q = 40 C
e = 1.6×10^-19 C
n = ?
n = q/e
n = 40/1.6×10^-19 C
= 2.6×10^20
A motorcyclist changes his speed from 20 km / h to 100 km / h in 3 seconds, maintaining a constant acceleration in that time interval. If the mass of the motorcycle is 200 kg and that of its rider is 80 kg, what is the value of the net force to accelerate the motorcycle? Help!
Answer:
2000 N
Explanation:
20 km/h = 5.56 m/s
100 km/h = 27.78 m/s
F = ma
F = m Δv/Δt
F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)
F = 2074 N
Rounded to one significant figure, the force is 2000 N.
1. A base-ball of mass 0.3kg approaches the bat at a speed of 30 miles/hour and when the ball hits the bat for 0.5 s, it started to move away from the bat at a speed of 60 miles/hour. Find the impulse
Answer:
I = 27kg.mi/h
Explanation:
In order to calculate the impulse of the ball, you use the following formula:
[tex]I=m\Delta v[/tex] [tex]=m(v-v_o)[/tex] (1)
m: mass of the ball = 0.3kg
v: speed of the ball after the bat hit it = 60mi/h
vo: speed of the ball before the bat hit it = 30mi/h
You replace the values of all parameters in the equation (1):
[tex]I=(0.3kg)(60mi/h-(-30mi/h))=27kg\frac{mi}{h}[/tex]
where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.
The imulpse of the ball is 27 kg.miles/hour
g If the interaction of a particle with its environment restricts the particle to a finite region of space, the result is the quantization of ____ of the particle.
Answer:
the result is the quantization of __Energy__ of the particle
Explanation:
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1) Which of the following describes the electricfield inside this conductor?
A) It is in the same direction as the original external field.
B) It is in the opposite direction from that of the original external field.
C) It has a direction determined entirely by the charge on its surface.
D) It is always zero.
2) The charge density inside the conductor is:______.
a) 0.
b) non-zero;but uniform.
c) non-zero;non-uniform.
d) infinite.
3) Assume that at some point just outside the surface of the conductor, the electric field has magnitudeE and is directed toward the surface of the conductor. What is the charge density eta on the surface of the conductor at that point?
Answer:
1) Option D is correct.
The electric field inside a conductor is always zero.
2) Option A is correct.
The charge density inside the conductor is 0.
3) Charge density on the surface of the conductor at that point = η = -E ε₀
Explanation:
1) The electric field is zero inside a conductor. Any excess charge resides entirely on the surface or surfaces of a conductor.
Assuming the net electric field wasn't zero, current would flow inside the conductor and this would build up charges on the exterior of the conductor. These charges would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
2) Since there are no charges inside a conductor (they all reside on the surface), it is logical that the charge density inside the conductor is also 0.
3) Surface Charge density = η = (q/A)
But electric field is given as
E = (-q/2πε₀r²)
q = -E (2πε₀r²)
η = (q/A) = -E (2πε₀r²)/A
For an elemental point on the surface,
A = 2πrl = 2πr²
So,
η = -E ε₀
Hope this Helps!!!
Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mule and refuses to continue on the walk. Cass yanks on the chain with a 67 N force at an angle of 30° above the horizontal. Determine the horizontal and vertical components of the tension force.
Answer:
Horizontal component: [tex]F_x = 58\ N[/tex]
Vertical component: [tex]F_y = 33.5\ N[/tex]
Explanation:
To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.
Therefore, for the horizontal component, we have:
[tex]F_x = F * cos(angle)[/tex]
[tex]F_x = 67 * cos(30)[/tex]
[tex]F_x = 58\ N[/tex]
For the vertical component, we have:
[tex]F_y = F * sin(angle)[/tex]
[tex]F_y = 67 * sin(30)[/tex]
[tex]F_y = 33.5\ N[/tex]
So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.
An object of mass 2 kg has a speed of 6 m/s and moves a distance of 8 m. What is its kinetic energy in joules?
Answer:
36 JoulesExplanation:
Mass ( m ) = 2 kg
Speed of the object (v) = 6 metre per second
Kinetic energy =?
Now,
We have,
Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]
Plugging the values,
[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]
Reduce the numbers with Greatest Common Factor 2
[tex] = {(6)}^{2} [/tex]
Calculate
[tex] = 36 \: joule[/tex]
Hope this helps...
Good luck on your assignment...
The Kinetic energy of the object will be "36 joules".
Kinetic energyThe excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.
