Answer:
Differences between freefall and weightlessness are as follows:
FreefallWhen a body falls only under the influence of gravity, it is called free fall.Freefall is not possible in absence of gravity.A body falling in a vacuum is an example of free fall.WeightlessnessWeightlessness is a condition at which the apparent weight of body becomes zero.Weightlessness is possible in absence of gravity.A man in a free falling lift is an example of weightlessness.Hope this helps....
Good luck on your assignment....
Consider the Earth and the Moon as a two-particle system.
A. Find an expression for the gravitational field g vector of this two-particle system as a function of the distance r from the center of the Earth.
B. G Plot the scalar component of g vector as a function of distance from the center of the Earth.
Answer:
a is the maybe I am not sure
Light bulb A is rated at 60 W and light bulb B is rated at 100 W. Both are designed to operate at 110 V. Which statement is correct?
Answer:
Option 5:
The 60W bulb has a greater resistance and a lower current than the 100 W bulb.
Explanation:
We have to compare the resistance and current of both bulbs.
Bulb A
Power = 60 W
Voltage = 110 V
Power is given as:
[tex]P = V^2 /R[/tex]
where V= voltage and R = resistance
[tex]=> 60 = 110^2 / R\\\\R = 201.6 \Omega[/tex]
Power is also given as:
P = IV
where I = current
=> 60 = I * 110
I = 60/110 = 0.54 A
Bulb B
Power = 100 W
Voltage = 110 V
To get resistance:
[tex]100 = 110^2 / R\\\\R = 121 \Omega[/tex]
To get current:
100 = I * 110
I = 100 / 110
I = 0.91 A
Therefore, by comparison, the 60W bulb has a greater resistance and a lower current.
A girl weighing 600 N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm under her weight. Find the spring constant and the total work done on it during the compression.
Answer:
The spring constant is 60,000 N
The total work done on it during the compression is 3 J
Explanation:
Given;
weight of the girl, W = 600 N
compression of the spring, x = 1 cm = 0.01 m
To determine the spring constant, we apply hook's law;
F = kx
where;
F is applied force or weight on the spring
k is the spring constant
x is the compression of the spring
k = F / x
k = 600 / 0.01
k = 60,000 N
The total work done on the spring = elastic potential energy of the spring, U;
U = ¹/₂kx²
U = ¹/₂(60000)(0.01)²
U = 3 J
Thus, the total work done on it during the compression is 3 J
A parallel-plate capacitor having circular plates of radius R and separation d is charged to a potential difference by a battery. It is then removed from the battery.If the plates are moved closer together (there may be more than one correct choice),
A. the amount of charge on each of them will increase
B. the amount of charge on each of them will decrease
C. the amount of charge on each of them will stay the same
D. the energy stored in the capacitor increases
A small charged bead has a mass of 3.6 g. It is held in a uniform electric field E = (200,000 N/C, up). When the bead is released, it accelerates upward with an acceleration of 24 m/s^2. What is the charge on the bead?
Answer:
The charge on the bead is [tex]q = 6.084 *10^{-7}\ C[/tex]
Explanation:
From the question we are told that
The mass of the bead is [tex]m = 3.6 \ g = 0.0036 \ kg[/tex]
The magnitude of the electric field is [tex]E = 200,000 \ N/C[/tex]
The acceleration of the bead is [tex]a = 24 m/s^2[/tex]
Generally, the electric force on the bead is mathematically represented as
[tex]F_ e = q E[/tex]
Where q is the charge on the bead
Now the gravitational force opposing the upward movement of the bead is mathematically represented as
[tex]F_g = mg[/tex]
Generally the net force on the bead is mathematically represented as
[tex]F = F_e - F_g = m* a[/tex]
=> [tex]qE - mg = ma[/tex]
Now substituting values
[tex]q * 200000 - 0.0036 *9.8 = 0.0036 * 24[/tex]
[tex]q = 6.084 *10^{-7}\ C[/tex]
A rotating flywheel can be used as a method to store energy. If it is required that such a device be able to store up to a maximum of 2.00 x 106 J when rotating at 443 rad/s, what moment of inertia is required
Answer:
Moment of inertia of the flywheel is equal to 10.19 kg-m^2
Explanation:
The maximum rotational energy to be stored by the flywheel [tex]E_{r}[/tex] = 2.00 x 10^6 J
Angular speed with which to store this energy ω = 443 rad/s
moment of inertia of the flywheel [tex]I[/tex] = ?
