Answer:
D
Explanation:
Which one of the following statements concerning the magnetic field inside (far from the surface) a long, current-carrying solenoid is true?
1) The magnetic field is zero.
2) The magnetic field is independent of the number of windings.
3) The magnetic field varies as 1/r as measured from the solenoid axis.
4) The magnetic field is independent of the current in the solenoid.
5) The magnetic field is non-zero and nearly uniform.
Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.
Answer:
[tex]0.0018833\ \text{m/s}[/tex]
Explanation:
[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]
[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
We have the relation
[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]
[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]
The required answer is [tex]0.0018833\ \text{m/s}[/tex].
A coil of resistance 100ohm is placed in a magnetic field of 1mWb the coil has 100 turns and a galvanometer of 400ohm resistance is connected in series with it. find the average emf and the current if the coil is moved in one tenth of a second from the given field to a field of 2.0mWb
Answer:
C
Explanation:
C
Momentum
Project: Egg Drop
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Answer:
get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it
Explanation:
10 POINTS! SPACE QUESTION!!
Answer; they are larger and made of rocky material
The half-life for a 400-gram sample of radioactive element X is 3 days. How much of element X remains after 15 days have passed?
A.
12.5 g
B.
25 g
C.
50 g
D.
100 g
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 x 108 m. Find the time it took for his voice to reach earth via radio waves. (b) Someday a person will walk on Mars, which is 5.6 x 1010 m from earth at the point of closest approach. Determine the minimum time that will be required for that person's voice to reach earth.
Answer:
a). 1.28333 seconds
b). 186.66 seconds
Explanation:
a). Given :
Distance between the earth and the moon, d = [tex]$3.85 \times 10^8$[/tex] m
Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s
Therefore the time required for the voice of Neil Armstrong to reach the earth via radio waves is given by :
[tex]$t=\frac{d}{c}$[/tex]
[tex]$=\frac{3.85 \times 10^8}{3 \times 10^8}$[/tex]
= 1.28333 seconds
b). Distance between Mars and the earth, d = [tex]$5.6 \times 10^{10}$[/tex] m
Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s
So, the time required for his voice to reach earth is :
[tex]$t=\frac{d}{c}$[/tex]
[tex]$=\frac{5.6 \times 10^{10}}{3 \times 10^8}$[/tex]
= 186.66 seconds
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
physics grade9 teacher guide
Answer:
huh
Explanation:
Philosophy: The Big Picture Unit 8
How does pragmatism differ from the utilitarianism of the previous era?
A. Utilitarianism did not consider rights and pragmatism did.
B. Pragmatism is concerned with function, utilitarianism with happiness.
C. Utilitarianism was created in England and pragmatism came from Germany.
D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
Describing a Wave
What does a wave carry?
Answer:
Waves carry energy from one place to another.
Explanation:
Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.
The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light.
a. True
b. False
Answer:
False
Explanation:
No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.
HURRY IM TIMED
How can you make people feel inspired?
By leading them on an emotional journey through various states to inspiration
By talking about something that interests you
By proving yourself to be a trustworthy speaker
By making them laugh and feel comfortable
Answer:
By talking about something that interesto you’
sorry if wrong
Explanation:
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
A candle is placed 50 cm from a diverging lens with a focal length of 28. What is the image distance in cm.
Answer:
v = -17.94 cm
Explanation:
Given that,
The candle is placed at a distance of 50 cm, u = -50 cm
The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)
We need to find the image distance. We know that the lens formula is as follows:
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]
So, the image distance is equal to 17.94 cm.
Select the correct answer.
Each square dance begins with what?
A. Handshake
B. Dosado
C. Bow or curtsy
D. Promenade
Answer:
C
Explanation:
The men bow to the women and the women curtsy to the men.
(took on test and got it right)
A winch is capable of hauling a ton of bricks vertically two stories (6.35 m ) in 24.5 s .
If the winch’s motor is rated at 5.80 hp , determine its efficiency during raising the load.
Answer: 84 %
Explanation:
Think of a hydropower dam . How is electrical energy produced from potential and kinetic energy ?
hydroelectric dam converts the potential energy stored in a water reservoir behind a dam to mechanical energy—mechanical energy is also known as kinetic energy. ... The generator converts the turbine's mechanical energy into electricity.
Hope this helps!
Answer:
Potential energy and kinetic energy are constituents of mechanical energy.
