Explanation:
some thing inside a resistor
Given that the temperature of a body is 527K determine the value in degree C
Answer:
253.85°C
Explanation:
Here is the formula for converting K to °C
527K − 273.15 = 253.85°C
g A small object of mass 2.5 g and charge 18 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field
Answer: [tex]1361.11\ N/C,\text{upward}[/tex]
Explanation:
Given
Mass of particle is [tex]m=2.5\ gm[/tex]
Charge of particle is [tex]q=18\ \mu C[/tex]
Electrostatic force must balance the weight of the particle
[tex]\lim_{n \to \infty} a_n \Rightarrow mg=qE\\\\\Rightarrow E=\dfrac{2.5\times 9.8\times 10^{-3}}{18\times 10^{-6}}\\\\\Rightarrow E=1361.1\ N/C[/tex]
Direction of the electric field is in upward direction such that it opposes the gravity force.
how many rings does saturn have
Answer:
From far away, Saturn looks like it has seven large rings. Each large ring is named for a letter of the alphabet. The rings were named in the order they were discovered.
Assume that the far point of a myopic (nearsighted) eye is 5.04 m in front of the eye. A lens is used to correct the vision, such that it can focus sharply an object at infinity. What is the power of the lens (in diopters; answer sign and magnitude)?
Answer:
[tex]P=-0.2D[/tex]
Explanation:
From the question we are told that:
Far-point [tex]v=-5.04m[/tex]
Where
u=-\alpha
Generally the equation for Lens is mathematically given by
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]P=\frac{1}{-5.04}-\frac{1}{\alpha}[/tex]
[tex]P=-0.2D[/tex]
.
A mass of 8.72 kg gains 446 J of gravitational potential energy. To what height was it lifted?
Answer:
[tex]\boxed {\boxed {\sf 5.22 \ m}}[/tex]
Explanation:
Gravitational potential energy is the energy an object possesses due to its position. It is calculated using the following formula:
[tex]E_P=mgh[/tex]
Where m is the mass, g is the acceleration due to gravity, and h is the height.
The object has a mass of 8.72 kilograms. Assuming this occurs on Earth, the acceleration due to gravity is 9.8 meters per second squared. The object gains 446 Joules of potential energy.
Let's convert the units of Joules. This makes the process of canceling units simpler later on. 1 Joule is equal to 1 kilogram meter squared per second squared. The object gains 446 J, which is equal to 446 kg *m²/s².
EP= 446 kg*m²/s²m= 8.72 kg g= 9.8 m/s²Substitute the values into the formula.
[tex]446 \ kg*m^2/s^2 = 8.72 \ kg * 9.8 \ m/s^2 *h[/tex]
Multiply on the right side of the equation.
[tex]446 \ kg*m^2/s^2 = 85.456 kg*m/s^2 *h[/tex]
We are solving for the height, so we must isolate the variable h. It is being multiplied by 85.456 kg*m/s². The inverse operation of multiplication is division, so we divide both sides by this value.
[tex]\frac{ 446 \ kg*m^2/s^2}{85.456 kg*m/s^2} = \frac{85.456 kg*m/s^2 *h}{85.456 kg*m/s^2}[/tex]
[tex]\frac{ 446 \ kg*m^2/s^2}{85.456 kg*m/s^2} =h[/tex]
The units of kg*m/s² cancel, leaving meters as our unit.
[tex]\frac{ 446 }{85.456 } \ m =h[/tex]
[tex]5.2190601011 \ m =h[/tex]
The original measurements of mass and potential energy have 3 significant figures, so our answer must have the same.
For the number we calculated, that is the hundredths place. The 9 in the thousandths place to the right tells us to round the 1 up to a 2.
[tex]5.22 \ m \approx h[/tex]
The object was lifted to a height of approximately 5.22 meters.
list at least types of motion
Answer:
These four are rotary, oscillating, linear and reciprocating. Each one moves in a slightly different way and each type of achieved using different mechanical means that help us understand linear motion and motion control.
(I got this off the web so credits to the rightful owner and I hope you have good day :)
The resistance of the light bulb changed as the voltage (and current) changed. Why does this resistance change occur?
A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?
