Answer:
By a factor of 9/4
Explanation:
Applying Coulomb's law,
F = kqq'/r²................... Equation 1
Assuming q and q' are the two point charges respectively.
Where k = coulomb's constant, r = distance between the charges.
When the point charges are 3.0 m apart,
F = kqq'/3²
F = kqq'/9.................... Equation 1
When they are moved to a new distance, 2.00 m
F' = kqq'/2²
F' = kqq'/4................. Equation 2.
Comparing equation 1 and equation 2.
F' = 9F/4
Hence the resulting mutual force change by a factor of 9/4
A ball is thrown at an angle 40.00 above the horizontal with an initial velocity of 22.0 m/s. What is the range of the ball?
Answer:
48.64 m
Explanation:
From the question,
Range(R) = (U²Sin2Ф)/g.................. Equation 1
Where U = initial velocity, Ф = Angle to the horizontal, g = acceleration due to gravity.
Given: U = 22 m/s, Ф = 40°, g = 9.8 m/s².
Substitute these values into equation 1
R = 22²Sin(40×2)/9.8
R = 484×0.9848/9.8
R = 48.64 m
Hence the range of the ball is 48.64 m
us
A 13.3 kg box sliding across the ground
decelerates at 2.42 m/s2. What is the
coefficient of kinetic friction?
(No unit)
Answer:
0.242
Explanation:
m = 13.3 kg
a =d= 2.42 m/s²
g = 10 m/s²
from the laws of friction F = ¶R
===> ¶ = F/R = ma/mg = a/g
¶ = a/g = 2.42/10 = 0.242
Run 2 17. Set # of slits to 2 18. Set Wave Length to 400nm 19. Set Slit width to 1600 nm 20. Set Slit spacing to 5000nm In row 18 21. Record distance to 1st bright fringe 22. Record distance to 2nd bright fringe 23. Record distance to 3rd bright fringe Knowing the screen distance to be 1m 24. Calculated the measured angle to 1st bright fringe 25. Calculated the measured angle to 2nd bright fringe 26. Calculated the measured angle to 3rd bright fringe Using sin(θ)=mλ/d 27. Calculate θ for 1st bright fringe
Answer:
a) m=1, y₁ = 0.08 m , θ₁ = 4.57º , b) m=2, y₂ = 0.16 m , θ₂ = 9.09º , c) m=3, y₃ = 0.24 m , θ₃ = 13.5º
Explanation:
After reading your strange statement, I understand that this is an interference problem, I transcribe the data to have it more clearly. Number of slits 2, distance between slits 5000 nm, wavelength 400 nm, distance to the screen 1 m.
They ask us to calculate the angles for the first, second and third interference, they also ask us to write down the distance from the central maximum.
The expression for constructive interference for two slits is
d sin θ = m λ
where d is the distance between the slits, λ is the wavelength used, m is an integer representing the order of interference
Let's use trigonometry to find the distance from the central maximum
tan θ = y / L
in all interference experiments the angle is small,
tan θ = sin θ / cos θ = sin θ
sint θ = y / L
let's replace
d y / L = m λ
y = m λ L / d
let's calculate
distance to the first maximum m = 1
y₁ = 1 400 10⁻⁹ 1/5000 10⁻⁹
y₁ = 0.08 m
distance to second maximum m = 2
y₂ = 2 400 10⁻⁹ 1/5000 10⁻⁹
y₂ = 0.16 m
distance to the third maximum m = 3
y₃ = 3 400 10⁻⁹ 1/5000 10⁻⁹
y₃ = 0.24 m
with these values we can search for each angle
tan θ = y / L
θ = tan⁻¹ y / L
for m = 1
θ₁ = tan⁻¹ (0.08 / 1)
θ₁ = 4.57º
for m = 2
θ₂ = tan⁻¹ (0.16 / 1)
θ₂ = 9.09º
for m = 3
θ₃ = tan⁻¹ (0.24 / 1)
θ₃ = 13.5º
A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when at terminal speed?
Answer:
The speed of the resistive force is 42.426 m/s
Explanation:
Given;
mass of skydiver, m = 75 kg
terminal velocity, [tex]V_T = 60 \ m/s[/tex]
The resistive force on the skydiver is known as drag force.
