Answer:
The voltage across the capacitor = 0.8723 V
Explanation:
From the question, it is said that the coil is quickly pulled out of the magnetic field. Therefore , the final magnetic flux linked to the coil is zero.
The change in magnetic flux linked to this coil is:
[tex]\delta \phi = \phi _f - \phi _i[/tex]
= 0 - BA cos 0°
[tex]\delta \phi =-BA \\ \\ \delta \phi =-B( \pi r^2) \\ \\ \delta \phi =(1.0 \ mT)[ \pi ( \frac{d}{2} )^2][/tex]
[tex]\delta \phi =(1.0 \ mT)[ \pi ( \frac{0.01}{2} )^2][/tex]
[tex]\delta \phi = -7.85*10^8 \ Wb[/tex]
Using Faraday's Law; the induced emf on N turns of coil is;
[tex]\epsilon = N |\frac{ \delta \phi}{\delta t} |[/tex]
Also; the induced current I = [tex]\frac{ \epsilon}{R}[/tex]
[tex]\frac{ \delta q}{ \delta t} = \frac{1}{R} (N|\frac{\delta \phi}{\delta t } |)[/tex]
[tex]\delta q = \frac{1}{R} N|\delta \phi |[/tex]
[tex]\delta q = \frac{1}{0.30} (10)| -7.85*10^8 \ Wb |[/tex]
[tex]\delta q = 2.617*10^{-6} \ C[/tex]
The voltage across the capacitor can now be determined as:
[tex]\delta \ V =\frac{ \delta q}{C}[/tex]
= [tex]\frac{ 2.617*10^{-6} \ C}{3.0 \ *10^{-6} F}[/tex]
= 0.8723 V
The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m^2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity.
A. What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity?
B. What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity?
Note: Complete Question:
The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is
β=10log(II0)dB,
where I0 is a reference intensity. For sound waves, I0 is taken to be 10−12W/m2. Note that log refers to the logarithm to the base 10.
Part A
What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity (i.e., I=10I0)?
Part B
What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity (i.e. I=100I0)?
Express the sound intensity numerically to the nearest integer.
Concepts and reason
The concept required to solve this problem is decibel scale of sound intensity.
Use the formula of sound intensity level in decibels and substitute the value of intensity to calculate decibels for all the parts.
Answer:
Find the given 2 attachments for complete solution. Thanks
g I can test a new wheel design by rolling it down a test ramp. I release a wheel of mass m=1.6 kg and radius r=0.37 m from rest at an initial height of h=6.7 m at the top of a test ramp. It rolls smoothly to the bottom without sliding. I measure the linear speed of the wheel at the bottom of the test ramp to be v=4.7 m/s. What is the rotational inertia of my wheel?
Answer:
The rotational inertia of my wheel is [tex]I =1.083 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the wheel is [tex]m = 1.6 \ kg[/tex]
The radius of the wheel is [tex]r = 0.37 \ m[/tex]
The height is [tex]h = 6.7 m[/tex]
The linear speed is [tex]v = 4.7 m/s[/tex]
According to the law of energy conservation
[tex]PE = KE + KE_R[/tex]
Where PE is the potential energy at the height h which is mathematically represented as
[tex]PE = mgh[/tex]
While KE is the kinetic energy at the bottom of height h
[tex]KE = \frac{1}{2} mv^2[/tex]
Where [tex]KE_R[/tex] is the rotational kinetic energy which is mathematically represented as
[tex]KE_R = \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
Where [tex]I[/tex] is the rotational inertia
So substituting this formula into the equation of energy conservation
[tex]mgh = \frac{1}{2} mv^2 + \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
=> [tex]I =[ \ mgh - \frac{1}{2} mv^2 \ ]* \frac{2 r^2}{v^2}[/tex]
substituting values
[tex]I =[ \ 1.6 * 9.8 * 6.7 - \frac{1}{2} * 1.6 *4.7^2 \ ]* \frac{2 * 0.37^2}{4.7^2}[/tex]
[tex]I =1.083 \ kg \cdot m^2[/tex]
An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the ground. A friend picks it up and throws it upward at 20 m/s at the instant the passenger is 5 m above the ground. (10 pts) a) Calculate the time for camera to reach passenger. b) Calculate the position and velocity of the camera when passenger catches it.
Answer:t=0.3253 s
Explanation:
Given
speed of balloon is [tex]u=3\ m/s[/tex]
speed of camera [tex]u_1=20\ m/s[/tex]
Initial separation between camera and balloon is [tex]d_o=5\ m[/tex]
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is
[tex]s=ut[/tex]
[tex]s=3\times t[/tex]
and distance travel by camera to reach balloon is
[tex]s_1=ut+\frac{1}{2}at^2[/tex]
[tex]s_1=20\times t-\frac{1}{2}gt^2[/tex]
Now
[tex]\Rightarrow s_1=5+s[/tex]
[tex]\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t[/tex]
[tex]\Rightarrow 5t^2-17t+5=0[/tex]
[tex]\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}[/tex]
[tex]\Rightarrow t=\dfrac{17\pm 13.747}{10}[/tex]
[tex]\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s[/tex]
There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is [tex]t=0.3253\ s[/tex]
velocity is given by
[tex]v=u+at[/tex]
[tex]v=20-10\times 0.3253[/tex]
[tex]v=16.747\ m/s[/tex]
and position of camera is same as of balloon so
Position is [tex]=5+3\times 0.3253[/tex]
[tex]=5.975\approx 6\ m[/tex]
Final answer:
The time for the camera to reach the passenger is 0.29 seconds. The position of the camera when the passenger catches it is 4.93 meters and the velocity of the camera at that moment is 17 m/s.