Answer:
Explanation:
(a)Solving for the acceleration of the bullet
acceleration = (vf^2 – vi^2) / 2d
acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)
acceleration = (78400 - 176400) / 0.24 m
acceleration = -98000 / 0.24
acceleration = -408333 m/s^2
(a)Solving for contact time with board
t^2 = 2d/a
t^2 = 2 * 0.12 m / 408333 m/s^2
t^2 = 0.24 m / 408333 m/s^2
t^2 = 5.8775558 x 10^-7
t = 0.0007666 s or 767 microseconds
When tightening a bolt, you push perpendicularly on a wrench a with force of 165 N at a distance of 0.140m from the center of the bolt. (A) How much torque are you exerting in newton x meters (relative to the center of the bolt)
Answer:
Part a)
23.1 Nm
..........
A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to be 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.
Required:
What is the stiffness of a typical interatomic bond in the alloy
Answer: hello some of your values are wrongly written hence I will resolve your question using the right values
answer:
stiffness = 1.09 * 10^-6 N/m
Explanation:
Given data:
Length ( l ) = 16 m
radius of wire ( r ) = 3.5 m
mass ( m ) = 5kg
Distance stretched ( Δl ) = 4 * 10^-3 m ( right value )
average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)
first step : calculate the area
area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2
γ = MgL / A Δl
= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]
= 784.8 / 0.165 = 4756.36 N/m^2
hence : stiffness = γ * bond length
= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m
Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )
1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )
1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) = [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ = [tex]\frac{V}{i+0.5}[/tex]
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ = [tex]\frac{V}{i+0.25}[/tex]
R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]
R₂ = 19.9 Ω
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
Data related to Meena’s and Malini’s journey is given below, plot a graph of their
respective journey on a graph paper. You have already plotted Meena’s Journey during the
summer vacation. On the same graph paper, now plot Malini’s Journey.
PDF task 2
Please do this for me urgent I can give you extra points if someone answers this in less than 1 hour.
Answer:
download the pdf
Explanation:
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event
Answer:
The rise in temperature is 0.06 K.
Explanation:
mass of bullet, m = 15 g
initial speed, u = 865 m/s
final speed, v = 534 m/s
mass of water, M = 13.5 kg
specific heat of water, c = 4200 J/kg K
The change in kinetic energy
[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]
According to the conservation of energy, the change in kinetic energy is used to heat the water.
K = m c T
where, T is the rise in temperature.
3473 = 13.5 x 4200 x T
T = 0.06 K
An astronaut on the moon drops a rock from rest. The rock falls 0.8m in one second of falling time. If the dropped rock fell for a total of two seconds of time instead of 1 second, then the distance traveled would be:____.
A) The same.
B) Doubled.
C) Tripled.
D) Quadruple.
E) None of the above.
Answer:
D) Quadruple.
Explanation:
We will use the second equation of motion to solve this problem:
[tex]s = v_it + \frac{1}{2}gt^2[/tex]
where,
s = distance travelled by the rock
vi = initial speed of rock = 0 m/s
t = time taken
g = acceleration due to gravity on the surface of the moon
Therefore,
[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)
Now, we double the time:
[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]
using equation (1):
s' = 4s
Hence, the correct option is:
D) Quadruple.
A bus moving on a straight road increases its speed uniformly from rest to 20m's over a time period of 1 min. The distance travelled during the time is (a) 150 m (b) 300 m (c) 600 m (d) 900 m
Explanation:
Given that,
Initial velocity (u) = 0 m/sFinal velocity (v) = 20 m/sTime taken (t) = 1 minute = 60 secondsIn order to find the distance travelled, firstly we need calculate the acceleration.
→ v = u + at
→ 20 = 0 + 60a
→ 20 = 60a
→ 20 ÷ 60 = a
→ ⅓ m/s² = a
Now, by using the 2nd equation of motion :
→ s = ut + ½at²
→ s = 0(60) + ½ × ⅓ × (60)²
→ s = ⅙ × 3600
→ s = 1 × 600
→ s = 600 m
Hence, the distance travelled is 600 m.
Which of these rotational quantities is analogous to mass in a linear system?
a.
Angle in radians
b.
Angular acceleration
c.
Torque
d.
Rotational inertia
Answer:
d
Explanation:
i think it is rotational inertia
because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.
hope it's right & helps !!!!!!!!!
the air pressure at the base of the mountain is 75.0cm of mercury while at the top is 60cm of mercury. Given that theaverage density of air is 1.25kg/m³ and the density of mercury is 13 600 kg/m³and gravity-10N/kg, calculate the height height of the mountain
Answer:
질문?