When a turbine is switched on, it rotates with mechanical energy.
Since a motor runs the turbine, it converts this mechanical energy to electrical energy.
A running Marites launched the egg she
stole as she was about to be caught with
a velocity of 25 m/s in a direction making
an angle of 20° upward with the
horizontal
a) What is the maximum height reached by
the egg?
b) What is the total flight time (between
launch and touching the ground) of the
egg?
c) What is the horizontal range (maximum
* above ground) of the egg?
d) What is the magnitude of the velocity
of the egg just before it hits the ground?
Answer:
a) y = 3.73 m, b) t = 1.74 s, c) R = 40.99 m,
d) vₓ = 23.49 m/s, v_y = -8.5 m / s
Explanation:
This is a projectile launching exercise, we start by breaking down the initial velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v_{oy} = v₀ sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 25 sin 20 = 8.55 m / s
v₀ₓ = 25 cos 20 = 23.49 m / s
a) when the egg reaches the maximum height its vertical speed is zero
v_y² = v_{oy}² - 2 g y
0 = v_[oy}² - 2g y
y = v_{oy}² / 2g
y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]
y = 3.73 m
b) flight time
y = v_{oy} t - ½ g t²
the time of flight occurs when the body reaches the ground y = 0
0 = (v_{oy} - ½ g t) t
The results are
t₁ = 0s this time is for using the body star
v_{oy} - ½ g t = 0
t = [tex]\frac{2v_{oy}^2}{g}[/tex]
t = 2 8.55 / 9.8
t = 1.74 s
c) the range
R = v₀² sin 2θ / g
R = 25² sin (2 20) / 9.8
R = 40.99 m
d) speed at the point of arrival
horizontal speed is constant
vₓ = v₀ₓ = 23.49 m/s
vertical speed is
v_y = Iv_{oy} - g t
v_y = 8.55 - 9.8 1.74
v_y = -8.5 m / s
How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 35.6 m and a propagation speed of 4.68 m/s? (the answer may not be a whole number)
Answer:
It will bob 7.887640449 times a minute
Explanation:
I hope this is correct!!
A 4 kg box is at rest on a table. The static friction coefficient u, between the box and table is 0.30, and
the kinetic friction coefficient Hi is 0.10. Then, a 10 N horizontal force is applied to the box.
Answer:
The box will not move from its position.
Explanation:
First, we will calculate the static frictional force that is stopping the box to move from its position:
[tex]f = \mu R = \mu W=\mu mg[/tex]
where,
f = static frictional force = ?
μ = coefficient of static friction = 0.3
m = mass of box = 4 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = (0.3)(4\ kg)(9.81\ m/s^2)\\f=11.77\ N[/tex]
Since the frictional force (11.77 N) is greater than the applied force (10 N).
Therefore, the box will not move from its position.
Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kgm2. The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable. (Hint: use what you know about relative velocity to help solve the problem
Answer:
[tex]0.3165\ \text{rad/s}[/tex]
Explanation:
m = Mass of person = 65 kg
d = Diameter of round table = 6.5 m
r = Radius = [tex]\dfrac{d}{2}=3.25\ \text{m}[/tex]
v = Velocity of person running = 3.8 m/s
[tex]I_t[/tex] = Moment of inertia of turntable = [tex]1850\ \text{kg m}^2[/tex]
Moment of inertia of the system is
[tex]I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2[/tex]
As the angular momentum of the system is conserved we have
[tex]L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}[/tex]
The angular velocity of the turntable is [tex]0.3165\ \text{rad/s}[/tex].
The diagram shows a charge moving into an electric field. The charge will most likely leave the electric field near which letter? OW OX OY OZ
you haven't attached the diagram, but i assume that this diagram is what you were talking about
Answer:
near Y
Explanation:
the electric field lines goes from a positive charge to a negative charge. This means that a positive charge would move in the same direction of the field lines, while a negative charge would move in the opposite direction of the field lines. the field lines are created from +vely charged plate to -vely charged plate so the negative charged particles moves towards the lower plate which is positively charged, and opposite to the direction of field lines.
The charge will most likely leave the electric field near Y letter. Hence option C is correct.