Explanation:
To calculate the time for the camera to reach the passenger, we first need to find the time it takes for the passenger to reach the height of the camera. The passenger is moving upward at a constant speed of 3 m/s, so it will take her 5 / 3 = 1.67 seconds to reach the height of the camera. Since the camera was thrown upward at 20 m/s, we can subtract the passenger's upward velocity to find the relative velocity of the camera with respect to the passenger: 20 - 3 = 17 m/s.
Using the relative velocity, we can calculate the time it takes for the camera to reach the passenger as follows: t = distance / relative velocity = 5 / 17 = 0.29 seconds.
b) To calculate the position of the camera when the passenger catches it, we can multiply the relative velocity by the time it takes for the camera to reach the passenger: position = relative velocity * time = 17 * 0.29 = 4.93 meters. The velocity of the camera when the passenger catches it will be the same as the relative velocity: 17 m/s.
atoms with electronegativity differences higher than 1.7 generally form ionic bonds. true or false
Answer: True
Explanation:
A covalent bond is defined as the bond which is formed when there is sharing of electrons and the atoms have electronegative difference between the elements less than 1.7. Example: [tex]H_2[/tex]
Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal. The electronegative difference between the elements is more than 1.7. Example: [tex]NaCl[/tex]
Which resistors in the circuit must always have the same current?
Answer:
Resistors in series in the circuit must always have the same current
Explanation:
Resistors are said to be connected in series if they are connected one after another.
The total resistance in the circuit with resistors connected in series is equal to the sum of individual resistances.
Individual resistors in series do not get the total source voltage. Total source voltage divide among them.
Answer:
it's a and d
Explanation:
because it all goes around if that makes sense.. and if u guys are here from apex
Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows a crystalline slab (refractive index 1.665) with a rectangular cross section. A ray of light strikes the slab at an incident angle of 1 = 37.0°, enters the slab, and travels to point P. This slab is surrounded by a fluid with a refractive index n. What is the maximum value of n such that total internal reflection occurs at point P?
Answer:
n = 1.4266
Explanation:
Given that:
refractive index of crystalline slab n = 1.665
let refractive index of fluid is n.
angle of incidence θ₁ = 37.0°
Critical angle [tex]\theta _c = sin^{-1} (\frac{n}{n_{slab}} )[/tex]
[tex]sin \theta _ c =\frac{n}{n_{slab}}[/tex]
According to Snell's law of refraction:
[tex]n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)[/tex]
At point P ; [tex]90 - \theta _2 \leq \theta _c[/tex]
[tex]\theta _2 = 90 - \theta _c[/tex]
Therefore:
[tex]n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}[/tex]
Then maximum value of refractive index n of the fluid is:
[tex]n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }[/tex]
[tex]n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }[/tex]
n = 1.4266
A 0.28-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 500 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = + 0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = -0.052 m, find the acceleration of the block.
Answer:
The block will accelerate at 92.86m/s²
Explanation:
The acceleration of a simple harmonic motion of a spring is expressed as
a= - kx/m
Where k = spring constant
x= displacement
m= mass of block
Given data
Spring constant k = 500N/m
Displacement x= - 0.052m
Mass of block m= 0.28kg
Pluging this parameters into the expression for acceleration we have
a= - 500*(-0.052)/0.28
a= 26/0.28
a= 92.86m/s²
Answer:
The block will accelerate at 92.86m/s²
Explanation:
QUESTIONS
A 800 kg person sprints up a 700 meter high stairway in 4.00 seconds flat. What is their power output in watts?
Answer:
Power;P = 1372000 W
Explanation:
We are given;
Mass of person;m = 800 kg
Distance sprinted;h = 700m
Time taken;t = 4 seconds
Now, the formula for Power is;
Power = Work done/Time taken
Where work done = Force x Distance
Now, Force is expressed as;
F = mg
Where m is mass and g is acceleration due to gravity = 9.8 m/s²
Thus, plugging in the relevant values into power equation gives;
Power;P = 800 x 9.8 x 700/4
Power;P = 1372000 W
The portion of a cello string between the bridge and upper end of the fingerboard (that part of the string that is free to vibrate) is of length 60.0 cm, and this length of the string has mass 2.00 g. The string sounds an A4 note (440 Hz) when played.
Where must the cellist put a finger (what distance x from bridge) to play a D5 note (587 Hz)? For both notes, the string vibrates in its fundamental mode.
Answer:
Explanation:
length of vibration l = .6 m
mass per unit length m = 2 x 10⁻³ / .6
= 3.33 x 10⁻³ kg/ m
n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]
n is frequency of vibration , l is length , T is tension in the string .
Apply this formula in the first case
440 = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]
Apply this formula for second case
n = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]
587 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]
Dividing
[tex]\frac{440}{587}[/tex] = [tex]\frac{l}{.6}[/tex]
l = .45 m .
The distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.
The given parameters;
length o the string at A4, l = 60 cmmass, m = 2.0 gfrequency of A4 note, f = 440 Hzfrequency of D5 note, f = 5587 HzThe frequency of a sound wave in a stretched string is calculated as;
[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
T is the tension in the stringμ is the mass per unit length[tex]f_1(2l_1) = f_2(2l_2)\\\\f_1l_1 = f_2l_2\\\\l_2 = \frac{f_1l_1}{f_2} \\\\l_2 = \frac{440 \times 0.6}{587} \\\\l_2 = 0.45 \ m\\\\l_2 = 45 \ cm[/tex]
Thus, the distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.