Recall that the energy of a rotating body is gotten from the equation
[tex]E_{r} = Iw^{2}[/tex]
Where [tex]E_{r}[/tex] is the rotational energy of the rotating body
[tex]I[/tex] = moment of inertia of the body
ω = angular speed of the rotating body
imputing the values into the equation, we'll have
2.00 x 10^6 = [tex]I[/tex] x [tex]443^{2}[/tex]
2.00 x 10^6 = [tex]I[/tex] x 196249
[tex]I[/tex] = (2.00 x 10^6) ÷ 196249 = 10.19 kg-m^2
A horizontal spring with spring constant 290 N/m is compressed by 10 cm and then used to launch a 300 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?
Answer:
Explanation:
check it out and rate me
. A mass m is traveling at an initial speed of 25.0 m/s. It is brought to rest in a distance of 62.5 m by a net force of 15.0 N. The mass is
Answer:
m = 3 kg
The mass m is 3 kg
Explanation:
From the equations of motion;
s = 0.5(u+v)t
Making t thr subject of formula;
t = 2s/(u+v)
t = time taken
s = distance travelled during deceleration = 62.5 m
u = initial speed = 25 m/s
v = final velocity = 0
Substituting the given values;
t = (2×62.5)/(25+0)
t = 5
Since, t = 5 the acceleration during this period is;
acceleration a = ∆v/t = (v-u)/t
a = (25)/5
a = 5 m/s^2
Force F = mass × acceleration
F = ma
Making m the subject of formula;
m = F/a
net force F = 15.0N
Substituting the values
m = 15/5
m = 3 kg
The mass m is 3 kg
The specific heat of a certain type of cooking oil is 1.75J/g°C. How much heat energy is needed to raise the temperature of 2.63kg of this oil from g
Complete question:
The specific heat of a certain type of cooking oil is 1.75J/g°C. How much heat energy is needed to raise the temperature of 2.63 kg of this oil from 23 °C to 191 °C?
Answer:
The heat energy required to raise the temperature of this cooking oil is 773220 J
Explanation:
Given;
mass of the cooking oil, m = 2.63 kg = 2630 g
specific heat capacity of the cooking oil, C = 1.75J/g°C
initial temperature of the cooking oil, T₁ = 23° C
final temperature of the cooking oil, T₂ = 191° C
The heat energy required to raise the temperature of this cooking oil;
Q = mcΔT
Q = 2630 x 1.75 x (191 - 23)
Q = 773220 J
Therefore, the heat energy required to raise the temperature of this cooking oil is 773220 J
A copper transmission cable 180 km long and 11.0 cm in diameter carries a current of 135 A.
Required:
a. What is the potential drop across the cable?
b. How much electrical energy is dissipated as thermal energy every hour?
Answer:
a) 43.98 V
b) E = 21.37 MJ
Explanation:
Parameters given:
Length of cable = 180 km = 180000 m
Diameter of cable = 11 cm = 0.11 m
Radius = 0.11 / 2 = 0.055 m
Current, I = 135 A
a) To find the potential drop, we have to find the voltage across the wire:
V = IR
=> V = IρL / A
where R = resistance
L = length of cable
A = cross-sectional area
ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm
Therefore:
V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)
V = 43.98 V
The potential drop across the cable is 43.98 V
b) Electrical energy is given as:
E = IVt
where t = time taken = 1 hour = 3600 s
Therefore, the energy dissipated per hour is:
E = 135 * 43.98 * 3600
E = 21.37 MJ (mega joules, 10^6)
When you stretch a spring 13 cm past its natural length, it exerts a force of 21
N. What is the spring constant of this spring?
A. 1.6 N/cm
B. 273 N/cm
C. 0.8 N/cm
D. 13 N/cm
Answer:
A. 1.6 N/cm
Explanation:
spring constant = 21/13 = 1.6 N/cm
Which characteristic gives the most information about what kind of element an atom is ?