Drag force is directly proportional to square of terminal velocity.
[tex]F_D = kV_T^2[/tex]
Where;
k is a constant
[tex]k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}[/tex]
When the new drag force is half of the original drag force;
[tex]F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s[/tex]
Therefore, the speed of the resistive force is 42.426 m/s
At terminal speed, the speed of the resistive force will be:
"42.426 m/s".
Force and speedAccording to the question,
Skydriver's mass, m = 75 kg
Terminal velocity, [tex]V_T[/tex] = 60 m/s
Constant = k
We know the relation,
→ [tex]F_D[/tex] = k[tex]V_T^2[/tex]
here, k = [tex]\frac{F_D_1}{V_T_1^2} = \frac{F_D_2}{V_T_2^2}[/tex]
Now,
[tex]F_D_2[/tex] = [tex]\frac{F_D_1}{2}[/tex]
[tex]\frac{F_D_1}{V_T_1^2}= \frac{F_D_2}{V_T_2^2}[/tex]
[tex]\frac{1}{V_T_1^2} = \frac{1}{2V_T_2^2}[/tex]
By applying cross-multiplication,
[tex]V_T_2^2 = \sqrt{\frac{V_T_1^2}{2} }[/tex]
By substituting the above values,
[tex]V_T_2[/tex] = 0.7071 ([tex]V_T_1[/tex])
= 0.7071 × 60
= 42.426 m/s
Thus the above response is correct.
Find out more information about velocity here:
https://brainly.com/question/742413
Objects floating in the water, like buoys, only bob up and down when waves pass. Why do they not get pushed all the way to wherever the wave goes
Answer:
Because as the waves propagates, the particles of the medium (molecules of water) vibrates perpendicularly (upward and downward) about their mean position and not in the direction of the waves.
Explanation:
A wave is a phenomena which causes a disturbance in a medium without any permanent deformation to the medium. Examples are; transverse wave and longitudinal wave. Waves transfer energy from one point in the medium to another.
The waves generated by water are transverse waves. Which are waves in which the vibrations of the particles of the medium is perpendicular to the direction of propagation of the waves.
Thus as the waves propagates, the molecules of water vibrates up and down and not along the direction of propagation of the waves. So that the floating objects do not get pushed in the direction of the waves every time.
Determine the magnitude of the force between two 11 m-long parallel wires separated by 0.033 m, both carrying 5.2 A in the same direction.
Answer:
[tex]F=1.8\times 10^{-3}\ N[/tex]
Explanation:
We have,
Length of wires is 11 m
Separation between wires is 0.033 m
Current in both the wires is 5.2 A
It is required to find the magnitude of force between two wires. The force between wires is given by :
[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]
So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]
A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matter it passes through. One such particle, initially traveling at 2.40 ✕ 106 m/s in a straight line, decreases in speed to 1.56 ✕ 106 m/s over a distance of 1.22 km.
(a) What is the magnitude of the force experienced by the muon?
(b) How does this force compare to the weight of the muon?
|F|/Fg =______
Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
An ac series circuit contains a resistor of 20 ohms, a capacitor of 0.75 microfarads of 120 x 10-3 H. If an effective (rms) voltage of 120 V is applied, what is the effective (rms) current when the circuit is in resonance
Answer:
The effective (rms) current when the circuit is in resonance is 6 A
Explanation:
Given;
resistance of the resistor, R = 20 ohms
capacitance of the capacitor, C = 0.75 microfarads
inductance of the inductor, L = 0.12 H
effective rms voltage, [tex]V_{rms}[/tex] = 120
At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).
The effective (rms) current, = [tex]V_{rms}[/tex] / R
= 120 / 20
= 6 A
Therefore, the effective (rms) current when the circuit is in resonance is 6 A
When a force of 20.0 N is applied to a spring, it elongates 0.20 m. Determine the period of oscillation of a 4.0-kg object suspended from this spring.
Answer:
1.26 secs.
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Extention (e) = 0.2 m
Mass (m) = 4 Kg
Period (T) =.?
Next, we shall determine the spring constant, K for spring.
