Answer:
current I = 38 mA
Explanation:
given data
distance r = 7.2 cm
repel each other force per unit length \frac{F}{l} = 4.2 nN/m
solution
we know 2 wire is parallel and when current flow through these wire they exert force each other due to magnetic field
and current I(1) = I(2) ................1
so
[tex]\frac{F}{l} = \frac{\mu _o}{2\pi } \times \frac{I(1) \times I(2)}{r}[/tex] ..................2
put here value
4.2 × [tex]10^{-9}[/tex] = [tex]\frac{4\pi \times 10^{-7}}{2\pi } \times \frac{I^2}{7.2\times 10^{-2}}[/tex]
solve it we get
I = 0.038884 A
current I = 38 mA
The current flow in the wires will be:
"38 mA".
Force, Current and DistanceAccording to the question,
Distance, r = 7.2 cm
Force per unit length, [tex]\frac{F}{l}[/tex] = 4.2 nN/m
Current passes, when wire is parallel:
→ I₁ = i₂
We know the relation,
→ [tex]\frac{F}{l}[/tex] = [tex]\frac{\mu_0}{2 \pi}\times \frac{I_1\times I_2}{r}[/tex]
By substituting the values, we get
4.2 × 10⁻⁹ = [tex]\frac{4 \pi\times 10^{-7}}{2 \pi}\times \frac{I^2}{7.2\times 10^{-2}}[/tex]
hence,
The current will be:
I = 0.038884 A or,
= 38 mA
Thus the above answer is appropriate.
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A non-reflective coating that has a thickness of 198 nm (n = 1.45) is deposited on top of a substrate of glass (n = 1.50). What wavelength of visible light is most strongly transmitted if it is illuminated perpendicular to its surface?
Answer:
The wavelength is [tex]\lambda_ 1 = 574.2 nm[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 198 nm = 198 *10^{-9 }\ m[/tex]
The refractive index of the non-reflective coating is [tex]n_m = 1.45[/tex]
The refractive index of glass is [tex]n_g = 1.50[/tex]
Generally the condition for destructive interference is mathematically represented as
[tex]2 * n_m * t * cos (\theta) = n * \lambda[/tex]
Where [tex]\thata[/tex] [tex]\theta[/tex] is the angle of refraction which is 0° when the light is strongly transmitted
and n is the order maximum interference
so
[tex]\lambda = \frac{2 * n * t * cos (\theta )}{n}[/tex]
at the point n = 1
[tex]\lambda _1 = \frac{2 * 1.45 * 198*10^{-9} * cos (0 )}{1}[/tex]
[tex]\lambda_1 = 574.2 *10^{-9}[/tex]
[tex]\lambda_1 = 574.2 nm[/tex]
at n =2
[tex]\lambda _2 = \frac{\lambda _1 }{2}[/tex]
[tex]\lambda _2 = \frac{574.2*10^{-9} }{2}[/tex]
[tex]\lambda _2 = 2.87 1 *10^{-9} \ m[/tex]
[tex]\lambda _2 = 287. 1 nm[/tex]
Now we know that the wavelength range of visible light is between
[tex]390 \ nm \to 700 \ nm[/tex]
So the wavelength of visible light that is been transmitted is
[tex]\lambda_ 1 = 574.2 nm[/tex]
A woman is listening to her radio, which is 174 m from the radio station transmitter. (a) How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz
Explanation:
It is given that,
The distance between the radio and the radio station is 174 m
We need to find how many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz.
f = 1540 kHz
Wavelength,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1540\times 10^3}\\\\\lambda=194.8\ m[/tex]
Let there are n wavelengths of the radio waves. So,
[tex]n=\dfrac{d}{\lambda}\\\\n=\dfrac{174}{194.8}\\\\n=0.89\ \text{wavelengths}[/tex]
There are 0.89 wavelengths.