What is electric charge ?Electric charge is the physical property of matter that experiences force when it is placed in electric field. F = qE where q is amount of charge, E = electric field and F = is force experienced by the charge. there are two types of charges, positive charge and negative charge which are generally carried by proton and electron resp. like charges repel each other and unlike charges attract each other. the flow charges is called as current. Elementary charge is amount of charge a electron is having, whose value is 1.602 x 10⁻¹⁹ C
The charge on the electron is negative, and the force is directed in the opposite direction as the electric field. When an electron is projected perpendicular to a uniform electric field, it experiences an electric force in the opposite direction of the field.
the trajectory of the charged particle in the electric filed is parabolic and in magnetic field it is circular. when this electron moves perpendicular to the electric field, electron experience force in opposite direction to the electric field and due to parabolic nature, it will leave at Y.
Hence option C is correct.
To know more about Electric field :
https://brainly.com/question/8971780
#SPJ7.
What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?
im confused hold on imma send you a link to the answerExplanation:
A toy car rolls down a ramp. Which force causes the car to move
Answer:
Gravity
Explanation:
Gravity pulls things down to earth and it is a force
1.) Calculate the mass of a solid gold rectangular bar that has dimensions lwh = 4.30 cm ✕ 14.0 cm ✕ 27.0 cm. (The density of gold is 19.3 ✕ 103 kg/m3.)
kg
2.)A brass ring of diameter 10.00 cm at 17.3°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 17.3°C. Assume the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to separate the two metals?
(b) What if the aluminum rod were 10.06 cm in diameter?
Answer:
1) m = 0.3137 kg
2a)T_f = -181.7°C
2b) T_f = -1176.97°C
Explanation:
1) We are given;
Length; l = 4.30 cm = 0.043 m
Width; w = 14.0 cm = 0.014 m
height; h = 27.0 cm = 0.027 m
density of gold; ρ = 19.3 × 10³ kg/m³
Formula for the density is known as;
ρ = mass/volume
Thus;
m =ρV
m = 19.3 × 10³ × (lwh)
m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)
m = 0.3137 kg
2a) We are given;
Diameter of brass; L_br = 10 cm
Diameter of aluminum; L_al = 10.01 cm
Now, to some for change in temperature we will use the formula;
L_f = L_i + αL_i(Δt)
Where α is coefficient of expansion.
Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.
Thus;
L_f(brass) = L_f(aluminium)
From table attached, α_brass ≈ 19 × 10^(-6) /°C
Also, α_aluminium ≈ 24 × 10^(-6) /°C
Thus;
L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))
Similarly,
L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Since L_f(brass) = L_f(aluminium), then;
10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Rearranging, we have;
10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.01 = Δt(-50.24 × 10^(-6))
Δt = 0.01/(-50.24 × 10^(-6))
Δt ≈ -199°C
Thus, temperature at which the combination must be cooled to separate the two metals is;
T_f = T_i + Δt
T_f = 17.3 + (-199)
T_f = -181.7°C
2b) Diameter of aluminum is now;
L_al = 10.06 cm
Thus;
10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.06 = Δt(-50.24 × 10^(-6))
Δt = 0.06/(-50.24 × 10^(-6))
Δt = -1194.27°C
T_f = 17.3 + (-1194.27)
T_f = -1176.97°C
what is friction and the types withe examples.
Explanation:
The answer is In the picture. Thanks.
You are locked inside the train car and want to get it moving to draw attention to your plight. There is effectively no friction between the axle and the car, and the train is on horizontal tracks. To try and get the car moving with respect to the ground, you run and slam with all your force against the wall at the front. What happens with the car after you slammed against the wall of the car
Answer:
the car movves briefly as you ran, however, it stops again after you ran in to the wall
Explanation:
Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.
If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.
i.e
[tex]m \times V_1 + M \times V_2 = 0[/tex] --- (1)
here;
m = mass of the boy
M = mass of the train
Thus;
[tex]m \times V_1 =- M \times V_2[/tex] --- (2)
As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.
Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is
Answer:
The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].
Explanation:
The work done is given as in terms of
[tex]W=\Delta TE[/tex]
Where ΔTE is the change in total energy.
This is given as
[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]
Rearranging it in terms of K_x gives
[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]
Which investigation BEST measures the gravitational force on a
toy car?
A. rolling the car down a steep ramp and measuring time
B. using a spring scale and measuring the weight of the car
C. pushing the car and measuring how far it travels before it stops
D. throwing the car in the air and measuring how far it goes before coming
down
Answer:
B : using a spring scale and measuring the weight of the car
Explanation: