Answer:
1620000J
Explanation:
Given parameters:
Resistance = 24ohm
Current = 5A
Time = 45min
Unknown:
Work done = ?
Solution:
Electrical work done is given as the product of power with time;
Electrical work done = power x time
Power = I²R
Electrical work done = I²R x time
Convert the given time to seconds;
1min = 60s
45 min = 45 x 60 = 2700s
Now,
Work done = 5² x 24 x 2700 = 1620000J
please help i’ll mark u branliest
Answer:
62
Explanation:
it doesn't need explanation
A jeweler's grinding wheel slows down at a constant rate from 185 rad/s to 105 rad/s while it rotates through 16.0 revolutions. How much time does this take?
Answer:
t = 0.6933 s
Explanation:
This is a rotational kinematics exercise, let's find the angular acceleration of the wheel
w² = w₀² + 2 α θ
α = (w² - w₀²) / 2 θ
Let's reduce the angles to the SI system
θ = 16 rev (2π rad / 1 rev) = 32π rad
let's calculate
α = (105² - 185²) / (2 32π)
α = -115.39 rad / s²
now let's use the relation
w = w₀ + α t
t = (w- w₀) /α
t = (105 - 185) / (- 115.39)
t = 0.6933 s
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
Required:
Find the minimum uncertainty in the momentum of the object. Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass=0.045 kg) and (c) an electron.
Answer:
[tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
[tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
[tex]0.31\times 10^{-4}\ \text{m/s}[/tex]
Explanation:
[tex]\Delta x[/tex] = Uncertainty in position = 1.9 m
[tex]\Delta p[/tex] = Uncertainty in momentum
h = Planck's constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]
m = Mass of object
From Heisenberg's uncertainty principle we know
[tex]\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}[/tex]
The minimum uncertainty in the momentum of the object is [tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
Golf ball minimum uncertainty in the momentum of the object
[tex]m=0.045\ \text{kg}[/tex]
Uncertainty in velocity is given by
[tex]\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
Electron
[tex]m=9.11\times 10^{-31}\ \text{kg}[/tex]
[tex]\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]0.31\times 10^{-4}\ \text{m/s}[/tex].
Jerry runs 60 meters east and then 20 meters west in 10
seconds. His average velocity is
m/s.
Answer: 8 meters per second
Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s
A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what isthe current in the primary?and WHy?
Answer:
The correct solution will be "12.0 A".
Explanation:
The given values are:
[tex]N_p= 500 \ turn[/tex]
[tex]N_s= 200 \ turn[/tex]
[tex]I_s= 3.0 \ A[/tex]
By using the transformer formula, we get
⇒ [tex]\frac{N_p}{N_s} =\frac{I_s}{I_p}[/tex]
⇒ [tex]I_p = I_s\times \frac{N_s}{N_p}[/tex]
On substituting the given values, we get
⇒ [tex]=3.0 \ A\times \frac{2000}{500}[/tex]
⇒ [tex]=12.0 \ A[/tex]
Can someone tell me what a free fall is
Answer:
free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it
Explanation:
hope this helped
Answer:A freefall is what you experience when you go skydiving and that is because you are falling out of the sky
Explanation:
Which is one of Edwin Hubble’s findings that supports the big bang theory?
Answer:
Edwin Hubble found that galaxies are constantly moving away from us. According to his observations with the Hubble Space Telescope, galaxies are moving at different speeds. This shows that the universe is expanding. The farther away a galaxy is, the faster it is moving away. Found this on google hope this helps.
Answer:
A) the universe started at a central point
Explanation:
taking the quiz on eg. :))
A pulley is in the form of a uniform solid cylinder of radius 7cm and mass 2kg. One end of a very light rope is fixed to wind the pulley and the other end is
free. When we pull the free end of the rope the pulley starts rotating from rest and accelerates uniformly. If the angular acceleration is 100rad/s so the
tangential acceleration of the rope is:
Chọn một:
a.