The spring constant, K can be obtained as follow:
Force (F) = 20 N
Extention (e) = 0.2 m
Spring constant (K) =..?
F = Ke
20 = K x 0.2
Divide both side by 0.2
K = 20/0.2
K = 100 N/m
Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:
Mass (m) = 4 Kg
Spring constant (K) = 100 N/m
Period (T) =..?
T = 2π√(m/K)
T = 2π√(4/100)
T = 2π x √(0.04)
T = 2π x 0.2
T = 1.26 secs.
Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.
Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what is the value of the spring constant if the spring is compressed from its rest length to a maximum depth of 2.4 m by the train
Answer:
187,500N/mExplanation:
From the question, the kinectic energy of the train will be equal to the energy stored in the spring.
Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².
Equating both we will have;
1/2 mv² = 1/2ke²
mv² = ke²
m is the mass of the train
v is the velocity of then train
k is the spring constant
e is the extension caused by the spring.
Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m
Substituting this values into the formula will give;
30000*4² = k*1.6²
[tex]k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}[/tex]
The value of the spring constant is 187,500N/m
in a certain region of space, the gravitational field is given by -k/r,where r=distance,k=const.if gravitational potential at r=r0 be v0,then what is the expression for the gravitational potential v?
options
1)k log(r/ro)
2)k log(ro/r)
3)vo+k log(r/ro)
4)vo+k log(ro/r)
plz help me out
I will mark u as brainliest if u answer correct
Answer:
The correct answer is option 3 .
Please check the answer once :)
Calculate the value of the translational partition function of O2 at 1000 K for a volume of 1 m3. Submit your answer multiplied by 10^−32 (i.e. if you calculate 4.2e33, input 42).
Answer:
The value of the function is [tex]q__{t }} = 1.878 *10^{35}[/tex]
Explanation:
From the question we are told that
The temperature is [tex]T = 1000 \ K[/tex]
The volume is [tex]V = 1 m^3[/tex]
Generally the transnational partition function is mathematically represented as
[tex]q__{t }} = [\frac{2 * \pi * m * k * T }{ N_a * h} ]^{\frac{3}{2} } * V[/tex]
Where m is the molar mass of oxygen with a constant value of [tex]m = 32 *10^{-3} \ kg/mol[/tex]
k is the Boltzmann constant with a value of [tex]k = 1.38 *10^{-23 } \ J/K[/tex]
[tex]N_a[/tex] is the Avogadro Number with a constant value of [tex]N_a = 6.022 *10^{23} \ atoms[/tex]
h is the Planck's constant with value [tex]h = 6.626 *10^{-34 } \ J\cdot s[/tex]
Substituting values
[tex]q__{t }} = [\frac{2 * 3.142 * 32*10^{-3} * 1.38 *10^{-23} * 1000 }{ 6.022 *10^{23} * [6.626 *10^{-34}] ^2 }]^{\frac{3}{2} } * 1[/tex]
[tex]q__{t }} = 1.878 *10^{35}[/tex]
7. Which statement is true about teens that are in Marcia’s final state of identity formation?
Answer:
D. All of the above
Explanation:
The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.
James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure, moratorium, and achievement.
What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube
Answer:
f = 357.29Hz
Explanation:
In order to calculate the fundamental frequency in the closed tube, you use the following formula:
[tex]f_n=\frac{nv}{4L}[/tex] (1)
n: order of the mode = 1
v: speed of sound = 343m/s
L: length of the tube = 24cm = 0.24m
You replace the values of the parameters in the equation (1):
[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]
The fundamental frequency of in the tube is 357.29Hz
A student stretches an elastic band by 0.8 m in 0.5 seconds. The spring constant of the elastic band is 40 N/m. What was the power exerted by the student
Answer:
The power exerted by the student is 51.2 W
Explanation:
Given;
extension of the elastic band, x = 0.8 m
time taken to stretch this distance, t = 0.5 seconds
the spring constant, k = 40 N/m
Apply Hook's law;
F = kx
where;
F is the force applied to the elastic band
k is the spring constant
x is the extension of the elastic band
F = 40 x 0.8
F = 32 N
The power exerted by the student is calculated as;
P = Fv
where;
F is the applied force
v is velocity = d/t
P = F x (d/t)
P = 32 x (0.8 /0.5)
P = 32 x 1.6
P = 51.2 W
Therefore, the power exerted by the student is 51.2 W
In Einstein's Thoery of Relativity. What did he believe was the relationship between energy and malter?
