Answer:
A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...
Explanation:
One example of a physical entity that is quantized is:
The amount of money in your pocket.
The amount can't have any fraction of 1 cent.
Its value must be an integer-multiple of cents, or 0.01 dollar.
When it increases or decreases, it jumps from one integer number of cents to the next integer number. It doesn't "slide" from one to the next. It can never have a value between two integer numbers of cents.
PLEASE HELP ME WITH THIS ONE QUESTION
Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)
A) 59 eV
B) 58 eV
C) 57 eV
D) 56 eV
Answer:
a. 59 ev. helpful answer
Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.
Answer:
The required ratio is 1.99.
Explanation:
We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.
We know that,
[tex]eV=\dfrac{1}{2}mv^2[/tex]
The LHS for both proton and an alpha particle is the same.
So,
[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]
So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.
Erica (37 kg ) and Danny (45 kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.7 m/s . At that instant he grabs hold of her. What is their speed just after he grabs her?
Answer:
V = 2.58 m/s
Explanation:
Below is the calculation:
Given the weight of Erica = 37 kg
The weight of Danny = 45 kg
Danny's speed to move upward = 4.7 m/s
Use below formula to find the answer.
m1 * u1 = (m1+m2) * V
V = m1*u1 / (m1+m2)
Here, m1 = 45
u1 = 4.7
m1 = 45
m2 = 37
Now plug the values in formula:
V = m1*u1 / (m1+m2)
V = (45*4.7)/(45+37)
V = 2.58 m/s
an artificial satellite is moving in a circular orbit of radius 36000 kilometre calculate its speed if it takes 24 hours to revolve around the earth
Explanation:
9420 km/hr is the correct answer
Hope this helps...☺
A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a rate of 202 Hz and spectral analysis was performed using FFT. What is the maximum frequency that will be displayed on the power spectrum plot
Answer:
Maximum frequency on power spectrum plot = 101 Hz
Explanation:
Given:
Time taken for output = 15 seconds
Frequency rate = 202 Hz
Find:
Maximum frequency on power spectrum plot
Computation:
Maximum frequency = Given frequency rate / 2
Maximum frequency on power spectrum plot = Frequency rate / 2
Maximum frequency on power spectrum plot = 202 / 2
Maximum frequency on power spectrum plot = 101 Hz
PLEASE HELP ME WITH THIS ONE QUESTION
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
k = [tex] \dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m} [/tex]
k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷
k = (1.12 × 10-³⁰)^2/3.346×10-²⁷
k = 1.25 × 10-⁶⁰ /3.346×10-²⁷
k = 0
ldk why, my answer is coming this :(
two particles woth each charge magnitude 2.0×10^-7 c but opposite signs are held 15cm apart.what are the magnitude and direction of the electric field E at tge point midway between charges
Answer:
The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.
Explanation:
Given;
charge of each particle, Q = 2 x 10⁻⁷ C
distance between the two charges, r = 15 cm = 0.15 m
distance midway between the charges = 0.075 m
The magnitude of the electric field is calculated as;
[tex]E_{net} = E_{+q} + E_{-q}\\\\E_{net} = \frac{kQ}{r_{1/2}^2} + \frac{kQ}{r_{1/2}^2}\\\\E_{net} = 2(\frac{kQ}{r_{1/2}^2})\\\\E_{net} = 2 (\frac{9\times 10^9 \ \times 2\times 10^{-7}}{0.075^2} )\\\\E_{net} = 6.4\times 10^5 \ N/C[/tex]
The direction of the electric field is from positive particle to negative particle.
You throw a football straight up. Air resistance can be neglected. When the football is 4.00 mm above where it left your hand, it is moving upward at 0.500 m/sm/s. What was the speed of the football when it left your hand
Answer:
u=8.868 m/s
Explanation:
The displacement of the ball is 4 meters
The final speed of the ball is 0.5 m/s
The initial speed of the ball is to be calculated. Using the equation of the rectilinear motion,
[tex]v^{2} =u^{2} +2as[/tex]
Plugging the values in the above expression,
[tex]\\0.5^{2} =u^{2} +2*(-9.8)*4\\\\u^{2} =78.4+0.25\\\\u^{2} =78.65\\\\u=8.868 m/s[/tex]
The correct formula for finding the relative velocity of an object is:
WILL MARK BRAINLIEST TO THE CORRECT ANSWER!!
Answer:
[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]
Explanation:
The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.
The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]
From the given options, the second option is the correct answer.
[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]
A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.3 m/s and at the top of the incline the mass is moving at 2.8 m/s. What is the work done by all non-conservative force in Joules?