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There are several types of drag on a car other than air resistance. Effects having to do with the squeezing of the tires (rolling resistance) and frictional forces in the drivetrain (the system that transfers energy from the engine to the rotation of the wheels) also must be taken into account. Engineers use the following equation to model the total force due to these different effects Fdrag=A+Bv+Cv2 For a Accord, these coefficients are estimated to be A=220.500 N, B=−5.930 N s/m, and C=0.611 N s2/m2. Suppose that the driver steadily accelerates the car from 0 km/hr to 100 km/hr over a 3.5 s. What is the magnitude of the work done by the drag forces?]
Answer:[tex]W=16.837\ kJ[/tex]
Explanation:
Given
Drag force is given by
[tex]F_{drag}=A+Bv+Cv^2[/tex]
for [tex]A=220.5\ N[/tex]
[tex]B=-5.93\ N-s/m[/tex]
[tex]C=0.611N-s/m^2[/tex]
car accelerate from 0 to [tex]100\ km/hr[/tex] in [tex]3.5\ s[/tex]
so acceleration is given by
[tex]v=u+at[/tex]
here u=initial velocity is zero
[tex]v=100\km/hr\approx 27.78\ m/s[/tex]
[tex]27.78=0+a(3.5)[/tex]
[tex]a=7.936\ m/s^2[/tex]
Now work done is given by
[tex]dW=F\cdot vdt[/tex]
[tex]\int_{0}^{W}dW=\int_{0}^{3.5}F\cdot vdt[/tex]
[tex]W=\int_{0}^{3.5}[Av+Bv^2+Cv^3]dt[/tex]
[tex]W=\int_{0}^{3.5}[220.5at-5.93a^2t^2+0.611a^3t^3]dt[/tex]
[tex]W=\int_{0}^{3.5}220.5\times 7.936tdt-\int_{0}^{3.5}5.93\times (7.936)^2t^2dt+\int_{0}^{3.5}0.611\times (7.936)^3t^3dt[/tex]
[tex]W=1749.88[\frac{t^2}{2}]_0^{3.5}-373.47[\frac{t^3}{3}]_0^{3.5}+305.389[\frac{t^4}{4}]_0^{3.5}[/tex]
[tex]W=10,718.015-5337.508+11,456.85[/tex]
[tex]W=16.837\ kJ[/tex]
The work done by drag forces on a car accelerating from 0 to 100 km/hr can be calculated using the equation for total force and its given coefficients, by first converting the velocity to m/s, then finding the average velocity to use it to find the drag force, and finally using the net force and distance covered to calculate the work done.
Explanation:The work done by the drag forces on a car accelerating steadily from 0 km/hr to 100 km/hr over 3.5 s can be calculated using the given equation for total force due to different effects, Fdrag=A+Bv+Cv2, and the coefficients for an Accord car. Please note that before calculating, the velocity needs to be converted from km/hr to m/s as SI units should be consistent.
First, find the average velocity of the car during its acceleration period, which is (0 + final velocity)/2. Then, substitute the values in the given equation Fdrag=A+Bv+Cv2 to find out the Net Force acting on the car.
Next, use this Net Force to find the Work done against the drag forces during the acceleration. The Work done (W) against a force is calculated by the equation W = F * d, where F is the net force (drag force) acting on the body and d is the total distance covered during the acceleration. To find the distance covered by the car, replace 's' in the equation s = ut + 1/2*a*t^2 with the average velocity of the car times the total time (t) it took to accelerate.
This will give you the total work done against drag forces on accelerating the car from 0 km/hr to 100 km/hr over 3.5 s.
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While Dr. Chesnutt is making breakfast, she turns on her 1060 W toaster, 500 W coffee pot, and 1230 W microwave at the same time. Part Description Answer Save Status A. If all three appliances are connected in parallel across a 120 V power source, what total current would they draw from the source? (include units with answer) Format Check Click here to check your answer 4 pts.100% 13% try penalty 8 try limit # tries: 0
Answer:23.25 A
Explanation:
Given
Rating of toaster [tex]P_1=1060\ W[/tex]
Coffee pot [tex]P_2=500\ W[/tex]
microwave [tex]P_3=1230\ W[/tex]
Voltage applied [tex]V=120\ V[/tex]
if they are connected in parallel then all three operates at same voltage
so their resistance are
[tex]P=\frac{V^2}{R}[/tex]
thus [tex]R_1=\frac{V^2}{P_1}=\frac{120^2}{1060}[/tex]
[tex]R_1=13.58\ \Omega[/tex]
[tex]R_2=\frac{V^2}{P_2}=\frac{120^2}{500}[/tex]
[tex]R_2=28.8\ \Omega[/tex]
[tex]R_3=\frac{V^2}{P_3}=\frac{120^2}{1230}[/tex]
[tex]R_3=11.707\ \Omega[/tex]
and [tex]V=IR[/tex]
where I=current
thus [tex]I_1=\frac{V}{R_1}=\frac{120}{13.58}[/tex]
[tex]I_1=8.83\ A[/tex]
[tex]I_2=\frac{V}{R_2}=\frac{120}{28.8}[/tex]
[tex]I_2=4.16\ A[/tex]
[tex]I_3=\frac{V}{R_3}=\frac{120}{11.707}[/tex]
[tex]I_3=10.25\ A[/tex]
Total current [tex]I=I_1+I_2+I_3=23.24\ A[/tex]
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.90 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
How fast are the waves traveling?
What is the amplitude A of each wave?
Answer:
Velocity=1.1m/s
Amplitude=0.35m
Explanation:
Given:
time 't' = 2.9s
wavelength 'λ'= 5.5m
distance 'd'=0.7m
The time period 't' is the time b/w two successive waves. Therefore, the time it takes from the boat to travel from its highest point to its lowest is a half period.
So, T = 2 x 2.9 => 5.8 s
As we know that frequency is the reciprocal of time period, we have
f= 1/T = 1/5.8 =>0.2 Hz
In order to find how fast are the waves traveling, the velocity is given by
Velocity = f λ
V= 0.2 x 5.5 =>1.1m/s
The distance between the boat's highest point to its lowest point is double the amplitude.
