A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d, place a block of mass m on top of the compressed spring, and then release the block. The spring launches the block upward, and the block rises to a maximum height some distance above the now-relaxed spring. Find the speed of the block just as it loses contact with the spring.
Answers
Answer:
[tex]v=d\sqrt{\frac{k}{m}} [/tex]
Explanation:
In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:
[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]
In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:
[tex]U_{0}=K_{f}[/tex]
The initial potential energy of the spring is given by the equation:
[tex]U_{0}=\frac{1}{2}kd^{2}[/tex]
the Kinetic energy of the block is then given by the equation:
[tex]K_{f}=\frac{1}{2}mv_{f}^{2} [/tex]
so we can now set them both equal to each other, so we get:
[tex]=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}[/tex]
This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:
[tex]kd^{2}=mv_{f}^{2}[/tex]
so now we can solve this for the final velocity, so we get:
[tex]v=d\sqrt{\frac{k}{m}} [/tex]
Related Questions
Find the net work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius. The coefficient of friction between the ground and the snake is 0.25, and the snake's weight is 54.0 N.
Answers
Answer:
The net work done by friction on the body of the snake is 333.35J
Explanation:
Work done is given by
W = F × s
Where W is the Work done
F is the force
and s is the distance covered
Since we are to determine the work done by friction, then we will determine the frictional force. The frictional force is given by
f = μN = μw
Where μ is the coefficient of friction
N is the normal reaction
and w is the weight
But, F = f
∴ W = μws
From the question
μ = 0.25
w = 54.0 N
Now, we will determine s
From the question,
We are to determine the work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius.
The distance s here is given by the circumference of the circle. Circumference of a circle is given by 2πr
∴ s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Hence,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
Hence, the net work done by friction on the body of the snake is 333.35J.
The net work done by friction on the body of the snake is :
-333.35J
FrictionFormulas:
Work done (W) = F × s
F = force
s = distance covered
f = μN = μw
μ = coefficient of friction
N = normal reaction
w = weight
Solution:
F = f
Weight is :
W = μws
μ = 0.25
w = 54.0 N
Distance covered:
s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Therefore,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
The net work done by friction on the body of the snake is 333.35J.
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Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, what is the weight of the object on the moon?
Answers
I have assumed a weight of 120 N on Earth.
Answer:
The object weighs 20 N on the moon
Explanation:
Weight
The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.
The formula to calculate the weight is:
W = m.g
If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:
[tex]g_m=1/6 g_e[/tex]
Dividing by ge:
[tex]g_m/g_e = 1/6[/tex]
An object of weight We=120 N on planet Earth has a mass of:
[tex]m = 120 / g_e[/tex]
Multiplying by gm:
[tex]m.g_m=120 g_m/g_e[/tex]
Substituting the ratio of accelerations of gravity:
[tex]m.g_m=120 * 1/6[/tex]
Since m.gm is the weight on the Moon Wm:
[tex]W_m=20~N[/tex]
The object weighs 20 N on the moon
Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, the weight of the object on the moon would be one-sixth of its weight on Earth.
Knowing an object's weight on Earth and applying the ratio of the acceleration caused by gravity on the moon to that on Earth will allow us to determine an object's weight on the moon.
Assume that the object's weight on Earth is W (in newtons).
On Earth, the acceleration due to gravity = 9.8 [tex]m/s^2[/tex].
On the moon, = (1/6) * 9.8 [tex]m/s^2[/tex] = 1.6333 [tex]m/s^2[/tex]
Weight = Mass * Acceleration due to gravity
Weight on the moon = W * (1/6)
Thus, the weight of the object on the moon would be one-sixth of its weight on Earth.
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A diffraction pattern is created on a screen when blue light is passed through a single slit. Does the central bright maximum in this pattern become wider or narrower, when the blue light is replaced by red light?
Answers
Answer:
becomes wider
Explanation:
From; asin θ= m λ
It was said in the question that in this particular instance blue light is replaced by red light. The wavelength of blue light is less than that of red light, it then follows that the central bright maximum in this pattern widens when red light is used.
Therefore the diffraction pattern becomes wider.
A car is accelerated at a constant rate from 15 m/s to 25 m/s. It takes the car 6 s to reach its final speed. What is the car’s acceleration?