Explanation:
Einsteins theory of relativity explains how space and time are linked for objects that are moving at a consistent speed in a straight line.
A hunter stands on a frozen pond (frictionless) and fires a 4.20g bullet at 965m/s horizontally.The mass of hunter + gun is 72.5kg. Find the recoil velocity of the hunter.
Answer:
The the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.
Explanation:
Given;
mass of bullet, m₁ = 4.2 g = 0.0042 kg
mass of hunter + gun = 72.5 kg
velocity of the bullet, u = 965 m/s
Momentum of the bullet when it was fired;
P = mv
P = 0.0042 x 965
P = 4.053 kg.m/s
Determine the recoil velocity of the hunter.
Total momentum = sum of the individual momenta
Total momentum = momentum of the bullet + momentum of the hunter
Apply the principle of conservation of momentum, sum of the momentum is equal to zero.
[tex]P_{hunter} + P_{bullet} = 0\\\\P_{hunter} = -P_{bullet}\\\\72.5v = -4.053\\\\v = \frac{-4.053}{72.5} \\\\v = - 0.056 \ m/s\\\\Thus, the \ recoil \ velocity \ of \ the \ hunter \ is \ 0.056 \ m/s, \ in \ opposite \ direction \ of \ the \ bullet.[/tex]
Therefore, the the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.
The two arms of a U-tube are not identical, one having 2.5 times the diameter of the other. A cork in the narrow arm requires a force of 16 N to remove it. The tube is filled with water and the wide arm is fitted with a piston. The minimum force that must be applied to the piston to push the cork out is:
Answer:
F₁ = 100 N
Explanation:
The pressure must be equally transmitted from the piston to the narrow arm. Therefore,
P₁ = P₂
F₁/A₁ = F₂/A₂
F₁/F₂ = A₁/A₂
where,
F₁ = Force Required to be applied to piston = ?
F₂ = Force to push cork at narrow arm = 16 N
A₁ = Area of wider arm = πd₁²/4
A₂ = Area of narrow arm = πd₂²/4
Therefore,
F₁/16 N = (πd₁²/4)/(πd₂²/4)
F₁ = (16 N)(d₁²/d₂²)
but, it is given that the diameter of wider arm is 2.5 times the diameter of the narrow arm.
d₁ = 2.5 d₂
Therefore,
F₁ = (16 N)[(2.5 d₂)²/d₂²]
F₁ = (16 N)(6.25)
F₁ = 100 N
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evporate
Answer:
Gain heat and evaporate
Explanation:
As water molecules are exposed to sunlight, they begin to heat up. This means that the molecules begin to jiggle faster, and as such take up less space. Since they take up less space they are less dense, and therefore more bouyant. This means that they begin to rise into the air, and evaporate. Hope this helps!
We can see Objects because of
A) reflection
B) refraction
C) transmission
D) diffraction
Please help a friend
Answer:
I believe it's A.)
Explanation:
Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.
Hope this helps you out : )
The position of an object moving along the x
axis is given by x(t) = 2t^2+t^3 +1, where x is
in meters and t in seconds. The acceleration of
the object at t = 2 seconds is:
4m/s?
Answer: 16 meters per second
Explanation:
The derivative of the position is the velocity.
The derivative of the velocity is the acceleration.
x(t) = 2t² + t³ + 1
x'(t) = v(t) = 4t + 3t²
x''(t) = v'(t) = a(t) = 4 + 6t
a(2) = 4 + 6(2)
= 4 + 12
= 16
A 15-m rope is pulled taut with a tension of 140 N. It takes 0.545 s for a wave to propagate along the rope. What is the mass of the rope
Answer:
Mass of the rope = 2.8 kg
Explanation:
The speed of waves travelling through a rope with linear density (μ) and under tension T is given as v = √(T/μ)
The speed of waves in the rope is also calculated as
v = (d/t)
d = L = length of the rope = 15 m
t = time taken for the wave to move through the rope = 0.545 s
Speed = v = (15/0.545) = 27.523 m/s
Speed = v = √(T/μ)
T = tension in the rope = 140 N
μ = linear density = ?