Therefore , we can write
Amplitude 'A'= d/2 =>0.7/2 =>0.35m
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.
a. Is angular momentum conserved?
b. Find the change in kinetic energy of the block, in J.
c. How much work was done in pulling the cord? in J.
Answer:
W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]
Explanation:
a) Yes, In the absence of external torques acting on the system, the angular momentum is conserved.
b) By the law of conservation of energy angular momentum
[tex]L_1=L_2[/tex]
[tex]I_1\omega_1=I_2\omega_2[/tex]
[tex]mr_1^2\omega_1=mr_2^2\omega_2\\\omega_2=(\frac{r_1}{r_2} )^2\omega_1[/tex]
[tex]\omega_2=(\frac{0.3}{0.15})^2\times2.85[/tex]
[tex]\omega_2=5.7\text{ rad/sec}[/tex]
c) work done in pulling the chord W= Final kinetic energy(K_2)-Initial Kinetic energy(K_1)
[tex]K_1=\frac{1}{2} mr_1^2\omega_1^2[/tex]
[tex]K_1=\frac{1}{2} \times0.025\times0.3^2\times2.85^2[/tex]
[tex]=9.12\times10^{-3} J[/tex]
Now,
[tex]K_2=\frac{1}{2} mr_2^2\omega_2^2[/tex]
[tex]K_2=\frac{1}{2} \times0.025\times0.15^2\times5.7^2[/tex]
[tex]K_2=18.24\times10^{-3}[/tex] J
Therefore, Work done W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]
a. Angular momentum is conserved in this system. b. The change in kinetic energy of the block can be calculated. c. No work is done in pulling the cord.
Explanation:a. Angular momentum is conserved when there are no external torques acting on the system. In this case, since the block is revolving on a frictionless surface, and there is no mention of any external torques, we can assume that angular momentum is conserved.
b. The change in kinetic energy of the block can be calculated using the equation ΔKE = KE_final - KE_initial. Since the block is modeled as a particle, its kinetic energy is given by KE = 1/2 * m * v^2, where m is the mass and v is the linear velocity. As the radius is changed, the linear velocity changes, and we can calculate the change in kinetic energy.
c. The work done in pulling the cord can be calculated using the equation W = ΔKE + ΔPE, where W is the work done, ΔKE is the change in kinetic energy, and ΔPE is the change in potential energy. In this case, since the block is on a frictionless surface, there is no change in potential energy, and we only need to calculate the change in kinetic energy.
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slab of ice floats on water with a large portion submerged beneath the water surface. The slab is in the shape of a rectangular solid. The volume of the slab is 20 m3 and the surface area of the top is 14 m2. The density of ice is 900 kg/m3 and sea water is 1030 kg/m3. No need to show work. a) Calculate the percentage of the volume of the ice that is submerged. b) Calculate the height, in meters, of the portion of the ice that is submerged. c) Calculate the buoyant force acting on the ice. d) Assume a polar bear has a mass of 400 kg. Calculate the maximum number of polar bears that could be supported by the slab without the slab sinking below the surface of the water.
Answer:
a) [tex]\%V = 87.36\,\%[/tex], b) [tex]x = 1.248\,m[/tex], c) [tex]F_{B} = 176488.341\,N[/tex], d) Six polar bears.
Explanation:
a) The slab of ice is modelled by the Archimedes' Principles and the Newton's Laws, whose equation of equilibrium is:
[tex]\Sigma F =\rho_{w}\cdot g \cdot A \cdot x-\rho_{i}\cdot g\cdot V = 0[/tex]
The height of the ice submerged is:
[tex]\rho_{w}\cdot A \cdot x = \rho_{i}\cdot V[/tex]
[tex]x = \frac{\rho_{i}\cdot V}{\rho_{w}\cdot A}[/tex]
[tex]x = \frac{\left(900\,\frac{kg}{m^{3}}\right)\cdot (20\,m^{3})}{\left(1030\,\frac{kg}{m^{3}} \right)\cdot (14\,m^{2})}[/tex]
[tex]x = 1.248\,m[/tex]
The percentage of the volume of the ice that is submerged is:
[tex]\%V = \frac{(1.248\,m)\cdot (14\,m^{2})}{20\,m^{3}} \times 100\,\%[/tex]
[tex]\%V = 87.36\,\%[/tex]
b) The height of the portion of the ice that is submerged is:
[tex]x = 1.248\,m[/tex]
c) The buoyant force acting on the ice is:
[tex]F_{B} = \left(1030\,\frac{kg}{m^{3}} \right)\cdot (1.248\,m)\cdot (14\,m^{2})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]F_{B} = 176488.341\,N[/tex]
d) The new system is modelled after the Archimedes' Principle and Newton's Laws:
[tex]\Sigma F = -n\cdot m_{bear}\cdot g-\rho_{i}\cdot V \cdot g + \rho_{w}\cdot V\cdot g = 0[/tex]
The number of polar bear is cleared in the equation:
[tex]n\cdot m_{bear} = (\rho_{w} - \rho_{i})\cdot V[/tex]
[tex]n = \frac{(\rho_{w}-\rho_{i})\cdot V}{m_{bear}}[/tex]
[tex]n = \frac{\left(1030\,\frac{kg}{m^{3}} - 900\,\frac{kg}{m^{3}} \right)\cdot (20\,m^{3})}{400\,kg}[/tex]
[tex]n = 6.5[/tex]
The maximum number of polar bears that slab could support is 6.
The slab of ice floating in water is analyzed in terms of the volume submerged, height of submerged portion, buoyant force, and maximum number of polar bears it can support.