Answers
Answer:
1.67 m/s²Explanation:
The car’s acceleration can be found by using the formula
[tex]a = \frac{v - u}{t} \\ [/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question we have
[tex]a = \frac{25 - 15}{6} = \frac{10}{6} = \frac{5}{3} \\ = 1.666666...[/tex]
We have the final answer as
1.67 m/s²Hope this helps you
what makes the ball stop on rolling after somtime
Answers
an object of mass 4kg moving with the initial velocity of 20m/s accelerates for 10s and attaind a final velocity of 60m/s calculate the initial momentum
Answers
Explanation:
m = 4kg
u = 20m/s
t = 10s
v = 60m/s
initial momentum = mass × initial velocity
= 4 × 20 = 80kgm/s
A rifle of mass M is initially at rest but free to recoil. It fires a bullet of mass m and velocity v (relative to the ground). After firing, the velocity of the rifle (relative to the ground) is:_____.a. –mv.
b. –Mv/m.
c. –mv/M.
d. –v.
e. mv/M.
Answers
Answer: Option C) -mv / M is the correct answer
Explanation:
given that;
A rifle of mass M is initially at rest but free to recoil
velocity = v
using conversation of moment
pi = pf
i.e initial moment = final moment
But initial moment was zero (0) since everything was rest but free to recoil
so
pi = pf
0 = mv + MV
V is the recoil velocity of the rifle
MV = -mv
V = -mv / M
Therefore Option C) -mv / M is the correct answer
A rock is thrown straight upward with a speed 6.000 m/s. If this rock takes 1.540 s to fall to the ground, from what height h was it released?
Answers
Answer:
3.8961038961038
Explanation:
6000/1540=3.8961038961038
X= v/t
24- What happens to a given mass of water as it is cooled from 4°C to zero?
(a) expands
(b) contracts
(c) vaporizes
(d) Neither expands, contracts, nor vaporizes.
Answers
Explanation:
A-It starts to expand. Hence ice and water of same weight has different volumes
Expansion happens to a given mass of water as it is cooled from 4°C to zero. Option A is correct.
What is temperature?Temperature directs to the hotness or coldness of a body. In straightforward terms,
it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles more the temperature.
Water loses density when it is cooled from 4°C to 0°C. It turns out that the temperature at which liquid water has the most significant density is 4 degrees Celsius.
It will increase in size when heated or cooled. Since most liquids shrink when they are chilled, it is rare for water to expand when cooled to lower temperatures.
As a mass of water is cooled from 4°C to zero, expansion occurs.
Hence, option A is correct.
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What average force is required to stop an 1100-kg car in 8.0s if the car is travelling at 95km/h?
Answers
Answer: 13062.5 N
What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h? i got the answer to be 13062.
Answer:
The answer is 13062.5 Newtons (N).
Explanation:
The mass of the car is 1100 kg.
The acceleration is 95 km/h.
Using this information, we can use Newton's 2nd Law, F=MA.
1100 kg * 95km/h = 104500 N
Because the answer wants average force, we need to divide the answer by 8 seconds, giving us 13062.5 N.
104500 N / 8 seconds = 13062.5 N (avg force).
A falling stone takes delta t = 0.32s to travel past a window 2.2m Tall. From what height above the top of the window did the stone fall?
Answers
Answer:
The height above the top of the window is 1.44 m
Explanation:
Given;
time of motion, t = 0.32 s
height traveled at the given time, h = 2.2m
determine the initial velocity of the stone;
h = ut + ¹/₂gt²
2.2 = u(0.32) + ¹/₂ x 9.8 x 0.32²
2.2 = 0.32u + 0.502
0.32u = 2.2 - 0.502
0.32u = 1.698
u = 1.698 / 0.32
u = 5.31 m/s
This initial velocity on top of the window becomes the final velocity from the height above the window.
v² = u² + 2gh
where;
u is the initial velocity of the stone from the height above the window;
5.31² = 0 + (2 x 9.8)h
19.6h = 28.196
h = 28.196/19.6
h = 1.44 m
Therefore, the height above the top of the window is 1.44 m
What happens when calcium reacts with chlorine?
A. Electrons become delocalized among the atoms
B. Electrons are shared between the calcium atoms and the chlorine
atoms
C. Electrons move from the chlorine atoms to the calcium atoms,
D. Electrons move from the calcium atoms to the chlorine atoms,
Answers
When calcium reacts with chlorine. Electrons move from the calcium atoms to the chlorine atoms. Option D is correct.
What is a chemical reaction?When one or more chemicals are transformed into one or more other compounds, a chemical reaction occurs.
In an ionic compound, calcium chloride is present. To build a complete outer shell of electrons, the calcium atom loses two electrons and each chlorine atom obtains one.
When calcium combines with chlorine, it forms calcium chloride. From the calcium atoms to the chlorine atoms, electrons travel.
Hence, option D is correct.
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If the velocity of a car changes from 0 meters per second (m/s) to 100 m/s in 10 seconds, what is the acceleration over that 10 second period?
Answers
Answer:
10m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 100m/s
Time taken = 10s
Unknown:
Acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time.
A = [tex]\frac{v - u}{t}[/tex]
v = final velocity
u = initial velocity
t = time taken
So, insert the parameters and solve;
A = [tex]\frac{100 - 0}{10}[/tex] = 10m/s²
A spaceship is accelerating at 1000 m/sec2 . How much force is required from the backthrusters to completely stop the spaceship?
Answers
Answer:
1000x Newton
Explanation:
Step one
given data
acceleration= 1000 m/s²
The question did not specify the mass of the mass of the space ship.
So, let's assume the mass is x kg
Step two:
Required is the force F in Newton
From Newtons first law, it states that a body will continue to be at rest or uniform motion unless acted upon by a force.
F=mass x Acceleration
F=ma
Substituting our given data we have
F=1000x Newton
A scientist decides to replicate an experiment completed by another scientist. Which statement describes something that would not affect the results of the replicated experiment
Answers
Hi, you've asked an incomplete question. However, I provided some explanation about the replication process that scientists do.
Explanation:
Replication in research involves carefully repeating an original experiment to see whether the same result would be arrived at as in previous research experiments.
For most scientists today, in other to avoid anything that would erroneously affect the results of the replicated experiment they usually follow the same procedures as carried by the previous researchers.
1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)10N/m
2. What will be the extension of this spring if the load is a) 4N and b) 75 g?
Answers
Answer:
1) k = 10 [N/m]
2) a-) x = 0.4 [m]
b) x = 0.075 [m]
Explanation:
To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.
F = k*x
where:
F = force [N] (units of Newtons]
k = spring constant [N/m]
x = distance = 10 [cm] = 0.1 [m]
Now, the weight is equal to the product of the mass by the gravity
W = m*g = F
where:
m = mass = 100 [g] = 0.1 [kg]
g = gravity acceleration = 10 [m/s²]
F = 0.1*10 = 1 [N]
Now clearing k
k = F/x
k = 1/0.1
k = 10 [N/m]
2)
a ) if the force is 4 [N]
clearing x
x = F/k
x = 4/10
x = 0.4 [m]
m = 75 [g] = 0.075 [kg]
W = m*g = F
F = 0.075*10 = 0.75 [N]
x = .75/10
x = 0.075 [m]
An ant can run at an average speed of 0.083 meters per second, calculate how long it would take an ant to run 100 meter dash in seconds and in minutes
Answers
Answer:
The time taken by the ant is 1204.82 seconds or 20.08 minutes
Explanation:
Given:
Speed of the ant (s) = 0.083 m/s
Distance traveled by the ant (D) = 100 m
We know that distance traveled by a body is equal to the product of the speed of the body and the time taken for its travel.
Let the time taken by the ant be 't' seconds.
Now, as per the formula:
Distance = Speed × Time
⇒ 100 m = 0.083 m/s × t
⇒ [tex]t = (100\ m)/(0.083\ m/s)[/tex]
[tex]\therefore t=1204.82\ s[/tex]
Now, we know that,
60 seconds = 1 minute
So, 1 second = [tex]\frac{1}{60}\ min[/tex]
Therefore, [tex]1204.82\ s = \frac{1204.82}{60}=20.08\ min[/tex]
Hence, the time taken by the ant is 1204.82 seconds or 20.08 minutes
Is Geothermal Energy renewable? Why or why not? Use in your own words.
Answers
Answer:
Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.
the resistance of a heater element is 1200 ohms and draws a circuit of 0.4 amperes calculate the voltage of the circuit
Answers
Answer:
0.00033333333
Explanation:
Two solid spheres are made from the same material, but one has twice the diameter of the other. Which sphere will have the greater bulk modulus?
Answers
Answer:
It will be the same for both
Explanation:
from this question we have one similarity between these two spheres.