Answer:
The atomic number
Explanation:
You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the base need to know the acceleration of a freely falling object at the surface of the planet in order to adequately design the structures. The Omicronians have told you that the value is gOP7=7.29 flurggrom2, but your architects use the units metersecond2, and from your previous experience you know that both the Omicronians and your architects are terrible at unit conversion. Thus, it's up to you to do the unit conversion. Fortunately, you know the unit equality relationships: 5.24flurg=1meter and 1grom=0.493second. What is the value of gOP7 in the units your architects will use, in meter-second2?
Answer:
5.724 meters / second^2
Explanation:
We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units, 5.25 flurg / meter, and 0.493 second / grom.
_____
We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.
_____
( 1 ) ( flurg / grom^2 ) / ( flurg / meters ) - first conversion unit
= flurg / grom^2 * meters /flurg
= ( meters * flurg ) / ( grom^2 * flurg )
= meters /grom^2,
7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2
( 2 ) ( meter / grom^2 ) / ( second / grom ) - second conversion unit
= meter / grom^2 * grom / second
= ( meter * grom ) / ( grom^2 * second )
= meter / ( grom * second ),
( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter / grom * second
( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2
( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )
The value of gOP7 in the units your architects will use is 5.724 [tex]m/s^2[/tex]
Given that 5.24 flurg = 1 meter,
1 grom = 0.493 second.
First, we will convert the length units:
[tex]7.29 flurg / grom^2 / 5.24 flurg / meter = 1.39 meter / grom^2[/tex]
Now we convert the time units:
[tex]1.39 meter / grom^2 / 0.493 second / grom = 2.82195 meter / grom * second[/tex]
[tex]2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2[/tex]
The value of gOP7 in the units the architects will use is [tex]5.724m/s^2[/tex]
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A mass of a spring vibrates in simple harmonic motion at a frequency of 8.0 Hz and an amplitude of 3.9 cm. If a timer is started with its displacement is a maximum, what is the speed of the mass when the timer reads 3 seconds
Answer:
e = 50.27 give / s
Explanation:
The expression for simple harmonic motion is
x = A cos (wt + Ф)
in this case they give us the amplitude A = 3.9 cm and frequency f = 8.0 Hz
The angular and linФear variables are related
e = 2π d
e = 2π 8
e = 50.27 give / s
let's look for the constant fi
so let's find the time to have the maximum displacement
v = dx / dt
v = -A w sin (wt +Ф)
for the point of maximum displacement the speed is I think
0 = - sin (0 + Ф)
therefore fi = 0
Let's put together the equation of motion
x = 0.039 sin (50.27 t)
v = 0.039 50.27 sin (50.27 3)
v = 1.96 50 0.01355
v = 0.0266 m / s
In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources was found to be equal to 7.5 x 10-4 m. Find the wavelength of the source of light if the eye-piece has to be moved transversely through a distance of 1.88 cm for 10 fringes.
Answer:
λ = 1.4 × 10^(-7) m
Explanation:
We are given;
distance of eye piece from the source;D = 1.5 m
distance between the virtual sources;d = 7.5 × 10^(-4) m
To find the wavelength, we will use the formula for fringe width;
X = λD/d
Where X is fringe width, λ is wavelength, while d and D remain as before.
Now, fringe width = eye-piece distance moved transversely/number of fringes
Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m
Thus,
Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m
Thus;
1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))
λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5
λ = 1.4 × 10^(-7) m
In an experiment to measure the acceleration due to gravity, g two values, 9.96 m/s2 and 9.72 m/s2 , are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value: g 5 9.80 m/s2 .)