An evacuated tube uses an accelerating voltage of 50 kV to accelerate electrons to hit a copper plate and produce X-rays. Non-relativistically, what would be the maximum speed (in m/s) of these electrons
Answer:
Explanation:
Accelerating voltage = 50 x 10³ V
energy created = q V where q is charge on electron and V is accelerating potential
So qV = 1/2 m v² , m is mass of electron and v is velocity
Putting values
1.6 x 10⁻¹⁹ x 50 x 10³ = 1/2 x 9.1 x 10⁻³¹ x v²
v² = 175.8 x 10¹⁴
v = 13.25 x 10⁷ m /s
A wheel 2.40 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.40 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.
A. What is the tangential speed?
B. Total acceleration
C. Angular position of point P.
Answer:
Explanation:
Radius of wheel = 1.2 m
A )
To know angular speed after t sec , we use the formula
ω = ω₀ + α t , where ω₀ is initial velocity , α is angular acceleration
ω = 0 + 4.4 x 2
= 8.8 rad / s
v= ωR , v is tangential speed , ω is angular speed , R is radius of wheel .
= 8.8 x 1.2 = 10.56 m /s
B )
radial acceleration
Ar = v² / R
= 10.56² / 1.2
= 92.93 m /s²
Tangential acceleration
At = angular acceleration x radius
= 4.4 x 1.2 = 5.28 m /s²
Total acceleration
= √ ( At² + Ar² )
=√ (5.28² +92.93²)
= 93 m /s²
C )
θ = ωt + 1/2 α t² where θ is angular position after time t .
= 0 + .5 x 4.4 x 2²
= 8.8 rad
= 180x 8.8/ 3.14 = 504.45 degree
initial position = 57.3°
final position = 504 .45 + 57.3
= 561.75 °
= 561.75 - 360
= 201.75 ° .
Position of radius vector of point P will be at angle of 201.75 from horizontal axis .
Wind gusts create ripples on the ocean that have a wavelength of 3.03 cm and propagate at 3.37 m/s. What is their frequency (in Hz)?
Answer:
Their frequency is 111.22 Hz
Explanation:
Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration and is expressed in units of length (m).
Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
The propagation speed of a wave is the quantity that measures the speed at which the wave's disturbance propagates throughout its displacement. The speed at which the wave propagates depends on both the type of wave and the medium through which it propagates. Relate wavelength (λ) and frequency (f) inversely proportional using the following equation:
v = f * λ.
Then the frequency can be calculated as: f=v÷λ
In this case:
λ=3.03 cm=0.0303 m (1m=100 cm)v= 3.37 m/sReplacing:
[tex]f=\frac{3.37 \frac{m}{s} }{0.0303 m}[/tex]
Solving:
f=111.22 Hz
Their frequency is 111.22 Hz
Each charge is equidistant from the origin. In which direction is the net electric field at the point P on the y-axis?
Answer:
"Upwards and towards the left" is the right answer.
Explanation:
The magnitude of the field will be:
⇒ [tex]E=\frac{kq}{r^2}[/tex]
And direction -> for negative charges, to positive charges, except charges.
Charging across the y-axis. It would be up to the aggregate field. Because the x-axis needs to charge. Total production is to the west.Thus the net field is upwards as well as to the left.
Total energy of a particle executing S.H.M of amplitude A is proportional to:
(a)A?
(b) A-2
(c) A
(d) A-
Answer:
If x = A sin w t where w is the angular frequency
then v = w A cos w t
Since KE = 1/2 m Vmax^2 and Vmax = w A maximum KE
the total energy is proportional to A^2
Also, since the maximum potential energy is
PEmax = 1/2 K A^2 where the KE is zero (maximum amplitude)
one can again see that the total energy is proportional to A^2
The bases of developing convective cumulus clouds will be relatively higher at a location with a relatively ______ difference between the surface temperature and surface dew point temperature.
Answer:
large
Explanation:
Cumulus clouds is a term in metrology that defines the type of clouds which are characterized by its low altitude, puffy appearance, and fair-weather nature. They are generally considered as low-level clouds, with less than than 2,000m in altitude except they are the more vertical cumulus congestus form.
Thus, it can be noted that, the difference between the surface dew point temperature and the surface temperature is related to relative humidity. Hence, in a situation when there is a LARGE difference between the surface temperature and the surface dewpoint temperature, then the relative humidity is very low (e.g., 10%).
Therefore, the bases of developing convective cumulus clouds will be relatively higher at a location with a relatively LARGE difference between the surface temperature and surface dew point temperature.
70 kg man walks out on a 10 kg beam that rests on, but is not attached to, two supports. When the beam just starts to tip, what is the force exerted on the beam by the right support
Answer:
The force is [tex]F = 784 \ N[/tex]
Explanation:
From the question we are told that
The mass of the man is [tex]m = 70 \ kg[/tex]
The mass of the beam is [tex]m_b = 10 \ kg[/tex]
Now from the question we can deduce that when this beam start to tip that both the force exerted by the weight of the man and that of the beam is been supported by the right support so
The force exerted on the right support is mathematically evaluated as
[tex]F = (m + m_b) * g[/tex]
substituting values
[tex]F = (70 + 10 ) * 9.8[/tex]
[tex]F = 784 \ N[/tex]
The force exerted on the beam by the right support is 784 Newton.
Given the data in the question;
Mass of the man; [tex]m_m = 70kg[/tex]Mass of the beam; [tex]m_b = 10kg[/tex]Force exerted on the beam by the right support; [tex]F = W = \ ?[/tex]
When the beam just starts to tip, the right support holds up the combined mass of the man and the beam.
Hence;
[tex]M_{net} = m_m + m_b\\\\M_{net} = 70kg + 10kg\\\\M_{net} = 80kg[/tex]
Now, To determine the force exerted on the beam by the right support, we use the general formula for weight or equation of force of gravity which is expressed as:
[tex]F = W = m * g[/tex]
Where m is mass and g represents the acceleration due to gravity( [tex]9.8m/s^2[/tex] )
We substitute our values into the equation
[tex]F = 80kg * 9.8m/s^2\\\\F = 784kg.m/s^2\\\\F = 784N[/tex]
Therefore, the force exerted on the beam by the right support is 784 Newton.
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A handheld glass rod can be charged by rubbing it with silk or a plastic bag while holding it in your hands. Would you conclude from this that glass is a conductor or an insulator? Why?
Answer:
a conductor is an object or type of material that allows the flow of charge (electrical current) in one or more directions.
Explanation:
. Materials made of metal are common electrical conductors.
To water the yard, you use a hose with a diameter of 3.0 cm. Water flows from the hose with a speed of 2.2 m/s. If you partially block the end of the hose so the effective diameter is now 0.50 cm, with what speed does water spray from the hose
Answer:
v₂ = 79.69 m/s
Explanation:
The initial diameter of the hose, d₁ = 3.0 cm = 0.03 m
Initial Cross Sectional Area, A₁ = πd₁²/4
A₁ = (π* 0.03²)/4
A₁ = 0.00071 m²
The initial speed of water from the hose, v₁ = 2.2 m/s
The diameter of the hose after blocking the end, d₂ = 0.50 cm = 0.005 m
Cross Sectional Area of the hose after blocking the end, A₂ = πd₂²/4
A₂ = (π* 0.005²)/4
A₂ = 0.0000196 m²
To get the speed, v₂, at which the water spray from the hose after blocking the end, we will use the continuity equation:
A₁v₁ = A₂v₂
0.00071 * 2.2 = 0.0000196 v₂
0.001562 = 0.0000196 v₂
v₂ = 0.001562/0.0000196
v₂ = 79.69 m/s
A small object with mass 3.80 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.70 m centered at the origin. It starts at the point with position vector 2.70 m. Then it undergoes an angular displacement of 8.70 rad.
(a) What is its new position vector?
in meters
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity?
in m/s
(d) In what direction is it moving?