According to the question,
Mass of object, m = 2 kg
Speed of object, v = 6 m/s
As we know the formula,
→ Kinetic energy (K.E),
= [tex]\frac{1}{2}[/tex] × m × v²
By substituting the values, we get
= [tex]\frac{1}{2}[/tex] × 2 × (6)²
= [tex]\frac{1}{2}[/tex] × 2 × 36
= 36 joule
Thus the above answer is appropriate.
Find out more information about Kinetic energy here:
https://brainly.com/question/25959744
A 25-kilogram object is placed on a compression spring, and it creates a displacement of 0.15 meters. What is the weight of an object that creates a displacement of 0.23 m on the same spring? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
I hope it is correct ✌️
1. A current of 0.001 A can be felt by the human body. 0.005 A can produce a pain response. 0.015 A can cause a loss of muscle control. In the procedures of this lesson, over 0.030 A of current traveled in the three-battery circuit. Why was this circuit safe to handle with dry hands?
Answer:
It was safe to handle the circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
Explanation:
Given;
Current of 0.001 A to be felt
Current of 0.005 A can produce a pain response
Current of 0.015 A can cause a loss of muscle control
Total current that traveled in the three-battery circuit = 0.03 A
Thus, we can conclude that, it was safe to handle the above mentioned circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
A 75kg passenger at the bottom of a roller coaster loop that has a radius of 20m. If the roller coaster car is moving 10m/s, what is the apparent weight of the passenger? g
Answer:
The apparent weight of the passenger is 360 N
Explanation:
Given;
The mass of the passenger, m = 75 kg
radius of the loop, r = 20 m
velocity of the roller coaster, v = 10 m/s
Centripetal force acting on this passenger is given as;
[tex]F = \frac{mv^2}{r}[/tex]
where;
F is the centripetal force acting on the passenger
m is the mass of the passenger
v is the velocity of the passenger
r is the radius of the track
[tex]F = \frac{mv^2}{r} \\\\F = \frac{75*10^2}{20} \\\\F = 375 \ N[/tex]
Real weight of the passenger,
W = mg
where;
g is acceleration due to gravity
W = 75 x 9.8
W = 735 N
Apparent weight of the passenger = Real weight - Centripetal force
Apparent weight of the passenger = 735 N - 375 N
Apparent weight of the passenger = 360 N
Therefore, the apparent weight of the passenger is 360 N
If two radio telescope dishes are wired together in the right way, the "D" used in determining the angular resolution is determined by
Answer:
D is determined by distance between the telescopes.
Explanation:
what is meaning of convection
Answer:
Convection is heat transfer through the movement of liquids and gases.
A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is:
Answer:
Total charge = 1800C
Explanation:
Q= IT
I = currentt = timeQ = chargeA car has a mass of 1200 kg and an acceleration of 4 m/s^2. If the friction on the car is 200 N, how much force is the thrust providing?
Answer:
5000N
Explanation:
According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e
∑F = m x a -------------(i)
From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e
∑F = F + Fₓ -------------(ii)
Where;
Fₓ = -200N [negative sign because the friction force opposes motion]
Combine equations(i) and (ii) together to get;
F + Fₓ = m x a
F = ma - Fₓ -------------(iii)
Where;
m = mass of car = 1200kg
a = acceleration of the car = 4m/s²
Now substitute the values of m, a and Fₓ into equation (iii) as follows;
F = (1200 x 4) - (-200)
F = 4800 + 200
F = 5000N
Therefore, the force the thrust is providing is 5000N
A car is traveling down a highway. It was moving with a velocity of 50m/s when the driver reads the speed limit and has to decelerate with an acceleration of -5m/s for 2 seconds. What is the momentum of this 500kg car after it decelerates?
Answer:
20,000 kg m/s
Explanation:
Given:
v₀ = 50 m/s
a = -5 m/s²
t = 2 s
Find: v
v = at + v₀
v = (-5 m/s²) (2 s) + (50 m/s)
v = 40 m/s
p = mv
p = (500 kg) (40 m/s)
p = 20,000 kg m/s
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves
O capture, amplify, and demodulate waves
change the amplitude and frequency of waves
O change the pulse and phase of waves
Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:
The diagram shows a device that uses radio waves.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves capture, amplify, and demodulate waves change the amplitude and frequency of waves change the pulse and phase of wavesAnswer:
2. capture, amplify, and demodulate waves
Explanation:
The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.
The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.
Hence, the correct option is 2.
Answer:
B
Explanation:
edge 2020
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12 + 35 t + 1
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1
Answer:
Velocity = 131 m/s
Speed = 131 m/s
Explanation:
Equation of motion, s = f(t) = 12t² + 35 t + 1
To get velocity of the particle, let us find the first derivative of s
v (t) = ds/dt = 24t + 35
At t = 4
v(4) = 24(4) + 35
v(4) = 131 m/s
Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s
Zuckerman’s test for sensation seeking measures which of the following characteristics?
dangerousness, antisocial traits, “letting loose,’ and intolerance for boredom
thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
adventurousness, physical prowess, creative morality, and charisma
dangerousness, adventurousness, creativity, and thrill and adventure seeking
The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
Explanation:
Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.
A noisy channel needs to transfer 87 kbps, but has a SNR of 11 dB (decibels). Calculate the minimum Bandwidth required , in kHz, according to Shannon.
Answer:
24KHz
Explanation:
See attached file
A street light is at the top of a pole that has a height of 17 ft . A woman 5 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?
Answer:
8 ft/s
Explanation:
This is a straight forward question without much ado.
It is given from the question that she walks with a speed of 8 ft/s
A brick is dropped with zero initial speed from the roof of a building and strikes the ground in 1.90 s. How tall is the building?
Answer:
17.69 m
Explanation:
The time it takes the brick to strike the ground is 1.90 seconds.
We can apply one of Newton's equation of linear motion to find the height of the building:
[tex]s = ut + 0.5gt^2[/tex]
where s = distance (in this case height)
u = initial velocity = 0 m/s
t = time = 1.90 s
g = acceleration due to gravity = 9.8 m/s^2
Therefore:
s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)
s = 0 + 17.68
s = 17.69 m
The height of the building is 17.69 m.
Earth orbiting the Sun The Earth is 1.5 ⋅ 10 8 km from the Sun (on average). How fast is the Earth orbiting the Sun in kilometers per second (on average)? You can assume the orbit of the Earth is a circle and that the circumference of a circle is equal to C = 2 π R where R is the radius of a circle (the distance between the center and the edge. Note that for our purposes, it is perfectly fine to assume π = 3 which allows for a pretty good approximation C = 6 R . Your answer does not need to be put into scientific notation, but if you choose to do so it will be marked correct! kilometers per second
Answer:
1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec
C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km circumference of orbit
v = C / t = 9 * 10E8 km / 3 * 10E7 sec = 30 km / sec = 18 mi/sec
A spherical balloon contains a charge +Q uniformly distributed over its surface. When it has a diameter D, the electric field at its surface has magnitude E. If the balloon is not blown up to thrice this diameter without changing the charge, the electric field at its surface is?
Answer:
E = 1/9 E₀
Explanation:
In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression
we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge
For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere
Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q
A = 4π r²
E 4π r² = Q /ε₀
E = 1 /4πε₀ Q / r²
the value of the indicated distance is 3 times the initial diamete
r = 3 D / 2
we substitute
E = 1/4 πε₀ Q (2/ 3D)²
for the initial conditions
E₀ = 1 / 4πε₀ Q (2/D)²
subtitled in the equation above
E = 1/9 E₀
The only factor connecting horizontal and vertical components of projectile motion is _____.
Answer:
VelocityExplanation:
When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.
Along the horizontal Ux = Ucos(theta)
Along the vertical Uy = Ucos(theta)
Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.
This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.
A hungry 177 kg lion running northward at 81.8 km/hr attacks and holds onto a 32.0 kg Thomson's gazelle running eastward at 59.0 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.
Answer:
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
Explanation:
Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:
[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]
Where:
[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.
[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.
[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.
After using the definition of momentum, the system is expanded:
[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]
Vectorially speaking, the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]
Where:
[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.
[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.
If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]
[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]
The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:
[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]
[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
The bulldog and skateboard have a combined mass of 20 kg. In case B (the middle of the three pictures of the bulldog and the well), the bulldog and skateboard have a KE of 380 J at the bottom of the well. How deep is the well in meters?
Answer:
h = 1.94 m
Explanation:
When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:
Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board
K.E = P.E
K.E = mgh
h = K.E/mg
where,
h = depth of well = ?
K.E = Kinetic Energy at bottom = 380 J
m = mass of bull dog and skate board = 20 kg
g = 9.8 m/s²
Therefore,
h = 380 J/(20 kg)(9.8 m/s²)
h = 1.94 m
A woman who weighs 500 N stands on an 8 m long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times bigger than the support force at the left end. How far from the right end is the woman standing
Answer:
[tex]1.6\; \rm m[/tex].
Explanation:
Let [tex]x[/tex] denote the distance (in meters) between the person and the right end of the board.
To keep the calculations simple, consider another unknown: let [tex]y[/tex] denote the support force (in Newtons) on the left end. The support force on the right end of this board would be [tex]3 \, y[/tex] (also in Newtons.)