Explanation:a) Percentage of the volume submerged:The volume of the submerged portion can be calculated using the formula:
Volume submerged = Volume of ice × (Density of ice / Density of water)
The percentage of the volume submerged can then be calculated by dividing the volume submerged by the total volume of the ice and multiplying by 100.
The height can be calculated by dividing the volume submerged by the surface area of the top.
The buoyant force is equal to the weight of the water displaced by the ice. It can be calculated using the formula:
Buoyant force = Volume of submerged portion × Density of water × Gravitational acceleration
The maximum number of polar bears that can be supported by the slab can be calculated by dividing the buoyant force by the weight of a single polar bear. To prevent the slab from sinking, the buoyant force must be equal to or greater than the weight of the polar bear.
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Sarah is 14 years old and skips school twice a week without any written explanation. What can she be charged with?
Answer:
Too many people are unaware or indifferent to that.” Fines can cost up to $500 per truancy, due within 30 days unless a judge gives an extension. For many students and families, it's another debt they can't pay. And if fines aren't paid, they can convert into an arrest warrant when a student turns 17.
Explanation:
A baseball is thrown a distance of 18m. What is its speed if it takes 0.5 seconds to cover the distance?
Answer:36m/s
Explanation:
Distance=18m
time=0.5 seconds
speed=distance ➗ time
Speed=18 ➗ 0.5
Speed=36m/s
The question is about calculating the speed at which a baseball was thrown given that it traveled a distance of 18 meters in 0.5 seconds. By using the speed formula (speed = distance/time), we find that the baseball was thrown at a speed of 36 meters per second.
Explanation:To calculate the speed of the baseball, we need to use the formula for speed which is speed = distance/time. Here, the distance covered by the baseball is 18m and the time taken is 0.5 seconds.
Substituting these values into the formula, we get speed = 18m / 0.5s = 36 m/s. So, the speed of the baseball is 36 meters per second.
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The electron transport chain (ETC), or respiratory chain, is linked to proton movement and ATP synthesis. Select the STATEMENTS that ACCURATELY describe the electron transport chain. (True/False)
a)Electron transfer in the ETC is coupled to proton transfer form the matrix to the intermembrane space.b)Electrons generated by he citric acid cycle in the mitochondrial matrix enter the ETC.c)The outer membrane of mitochondria is readily permeable to small molecules and hydrogen ions.d)Electron carriers in the mitochondrial matrix include ubiquinone (coenzyme Q), FMN, and cytochrome c.e)Prosthetic groups, such as iron-sulfer centers, are directly involved with electron transfer.f)Electron carriers are organized into four complexes of proteins and prosthetic groups.g)The reactions of the ETC take place in the outer membrane of mitochondria.
Final answer:
The electron transport chain (ETC) occurs in the inner mitochondrial membrane and involves protein complexes and mobile carriers to produce ATP. Electrons from the citric acid cycle enter the ETC, leading to proton movement and ATP synthesis. The outer membrane is permeable to small molecules, but the ETC components, including carriers such as ubiquinone and cytochrome c, are located in the inner membrane.
Explanation:
The electron transport chain (ETC) is a critical step in cellular respiration, taking place in the inner mitochondrial membrane of eukaryotic cells. The process involves several protein complexes and mobile carriers that facilitate the transfer and stepwise release of energy from reduced substrates like NADH and FADH₂ to produce ATP via oxidative phosphorylation.
a) True: Electron transfer in the ETC is indeed coupled to proton transfer from the matrix to the intermembrane space.b) True: Electrons generated by the citric acid cycle in the mitochondrial matrix do enter the ETC.c) True: The outer membrane of mitochondria is permeable to small molecules and ions, although hydrogen ions' passage is more specifically regulated by the ETC.d) False: Electron carriers such as ubiquinone and cytochrome c are embedded in the inner mitochondrial membrane, not in the matrix.e) True: Prosthetic groups such as iron-sulfur centers are indeed directly involved with electron transfer within the ETC.f) False: The ETC consists of four, not three, complexes of proteins and prosthetic groups.g) False: The reactions of the ETC take place in the inner mitochondrial membrane, not the outer.The correct statements about the electron transport chain (ETC) are a,b,e,f which include its coupling with proton transfer, the role of electrons from the citric acid cycle, the significance of prosthetic groups, and the organization into four complexes.
To accurately describe the Electron Transport Chain (ETC) and assess the statements given, we will evaluate each statement based on what we know about the ETC.
True. As electrons move through the ETC, energy is used to pump hydrogen ions (H⁺) from the mitochondrial matrix into the intermembrane space, creating a proton gradient.True. High-energy electrons carried by NADH and FADH₂, produced in the citric acid cycle, enter the ETC.False. While the outer membrane is permeable to small molecules due to porins, it is not freely permeable to protons (H+). The inner mitochondrial membrane tightly regulates the transfer of protons.False. Ubiquinone (coenzyme Q) and cytochrome c are located within the inner mitochondrial membrane, not the matrix, though they are involved in the ETC.True. These groups are essential components of the protein complexes and play a crucial role in electron transfers.True. The ETC consists of four main protein complexes (I, II, III, IV) that facilitate the transfer of electrons.False. The reactions occur in the inner mitochondrial membrane, where the protein complexes are embedded.Therefore the correct statements out of the given ones are a, b, e, f.
our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three on the right (total mass mp=10.1 kgmp=10.1 kg). The distance from the fulcrum to the center of the pile of rocks is rp=0.360 m.rp=0.360 m. Answer the two questions below, using three significant digits. Part A: What is the value of the torque (????pτp) produced by the pile of rocks? (Enter a positive value.)