200 N
b.
49 N
C.
0.49 N
d.
7N
To remove a stain using a solvent the stain has to become dissolved in the solvent
True
False
Answer:
True
Explanation:
have a good day:)
Answer: This statement is True
A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How long does it take to turn the 5.5 revolutions?
Answer:
(a) The final angular speed is 12.05 rad/s
(b) The time taken to turn 5.5 revolutions is 5.74 s
Explanation:
Given;
number of revolutions, θ = 5.5 revolutions
acceleration of the wheel, α = 20 rpm/s
number of revolutions in radian is given as;
θ = 5.5 x 2π = 34.562 rad
angular acceleration in rad/s² is given as;
[tex]\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2[/tex]
(a)
The final angular speed is given as;
[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s[/tex]
(b) the time taken to turn 5.5 revolutions is given as
[tex]\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s[/tex]
How much energy is required to move a 1000 kg object from the Earth's surface to an altitude twice the Earth's radius?
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
Since the object must be moved away to a distance greater than the radius of the Earth, then change in gravitational potential energy must be based on Newton's Law of Gravitation.
By the Work-Energy Theorem, the work ([tex]W[/tex]), in joules, done on the object is equal to the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]W = U_{g}[/tex] (1)
[tex]W = -G\cdot m\cdot M\cdot \left(\frac{1}{r_{f}}-\frac{1}{r_{o}} \right)[/tex] (1b)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.[tex]m[/tex] - Mass of the object, in kilograms.[tex]M[/tex] - Mass of the Earth, in kilograms.[tex]r_{o}[/tex] - Initial distance, in meters.[tex]r_{f}[/tex] - Final distance, in meters.If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 1000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]r_{o} = 6.371\times 10^{6}\,m[/tex] and [tex]r_{f} = 19.113\times 10^{6}\,m[/tex], then the energy required to move the object from the Earth's surface is:
[tex]W = -\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1000\,kg)\cdot (5.972\times 10^{24}\,kg)\cdot \left[\frac{1}{19.113\times 10^{6}\,m} - \frac{1}{6.371\times 10^{6}\,m} \right][/tex][tex]W = 4.171\times 10^{10}\,J[/tex]
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
We kindly invite to check this question on gravitational potential energy: https://brainly.com/question/19768887
A boy on a bicycle rides in a circle of radius ro at speed vo. If the boy now rides at a radius equal to half the initial radius ro, by what approximate factor must he change his speed in order to have the same radial acceleration
Answer:
The speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
Explanation:
The radial or centripetal acceleration is given by:
[tex] a_{c} = \frac{v^{2}}{r} [/tex]
Where:
v: is the speed = v₀
r: is the radius = r₀
[tex] a_{c} = \frac{v_{0}^{2}}{r_{0}} [/tex] (1)
If the radius is now equal to half the initial radius the speed must be:
[tex]a_{c} = \frac{v^{2}}{r_{0}/2}[/tex] (2)
By equating equation (1) and (2):
[tex] \frac{v_{0}^{2}}{r_{0}} = \frac{v^{2}}{r_{0}/2} [/tex]
[tex]v^{2} = \frac{v_{0}^{2}}{2}[/tex]
[tex] v = \frac{v_{0}}{\sqrt{2}} [/tex]
Therefore, the speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
I hope it helps you!
where does a solid material go when a solution is made
Answer:
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Explanation:
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please answer this question
Which best describes a characteristic of an adiabatic process?
Answer:
Can you please provide the choices? For now this is all I can give you.Explanation:
In thermodynamics, an adiabatic process is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings.Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work. It also conceptually supports the theory used to explain the first law of thermodynamics and is therefore a key thermodynamic concept.
The system takes in heat but does not release it. If an adiabatic process is reversible, there is no change in entropy; otherwise, there is a rise in entropy, or degree of disorder.
What is adiabatic process?