27.523 = √(140/μ)
27.523² = (140/μ)
(140/μ) = 757.512
μ = (140/757.512) = 0.1848155556 = 0.1848 kg/m
Linear density = μ = (m/L)
m = mass of the rope = ?
L = length of the rope = 15 m
0.1848 = (m/15)
m = 0.1848 × 15 = 2.77 kg = 2.8 kg to 1 d.p.
Hope this Helps!!!
I WILL MARK YOU AS BRAINLIEST!!! An object is launched straight up into the air with an initial velocity of 40 meters per second, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 /2, after how many seconds will the object hit the ground? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
8.9 seconds
Explanation:
The height of the object at time t is:
y = h + vt − 4.9t²
where h is the initial height, and v is the initial velocity.
Given h = 30 and v = 40:
y = 30 + 40t − 4.9t²
When y = 0:
0 = 30 + 40t − 4.9t²
4.9t² − 40t − 30 = 0
Solving with quadratic formula:
t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)
t = [ 40 ± √(1600 + 588) ] / 9.8
t = 8.9
It takes 8.9 seconds for the object to land.
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
a-velocity
b-mass
c-momentum
d-direction
Answer:
b. Mass
Explanation:
This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.
As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.
Therefore, the correct answer is B. Mass.
This question involves the concept of the law of conservation of momentum.
Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.
The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.
Mathematically,
[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]
where,
m₁ = mass of the first object
m₂ = mass of the second object
u₁ = velocity of the first object before the collision
u₂ = velocity of the second object before the collision
v₁ = velocity of the first object after the collision
v₂ = velocity of the second object after the collision
Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.
Learn more about the law of conservation of momentum here:
brainly.com/question/1113396?referrer=searchResults
The attached picture illustrates the law of conservation of momentum.
Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response with at least 3 reasons with details regarding concepts from the units learned in this course.
Answer:
Improvement in observational, and exploratory technology
Rapid increase in knowledge
International collaboration
Explanation:
Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.
Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.
Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.
International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.
In one of the classic nuclear physics experiments performed by Ernest Rutherford at the beginning of the 20th century, alpha particles (helium nuclei) were shot at gold nuclei and their paths were substantially affected by the Coulomb repulsion from the nuclei. If the energy of the (doubly charged) alpha nucleus was 5.1 MeV, how close to the gold nucleus (79 protons) could it come before being deflected? r =
Answer:
r = 3.8 × 10 ⁻¹⁴ m
Explanation:
given data
alpha nucleus = 5.1 MeV
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C
Kinetic energy of the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) = 9.564 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
solution
we know that when its kinetic energy is equal to the potential energy than alpha particle will deflect \
so
Kinetic energy = potential energy = k q₁q₂ ÷ r ..................1
here r is close distance the alpha particle
so r will be put here value
r = ( 9 × 10⁹ × 3.2 × 10⁻¹⁹ × 1.264 × 10⁻¹⁷ ) ÷ ( 9.564 × 10⁻¹³ )
r = 3.8 × 10 ⁻¹⁴ m
Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?
Answer:
a) E = 2.7x10⁶ N/C
b) F = 54 N
Explanation:
a) The electric field can be calculated as follows:
[tex] E = \frac{Kq}{d^{2}} [/tex]
Where:
K: is the Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
d: is the distance
Now, we need to find the electric field due to charge 1:
[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]
The electric field due to charge 2 is:
[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]
The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):
[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]
Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.
b) The force on a charge q₃ situated there is given by:
[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]
[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]
Therefore, the force on a charge q₃ situated there is 54 N.
I hope it helps you!
(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].
(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.
The answer can be explained as follows.
Electric FieldGiven that the two charges are;
[tex]q_1 = 50\times 10^{-6}\,C[/tex] and [tex]q_2 = -25\times 10^{-6}\,C[/tex](a) At the midpoint; [tex]r = 0.5\,m[/tex].
We know that the electric field due to charge [tex]q_1[/tex].