Answer:
499.7 J
Explanation:
Since total mechanical energy is conserved,
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.
So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
Substituting the values of the variables into the equation, we have
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)² + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)² + W₂
0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s² + W₂
907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s² + W₂
907.38 kgm²/s² = 407.68 kgm²/s² + W₂
W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²
W₂ = 499.7 kgm²/s²
W₂ = 499.7 J
Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J
Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the beam, and moves at 314 m/s. What is the maximum magnetic force it can feel
Answer:
The maximum magnetic force is 2.637 x 10⁻¹² N
Explanation:
Given;
Power, P = 8.25 m W = 8.25 x 10⁻³ W
charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C
speed of the charge, v = 314 m/s
area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²
The intensity of the radiation is calculated as;
[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]
The maximum magnetic field is calculated using the following intensity formula;
[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]
The maximum magnetic force is calculated as;
F₀ = qvB₀
F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)
F₀ = 2.637 x 10⁻¹² N
If a magnifying glass has a power of 10.0 D, what is the magnification it produces when held 6.55 cm from an object?
Answer: The magnification of the magnifying glass is -2.9
Explanation:
The equation for power is given as:
[tex]P=\frac{1}{f}[/tex]
where,
P = Power = 10 D
f = focal length
Putting values in above equation, we get:
[tex]f=\frac{1}{10}=0.1m=10 cm[/tex] (Conversion factor: 1 m = 100 cm)
The equation for lens formula follows:
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]
where,
v = image distance
u = Object distance = 6.55 cm
Putting values in above equation, we get:
[tex]\frac{1}{v}=\frac{1}{10}-\frac{1}{(6.55)}\\\\\frac{1}{v}=\frac{6.55-10}{10\times 6.55}\\\\v=\frac{65.5}{-3.45}=-18.98cm[/tex]
Magnification (m) can be written as:
[tex]m=\frac{-v}{u}[/tex]
Putting values in above equation, we get:
[tex]m=\frac{-18.98}{6.55}\\\\m=-2.9[/tex]
Hence, the magnification of the magnifying glass is -2.9
Is the following chemical reaction balanced?
2H202-H2O + O2
yes
no
A ball is thrown vertically upward at 24.0 ms can reach a height of 28.8m ( neglecting air resistance).The speed,in m/s,when it is halfway to its highest point is (using g= 10 ms ^2)
Answer:
The answer is "[tex]16.79\ \frac{m}{s}[/tex]"
Explanation:
In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]
Using formula:
[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]
[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]
. A car increases velocity from 20 m/s to 60 m/s in a time of 10 seconds. What was the acceleration of the car?
Answer:
0.3333
Explanation:
Acceleration = change in velocity/time
a = 20 m/s / 60 m/s
a = 0.3333 m/s^2
A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after landing on the pad.
Required:
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision
a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]
b. The constant force exerted on the pole vaulter is 6799.52 N
a. We use Newton's equation of motion,
[tex]v=u+at\\\\S=ut+\frac{1}{2}at^{2}[/tex]
Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.
Given that, s = 5.1 m , t = 0.29s, u = 0
Substitute in above equation.
[tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]
the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]
b. The constant force exerted on the pole vaulter due to the collision is given as, [tex]Force=mass*acceleration[/tex]
[tex]Force=56*121.42=6799.52N[/tex]
Learn more:
https://brainly.com/question/13532462
B. Complete the lists:
Things that I must do for my family
Things I must never do to my family
1.
2.
2.
3.
3.
4.
5.
5.
Answer:
Things you should do for your family
help your parentstreat them kindlylisten and obey themappreciate them for anything they do for you talk softlythings you shouldn't
backanswering them Disobey And anything that's harsh or make it parents sadWhich statement describes an action-reaction pair?
O A. You push on a car, and the car pushes back on you.
B. A book pushes down on a table, and the table pushes down on the
Earth.
C. The Moon pulls on Earth, and Earth pulls on the Sun.
D. You push down on your shoe, and Earth's gravity pulls down on the
shoe.
Answer:
A
Explanation:
a pex
A gymnast of mass 70.0 kgkg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81m/s29.81m/s2 for the acceleration of gravity.
PART A Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.
PART B Calculate the tension TTT in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.00 m/s2
PART C Calculate the tension TTT in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.00 m/s2m/s2 .
Answer:
43994
Explanation:
Hope this helps!
Calculate the current flowing when the voltage across is 35V and the resistance is 7ohms.
Explanation:
V= IR
35=I×7
I=35/7
I=5amperes
pls give brainliest
Hydrogen carried in light phase
Answer:
because it is helpful to human beings I think
A 12 kg hanging sculpture is suspended by a 80-cm-long, 6.0 g steel wire. When the wind blows hard, the wire hums at its fundamental frequency. What is the frequency of the hum
Answer:
[tex]F=78.3hz[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=12[/tex]
Length [tex]l=80cm=0.8m[/tex]
Linear density [tex]\mu= 6.0g[/tex]
Generally the equation for Frequency is mathematically given by
[tex]F=\frac{1}{2l}\sqrt{\frac{T}{K}}[/tex]
[tex]F=\frac{1}{2(0.8)}\sqrt{\frac{12*9.8*0.8}{6*10^{-3}}}[/tex]
[tex]F=78.3hz[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
The color orange has a wavelength of 590 nm. The energy of an orange photon is approximately 0.337 eV.
The correct answer is option E.
To calculate the energy of a photon, we can use the equation:
E = (hc) / λ
where E is the energy of the photon, h is the Planck's constant (6.626 x [tex]10^-^3^4[/tex]J·s or 6.626 x[tex]10^-^1^9^[/tex] eV·s), c is the speed of light (3.00 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light.
Given that the wavelength of orange light is 590 nm (or 590 x [tex]10^-^9[/tex]m), we can substitute the values into the equation:
E = [(6.626 x[tex]10^-^1^9^[/tex] eV·s) x (3.00 x [tex]10^8[/tex] m/s)] / (590 x[tex]10^-^9[/tex]m)
E = (1.9878 x [tex]10^-^1^0[/tex]eV·m) / (590 x [tex]10^-^9[/tex] m)
E = 3.3695 x [tex]10^-^1[/tex] eV
For more such information on: wavelength
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The question probable may be:
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x [tex]10^-^1^9^[/tex], 1 eV = 1.6 x[tex]10^-^1^9^[/tex]J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
E) 0.337 eV
Using pascals principle, F1/A1=F2/A2, solve this like a proportion.
A force of 50 N is applied to an area of 200 sq feet, how much force will be applied of the area to be covered is 50 sq feet?
Answer:
NO CLUE
Explanation:
GOOD LUCK THOUGH
A 35 kg child slides down a playground slide at a constant speed. The slide has a height of 3.8 m and is 8.0 m long. Find the magnitude of the kinetic friction force acting on the child.
Answer:
The magnitude of the kinetic frictional force acting on the child is 162.93 N
Explanation:
Given;
mass of the child, m = 35 kg
height of the slide, h = 3.8 m
length of the slide, d = 8.0 m
The change in thermal energy associated with the kinetic frictional force is calculated as follows;
[tex]\Delta E_{th} + \Delta K.E + \Delta U = 0\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i) =0\\\\since \ the \ speed \ is \ constant, \ v_f = v_i \ and \ \Delta K.E = 0\\\\Also, \ final \ height \ , h _f= 0\\\\\Delta E_{th} - mgh_i = 0\\\\\Delta E_{th} = mgh_i\\\\\Delta E_{th} = 35 \times9.8 \times 3.8\\\\\Delta E_{th} = 1303.4 \ J[/tex]
The magnitude of the kinetic frictional force that produced this thermal energy is calculated from the work done by frictional force;
[tex]\Delta E_{th} = F \times d\\\\F = \frac{\Delta E_{th} }{d} \\\\F = \frac{1303.4}{8} \\\\F = 162.93 \ N[/tex]
Therefore, the magnitude of the kinetic frictional force acting on the child is 162.93 N
why does a spherometer have three legs?
spherometer is a device used to measure curved in surface
it have 3 legs which form equivalent triangle.
geometry says that 3 point determine a plane that's why it have 3 legs
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
26.
Which one of the following is not a vector quantity?
(2)
A) acceleration
C) displacement
E) instantaneous velocity
B) average speed
D) average velocity
Answer:
Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.
The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2 pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2
Answer:
Explanation:
a)
Given that:
V = 25 mi/hr
To ft/sec, we have:
[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{110}{3} ft/s[/tex]
[tex]\rho = 0.075 \ lb/ft^3[/tex]
[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]
[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]
[tex]C_d = 0.28[/tex]
A = 25ft²
Recall that:
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]
[tex]F_d =10.967 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]
For 70 miles per hour, we have:
[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{308}{3} ft/s[/tex]
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]
[tex]F_d =85.99 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]
How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?
Answer:
Power = 70 W
Explanation:
Given that,
Force, F = 70 N
Height, h = 5 m
Time, t = 5 s
We need to find the power of the object. We know that,
Power = work done/time
Put all the values,
[tex]P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W[/tex]
So, the required power is 70 W.