A thermodynamic process known as an adiabatic process takes place when there is no exchange of mass or heat between the system and its surroundings.
An adiabatic process only transfers energy to the environment as work, in contrast to an isothermal process.
The system's overall heat remains constant since there is no heat exchange with the environment.
It is a fundamental idea in thermodynamics and theoretically supports the theory that underlies the first law of thermodynamics.
Therefore, no heat flow in and out of system, which is a characteristic of an adiabatic process.
Learn more about adiabatic process here:
https://brainly.com/question/11938296
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Help!!
A 30-N force is applied to a 4-kg object to move it with a constant
velocity of 2 m/s across a level surface. The coefficient of friction
between the object and the surface is approximately (Use the
approximation: g - 10 m/s/s.)
A 0.20
B O 0.50
C 0.55
D 0.75
Answer:
[tex]\mu=0.75[/tex]
Explanation:
Given that,
Force acting on an object, F = 30 N
Mass of the object, m = 4 kg
It is moving with a constant velocity of 2 m/s across a level surface.
We need to find the coefficient of friction between the object and the surface. Let it is μ. Force in terms of coefficient of friction is given by :
F = μ N, Where N is normal force, N = mg
[tex]\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{30}{4\times 10}\\\\\mu=0.75[/tex]
So, the coefficient of friction between the object and the surface is 0.75.
Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time interval?
Answer:
answer is 11.76 meter
Explanation:
use 2nd equation of motion
S=ut+1/2at^2
An object that is oscillating on a spring is given by the equation x = (10.0 cm) cos[(6.00 s-1)t]. At what value of t after t = 0.00 s is the cart first located at x = 8.00 cm?
Answer:
[tex]t=0.0107\ \text{s}[/tex]
Explanation:
[tex]x=10\cos(6t)[/tex]
Now [tex]x=8\ \text{cm}[/tex]
Substituting the value of [tex]x[/tex] in the equation we get
[tex]8=10\cos6t[/tex]
[tex]\Rightarrow 0.8=\cos6t[/tex]
[tex]\Rightarrow \cos^{-1}0.8=6t[/tex]
[tex]\Rightarrow t=\dfrac{\cos^{-1}0.8}{6}[/tex] the values here are used found in radians
[tex]\Rightarrow t=0.0107\ \text{s}[/tex]
So, at [tex]t=0.0107\ \text{s}[/tex] the value of [tex]x=8\ \text{cm}[/tex].
As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65 degrees above the horizontal 45 N at an angle of 65 degrees above the horizontal.
500 kg
The horizontal component of the force acting on the crate is?
Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
[tex]\theta[/tex] = Angle at which the force is being applied = [tex]65^{\circ}[/tex]
Horizontal component of force is given by
[tex]F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}[/tex]
The horizontal component of the force acting on the crate is 19.01 N.
Someone help me pls
Explanation:
The dog was stationary at segment c
Answer:
I don't know if I'm wrong but I'm pretty sure stationary means that the thing is still. I would go with C And maybe D????
Two tectonic plates moving toward one another are at a
ANSWER CHOICES
convergent boundary.
divergent boundary.
subduction boundary.
transform boundary.
Answer:
A. cause i just took the test
Explanation:
Answer:
its A
Explanation:
no explanation is needed, just trust me.
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4. Determine the magnitude of force at point and determine if the ladder will slip.
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
answer this plzzzzzzzzzzzzz
Helicopters rotor blades, could spin at high speed of 510 rpm. Find the angular displacement in radian for 3 hour(s) operation
Answer:
The angular displacement of the blade is 576,871.2 radians
Explanation:
Given;
angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)
time of motion, t = 3 hours
The angular speed of the Helicopters rotor blades in radian per second is given as;
[tex]\omega = \frac{510 \ rev}{mins} *\frac{2 \pi \ rad}{1 \ rev} *\frac{1 \ min}{60 \ s}\\\\\omega = 53.414 \ rad/s[/tex]
The angular displacement in radian is given as;
θ = ωt
where;
t is time in seconds
θ = (53.414)(3 x 60 x 60)\\
θ = 576,871.2 radians
Therefore, the angular displacement of the blade is 576,871.2 radians
A 150 kg boy and his bike are traveling 12 m/s when he slams on his breaks and stop at his friend’s house. How much impulse is required to produce this change in momentum?