Answer:
(1) Percent Difference = 2.47%
(2) Percent Error (9.96 m/s²) = 1.63 %
Percent Error (9.72 m/s²) = 0.82 %
(3) Percent Error (Mean) = 0.41 %
Explanation:
(1)
Percent Difference = [(9.96 m/s² - 9.72 m/s²)/(9.72 m/s²)]*100 %
Percent Difference = 2.47%
(2)
Percent Error = (|Measured Value - Original Value|/Original Value)*100%
Therefore,
Percent Error (9.96 m/s²) = (|9.96 m/s² - 9.8 m/s²|/9.8 m/s²)*100%
Percent Error (9.96 m/s²) = 1.63 %
Now,
Percent Error (9.72 m/s²) = (|9.72 m/s² - 9.8 m/s²|/9.8 m/s²)*100%
Percent Error (9.72 m/s²) = 0.82 %
(3)
First we need to find the mean of values:
Mean = (9.96 m/s² + 9.72 m/s²)/2
Mean = 9.84 m/s²
Therefore,
Percent Error (Mean) = (|9.84 m/s² - 9.8 m/s²|/9.8 m/s²)*100%
Percent Error (Mean) = 0.41 %
If a water wave completes one cycle in 2 seconds, what is
the period of the wave?
0.5 seconds
O4 seconds
2 seconds
0.2 seconds
Done
The period of a wave is the time it takes the wave to complete one cycle (at a fixed location).
So if a wave completes one cycle in 2 seconds, then that is its period.
A 54.0 kg ice skater is moving at 3.98 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.802 m around the pole.
(a) Determine the force exerted by the horizontal rope on her arms.N
(b) What is the ratio of this force to her weight?(force from part a / her weight)
Answer:
(a) force is 1066.56N
Explanation:
(a) MV²/R
WILL MARK THE BRAINLIEST!!! The diagram shows a carrier wave that is used to transmit information. Which best illustrates how the carrier wave would likely appear after pulse modulation?
Answer:
Second image in your list of possible answers
Explanation:
The second option is what you would expect from modulating a sinusoidal carrier wave of higher frequency after being modulated by a square pulse of lower frequency that allows part of the carrier signal to travel during the time the square signal is constant different from zero, and be absent (flat) during the time the square pulse signal has amplitude zero.
The second line is the best picture of a pulse-modulated carrier.
It would be easy to build a circuit where each pulse ... when it comes along ... just switches the carrier OFF for as long as the pulse lasts.
If the set W is a vector space, find a set S of vectors that spans it. Otherwise, state that W is not a vector space. W is the set of all vectors of the form [a - 4b 5 4a + b -a - b], where a and bare arbitrary real numbers.
a. [1 5 4 -1], [-4 0 1 -1]
b. [1 0 4 -1], [-4 5 1 -1]
c. [1 0 4 -1], [-4 0 1 -1], [0 5 0 0]
d. Not a vector space
Answer:
Choice d. The set of vectors: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex] isn't a vector space over [tex]\mathbb{R}[/tex].
Explanation:
Let a set of vectors [tex]V[/tex] to be a vector field over some field [tex]\mathbb{F}[/tex] (for this question, that "field" is the set of all real number.) The following must be true:
The set of vectors [tex]V[/tex] includes the identity element [tex]\mathbf{0}[/tex]. In other words, there exists a vector [tex]\mathbf{0} \in V[/tex] such that for all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].[tex]V[/tex] should be closed under vector addition. In other words, for all [tex]\mathbf{u},\, \mathbf{v} \in V[/tex], [tex]\mathbf{u} + \mathbf{v} \in V[/tex].[tex]V[/tex] should also be closed under scalar multiplication. In other words, for all [tex]\mathbf{v} \in V[/tex] and all "scalar" [tex]m \in \mathbb{F}[/tex] (in this question, the "field" is the set of all real numbers, so [tex]m[/tex] can be any real number,) [tex]a\,\mathbf{v} \in V[/tex].Note that in the general form of a vector in [tex]V[/tex], the second component is a always non-zero. Because of that non-zero component,
Assume by contradiction that [tex]V[/tex] is indeed a vector field. Therefore, it should contain a zero vector. Let [tex]\mathbf{0}[/tex] denote that zero vector. For all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].
Using the definition of set [tex]V[/tex]: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex], there exist real numbers [tex]a[/tex] and [tex]b[/tex], such that:
[tex]\displaystyle \mathbf{v} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].
Hence, [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex] is equivalent to:
[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} + \mathbf{0} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].
Apply the third property that [tex]V[/tex] is closed under scalar multiplication. [tex]-1[/tex] is indeed a real number. Therefore, if [tex]\mathbf{v}[/tex] is in
Therefore:
[tex]\displaystyle -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} \in V[/tex].
Apply the second property and add [tex]\displaystyle - \mathbf{v} = -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex] to both sides of [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex]. The left-hand side becomes:
[tex]\mathbf{v} - \mathbf{v} + \mathbf{0} = \mathbf{0}[/tex].
The right-hand side becomes:
[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} = \begin{bmatrix}a - 4\, b - (a - 4\, b) \\ 5 - 5 \\ 4\, a+ b-(4\, a+ b)\\ -a -b - (-a -b)\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].
Therefore:
[tex]\displaystyle \mathbf{0} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].
However, [tex]\mathbf{0} = \displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex] isn't a member of the set [tex]\displaystyle V = \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex]. That's a contradiction, because [tex]\mathbf{0}[/tex] was supposed to be part of [tex]V[/tex].
Hence, [tex]V[/tex] isn't a vector space by contradiction.
Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.547 T.
Answer:
Wavelength is 0.359 m
Explanation:
Given that,
Magnetic field, B = 0.547 T
We need to find the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.547 T.
The frequency of revolution of proton in the cyclotron is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
m is mass of proton
q is charge on proton
So,
[tex]f=\dfrac{1.6\times 10^{-19}\times 0.547}{2\pi \times 1.67\times 10^{-27}}\\\\f=8.34\times 10^6\ Hz[/tex]
We know that,
Speed of light, [tex]c=f\lambda[/tex]
[tex]\lambda[/tex] = wavelength
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{834\times 10^6}\\\\\lambda=0.359\ m[/tex]
So, the wavelength of the radiation produced by a proton is 0.359 m.
A parallel-plate capacitor has a plate separation of 1.5 mm and is charged to 450 V. 1) If an electron leaves the negative plate, starting from rest, how fast is it going when it hits the positive plate
Answer:
Explanation:
this is the answer to your question
The electron is going with a velocity of 1.25 × 10⁷ m/s when it hits the positive plate.
What is law of the conservation of mechanical energy?According to the law of the conservation of mechanical energy, the total mechanical energy is always conserved by an electron. We can say that the sum of potential energy (U) and kinetic energy (K) is always constant.
K + U = E
Given, the distance between the two parallel plates = 1.5 mm
The potential difference between the plates, V = 450V
The charge on an electron, q = [tex]-1.6\times 10^{-19} C[/tex]
The mass of an electron, m = 9.1× 10⁻³¹ Kg
The change in the potential energy of the charge moving through the potential difference of 450V.
ΔU = qΔV = (-1.6× 10⁻¹⁹)(450) = -7.2 × 10⁻¹⁷J
From the law of the conservation of mechanical energy, we can write:
K + U = E
ΔK + ΔU = 0
ΔK = -ΔU
1/2mv² = -ΔU
v² = -2ΔU/m
[tex]v^2 =\frac{-2\times (-7.2\times 10{-17})}{9.1\times 10^{-31}}[/tex]
[tex]v=\sqrt{1.58\times 10^{14}}[/tex]
v = 1.25 × 10⁷ m/s
Therefore, the electron is going with the speed of 1.25 × 10⁷ m/s when it hits the positive plate.
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A wire has an electric field of 6.2 V/m and carries a current density of 2.4 x 108 A/m2. What is its resistivity
Answer:
The resistivity is [tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 6.2 V/m[/tex]
The current density is [tex]J = 2.4 *10^{8} \ A/m^2[/tex]
Generally the resistivity is mathematically represented as
[tex]\rho = \frac{E}{J}[/tex]
substituting values
[tex]\rho = \frac{6.2}{2.4 *10^{8}}[/tex]
[tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]
"KATZPSEF1 7.P.053.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Two black holes (the remains of exploded stars), separated by a distance of 10.0 AU (1 AU = 1.50 ✕ 1011 m), attract one another with a gravitational force of 7.70 ✕ 1025 N. The combined mass of the two black holes is 6.00 ✕ 1030 kg. What is the mass of each black hole?"
Answer:
There are two possible solutions.
M1 = 4.68*10^30kg, M2 = 5.53*10^30kg
M1 = 5.53*10^30kg, M2 = 4.7*10^29kg
Explanation:
In order to find the mass of each black hole, you take into account the gravitational force between them and the sum of their masses.
You use the formula for the gravitational force between two masses:
[tex]F_g=G\frac{M_1M_2}{r^2}[/tex] (1)
G: Cavendish's constant = 6.674*10^-11 m^3/kg.s^2
M1, M2: mass of each black hole = ?
r: distance between the black holes = 10.0 AU = 10.0(1.50*10^11m) = 1.5*10^12m
Fg: gravitational force between the black holes = 7.70*10^25N
Furthermore, you take into account that the sum of the masses M1 and M2 is:
M1 + M2 = 6.00*10^30 kg (2)
You solve the equation (2) for M2.
[tex]M_2=6.00*10^{30}-M_1[/tex]
Next, you replace the obtained expression for M2 into the equation (1) and solve for M1, as follow (for simplicity, you do not add the units):
[tex]F_g=G\frac{M_1(6.00*10^{30}-M_1)}{r^2}\\\\\frac{r^2F_g}{G}=6.00*10^{30}M_1-M_1^2\\\\\frac{(1.5*10^{12})^2(7.70*10^{25})}{(6.674*10^{-11}}=6.00*10^{30}M_1-M_1^2\\\\2.59*10^{60}=6.00*10^{30}M_1-M_1^2\\\\M_1^2-6.00*10^{30}M_1+2.59*10^{60}=0[/tex]
Then, you have obtained a quadratic polynomial. You solve it with the quadratic formula:
[tex]M_1=\frac{-(-6.00*10^{30})\pm \sqrt{(-6.00*10^{30})^2-4(1)(2.59*10^{60}))}}{2(1)}\\\\M_1=\frac{6.00*10^{30}\pm 5.06*10^{30}}{2}\\\\M_1=4.68*10^{29}\\\\M_1=5.53*10^{30}[/tex]
Both results are consistent, then the mass of one black hole can be 4.68*10^30kg and also 5.53*10^30kg.
The other black hole has a mass of:
[tex]M_2=6.00*10^{30}kg-4.68*10^{29}kg=5.53*10^{30}kg\\\\M_2=6.00*10^{30}kg-5.53*10^{30}kg=4.7*10^{29}kg[/tex]
Hence, you have a pair of solutions:
M1 = 4.68*10^30kg, M2 = 5.53*10^30kg
M1 = 5.53*10^30kg, M2 = 4.7*10^29kg
a ring with a clockwise current is situated with its center directly above another ring. The current in the top ring is decreasing. What is the directiong of the induced current in the bottom ring
Answer:
clockwise
Explanation:
when current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil.
Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.
The direction of the induced current in the bottom ring is in the clockwise direction.
The given problem is based on the concept and fundamentals of the induced current and the direction of flow of the induced current.
When current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil. Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.Thus, we can conclude that the direction of the induced current in the bottom ring is in the clockwise direction.
