex]
Velocity = [tex]\frac{8}{12}[/tex] = 0.67m/s due east
If the speed of an object is doubled, what happens to its kinetic energy?
Answer:
Kinetic energy is energy of motion. Doubling the speed will quadruple the kinetic energy. The relationship between speed and kinetic energy is: This means that the factor by which kinetic energy increases is the square of the factor by which speed or velocity increases for a given object.
Explanation:
PLS HELP WILL MARK AS BRAINLIEST ANSWER
any shot in front of the 3
point line is worth how
many points? *
Answer:
1 point?
Explanation:
prove the identity
Trigonometry grade 10
Answer:
and is in photo given.I didn't get time to type.
Grapevine Co. paid a dividend of $3 on its common stock yesterday. The company's dividends are expected to grow at a constant rate of 9% indefinitely. If the required rate of return on this stock is 13%, compute the current value per share of Grapevine Co. stock.
A. $81.75
B. $47.90
C. $75
D. $56.25
Answer:
the current value per share of Grapevine Co. stock is $ 81.75
Option A) $ 81.75 is the correct answer
Explanation:
Given that;
paid dividend D₀ = $ 3
company's dividends are expected to grow at a constant rate of g = 9% = 0.09
return of stocks r = 13% = 0.13
current value per share of Grapevine Co. stock = ?
Now, to calculate the current value per share of Grapevine Co. stock; we use the expression;
D₀( 1 + g) / ( r - g)
we substitute
current value per share of Grapevine Co. stock = 3( 1 + 0.09) / ( 0.13 - 0.09)
= 3(1.09) / 0.04
= 3.27 / 0.04
= 81.75
Therefore, the current value per share of Grapevine Co. stock is $ 81.75
Option A) $ 81.75 is the correct answer
define what is false
Answer: not according with truth or fact/ incorrect.
Explanation:
5. Did objects have to touch to interact? What causes this?
Answer:
★ For example, a useful analogy for explaining the Earth's gravity force is that the Earth can pull on objects without touching them just like a magnet can affect other objects without touching them. In Addition, the main notion to convey here is that forces can act at a distance with no perceivable substance in between.
Explanation:
Hope you have a great day :)
No, objects have no need to touch for interaction.
Forces that interact with objects without touchingWe know that there are some forces which interact with the objects from far distance without touching such as gravity of the earth. The Earth's gravitational force is that the Earth can pull on objects without touching them so we can conclude that objects have no need to touch for interaction with one another.
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