_____° from the +x direction.
(e) What is its acceleration?
in m/s2
(f) What total force is exerted on the object?
in N
Answer:
Explanation:
angular velocity
ω = 1.65 rad /s
radius R = 2.70 m
angular displacement = 8.70 rad
a )
New position vector in vector form
= R cos8.7 i + R sin8.7 j
= 2.7 cos8.7 i + 2.7 sin8.7 j
= 2.7 x .748 i + 2.7 x .663 j
= 2.01 i + 1.79 j
b )
8.7 radian = 180/π x 8.7 degree
= 498.72 degree
= 498.72 - 360
= 138.72 degree
It will be in second quadrant .
angle made with positive x - axis
= 138.72 degree .
c )
velocity
v = ω R
= 1.65 x 2.7
= 4.455 m /s
d )
It is moving in a direction making 138.72° with positive x direction .
e )
acceleration will be centripetal acceleration
= v²/ R
= 4.455² / 2.7
= 7.35 m /s²
f ) force = mass x acceleration
= 3.8 x 7.35
= 27.93 N .
4. The capacitance of a capacitor is increased by a factor of 1.5 when it is completely filled
with a certain dielectric material. Find the dielectric constant of the material and its
electric susceptibility
Answer:
a. Dielectric constant, ε = 1.5 b. Electric susceptibility, χ = 0.5
Explanation:
a. Dielectric constant
Since the capacitance of the capacitor is increased by a factor of 1.5. Let its initial capacitance be C and its final capacitance after adding the material be C'.
Since C' = εC where ε = relative permittivity,
Also, C' = 1.5C
Comparing both equations for C', ε = 1.5.
Since ε = relative permittivity = dielectric constant,
dielectric constant = 1.5
So, the dielectric constant = 1.5
b. Electric susceptibility
The electric susceptibility χ is given by
χ = ε - 1 where ε = dielectric constant
Since ε = 1.5,
χ = ε - 1
χ = 1.5 - 1
χ = 0.5
So the electric susceptibility χ = 0.5
Consider two coils, with the first coil having twice as many loops as the second. Given the flux Φ though each loop of the first coil due to current in the second coil, what can be said about the flux through each loop of the second coil due to an equal current in the first coil?
Answer:
[tex]$ \phi_{21} = \frac{\phi_{12}}{2} $[/tex]
Which means that the flux through each loop of the second coil is half as much as flux through each loop of first coil.
Explanation:
The flux through each loop of the first coil due to current in the second coil is,
[tex]\phi_{12} = \phi[/tex]
The number of loops in the first coil is
no. of loops = 2N
Total flux passing through the first coil is
[tex]\phi_{12} = 2N\phi[/tex]
The flux through each loop of the second coil due to current in the first coil is,
[tex]\phi_{21} = \phi[/tex]
The number of loops in the second coil is
no. of loops = N
Total flux passing through the second coil is
[tex]\phi_{21} = N\phi[/tex]
Comparing both
[tex]\phi_{12} = \phi_{21} \\\\ 2N\phi = N\phi\\\\\phi_{21} = \frac{\phi_{12}}{2}[/tex]
Which means that the flux through each loop of the second coil is half as much as flux through each loop of first coil.
A spring with k = 35.5 N/m has a mass of 5.50 kg attached to it. An external force F = (4.40 N)sin[(6.80 s−1)t] drives the spring mass system so that it oscillates without any resistive forces. What is the amplitude of the oscillatory motion of the spring-mass system?
Answer:
A = 0.02 m
Explanation:
The spring constant, k = 35.5 N/m
The attached mass, m = 5.50 kg
The expression for the external force, F = (4.40 N)sin[(6.80 s⁻¹)t].....(1)
The general expression for the external force, F = F₀ sin (wt).............(2)
Comparing equations (1) and (2):
The forced frequency, [tex]\omega = 6.80 rad/s[/tex]
F₀ = 4.40 N
The natural frequency can be calculated using the formula:
[tex]\omega_0 = \sqrt{\frac{k}{m} } \\\\\omega_0 = \sqrt{\frac{35.5}{5.5} } \\\\\omega_0 = 2.54 rad/s[/tex]
The amplitude of oscillation of a spring-mass system in the steady state:
[tex]A = \frac{F_0}{m(\omega^2 - \omega_o^2)} \\\\A = \frac{4.4}{5.5(6.8^2 - 2.54^2)} \\\\A = 0.02 m[/tex]
A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential of the surface of this sphere and find it to be 14 kV relative to infinity. Find the excess charge on this sphere.
Answer:
The excess charge is [tex]Q = 3.5 *10^{-7} \ C[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 45 \ cm = 0.45 \ m[/tex]
The potential of the surface is [tex]V = 14 \ kV = 14 *10^{3} \ V[/tex]
The radius of the sphere is
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.45}{2}[/tex]
[tex]r = 0.225 \ m[/tex]
The potential on the surface is mathematically represented as
[tex]V = \frac{k * Q }{r }[/tex]
Where k is coulomb's constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
given from the question that there is no other charge the Q is the excess charge
Thus
[tex]Q = \frac{V* r}{ k}[/tex]
substituting values
[tex]Q = \frac{14 *10^{3} 0.225}{ 9*10^9}[/tex]
[tex]Q = 3.5 *10^{-7} \ C[/tex]
How did the horizontal velocity vector component change during the flight of the cannonball in the simulation
Answer:
The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight
Consider the waveform expression. y(x,t)=ymsin(2.39+693t+0.197x) The transverse displacement ( y ) of a wave is given as a function of position ( x in meters) and time ( t in seconds) by the expression. Determine the wavelength, frequency, period, and phase constant of this waveform.
Answer:
- λ = 31.89
- f = 110.29Hz
- Ф = 2.39
Explanation:
You have the following waveform expression:
[tex]y(x,t)=ym\ sin(2.39+693t+0.197x)[/tex] (1)
The general expression for a wave can be written as:
[tex]y(x,t)=y_o\ sin(kx\pm \omega t+\phi)[/tex] (2)
The sign of the term wt determines the direction of the motion of the wave.
In comparison with the equation (1) you have:
k: wavenumber = 0.197
w: angular frequency = 693
Ф: phase constant of the wave = 2.39
- The wavelength of the wave is given by the following formula:
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{0.197}=31.89m[/tex]
The wavelength of the wave is 31.89m
- The frequency is:
[tex]f=\frac{\omega}{2\pi}=\frac{693}{2\pi}=110.29Hz[/tex]
The frequency of the wave is 110.29Hz
- The phase constant is 2.39
What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m2
Answer:
The transmitted intensity is [tex]I = 40 \ W/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light [tex]I_o = 80 \ W /m^2[/tex]
Generally for a single filter the transmitted intensity of light is mathematically evaluated as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{80}{2}[/tex]
[tex]I = 40 \ W/m^2[/tex]
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. The dipole is initially oriented so that the charge Q is located in the plane that bisects the dipole. Assume that r>>s
Immediately after the dipole is released:
a. What is the magnitude of the force on the dipole?
b. What is the magnitude of the torque on the dipole?
Answer:
a) the magnitude of the force is
F= Q([tex]\frac{kqs}{r^3}[/tex]) and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E = [tex]\frac{kq}{r^{2} }[/tex]
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = [tex]\frac{kp}{r^{3} }[/tex] , where p = q × s
E(r) = [tex]\frac{kqs}{r^{3} }[/tex] where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q([tex]\frac{kqs}{r^3}[/tex])
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Part A: The expression of the force on the dipole is [tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex].
Part B: The expression of the torque on the dipole is [tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex].
How do you calculate the force and torque on the dipole?Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. Also, r>>s.
Part A
The electric field due to dipole at a distance r is given below.
[tex]E_r = \dfrac {qs}{4\pi \epsilon_0 r^3}[/tex]
The magnitude of the force can be given as below.
[tex]F = QE_r[/tex]
[tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex]
Hence the expression of the force on the dipole is [tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex].
Part B
The torque on the dipole will dependent on the perpendicular forces on the dipole. The expression of the torque is given below.
[tex]\tau = F \times r \times sin \theta[/tex]
For the perpendicular forces, θ = 90°. Hence the torque is given below.
[tex]\tau = F\times r[/tex]
[tex]\tau = \dfrac {Qqs}{4 \pi \epsilon_0 r^3} \times r[/tex]
[tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex]
Hence the expression of the torque on the dipole is [tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex].