- they are both made from the same material,
The difference between both spheres is that:
- one of the spheres has its diameter to be twice as large as that of the other one.
We are to say the sphere with the greater bulk modulus.
If the material is the same thenthe Bulk modulus is also the same. It is not dependent on the material since it is a constant for that materia
Therefore the correct answer is:
It will be the same for both spheres.
How do the magnetic fields of Uranus and Neptune suggest that the mantles inside those planets are fluid?
Answers
Answer:
The surface must be a good absorber of light. Spectra of the planets proved that methane in the atmospheres of these planets is what absorbs red light.
Explanation:
A projectile is fired with an initial speed of 36.6 m/s at an angle of 42.2° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the speed of the projectile 1.50 s after firing.
Answers
Answer:
Explanation:
a) Maximum height id expressed as;
H = u²sin² theta/2g
H = 36.6²(sin42.2)²/2(9.8)
H = 1,339.56(0.6717)/19.6
H = 899.81/19.6
H = 45.91m
Hence the maximum height is 45.91m
b) The Time of flight is the total time in air expressed as;
T = 2usin theta/g
T = 2(36.6)sin42.2/9.8
T = 73.2sin42.2/9.8
T = 49.17/9.8
T = 5.017secs
Hence the total time in air is 5.017scs
c) Range = U²sin2(theta)/g
Range = 36.6²sin2(42.2)/9.8
Range = 1,339.5(0.9952)/9.8
Range = 1,333.11/9.8
Range = 136.03m
Hence the range is 136.03m
d) USing the rime of flight formula;
T =2Usintheta/g
1.5 = 2Usin42.2/9.8
2Usin42.2 =1.5*9.8
2Usin42.2 = 14.7
U = 14.7/2sin42.2
U = 14.7/1.3434
U = 10.94m/s
Hence the speed of the projectile is 10.94m/s
At high noon, the sun is almost directly above (about 2.0 degrees from the vertical) and a tall redwood tree casts a shadow that is 10m long. How tall is the redwood tree?
Answers
290m
The light always travels in a straight line.
At high noon, the ray from the sun is 2° from the vertical axis.
tan θ = (Opposite side)/(Adjacent side)
On applying above trigonometric formula, we get,
tan 2 = 10/h
0.035 = 10/h
∴ h = 10/0.035 = 290 m
If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?
Answers
Answer:
p₂ / p₁ = 2 (v₁ / v₂)
Explanation:
The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,
p = m v / √[1+ (v/c)² ]
for the case of speeds much lower than the speed of light this expression is close to
p = m v
In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small
p₂ = 2 p₁
p₂/p₁ = 2
in consecuense
m v₂ = 2 m v₁
v₂ = 2 v₁
consider particles of equal mass.
By the time their speeds increase they enter the relativistic regime
p₂ = mv₂ /√(1 + v₂² /c²)
p₁ = m v₁ /√(1 + v₁² / c²)
let's look for the relationship between these two moments
p₂ / p₁ = mv₂ / mv₁ [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)
from the initial statement
p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)
we take c from the root
p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]
this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1
p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]
p₂ / p₁ = 2 (v₁ / v₂)
we see the value of the moment depends on the speed of the particles
A typical sheet of printer paper is 8.5 inches wide and 11 inches long.
a.) Calculate the length of one sheet in cm
b.) Calculate the width of one sheet in cm?
Answers
Answer:
Explanation:
Given the length and width in inches
Length = 8.5in
Width = 11in
b) Convert to centimeters
1in = 2.54cm
8.5in = x
Cross multiply
x = 8.5 × 2.54
x = 21.59cm
Hence the width in cm is 21.59cm
a) 11in = y
1in = 2.54cm
Cross multiply
x = 11 × 2.54
x = 27.94cm
Hence the length in cm is 27.94cm
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.90 m above the pool.
Required:
a. What is her highest point above the board?
b. How long a time are her feet in the air?
c. What is her velocity when her feet hit the water?
Answers
Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
In Figure 1 , an engine that delivers power P is attached to a pulley that lifts a box of mass m0 a vertical distance of y0 at a constant speed in time t0 . If, as shown in Figure 2 the engine again delivers power P , how long would it take the engine to raise a box of mass 3m0 the same vertical distance y0 at a constant speed?