Learn more about the concept of induced current here:
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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 60.0 cm. An electron is released from rest at a point midway between the charges and moves along the line connecting them. Part A What is the electric potential energy of the electron when it is at the midpoint
Answer:
U =-2.39*10^-18 J
Explanation:
In order to calculate the electric potential energy of the electron you use the following formula:
[tex]U=k\frac{q_1q_2}{r}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r: distance between charges
In this case the electron is at point midway between two charges, then the electric potential energy is the sum of two contributions:
[tex]U=U_1+U_2=k\frac{eq_1}{r}+k\frac{eq_2}{r}=\frac{ke}{r}[q_1+q_2][/tex]
e: charge of the electron = 1.6*10^-19C
q1: charge 1 = 3.00nC = 3.00*10^-9C
q2: charge 2 = 2.00nC = 3.00*10^-9C
r: distance to each charge = 60.0cm/2 = 30.0cm = 0.3m
If you consider that the electron is at the origin of coordinates, with the first charge in the negative x axis, and the other one in the positive x axis, you have:
[tex]U=\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{0.6m}[-3.0*10^{-9}C+2.0*10^{-9}C]\\\\U=-2.39*10^{-18}J[/tex]
The electric potential energy of the electron is -2.39*10^-18 J
The process by which energy is realized is known as?
Answer:
did you mean released?
Explanation:
If so the process is called respiration
a beam of 1mev electrons strike a thick target. for a beam current of 100 microampere, find the power dissipated in the target
Answer:
power dissipated in the target is 100 W
Explanation:
given data
electrons = 1 mev = [tex]10^{6}[/tex] eV
1 eV = 1.6 × [tex]10^{-19}[/tex] J
current = 100 microampere = 100 × [tex]10^{-6}[/tex] A
solution
when energy of beam strike with 1 MeV so energy of electron is
E = e × v ...................1
e is charge of electron and v is voltage
so put here value and we get voltage
v = 1 ÷ 1.6 × [tex]10^{-19}[/tex]
v = [tex]10^{6}[/tex] volt
so power dissipated in target
P = voltage × current ..............2
put here value
P = [tex]10^{6}[/tex] × 100 × [tex]10^{-6}[/tex]
P = 100 W
so power dissipated in the target is 100 W
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F ’= F 0.25
Explanation:
This problem refers to the electric force, which is described by Coulomb's law
F = k q₁ q₂ / r²
where k is the Coulomb constant, q the charges and r the separation between them.
The initial conditions are
F = k q_A q_B / d²
they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d
F ’= k (q / 4 q / 4) / (0.5 d)²
F ’= k q / 16 / 0.25 d²
F ’= k q² / d² 0.0625 / 0.25
F ’= F 0.25
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.
What is Coulomb's law?Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]
where,
[tex]q_A [/tex] and [tex]q_B[/tex] are the charges of A and B (and equal to q).k is the Coulomb's constant.If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:
[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
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1.3kg of gold at 300K comes in thermal contact with 2.4kg copper at 400K. The specific heats of Au and Cu are 126 J/kg-K and 386 J/kg-K respectively. What equilibrium temperature do they reach
Answer:
The final temperature of the metals will be 384.97 K
Explanation:
For the gold;
mass = 1.3 kg
temperature = 300 K
specific heat = 126 J/kg-K
For the copper;
mass = 2.4 kg
temperature = 400 K
specific heat = 386 J/kg-K
Firstly, we will have to calculate for the thermal energy possessed by each of the metal.
The heat possessed by a body = mcT
Where,
m is the mass of the body
c is the specific heat of the body, and
T is the temperature of the body at that instance
so we calculate for the thermal energy of the gold and the copper below
For gold;
heat energy = mcT = 1.3 x 126 x 300 = 49140 J
For copper;
heat energy = mcT = 2.4 x 386 x 400 = 370560 J
When the two metal come in thermal contact, this heat is evenly distributed between them.
The total heat energy = 49140 J + 370560 J = 419700 J
At thermal equilibrium, the two metals will be at the same temperature, to get this temperature, we equate the total thermal energy to the heat energy that will be possessed by the metals at equilibrium.
419700 = (1.3 x 126 x T) + (2.4 x 386 x T) = 163.8T + 926.4T
419700 = 1090.2T
T = 419700/1090.2 = 384.97 K
The final temperature of the metals will be 384.97 K