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At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spaceship's diameter is 35 mm , and give your answer as the time needed for one revolut
Answer:
Time needed for one revolution is 0.38 s
Explanation:
The formula for the frequency of rotation of a spaceship, to create the desired artificial gravity, is as follows:
f = (1/2π)√(a/r)
where,
f = frequency of rotation = ?
a = artificial gravity required = 0.5 g
g = acceleration due to gravity on surface of Earth = 9.8 m/s²
r = radius of ship = 35 mm/2 = 17.5 mm = 17.5 x 10⁻³ m
Therefore,
f = (1/2π)√[(0.5)(9.8 m/s₂)/(17.5 x 10⁻³ m)]
f = 2.66 Hz
Now, for the time required for one revolution, is given as:
Time Period = T = 1/f
T = 1/2.66 Hz
T = 0.38 s
The time required for one revolution to simulate the desired gravity is 0.38 s.
The frequency can be calculate by the formula
[tex]\bold {f = (\dfrac {1}{2\pi})\sqrt{ar}}[/tex]
where,
f - frequency of rotation = ?
a- artificial gravity required = 0.5 g
g - gravitational acceleration on surface of Earth = 9.8 m/s²
r - radius of ship = 35 mm/2 = 17.5 mm = 17.5 x 10⁻³ m
Put the value in the equation,
[tex]\bold {f = \dfrac {1}{2\pi}\squrt {(0.5)(9.8\ m/s^2)}{(17.5 x 10^{-3} m)}}\\\\\bold {f = 2.66\ Hz}[/tex]
the time required for one revolution can be calculated as
[tex]\bold {T =\dfrac 1f}\\\\\bold {T = \dfrac 1{2.66}\ Hz}\\\\\bold {T = 0.38\ s}[/tex]
Therefore, the time required for one revolution to simulate the desired gravity is 0.38 s.
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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge at point P, which is on the x-axis, 9.83 mm from the origin
Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
[tex]\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}][/tex] (1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
[tex]\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°[/tex]
The distance r is:
[tex]r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m[/tex]
You replace the values of all parameters in the equation (1):
[tex]\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}[/tex]
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
A piece of iron rests on top of a piece of wood floating in a bathtub. If the iron is removed from the wood, and kept out of the water, what happens to the water level in the tub?
a. It goes up.
b. It does not change.
c. It goes down.
d. It is impossible to determine from the information given.
Answer:
It goes down.
The water level remain the same.
Explanation:
This can be explained using Archimedes principle which states that a body fully or partially submerged in a fluid is acted by an upward bouyant force which is equal to the weight of the fluid the body displaced.
The wood will only sink if the weight of the wood is greater than the weight of the fluid the wood displaced, but the weight of the wood is equal to the weight of fluid displaced, therefore the wood will float.
Therefore, the weight of the wood is the same as the weight of the fluid displaced, so the wood will be at the same level as the water.
If the iron is removed, the level of the water goes down because iron weight is bigger than the water displaced and it tends to increase the water level but since it is removed, the water level will decrease.
The study of charges in motion and their
interaction with magnetic fields is known as....?
(A) electrostatics
(B) electrodynamics
(C) electromotive
(D) electrons
(E) none of the above
2. Any material with an unequal number of
electrons and or protons could generally be
termed as .....?
Answer:
The study of charges in motion and their interaction with magnetic fields is known as electromagnetism.
Any material with an unequal number of electrons and or protons could generally be termed as ion.
Explanation:
i hope this will help you :)
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is
Answer:
Torque τ =w ×0 = 0
Explanation:
We know that the torque is given by the product of the force and perpendicular distance between the force and the axis.
Here the gravity force act at the center and the rotational axis is also passing through the center.
Therefore the perpendicular distance between the force and the rotational axis would be zero.