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The torque produced by the pile of rocks is [tex]\tau = 35.63\ N \cdot m[/tex]
b
The distance of the single for equilibrium to occur is [tex]r_s =1.62 \ m[/tex]
Explanation:
From the question we are told that
The mass of the left rock is [tex]m_s = 2.25 \ kg[/tex]
The mass of the rock on the right [tex]m_p = 10.1 kg[/tex]
The distance from fulcrum to the center of the pile of rocks is [tex]r_p = 0.360 \ m[/tex]
Generally the torque produced by the pile of rock is mathematically represented as
[tex]\tau = m_p * g * r_p[/tex]
Substituting values
[tex]\tau = 10.1 * 9.8 * 0.360[/tex]
[tex]\tau = 35.63\ N \cdot m[/tex]
Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows
The torque due to the single rock is
[tex]\tau = m_s * g * r_s[/tex]
At equilibrium the both torque are equal
[tex]35.63 = m_s * r_s * g[/tex]
Making [tex]r_s[/tex] the subject of the formula
[tex]r_s = \frac{35.63 }{m_s * g}[/tex]
Substituting values
[tex]r_s = \frac{35.63 }{2.25 * 9.8}[/tex]
[tex]r_s =1.62 \ m[/tex]
A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of radius R. The small and large cars have accelerations as and aL respectively, as they travel around the curve. The magnitude of as is twice of that of aL. How does the speed of the small car VS compare to the speed of the large car VL as they move around the curve
Answer:
[tex]v_S=\sqrt{2}v_L[/tex]
Explanation:
The acceleration experimented while taking a curve is the centripetal acceleration [tex]a=\frac{v^2}{r}[/tex]. Since [tex]a_S=2a_L[/tex], we have that: [tex]\frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}[/tex]
They take the same curve, so we have: [tex]r_S=r_L=R[/tex]
Which means: [tex]v_S^2=2v_L^2[/tex]
And finally we obtain: [tex]v_S=\sqrt{2}v_L[/tex]
A 12 A fuse is placed in a parallel circuit that has two branches. 8 A flows in branch 1 and 6 A flows in branch 2. This fuse
1. will blow because the total current in this circuit is 14 A which is greater than 12 A.
2. will blow because a 12 A fuse only allows 6 A to flow in each branch.
3. will not blow because each branch is less than 12 A.
4. will not blow because the average current is 7 A which is less than 12 A.
Answer:
1. will blow because the total current in this circuit is 14 A which is greater than 12 A.
Explanation:
According to Kirchoff current law (KCL) which states that the total current flowing in a circuit is equal to the sum of the individual branch current.
If the supply current is greater than the sum of the individual branch current, then the load will collapse or blow off.
In the question given, the total current of the fuse is 12A
Sum of branch currents = current in branch 1 + current in branch 2
= 8A+6A
= 14A
As we can see that the supply current is lower than the sum of the branch current, this will cause the fuse to blow because some of the branch current will be sent back on the fuse and thereby causing the fuse to blow.
Albert and Emmy purchase identical state-of-the-art atomic watches to take on their interplanetary entomology expedition. After a busy day collecting spacebug specimens, Albert waits in the main spaceship while Emmy flies a shuttle to the nearest food court to pick up dinner. As Emmy flies by at a very high but constant velocity, Albert measures the rate at which Emmy's watch is ticking and compares it with his watch.
Albert observe that Emmy's watch is ticking ________ his own watch.
A. more slowly than
B. more quickly than
C. at the same rate as
Answer:
Option is A, Albert observes that Emmy's watch is ticking more slowly than his watch.
Explanation:
According to Einstein's theory of relativity, the time between two observers, as one of them is moving respect to the other, is different, occurring the known time dilation. This theory indicates that, for the observer that is in an inertial frame of reference, he will measure a clock that is moving relative to him, to tick slower than his clock that is at rest. The faster the relative velocity, the greater the time dilation between the two of them.
In our case, the person that is at rest (Albert) will measure the clock of the observer that is moving (Emmy) as ticking slower than his own watch, because Emmy is moving at high speed in relation to Albert (that is at rest), thus, time dilation is occurring for Emmy.
Therefore, the correct option is A, Albert observes that Emmy's watch is ticking more slowly than his watch.
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Two speakers produce waves of the same wavelength that are in phase. 1) At a point midway between the speakers, you would expect to hear: Louder sound Softer sound Alternating louder and softer sounds Louder or softer sounds depending on the wavelength No interference You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. (Survey Question) 2) Briefly explain your answer to the previous questio
Answer:
louder
Explanation:
Since the two speakers producing same wavelength that are in phase,at the midpoint, the waves travel the same distance and hence path difference is zero
hence constructive interference takes place , due to this a louder sound is observed .
hence the answer is a) louder
When two speakers produce waves of the same wavelength that are in phase at a point midway between the speakers, constructive interference occurs. This results in the overlapping and combination of the waves to form a wave with higher amplitude, creating a louder sound.
Explanation:The subject of this question involves the principle of wave interference in physics. This is a phenomenon that occurs when two waves come together while traveling through the same medium. At a point midway between the speakers, when two speakers produce waves of the same wavelength that are in phase, taking into account the path lengths traveled by the individual waves, you would expect to hear a louder sound.
This is a case of constructive interference, where the two sound waves, being in phase and of the same wavelength, will overlap and combine to form a wave with a greater amplitude, leading to a louder sound. This is explained in Figure 17.17 and 16.36, where the difference in the path lengths is one wavelength, resulting in total constructive interference and a resulting amplitude equal to twice the original amplitude.
However, it is worthy to note that in the real world recognition of this increased amplitude or louder sound will depend on the specific frequency of the sound, as sonic perception can vary with frequencies. This explanation is in reference to a single tone or frequency. When discussing music which is composed of many frequencies, the actual perception might be a bit more complex.
