Answer:
5.5 km
Explanation:
First, we convert the distance from km/h to m/s
910 * 1000/3600
= 252.78 m/s
Now, we use the formula v²/r = gtanθ to get our needed radius
making r the subject of the formula, we have
r = v²/gtanθ, where
r = radius of curvature needed
g = acceleration due to gravity
θ = angle of banking
r = 252.78² / (9.8 * tan 50)
r = 63897.73 / (9.8 * 1.19)
r = 63897.73 / 11.662
r = 5479 m or 5.5 km
Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km
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:
https://brainly.com/question/3712635
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.
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
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.
Find the force on a proton moving with velocity (2i+3j+4k)10^5m/s in a uniform magnetic field of 0.5k T. What is the angle between the magnetic field lines and the velocity?
Answer:
Explanation:
Force on charge particles
F = q ( v x B )
= 1.6 x 10⁻¹⁹ x [ ( 2i+3j+4k) x .5k ] x 10⁵
= 1.6 x 10⁻¹⁴ x ( 1.5 i - j )
= (2.4 i - 1.6 j ) x 10⁻¹⁴ N
magnitude of this vector
= 2.88 x 10⁻¹⁴ N
Angle between B and v
cosθ = [tex]\frac{(2i+3j+4k).(.5k)}{\sqrt{2^2+3^2+4^2}\times .5 }[/tex]
= [tex]\frac{2}{2.69}[/tex]
cosθ = .74
θ = 42° .
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.
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)
Which coefficient of O2 is needed to balance the equation C3H8 + O2 → 3CO2 + 4H2O?
Answer:
5
Explanation:
Count the number of moles of O in the product side:
3x2 + 4
= 10
Divide it by 2 since O exists as diatomic molecule (O2)
10/2 = 5
Therefore, the coefficient required is 5.
Answer:
c.) 5
Explanation:
A wire of radius 0.6 cm carries a current of 94 A that is uniformly distributed over its cross-sectional area. Find the magnetic field B at a distance of 0.2 cm from the center of the wire.
Answer:
0.00299T
Explanation:
The magnetic field B = 0.00299T
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]
You measure the current through a 27.7 Ω resistor to be 753 mA . What is the potential difference across the contacts of the resistor?
Answer:
20.9 volts
Explanation:
R = 27.7 Ω
I = 753 mA = 0.753 A
V = ?
From Ohms law, V = IR
V = 0.753×27.7
V = 20.8581
V = 20.9 volts
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat frequency is generated by the first overtone of the closed pipe with the fundamental of the open pipe
Answer:
fb = 240.35 Hz
Explanation:
In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.
Open tube:
[tex]f=\frac{v_s}{2L}[/tex] (1)
vs: speed of sound = 343m/s
L: length of the open tube = 0.47328m
You replace in the equation (1):
[tex]f=\frac{343m/s}{2(0.47228m)}=362.36Hz[/tex]
Closed tube:
[tex]f'=\frac{v_s}{4L'}[/tex]
L': length of the closed tube = 0.702821m
[tex]f'=\frac{343m/s}{4(0.702821m)}=122.00Hz[/tex]
Next, you use the following formula for the beat frequency:
[tex]f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz[/tex]
The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz
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
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]
Learn more :
https://brainly.com/question/17192728
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.
Learn more about Coulomb's law here: https://brainly.com/question/506926
A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initial
speed of 32 m/s, what is its new speed?
A. 17 m/s
B. 15 m/s
C. 47 m/s
D. 32 m/s
Answer:
17 m/s
Explanation:
Using formula a = (v-u) /t
acceleration a = -1.5 m/s2
final velocity v = unknown
initial velocity u = 32 m/s
time t = 10s
-1.5 = (v-32)/10
-15 = v - 32
-15 + 32 = v
v = 17 m/s
What is the one single most important reason that human impact on the planet has been so great?
Answer:
Increasing population
Explanation:
As we can see that the death rate is decreasing while at the same time the birth rate is increasing due to which it increased the population that directly impact the planet so great
Day by day the population of the villages, cities, states, the country is increasing which would create a direct human impact on the planet
Therefore the increasing population is the one and single most important reason
a fixed amount of ideal gas is held in a rigid container that expands negligibly when heated. at 20 the gas pressure is p. if we added enough heat to increase the temperature from 20 to 40, the pressure will be
Answer:
When the temperature of the gas is increased from 20 to 40, the pressure will be 2p
Explanation:
Given;
initial temperature of the gas, T₁ = 20 K
final temperature of the gas, T₂ = 40 k
initial pressure of the gas, P₁ = P
final pressure of the gas, P₂ = ?