Hence the torque will be
Torque = Force × Perpendicular distance
Torque = mg×0 = 0
Therefore the torque would be zero.
following are the response to the given question:
The torque is calculated as the product of the pressure and also the perpendicular distance between both the force and the axis.In this case, gravity acts at the center, as well as the rotational axis likewise passes through it.As a result, the perpendicular distance between both the force and the rotational axis is 0.Calculating the torque:
[tex]\text{= Force} \times \text{Perpendicular distance}[/tex]
Therefore, the final torque would be '0'.
Learn more:
brainly.com/question/24630646
The net torque acting on an object is zero. What can you say about the angular momentum (about the same axis) of the object?
Answer:
The angular momentum (about the same axis) of the object is constant
Explanation:
Torque is the time time derivative of angular momentum.
τ = dL / dt
where;
dL is the change in angular momentum
τ is torque acting on the object
dt is change in time
when net torque acting on an object is zero, then
dL / dt = 0
Change in angular momentum (ΔL) is zero, therefore we can say that the angular momentum (L) is constant.
Thus, the angular momentum (about the same axis) of the object is constant.
79. An example of an electrical insulator is
a. graphite.
b. glass.
c. aluminum
d. tungsten
Answer:
B. Glass
Explanation:
An electrical insulator is a substance that does not conduct electricity.
Glass has tightly bounded electrons, that is why it is an insulator of electricity.
Hey there! The answer to your question is below.
The correct answer would be B.GLASS
Glass is a insulator and rubber too
Glass, wood, plastic and more are all insulators
So the answer would be B. Glass
Hope this helps!
By: xBrainly
Which one of the following is the shortest length?
A)
100 meters
C)
104 millimeters
E)
10 nanometers
B)
10² centimeters
D)
105 micrometers
Answer:
Option E (10 nanometers) is the shortest lengthExplanation:
From,
1cm = [tex]10^{-2}m[/tex]
1mm = [tex]10^{-3}m[/tex]
1nanometer = [tex]10^{-9[/tex]
1micrometer = [tex]10^{-6[/tex]
Therefore,
A) [tex]10^0[/tex] meters = 1meter
B) [tex]10^2[/tex] cm = [tex]10^2 * 10^{-2} = 1meter[/tex]
C) [tex]10^4[/tex] mm = [tex]10^4 * 10^{-3} = 10meter[/tex]
D) [tex]10^5[/tex] micrometer = [tex]10^5 * 10^{-6} = 0.1meter[/tex]
E) [tex]10[/tex] nanometer = [tex]10 * 10^-9 = 1*10^{-8}[/tex]
Therefore 10nanometers is the shortest length
For more information on length conversions, visit
https://brainly.com/subject/physics
A brass rod with a mass of 0.300 kg slides on parallel horizontal iron rails, 0.440 m apart, and carries a current of 15.0 A. The coefficient of friction between the rod and rails is 0.300. What vertical, uniform magnetic field is needed to keep the rod moving at a constant speed
Answer:
The magnitude of the magnetic field is [tex]B = 0.0890 \ T[/tex]
Explanation:
From the question we are told that
The mass of the rod is [tex]m =0.300 \ kg[/tex]
The distance of separation is [tex]d = 0.440 \ m[/tex]
The current is [tex]I = 15.0 \ A[/tex]
The coefficient of friction is [tex]\mu = 0.300[/tex]
Generally for the rod the rod to continue moving at a constant speed
The frictional force must equal to the magnetic field force so
[tex]F_m = F_f[/tex]
Where [tex]F_m = B* I * d[/tex]
and [tex]F_f = \mu * m * g[/tex]
[tex]B*I *d = \mu * m * g[/tex]
=> [tex]B = \frac{\mu * m * g }{I * d }[/tex]
substituting values
[tex]B = \frac{0.2 * 0.300 * 9.8 }{ 15 * 0.440 }[/tex]
[tex]B = 0.0890 \ T[/tex]
g The force of kinetic friction for a particular pair of interacting objects is always _____ the force of static friction. less than greater than equal to None of the above
Answer:
less than
Explanation:
The force of kinetic friction for a particular pair of interacting objects is always less than the force of static friction.
The force of static friction between two surfaces is always higher than the force of kinetic friction.