Answer:
A- to simplify the study of things
Explanation:
a visual reference
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
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.
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
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!! :))
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.
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
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
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
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.
Learn more about electromagnetic waves at: https://brainly.com/question/8553652
Answer:
gamma rays and X-rays
Explanation:
d on edge I got 100%
"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
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
Learn more about the capacitors here:
https://brainly.com/question/15052170
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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]
A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 4.10 cm.
Calculate:
a. The capacitance
b. The radius of the inner sphere.
c. The electric field just outside the surface of the inner sphere.
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
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.
A 2.0 kg handbag is released from the top of the Leaning Tower of Pisa, and 55 m before reaching the ground, it carries a speed of 29 m / s. What was the average force of air resistance?
Answer:
4.31 N
Explanation:
Given:
Δy = -55 m
v₀ = 0 m/s
v = -29 m/s
Find: a
v² = v₀² + 2aΔy
(-29 m/s)² = (0 m/s)² + 2a (-55 m)
a = -7.65 m/s²
Sum of forces in the y direction:
∑F = ma
R − mg = ma
R = m (g + a)
R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)
R = 4.31 N
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
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
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:
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.
Answer:
The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 11^o[/tex]
The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as
[tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]
Where [tex]\beta[/tex] is mathematically represented as
[tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]
[tex]\beta = 708.1 \ rad[/tex]
So
[tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
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
An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?
Answer:
The highest rms voltage will be 8.485 V
Explanation:
For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load
If the peak or maximum voltage should not exceed 12 V, then from the relationship
[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]
where [tex]V_{rms}[/tex] is the rms voltage
[tex]V_{p}[/tex] is the peak or maximum voltage
substituting values into the equation, we'll have
[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V
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
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
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.
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
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].
To know more air bubble volume,
https://brainly.com/question/10509397
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
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°
If you want to learn more about vectors, you can read:
https://brainly.com/question/7802337
what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
[tex]F_{e}[/tex] - W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium
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