Apply pressure law of gases;
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} \\\\P_2 = \frac{40P}{20} \\\\P_2 = 2P[/tex]
Therefore, when the temperature of the gas is increased from 20 to 40, the pressure will be 2p
An electric current through neon gas produces several distinct wavelengths of visible light. What are the wavelengths (in nm) of the neon spectrum, if they form first-order maxima at angles of 49.67°, 50.65°, 52.06°, and 52.89° when projected on a diffraction grating having 11,000 lines per centimeter? (Round your answers to the nearest nanometer. Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign. Enter your answers from smallest to largest.)
Answer:
Explanation:
If a be grating element
a = 1 x 10⁻² / 11000
= .0909 x 10⁻⁵
= 909 x 10⁻⁹ m
for first order maxima , the condition is
a sinθ = λ where λ is wavelength
909 x 10⁻⁹ sin 49.67 = λ₁
λ₁ = 692.95 nm .
λ₂ = 909 x 10⁻⁹ sin 50.65
= 702.91 nm
λ₃ = 909 x 10⁻⁹ sin 52.06
= 716.88 nm
λ₄ = 909 x 10⁻⁹ sin 52.89
= 724.90 nm
692.95 nm , 702.91 nm , 716.88 nm , 724.90 nm .
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]
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
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
"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
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
Quadrupling the power output from a speaker emitting a single frequency will result in what increase in loudness (in units of dB)
Answer:
6.02 dB increase
Explanation:
Let us take the initial power from the speaker P' = P Watt
then, the final power P = 4P Watt
for a given unit area, initial intensity (power per unit area) will be
I' = P Watt/m^2
and the final quadrupled sound will produce a sound intensity of
I = 4P Watt/m^2
Increase in loudness is gotten from the relation
ΔL = [tex]10log_{10} \frac{I}{I'}[/tex]
where
I = final sound intensity
I' = initial sound intensity
imputing values of the intensity into the equation, we have
==> [tex]10log_{10} \frac{4P}{P}[/tex] = [tex]10log_{10} 4[/tex] = 6.02 dB increase
Four forces act on bolt A as shown; F1 150N, F2 80N, F3 110N and F4 100N. Determine the magnitude and direction of the resultant of the forces of the bolt, A.
Complete Question
The complete question(reference (chegg)) is shown on the first uploaded image
Answer:
The magnitude of the resultant force is [tex]F = 199.64 \ N[/tex]
The direction of the resultant force is [tex]\theta = 4.1075^o[/tex] from the horizontal plane
Explanation:
Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be [tex]Fcos(\theta )[/tex] while if the force is moving away from the angle then the resolved force is [tex]Fsin (\theta )[/tex]
Now from the diagram let resolve the forces to their horizontal component
So
[tex]\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)[/tex]
[tex]\sum F_x = 199.128 \ N[/tex]
Now resolving these force into their vertical component can be mathematically evaluated as
[tex]\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)[/tex]
[tex]\sum F_{y} = 14.30[/tex]
Now the resultant force is mathematically evaluated as
[tex]F = \sqrt{F_x^2 + F_y^2}[/tex]
substituting values
[tex]F = \sqrt{199.128^2 + 14.3^2}[/tex]
[tex]F = 199.64 \ N[/tex]
The direction of the resultant force is evaluated as
[tex]\theta = tan^{-1}[\frac{F_y}{F_x} ][/tex]
substituting values
[tex]\theta = tan^{-1}[\frac{ 14.3}{199.128} ][/tex]
[tex]\theta = 4.1075^o[/tex] from the horizontal plane
Wind erosion can be reduced by _____.
. 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
Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm, 235U, 238U, 40K, or 14C? There may be more than one answer that is appropriate. Explain your reasoning for why the remaining scenario(s) would be inappropriate/impossible to use that particular isotope. Answers should include a discussion on usable ages for each system and whether the necessary isotopes would be found in the material to be dated.
a. A meteorite that formed early in the formation of the solar system.
b. A rock formed through a mountain building event around 420 million years ago.
c. Volcanic ash from an eruption 60 million years ago.
d. An earthquake scarp that formed along the San Andreas Fault 50 years ago.
e. An Incan archaeological dig site in the highlands of Peru.
f. A tree from a forest in England that is suspected to be the oldest in the British Isles.
Answer:
a) 238U, 40K and 87Rb, b) 235U and to a lesser extent 40K , c) he 235U,
d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful
Explanation:
One of the applications of radioactive decay is the dating of different systems.
To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.
Let's write the half-life times of the given materials
87Rb T ½ = 4.75 1010 years
147Sm T ½ = 1.06 1011 years
235U = 7,038 108 years
238U = 4.47 109 years
40K = 1,248 109 years
14C = 5,568 103 years
we already have the half-life of the different elements given
a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate
b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed
c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent
d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed
e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed
f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed
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