Options:
1/3 t0
t0
root(3 t0
3t0
9t0
Answers
Answer:
t = 3 t₀
Explanation:
The formula for the power delivered by the engine can be given in terms of the change in the potential energy of the box as follows:
P = P.E/t = mgy/t
where,
P = Power delivered by the engine
m = mass of box lifted = m₀
g = acceleration due to gravity
y = vertical distance or height box covered = y₀
t = time taken = t₀
Therefore,
P = m₀gy₀/t₀
t₀ = m₀gy₀/P -------------- equation (1)
Now, if we change the mass of box as:
m = 3m₀
y = y₀ (constant)
P = Constant
Then,
P = 3m₀gy₀/t
t = 3m₀gy₀/P
using equation (1):
t = 3 t₀
Answer:
t=3to
Explanation:
I took the quiz it is right
You are standing on a circular track that is 332 m long. You begin jogging at the start line and keep jogging until you complete a full lap and stop at the same point you
started at.
ard
• What is your final displacement?
Answers
Answer:
Zero.
Explanation:
By definition, the displacement is the difference between the final position and the initial position, so independent from the distance traveled (in this case a full lap), since the initial and final position are the same, the net displacement is just Zero.A copper collar is to fit tightly about a steel shaft whose diameter is 6.0000 cm at 19°C. The inside diameter of the copper collar at that temperature is 5.9800 cm. To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the temperature of both the steel shaft and copper collar are raised simultaneously?
Answers
Answer:
T' = 865.15 °C
Explanation:
In order for the copper collar to just slip on the steel shaft the, assuming are heated simultaneously, we must find the final parameters of both and equate them. Because the final diameters of both must be same for the slipping to occur.
FOR COPPER COLLAR:
dc' = dc(1 + ∝c*ΔT)
where,
dc' = final diameter of copper ring
dc = initial diameter of copper ring = 5.98 cm
∝c = coefficient of linear expansion for copper = 16 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (1)
FOR STEEL SHAFT:
ds' = ds(1 + ∝s*ΔT)
where,
ds' = final diameter of steel shaft
ds = initial diameter of steel shaft = 6 cm
∝s = coefficient of linear expansion for steel = 12 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (2)
Comparing equation (1) and equation (2):
(5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
(5.98 cm/6 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = [1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
0.9967 + (1.59 x 10⁻⁵ °C⁻¹)ΔT = 1 + (12 x 10⁻⁶ °C⁻¹)ΔT
1 - 0.9967 = [(15.9 -12) x 10⁻⁶ °C⁻¹]ΔT
0.0033/3.9 x 10⁻⁶ °C⁻¹ = ΔT
ΔT = 846.15 °C
but,
ΔT = T' - T = T' - 19°C = 846.15°C
T' = 846.15 °C + 19 °C
T' = 865.15 °C
2. A solenoid. Suppose the south end of a bar magnet was introduced to the right end of this solenoid at a constant velocity. What direction would you expect the induced magnetic field to be in?4. To decrease the magnitude of current induced in the inductor using a bar magnet, you could (select all that apply):a. Decrease the strength of the magnet.b. Decrease the velocity of the magnet going into the solenoid.c. Decrease the number of coils in the solenoid.d. Increase the cross sectional area of the solenoid.6. List the TWO values you need to record to measure a voltage on the oscilloscope
Answers
Answer:
2) deflection must be towards the negative side of the voltage.
4) the correct statements are: b and c
Explanation:
2) This question is based on Faraday's law of induction, when we introduce a magnet in a solenoid an induced current is produced that generates a voltage that is given by
E = - N d [tex]\phi_{B}[/tex] / dt
where [tex]\phi_{B}[/tex] = B. A
The bold are vectors
Therefore, when applying this formula to our case, the induction lines of the magnetic field increase as we approach the solenoid, as the South pole approaches the lines are in the direction of the magnet, therefore the normal to the solenoid that has an outgoing direction and the magnetic field has 180º between them and the cos 180 = -1; consequently the deflection must be towards the negative side of the voltage.
4) From the Faraday equation we can see that the inductive electromotive force depends
* The magnitude of B that changes over time
* The area of the loop that changes over time
* The angle between B and the area that changes over time
* A combination of the above
With this analysis we will review the different alternatives given
a) False. It takes a temporary change and an absolute value of B
b) True. As the speed decreases, the change in B decreases, that is, dB / dt decreases
c) True. The current is induced in each turn, if there is a smaller number the total current will be smaller
d) False. A temporary change of area is needed, in addition to increasing the area the current increases
We can see that the correct statements are: b and c