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A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the dry steel frying pan is only 0.100 N. Knowing the coefficient of kinetic friction between the two materials (0.3), he quickly calculates the normal force in newtons. What is it
Answer:
Normal force = 3N
Explanation:
We are given;
Coefficient of kinetic friction friction; μ_k = 0.3
Frictional force;F_f = 0.1N
Now,the formula for frictional force is;
F_f = μ_k*N
Where μ_k is coefficient of friction and N is the normal force.
So, making N the subject, we have;
N = μ_k/F_f
Plugging in the relevant values to obtain;
N = 0.3/0.1
N = 3N
A 1.50 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 C) the ammeter reads 18.8 A, while at 92.0 C it reads 17.4 A. You can ignore any thermal expansion of the rod.
1.Find the resistivity and for the material of the rod at 20 C. (rho= ? Ω*m)2.Find the temperature coefficient of resistivity at 20 C for the material of the rod.α= ? (C)^-1)
The answer to the questions are:
1. The resistivity for the material of the rod at 20 °C (ρ) is
[tex]1.26378\times 10^{-5} \Omega m[/tex] .
2. The temperature coefficient of resistivity at 20 °C for the material of the
rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].
Given to us:
Voltage across the rod, V = 15.0 V
Length of rod, L = 1.5 m = 150 cm,
Diameter, d = 0.55 cm,
[tex]Radius, r= \frac{d}{2}=\frac{0.550}{2}=0.275\ cm[/tex]
[tex]\begin{aligned}Area, A&= \pi r^2\\&=\pi\times (0.275)^2\\&=0.075625\pi\ cm^2\\\end{aligned}[/tex]
Initial temperature, [tex]T_o=20.0\ ^oC[/tex]
current at [tex]T_o[/tex], [tex]I_o= 18.8\ A[/tex]
Final temperature, [tex]T_1=92.0\ ^oC[/tex]
current at [tex]T_1[/tex], [tex]I_1=17.4\ A[/tex]
1.) To find out the resistivity of the rod(ρ),
Resistant of the rod(R),
[tex]\begin{aligned}\\R_o&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{18.8} \\&=0.7979\ \Omega \\\end{aligned}[/tex]
Resistivity of the rod at [tex]20^o\ C[/tex](ρ),
[tex]\begin{aligned}\\\rho&=\frac{RA}{L}\\&=\frac{0.7979\times 0.075625\pi}{150}\\&=0.00126378\ \Omega cm\\&=1.26378\times 10^{-3} \Omega cm\\&=1.26378\times 10^{-5} \Omega m\\\end{aligned}[/tex]
Hence, the resistivity for the material of the rod at 20 °C (ρ) is [tex]1.26378\times 10^{-5} \Omega m[/tex] .
2.) To find out the temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation,
[tex]{R_1} ={R_o}[ 1+\alpha(T_1-T_o)][/tex]
we need resistant of the rod([tex]R_1[/tex]),
[tex]\begin{aligned}\\R_1&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{17.4} \\&=0.862\ \Omega \\\end{aligned}[/tex]
Now, solving to get the value of α
[tex]\begin{aligned}{R_1} &={R_o}[ 1+\alpha(T_1-T_o)]\\0.862&=0.7979[1+\alpha(92-20)]\\\frac{0.862}{0.7979}&= [1+\alpha(72)]\\1.0804&=[1+\alpha(72)]\\0.0804&=\alpha(72)\\\alpha&=0.0011169\ ^oC^{-1}\\\alpha&=1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex]
Hence, the temperature coefficient of resistivity at 20 °C for the material of the rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].
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To find the resistivity and temperature coefficient of resistivity of the rod material, we need to use Ohm's Law and the formula for resistivity. The resistivity can be calculated using the resistance, area, and length of the rod, while the temperature coefficient of resistivity can be found by comparing resistivities at different temperatures.
Explanation:To find the resistivity of the material of the rod at 20°C, we can use the formula:
Resistivity (ρ) = (Resistance × Area) / (Length)
First, we need to find the resistance of the rod using Ohm's Law:
Resistance (R) = Voltage (V) / Current (I)
Next, we can substitute the values into the formula to find the resistivity. The temperature coefficient of resistivity can be calculated using the equation:
α = (ρ₂ - ρ₁) / (ρ₁ × (T₂ - T₁))
Where ρ₁ and ρ₂ are the resistivities at temperatures T₁ and T₂ respectively.
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distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. (a) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. (b) Find the speed of the galaxy relative to the earth.
Answer:
The speed of the galaxy relative to the Earth is [tex]3.09\times 10^6\ m/s[/tex].
Explanation:
We have,
(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.
(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :
[tex]v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s[/tex]
So, the speed of the galaxy relative to the Earth is [tex]3.09\times 10^6\ m/s[/tex].
The galaxy is receding from the Earth due to redshift. Using the redshift value, we can calculate the speed of the galaxy.
Explanation:The discrepancy in the measured wavelengths of light from a distant galaxy compared to its original wavelength on Earth is indicative of the galaxy moving away from Earth. This phenomenon, known as redshift, occurs when an object is moving away from the observer, causing the wavelength of light to stretch.
To calculate the speed of the galaxy relative to Earth, we can use the equation v = zc, where v is the speed, z is the redshift, and c is the speed of light. By plugging in the given values of 438.6 nm and 434.1 nm for the measured and original wavelengths respectively, we can solve for z. Once we know z, we can calculate the speed of the galaxy.
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A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B field (B is constant in time) and then back into a field free region to the right. The self inductance of the loop is negligible
True/False:
a. When leaving the field the coil experiences a magnetic force to the left.
b. Upon entering the field, a clockwise current flows in the loop.
c. Upon leaving the field, a clockwise current flows in the loop.
d. When entering the field the coil experiences a magnetic force to the right.