Please someone help me with this I’ll give brainliest
Answer:
J = 1800 kg-m/s
Explanation:
Given that,
Mass of a boy, m = 150 kg
Initial velocity of a boy, u = 12 m/s
Finally, it stops, v = 0
We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,
[tex]J=m(v-u)\\\\=150\times (0-12)\\\\=-1800\ kg-m/s\\\\|J|=1800\ kg-m/s[/tex]
So, the impulse is equal to 1800 kg-m/s
A 2 kg toy car moves at a speed of 5 m/s.
What is the kinetic energy of the car?
Answer:
The answer is 25 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]KE = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]KE = \frac{1}{2} \times 2 \times {5}^{2} \\ = 1 \times {5}^{2} [/tex]
We have the final answer as
25 JHope this helps you
A circular coil of 185 turns and radius 4.50 cm is placed inside and coaxial to a solenoid with 350 turns/m. If the current in the solenoid is represented by the equation I = a(1 − e−bt) where a = 12.5 A and b = 2.10 s−1, determine the magnitude of the induced emf in the coil when t = 1.50 s.
Answer:
Magnitude of the induced emf is 11.62 V
Explanation:
Given:
No. of turns of the circular coil, N=185
Radius, R=4.50 cm=0.045 m
No. of turns of solenoid per meter(m), n=350
a=12.5 A
b=2.10 [tex]s^{-1}[/tex]
Now,
To determine the emf induced in the coil at t = 1.50 s:
The given equation is:
[tex]I=a(1-e^{-bt})[/tex]
Now,
[tex]B=\mu_{o}nI=\mu_{o}na(1-e^{-bt})[/tex]
[tex]\frac{dB}{dt}=\mu_{o}nabe^{-bt}[/tex]
Now, substituting the respective values:
[tex]\frac{dB}{dt}=-4\pi\times 10^{-7}\times 350\times 12.5\times 2.10e^{-2.10\times 1.50}[/tex]
[tex]\frac{dB}{dt}=0.988\ T/s[/tex]
Now,
[tex]emf=NA\frac{dB}{dt}[/tex]
where,
A=Area=[tex]\pi R^{2}[/tex]
Thus,
[tex]emf=185\times \pi\times (4.50\times 10^{-2})^{2}\times 11.545e^{-3.15}[/tex]
[tex]emf=11.62 V[/tex]
A hunter aims directly at a target (on the same level) 120 m away. If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target
Answer:
The distance the bullet will miss the target is 1.13 m.
Explanation:
Given;
Initial velocity of the bullet = 250 m/s
Distance of the target = 120 m
Time of motion;
t = 120 / 250
t = 0.48 s
During this time the bullet is under the gravitational pull and the distance it will miss the target is given by;
Y = V₀y + ¹/₂gt²
where;
V₀y is the initial vertical velocity = 0
Y = 0+ ¹/₂gt²
Y = ¹/₂(9.8)(0.48)²
Y = 1.13 m
Therefore, the distance the bullet will miss the target is 1.13 m.
what happens to the grvatational fore between two objects as they move fartger apart
Answer:
♡ madeline here ♡
heres your answer
This force of gravitational attraction
is depending on the masses of both
objects and contrarily corresponding
to the square of the distance that divided
their middle parts. This means that as you
move away from an object the gravitational
force lowers. i hope i could help you! ☆
have a great day!
- madeline/madi ✧･ﾟ: *✧･ﾟ:･ﾟ✧*:･ﾟ✧･ﾟ
Explanation:
I need help with this science work