Now there are two unknowns. At least two equations will be required for finding the exact solutions. For that, consider this board as a lever, but with two possible fulcrums. Refer to the two diagrams attached. (Not to scale.)
In the first diagram, the support at the left end of the board is considered as the fulcrum. In the second diagram, the support at the right end of the board is considered as the fulcrum.Calculate the torque in each situation. Note that are four external forces acting on this board at the same time. (Two support forces and two weights.) Why does each of the two diagrams show only three? In particular, why is the support force at each "fulcrum" missing? The reason is that any force acting on the lever at the fulcrum will have no direct impact on the balance between torques elsewhere on the lever. Keep in mind that the torque of each force on a lever is proportional to [tex]r[/tex], the distance between the starting point and the fulcrum. Since that missing support force starts right at the fulcrum, its [tex]r[/tex] will be zero, and it will have no torque in this context.
Hence, there are three (non-zero) torques acting on the "lever" in each diagram. For example, in the first diagram:
The weight of the board acts at the center of the board, [tex](1/2) \times 8\; \rm m = 4\; \rm m[/tex] from the fulcrum. This force will exert a torque of [tex]\tau(\text{weight of board}) = 4\; \rm m \times (-100\; \rm N) = (-400\; \rm N \cdot m)[/tex] on this "lever". The negative sign indicates that this torque points downwards.The weight of the person acts at [tex]x\;\rm m[/tex] from the right end of the board, which is [tex](8 - x)\; \rm m[/tex] from the fulcrum at the other end of this board. This force will exert a torque of [tex]\tau(\text{weight of person}) = (8 - x)\; {\rm m \times (-500\; \rm N)} = (-500\, \mathnormal{(8 - x)})\; \rm N \cdot m[/tex] on this "lever". This torque also points downwards.The support on the right end of the board acts at [tex]8\; \rm m[/tex] from the fulcrum (i.e., the left end of this board.) This force will exert a torque of [tex]\tau(\text{support, right}) = 8\; {\rm m} \times (3\, \mathnormal{y})\; {\rm N} = (24\, y)\; \rm N \cdot m[/tex] on the "lever". This torque points upwards.If the value of [tex]x[/tex] and [tex]y[/tex] are correct, these three torques should add up to zero. That is:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\, (8 - x))}_{\text{person}} + \underbrace{24\, y}_{\text{support}} = 0[/tex].
That gives the first equation of this system. Similarly, a different equation can be obtained using the second diagram:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\,x)}_{\text{person}} + \underbrace{8\, y}_{\text{support}} = 0[/tex].
Combine these two equations into a two-by-two system. Solve the system for [tex]x[/tex] and [tex]y[/tex]:
[tex]\left\lbrace\begin{aligned}&x = 1.6\\ &y = 150\end{aligned}\right.[/tex].
In other words, the person is standing at about [tex]1.6\; \rm m[/tex] from the right end of the board. The support force at the left end of the board is [tex]150\; \rm N[/tex].
A diver wants to jump from a board, the initial height is 10 meters and he wants to reach a horizontal distance of 2 meters. What minimum speed must he have when jumping from the board to achieve his goal?
Answer:
1.4 m/s
Explanation:
The minimum speed will be when the diver's initial velocity is horizontal.
First, find the time it takes for the diver to fall 10 meters.
Given:
Δy = 10 m
v₀ᵧ = 0 m/s
aᵧ = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
10 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.43 s
Now find the initial horizontal velocity.
v = (2 m) / (1.43 s)
v = 1.4 m/s
Two small charged spheres are 7.59 cmcm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now?
Answer:
The two small charged spheres are now 4.382 cm apart
Explanation:
Given;
distance between the two small charged sphere, r = 7.59 cm
The force on each of the charged sphere can be calculated by applying Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
F is the force on each sphere
q₁ and q₂ are the charges of the spheres
r is the distance between the spheres
[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]
Therefore, the two small charged spheres are now 4.382 cm apart.
Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
W = 12.96 J
A student in her physics lab measures the standing-wave modes of a tube. The lowest frequency that makes a resonance is 30 Hz. As the frequency is increased, the next resonance is at 90 Hz.
What will be the next resonance after this?
Answer:
The next resonance will be 150 Hz.
Explanation:
The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.
The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.
Frecuency of the standing sound wave modes in a open-closed tube is:
fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)
The next resonance is at 90 Hz. This means that it occurs when n = 3:
f₃=3*30 Hz= 90 Hz
This means that the next resonance occurs when n = 5:
f₅=5*30 Hz= 150 Hz
The next resonance will be 150 Hz.