The loop experiences a magnetic force to the left when leaving the field, a counterclockwise current flows when entering, a clockwise current flows when leaving, and a magnetic force to the left when entering the field.
Explanation:The statement a is true. When leaving the magnetic field, the loop experiences a magnetic force to the left. This is because the magnetic field lines are directed from right to left in the field and the loop opposes the change in magnetic flux by generating a current that creates its own magnetic field, according to Lenz's law.
The statement b is false. Upon entering the field, a counterclockwise current flows in the loop. This is because the increasing magnetic flux induces a current that opposes the increase, as stated by Lenz's law.
The statement c is true. Upon leaving the field, a clockwise current flows in the loop. This is because the decreasing magnetic flux induces a current that opposes the decrease, according to Lenz's law.
The statement d is false. When entering the field, the coil experiences a magnetic force to the left. This is because the magnetic field lines are directed from left to right in the field and the loop opposes the change in magnetic flux by generating a current that creates its own magnetic field, according to Lenz's law.
Two football players collide head-on in midair, moving along the same horizontal direction, while trying to catch a thrown football. The first player is 89.5 kg and has an initial velocity of 6.05 m/s (in the positive direction), while the second player is 111 kg and has an initial velocity of –3.55 m/s. What is their velocity just after impact if they cling together?
Final answer:
The final velocity of the two football players after the collision is 0.738 m/s in the positive direction, calculated using the conservation of momentum.
Explanation:
The scenario described involves a conservation of momentum problem, where two football players collide and cling together. We can solve for the final velocity by using the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces are acting on it.
To find the final combined velocity of the players after the collision, we use the formula:
Momentum before collision = Momentum after collision
(m1 × v1) + (m2 × v2) = (m1 + m2) × v_final
Plugging in the given values:
(89.5 kg × 6.05 m/s) + (111 kg × (-3.55 m/s)) = (89.5 kg + 111 kg) × v_final
After calculating both sides,
541.475 kg·m/s - 393.55 kg·m/s = 200.5 kg · v_final
We get the final combined velocity,
v_final = 147.925 kg·m/s / 200.5 kg
v_final = 0.7377 m/s (to four significant figures)
The players will be moving with a velocity of 0.738 m/s in the positive direction.
Two students are holding opposite ends of a spring in a classroom. One student stands on the left end of the classroom and the other stands at the right end. They shake the spring so that a longitudinal wave travels along the spring. In which directions will the longitudinal wave oscillate?
Answer:
Up And Down
Explanation:
In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave. Longitudinal waves are always characterized by particle motion being parallel to wave motion.
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Final answer:
In a longitudinal wave on a horizontal spring, the medium oscillates back and forth horizontally, parallel to the direction of the wave's travel.
Explanation:
The question pertains to the behavior of longitudinal waves in a spring. When one student shakes the spring to create a longitudinal wave, the oscillation of the spring occurs in the same direction as the wave's propagation. This means that in a longitudinal wave, the medium—the spring in this case—oscillates parallel to the wave's direction of motion. Therefore, if the spring is held horizontally and the wave travels from left to right, the individual coils of the spring will move left and right along the same horizontal line, compressing and expanding as the wave passes through.
In contrast, with transverse waves, the medium moves perpendicularly to the direction of the wave's travel, such as in the motion of a rope being moved up and down while the wave travels horizontally.
An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnetic field of 0.508 T at right angles to the direction of the electron’s motion. The mass of the electron is 9.11 × 10−31 kg and its charge is 1.60218 × 10−19 C. What is the magnitude of the force on the electron due to the magnetic field? Answer in units of N.
Answer:
Magnetic force is equal to [tex]1.37\times 10^{-11}N[/tex]
Explanation:
We have given electron is accelerated with a potential difference of 81700 volt.
Magnetic field B = 0.508 T
Angle between magnetic field and velocity [tex]\Theta =90^{0}[/tex]
Mass of electron [tex]m=9.11\times 10^{-31}kg[/tex]
Charge on electron [tex]e=1.6\times 10^{-19}C[/tex]
By energy conservation.
[tex]\frac{1}{2}mv^2=qV[/tex]
[tex]\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=1.6\times 10^{-19}\times 81700[/tex]
[tex]v=169.4\times 10^6m/sec[/tex]
Magnetic force on electron
[tex]F=qvBsin\Theta[/tex]
[tex]F=1.6\times 10^{-19}\times 169.4\times 10^6\times 0.508\times sin90^{\circ}[/tex]
[tex]=1.37\times 10^{-11}N[/tex]
Answer:
Explanation:
After acceleration under potential difference , velocity v acquired can be calculated by the following expression
V e = 1/2 m v² ;
V is potential under which electron with mass m and charge e is accelerated to velocity v .
81700 x 1.60218 x 10⁻¹⁹ = .5 x 9.11 x 10⁻³¹ x v²
v² = 28737 x 10¹²
v = 169.52 x 10⁶ m /s
Force = Bev , B is magnetic field , e is charge on lectron and v is its velocity
= .508 x 1.60218 x10⁻¹⁹ x 169.52 x 10⁶
= 128 x 10⁻¹³ N.
Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is the electric force between them (in N)?
Explanation:
Th electric force between charges is inversely proportional to the square of distance between them. It means,
[tex]F\propto \dfrac{1}{r^2}[/tex]
Initial distance, r₁ = 2 cm
Final distance, r₂ = 0.25 cm
Initial force, F₁ = 1 N
We need to find the electric force between charges if the new separation of 0.25 cm. So,
[tex]\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N[/tex]
So, the new force is 64 N if the separation between charges is 64 N.