Answer:
the correct answer is c v₁> 12.5 m / s
Explanation:
This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.
v² = v₀² + 2 a₁ x
as part of rest v₀ = 0
a₁ = v² / 2x
a₁ = 25² / (2 120)
a₁ = 2.6 m / s²
now we can find the velocity for the distance x₂ = 60 m
v₁² = 0 + 2 a1 x₂
v₁ = Ra (2 2,6 60)
v₁ = 17.7 m / s
these the speed at 60 m
we see that the correct answer is c v₁> 12.5 m / s
Find the distance to a Sun-like star (L=3.8x1026 watts) whose apparent brightness at Earth is 1.0 x10-10 watt/m2.
Answer:
5.49 x 10^17 m is the distance between the sun-like star to the earth
Explanation:
Radiation intensity on Earth = 1.0 x 10^-10 W/m^2
Power of radiation of the star = 3.8 x 10^26 W
Recall that the intensity of radiation is given as
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where
[tex]I[/tex] = intensity of radiation
P = power of radiation
A is the area through which the radiation spreads out in all three dimensional direction.
A = [tex]\frac{P}{I}[/tex] = [tex]\frac{3.8*10^{26} }{1.0*10^{-10} }[/tex] = 3.8 x 10^36 m^2
This area is spread out in the form of a sphere of area
A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]r^{2}[/tex]
3.8 x 10^36 = 12.568[tex]r^{2}[/tex]
[tex]r^{2}[/tex] = (3.8 x 10^36)/12.568 = 3.02 x 10^35
r = [tex]\sqrt{3.02*10^{35} }[/tex] = 5.49 x 10^17 m this is the distance of the star to the Earth
Which of the following represents a concave mirror? +f,-f,-di,+di
Answer:
fully describes the concave mirror is + f
Explanation:
A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f
This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di
Answer:
+f
Explanation:
because you have to be really dumb to get an -f
7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
9.09 × 10³ capacitors must be connected in parallel.
How to calculate the number of capacitors connected in parallel?
Given C = 1.00μF = 1 × 10⁻⁶ F
q = 1.00 C
V = 110 V
The equivalent capacitance is given by
Ceq = q/V
where q = total charge on all the capacitors
V = potential difference
For N number of identical capacitors in parallel,
Ceq = NC
Therefore,
NC = q/V
N = q/VC
Putting on the values in the above formula,
N = 1/ (110)(1 × 10⁻⁶)
= 1 / 110 × 10⁻⁶
= 9.09 × 10³
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23.15. Can an object carry a charge of 2.0 10-19 C?
Answer:
Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)
Where the charge of a single proton is:
C = 1.6x10^-19 C
Now, you need to remember that when we are working with charges, we are working with discrete math:
What means that?
If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)
So the consecutive charge will be:
C = 2*1.6x10^-19 C = 3.2x10^-19 C.
So, because we are working in discrete math, we can not have any object that has charge between 1.6x10^-19 C and 3.2x10^-19 C.
Particularly, 2.0x10^-19 C is in that range, so we can conclude that:
No, an object can not carry a charge of 2.0x10^-19 C.
A copper telephone wire has essentially no sag between poles 36.0 m apart on a winter day when the temperature is −20.0°C. How much longer is the wire on a summer day when the temperature is 34.0°C?
Answer:
The extension is [tex]\Delta L = 0.033 \ m[/tex]
Explanation:
From the question we are told that
The length of the wire on a winter day is [tex]L_w = 36.0 \ m[/tex]
The temperature on the winter day is [tex]T_w = -20.0 ^o C[/tex]
The temperature on a summer day is [tex]T_s = 34.0 ^0 C[/tex]
The the extension of the wire on a summer day is mathematically represented as
[tex]\Delta L = \alpha L_w [T_s - T_w][/tex]
Where
[tex]\alpha[/tex] is the coefficient of linear expansion of copper with a values [tex]\alpha = 17 *10^{-6}[/tex]
substituting value
[tex]\Delta L = 17 *10^{-6} * 36.0 [34 - [-20]][/tex]
[tex]\Delta L = 0.033 \ m[/tex]
What is the average flow rate in cm3 /s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L
Answer:
2.78 cm³/s
Explanation:
From the question,
Q = v/A'.................... Equation 1
Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.
Given: v = 100 km/h, A' = 10 km/L
Substitute this value into equation 1
Q = 100/10
Q = 10 L/h.
Now, we convert L/h to cm³/s.
Since,
1 L = 1000 cm³, and
1 h = 3600 s
Therefore,
Q = 10(1000/3600) cm³/s
Q = 2.78 cm³/s
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
λ = 5.2 x 10⁻⁷ m = 520 nm
g An object with mass 1kg travels at 3 m/s and collides with a stationary object whose mass is 0.5kg. The two objects stick together and continue to move. What is the velocity of the two objects together after collision
Answer:
2
Explanation:
since the second object was in it stationary, it velocity is 0 m/s
What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
c. The incident light must have at least as much energy as the electron work function
A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the friction force between the car and the track, calculate how much the car travels in 10 s
Answer:
50 m
Explanation:
F = ma
10 N = (10 kg) a
a = 1 m/s²
Given:
v₀ = 0 m/s
a = 1 m/s²
t = 10 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²
Δx = 50 m
An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 75.2 ∘C How much ice at a temperature of -22.8 ∘C must be dropped into the water so that the final temperature of the system will be 32.0 ∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kg
Answer:
Explanation:
We shall apply the theory of
heat lost = heat gained .
heat lost by water = mass x specific heat x temperature diff
= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J
heat gained by ice to attain temperature of zero
= m x 2100 x 22.8 = 47880 m
heat gained by ice in melting = latent heat x mass
= 334000m
heat gained by water at zero to become warm at 32 degree
= m x 4190 x 32 = 134080 m
Total heat gained = 515960 m
So
515960 m = 51587.28
m = .1 kg
= 100 gm
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
38.A student pushes a 0.15 kg box down against a spring doing 25 J of work on the spring. The student releases the box which launches the box into the air. What is the maximum height reached by the box assuming negligible frictional forces
Answer:
Explanation:
Potential energy stored in the spring = 25 J
This energy is converted into gravitational potential energy . If h be the height attained
gravitational potential energy = mgh
mgh = 25
.15 x 9.8 x h = 25
h = 17 m
An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is the volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent.
Answer:
V2 = 21.44cm^3
Explanation:
Given that: the initial volume of the bubble = 1.3 cm^3
Depth = h = 160m
Where P2 is the atmospheric pressure = Patm
P1 is the pressure at depth 'h'
Density of water = ρ = 10^3kg/m^3
Patm = 1.013×10^5 Pa.
Patm = 101300Pa
g = 9.81m/s^2
P1 = P2+ρgh
P1 = Patm +ρgh
P1 = 1.013×10^5+10^3×9.81×160.
P1 = 101300+1569600
P1 = 1670900 Pa
For an ideal gas law
PV =nRT
P1V1/P2V2 = 1
V2 = ( P1/P2)V1
V2 = (P1/Patm)V1
V2 = ( 1670900 /101300 Pa) × 1.3
V2 = 1670900/101300
V2 = 16.494×1.3
V2 = 21.44cm^3
The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
The formula of the pressure of the static fluid
P1 = P2+ρgh
Where,
P1 - pressure at depth 'h'
P2 - atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] = 1670900 Pa
h - Depth = 160m
ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]
g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]
The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]
[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]
For an ideal gas,
PV =nRT
[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]
[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]
So,
[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]
Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
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Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)
A. Their electric potential energy keeps decreasing.
B. Their acceleration keeps decreasing.
C. Their kinetic energy keeps increasing.
D. Their kinetic energy keeps decreasing.
E. Their electric potential energy keeps increasing.
Answer:
(A)
Explanation:
We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.
Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.
What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?
Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.
A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter of 50 cm. The speed of the top of the wheel is
Answer:
The speed of the top of the wheel is twice the speed of the car.
That is: 72 m/s
Explanation:
To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,
plus the velocity due to the translation of the center of mass (v = 36 m/s).
The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]
Then the addition of these two velocities equals:
[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]
Two electric force vectors act on a particle. Their x-components are 13.5 N and −7.40 N and their y-components are −12.0 N and −4.70 N, respectively. For the resultant electric force, find the following.
(a) the x-component N
(b) the y-component N
(c) the magnitude of the resultant electric force N
(d) the direction of the resultant electric force, measured counterclockwise from the positive x-axis ° counterclockwise from the +x-axis
Answer:
Explanation:
Given two vectors as follows
E₁ = 13.5 i -12 j
E₂ = -7.4 i - 4.7 j
Resultant E = E₁ + E₂
= 13.5 i -12 j -7.4 i - 4.7 j
E = 6.1 i - 16.7 j
a ) X component of resultant = 6.1 N
b ) y component of resultant = -16.7 N
Magnitude of resultant = √ ( 6.1² + 16.7² )
= 17.75 N
d ) If θ be the required angle
tanθ = 16.7 / 6.1 = 2.73
θ = 70° .
counterclockwise = 360 - 70 = 290°
By working with the vector forces, we will get:
a) The x-component is 1.5 Nb) The y-component is -12.2 Nc) The magnitude is 12.9 Nd) The direction is 277.01°.How to find the resultant force?
Remember that we can directly add vector forces, so if our two forces are:
F₁ = <13.5 N, -7.5 N>
F₂ = < -12 N, -4.70 N>
Then the resultant force is:
F = F₁ + F₂ = <13.5 N + (-12 N), -7.5 N + ( -4.70 N) >
F = < 1.5 N, -12.2 N>
so we have:
a) The x-component is 1.5 N
b) The y-component is -12.2 N
c) The magnitude will be:
|F| = √( (1.5 N)^2 + (-12.2 N)^2) = 12.29 N
d) The direction of a vector <x, y> measured counterclockwise from the positive x-axis is given by:
θ = Atan(y/x)
Where Atan is the inverse tangent function, then here we have:
θ = Atan(-12.2 N/1.5 N) = 277.01°
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Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum
Answer:
a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.
Explanation:
a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:
[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]
Where:
[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.
[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.
The impact experimented by the ball due to collision is:
[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]
By using the definition of momentum, the expression is therefore expanded:
[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]
Where:
[tex]m[/tex] - Mass of the ball, measured in kilograms.
[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.
If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:
[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]
The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.
b) The magnitudes of initial and final momentums of the ball are, respectively:
[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is:
[tex]\Delta p = p_{f}-p_{o}[/tex]
[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.
c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.
In a nutshell, the right choice is option D.
21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
I got it right on my quiz!
I hope this helps!! :))
The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object on this interval by sub-dividing the interval into the indicated number of sub-intervals. Use the left endpoint of each sub-interval to compute the height of the rectangles.
v= 4t + 5(m/s) for 3 < t < 7; n = 4
The approximate displacement of the object is______m.
Answer:
The approximate displacement of the object is 23 m.
Explanation:
Given that:
v = 4t + 5 (m/s) for 3< t< 7; n= 4
The approximate displacement of the object can be calculated as follows:
The velocities at the intervals of t are :
3
4
5
6
the velocity at the intervals of t = 7 will be left out due the fact that we are calculating the left endpoint Reimann sum
n = 4 since there are 4 values for t, Then there is no need to divide the velocity values
v(3) = 4(3)+5
v(3) = 12+5
v(3) = 17
v(4)= 4(4)+5
v(4) = 16 + 5
v(4) = 21
v(5)= 4(5)+5
v(5) = 20 + 5
v(5) = 25
v(6) = 4(6)+5
v(6) = 24 + 5
v(6) = 29
Using Left end point;
[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]
= 23 m
Which two types of electromagnetic waves have higher frequencies than the waves that make up ultraviolet light?
radio waves and infrared light
visible light and X-rays
microwaves and gamma rays
gamma rays and X-rays
The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.
WHAT ARE ELECTROMAGNETIC WAVES?Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:
Radio wavesInfraredUltravioletVisible lightX-raysGamma raysmicrowaveEach electromagnetic wave have a specific frequency and wavelength.
However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.
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Answer:
gamma rays and X-rays
Explanation:
d on edge I got 100%
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.
Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑ [tex]q_{i} / r_{i}[/tex]
where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resistance r/9 what is the diameter of the second wire g
Answer:
d₂ = 3dThe diameter of the second wire is 3 times that of the initial wire.Explanation:
Using the formula for calculating the resistivity of an object to find the diameter.
Resistivity P = RA/L
R is the resistance of the material
A is the cross sectional area
L is the length of the material
Since A = πd²/4
P = R( πd²/4)/L
P = Rπd²/4L ... 1
If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;
P₂ = (R/9)A₂/L₂
P₂ = (R/9)(πd₂²/4)/L₂
P₂ = (Rπd₂²/36)/L₂
P₂ = (Rπd₂²)/36L₂
Since the length and resistivity are the same;
P = P₂ and L =L₂
Equating 1 and 2;
Rπd²/4L = (Rπd₂²)/36L₂
Rπd²/4L = (Rπd₂²)/36L
d² = d₂²/9
d₂² = 9d²
Taking the square root of both sides;
√d₂² = √9d²
d₂ = 3d
Therefore the diameter of the second wire is 3 times that of the initial wire
An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
with a speed of 75 m sl. Determine
1) A vehicle of mass 1600 kg moves
the magnitude of its momentum.
Answer:
120000 kgxm/s
Explanation:
momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000 kgxm/s
A source containing a mixture of hydrogen and deuterium atoms emits light at two wavelengths whose mean is 540 nm and whose separation is 0.170 nm. Find the minimum number of lines needed in a diffraction grating that can resolve these lines in the first order.
Answer:
N=3176.5rulling
Explanation:
We were told that the source containing a mixture of hydrogen and deuterium atoms emits light with
wavelengths whose mean is 540 nm
Then λ= 540 nm, but we need to convert to metre which = (540× 10⁻⁹m)
Also whose separation is 0.170 nm, which mean the difference between the wavelength is 0.170 nm then
Δ λ = 0.170 nm the we convert to metre we have. Δλ= 0.170 nm= (0.170×10⁻⁹m)
the formular below can be used to can be used to calculate our minimum number of lines
N= λ /(m Δλ)
Where N is number of fillings i.e number of lines
λ= wavelength
Δλ= difference in wavelength
m=1
Then if we substitute the values we have
,N= (540× 10⁻⁹ m)/[(1)*(0.170× 10⁻⁹m)]
N =3176.5rulling
Therefore, minimum number of lines = =3176.5rulling
Will give brainliest ASAP! Please help (1/10 questions, will mark 5 stars and brainliest for all answers if correct)
Answer:
Option (A)
Explanation:
A 20 kg boy chases the butterfly with a speed of 2 meter per second.
Angle at which he runs is 70° North of West.
Therefore, Horizontal component (Vx) directing towards West will be,
Vx = v(Cos70°)
Vy = v(Sin70°)
Since momentum of a body is defined by,
Momentum = Mass × Velocity
Therefore, Westerly component of the momentum will be,
Momentum = 20 × (v)(Cos70°)
= 20 × 2Cos70°
= 13.68
≈ 13.7 kg-meter per second
Therefore, Option (A) will be the answer.