Explanation:
평균 공기 밀도가 1.25kg/m³이고 수은 밀도가 13600kg/m³이고 g=10N/kg인 경우 산 기슭의 기압은 수은의 75.0cm이고 정상의 수은은 60cm입니다. 산의 높이를 계산?
68.6 bags are required to fill the specified volume of 10"x20"x20".
It is 40 lbs/1.9 gallons, assuming US gallons, for the topsoil.
US gallons equal 3785 ml. 1 lb = 1/2.2 = 0.454 kg or 454 grammes since 1 kilogramme equals 2.2 lbs. 3785 x 1.9 = 7192 ml is equal to 1.9 gallons. 40 lbs = 18160 g. Therefore, the topsoil density is 2.52 g/cc (18160/7192).
The volume of the peat bag is 14x20x30 inches, which is equal to 35.6x50.8x76.2 cm (1 inch = 2/54 cm)=137806 ccs.
In other words, for the 40 lb again, 18160 grams/137806 equals 0.13 g/cc. The peat is therefore significantly lighter than topsoil.
The volume of the latter volume is 120"x240"x20" or 576,000 cubic inches, and the volume of a bag is 14"x20"x30" or 8400 cubic inches, hence 576,000/8400 = 68.6 bags are required to fill the specified volume of 10"x20"x20".
Learn more about density herebrainly.com/question/4970627
SPJ1
Which of the following would most likely produce the strongest magnetic
field?
A. A single moving electron
B. A stationary electric charge
C. A current in a straight wire
D. A current in a coil
Answer:
I current in a coil,,,,,,
Answer:D?
Explanation:
Sorry if i'm wrong....
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hanging vertically, the mass is pulled aside a small distance of 7.6 cm and released from rest. While the mass is swinging the cord exerts an almost-constant force on it. For this problem, assume the force is constant as the mass swings. How much work in J does the cord do to the mass as the mass swings a distance of 8.0 cm
Answer:
work done is -2.8 × 10⁻⁶ J
Explanation:
Given the data in the question;
mass of the pendulum m = 6 kg
Length of core = 1.7 m
Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad
In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.
Now, the required work done will be;
[tex]W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta[/tex]
[tex]W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta[/tex]
[tex]W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta[/tex]
W = [tex]-mgl[[/tex] -cosθ [tex]]^{0.047}_{0.045 }[/tex]
W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]
W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]
W = -2.8 × 10⁻⁶ J
Therefore, work done is -2.8 × 10⁻⁶ J
how can you convert galvanometer into ammeter?
Answer:
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
Explanation:
This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.
Uuse Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire loop. You will need to use the right-hand rule to find the direction of the induced current.
Answer:
Explanation:
According to the Fleming's right hand rule, if we spread our right hand such that the thumb, fore finger and the middle finger are mutually perpendicular to each other, then the thumb indicates the direction of force, fore finger indicates the direction of magnetic field, then the middle finger indicates the direction of induced current.
According to the Lenz's law, the direction of induced emf is such that it always opposes the cause due to which it is produced.
Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stream is 6.8 m wide and 3.2 m deep, and flows at 2.4 m/s. If the river has width 10.4 m and speed 2.8 m/s, what is its depth?
Answer:
The depth of the resulting stream is 3.8 meters.
Explanation:
Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:
[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)
All volume flows are measured in cubic meters per second.
Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:
[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]
[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)
[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.
[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.
[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.
[tex]w_{3}[/tex] - Width of the resulting stream, in meters.
[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.
If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:
[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]
[tex]h_{3} = 3.8\,m[/tex]
The depth of the resulting stream is 3.8 meters.
Assume the speed of sound is 343 m/s. You are sitting 150 m away from home plate at a baseball game. How much time in seconds elapses between the batter hitting a home run and the moment you actually hear the batter hitting the ball
Answer:
t = 0.437 s
Explanation:
Sound is a wave so its speed is constant
v = x / t
t = x / v
indicates that the distance is x = 150 m
t = 150/343
t = 0.437 s
this is the time it takes to hear the hit
To see the blow it is almost instantaneous since the speed of light is much greater c = 3 10⁸ m / s
b. Projectile on cliff (range)
An object of mass 5 kg is projected at an angle of 25° to the horizontal with a speed of 22 ms-1 from the top of the cliff.
The height of the cliff is 21 m. Take g, the acceleration due to gravity, to be 9.81 ms2
How far horizontally (to 1 decimal place) from the base of the cliff does the object land?
Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]
t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m
A parallel-plate capacitor is constructed of two horizontal 16.8-cm-diameter circular plates. A 1.8 g plastic bead, with a charge of -4.4 nC is suspended between the two plates by the force of the electric field between them.
a. Which plate, the upper or the lower, is positively charged?
b. What is the charge on the positive plate?
Answer:
Explanation:
Given that:
diameter of the circular plates = 16.8 cm
mass of the plastic bead = 1.8 g
charge q = -4.4 nC
From above, the area of the circular plates is:
[tex]Area = \pi r^2[/tex]
[tex]Area = \pi (\dfrac{d}{2})^2[/tex]
[tex]Area = \pi (\dfrac{16.8}{2*100} m)^2 \[/tex]
Area = 0.022 m²
Thus, as the plastic beads glide between the two plates of the capacitor, there exists a weight acting downwards while the weight is balanced by the force of the plates acting upwards.
Hence, this can be achieved only when the upper plate is positively charged.
b)
Recall that
Force (F) = qE
where;
F = mg
mg = qE
[tex]E = \dfrac{mg}{q}[/tex]
[tex]E = \dfrac{1.8*10^{-3}*9.8}{4.4*10^{-9}}[/tex]
E = 4.0 × 10⁶ N/C
From the electric field;
[tex]E = \dfrac{\Big(\dfrac{Q}{A}\Big)}{e_o}[/tex]
[tex]4.0*10^{6} = \dfrac{\Big(\dfrac{Q}{0.022}\Big)}{8.85*10^{-12}}[/tex]
[tex]4.0*10^{6}*8.85*10^{-12} = {\Big(\dfrac{Q}{0.022}\Big)}{}[/tex]
[tex]Q= 4.0*10^{6}*8.85*10^{-12}*0.022[/tex]
Q = 7.788 × 10⁻⁷ C
Q = 779 nC
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.0 cm and a current of 12 A. The bigger loop has a current of 20 A. The magnetic field at the center of the loops is found to be zero.
Required:
What is the radius of the bigger loop?
Answer:
the radius of the bigger loop is 5 cm.
Explanation:
Given;
current in the smaller loop, I₁ = 12 A
current in the larger loop, I₂ = 20 A
radius of the smaller loop, r₁ = 3 cm
let the radius of the larger loop, = r₂
Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.
[tex]B= \frac{\mu_0 I}{2r}[/tex]
The magnetic field at the center of the smaller loop;
[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]
The magnetic field at the center of the bigger loop;
[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]
If the magnetic field at the center is zero, then B₁ = B₂
[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]
Therefore, the radius of the bigger loop is 5 cm.
A 69.0-kg astronaut is floating in space, luckily he has his trusty 28.0-kg physics book. He throws his physics book and accelerates at 0.0130 m/s2 in the opposite direction. What is the magnitude of the acceleration of the physics book?
Answer:
0.032 [tex]m/s^2[/tex]
Explanation:
Given :
Weight of the astronaut = 69 kg
Weight of the physics book = 28 kg
Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]
The force that is applied on the astronaut :
[tex]F=ma[/tex]
[tex]$=69 \times 0.013$[/tex]
= 0.897 N
Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N
Therefore, the acceleration of the physics book is given by :
[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]
[tex]$a = \frac{0.897}{28}$[/tex]
a = 0.032 [tex]m/s^2[/tex]
Hence, the acceleration of the physics book is 0.032 [tex]m/s^2[/tex].
Answer:
The acceleration of astronaut is 5.27 x 10^-3 m/s^2.
Explanation:
mass of astronaut, M = 69 kg
Mass of book, m = 28 kg
acceleration of book, a = 0.013 m/s^2
Let the acceleration of astronaut is A.
According to the Newton's third law, for every action there is an equal and opposite reaction.
So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.
M A = m a
69 x A = 28 x 0.013
A = 5.27 x 10^-3 m/s^2
Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?
Answer:
Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.
En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.
La persona A aplica una fuerza:
Fa = -3N
La persona B aplica una fuerza:
Fb = 5N
La persona C aplica una fuerza Fc, la cual aún no conocemos.
Pero sabemos que la caja está en equilibrio físico, por lo que:
Fa + Fb + Fc = 0N
reemplazando los valores que conocemos, obtenemos:
-3N + 5N + Fc = 0N
Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.
Fc = 0N + 3N - 5N
Fc = -2N
Podemos concluir que la persona C aplica una fuerza horizontal de -2N