[tex]E_1 = k\,\frac{q_1}{r^2}[/tex]Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]
[tex]E_1 = (9\times 10^9) \times\frac{(50 \times 10^{-6})}{(0.5)^2}=1.8\times 10^6N/C[/tex]The electric field due to charge [tex]q_2[/tex] is given by;
[tex]E_2 = (9\times 10^9) \times\frac{(-25 \times 10^{-6})}{(0.5)^2}=-9\times 10^5\,N/C[/tex]Therefore, the net electric field in the midpoint is given by;
[tex]E_{net} =E_2+E_1[/tex][tex]\implies E_{net}=1.8 \times 10^6 N/C + 9 \times 10^5\,N/C=2.7\times 10^6\,N/C[/tex]The direction is towards the right side.
Electrostatic Force(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.
So the force on the charge is ;
[tex]F=E_{net} \times q_3=(2.7 \times 10^6\,N/C) \times (20\times 10^{-6}\,C)=54\,N[/tex]Find out more about electrostatic force and fields here:
https://brainly.com/question/14621988
"On earth, you have a pendulum of length L that oscillates with period T. Your friend lives on a planet where the acceleration of gravity is four times as big as it is on the earth. What should be the length of a pendulum on your friend s planet so that it also oscillates with the same period T
Answer:
4L
Explanation:
Data provided in the question according to the question is as follows
Length = L
Gravity = G
For friend
Length = ?
Growth = 4G
Moreover,
[tex]T_1 = T_2[/tex]
Based on the above information ,
Now we have to apply the simple pendulum formula which is shown below:
[tex]T = 2\pi \frac{L}{G}[/tex]
Now equates these equations in both sides
[tex]2\pi \frac{L}{G} = 2\pi \frac{L}{4G}[/tex]
So after solving this, the length of the pendulum is 4L
Answer:
the length of a pendulum on your friend s planet should be 4 times than that on earth
Explanation:
We know that time period of simple pendulum is given by
[tex]T= 2\pi\sqrt{\frac{L}{g} }[/tex]
L= length of pendulum
g= acceleration due to gravity
therefore, Let T_1 and T_2 be the time period of the earth and other planet respectively.
[tex]\frac{T_1}{T_2} =\sqrt(\frac{L_1}{L_2}\times\frac{g_2}{g_1})[/tex]
ATQ
T_1=T_2=T, g_2=4g_1
Putting the values in above equation and solving we get
[tex]\frac{L_1}{L_2} =\frac{1}{4}[/tex]
A generator is connected to a resistor and a 0.049-H inductor in series. The rms voltage across the generator is 7.9 V. When the generator frequency is set to 100 Hz, the rms voltage across the inductor is 2.8 V. Determine the resistance of the resistor in this circuit
Answer:
56.04 ohmsExplanation:
The voltage across the inductor VL = IXL
I is the total current flowing in the circuit and XL is the inductive reactance.
First we need to get the current flowing in the circuit.
From the expression above;
I = VL/XL
Since XL = 2πfL
I = VL/ 2πfL
Given VL = 2.8V, frequancy f = 100Hz and inductance L = 0.049-H
I = 2.8/2π*100*0.049
I = 2.8/30.79
I = 0.091A
Also;
Vrms = VL + VR
VR is the voltage across the resistor.
VR = Vrms - VL
VR = 7.9 - 2.8
VR = 5.1V
Then we can calculate the resistance of the resistor
According to ohms law VR = IR
Since the inductance and resistance ar connected in series, the same current will flow through them.
R = VR/I
R = 5.1/0.091
R = 56.04 ohms
Hence the resistance of the resistor in this circuit is 56.04 ohms
suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the string when the ball is at the highest point
The complete question is missing, so i have attached the complete question.
Answer:
A) FBD is attached.
B) The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
Explanation:
A) I've attached the image of the free body diagram.
B) The formula for the net force is given as;
F_net = mv²/r
We know that angular velocity;ω = v/r
Thus;
F_net = mω²r
Now, the minimum downward force is the weight and so;
mg = m(ω_min)²r
m will cancel out to give;
g = (ω_min)²r
(ω_min)² = g/r
ω_min = √(g/r)
The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).