An elevator, in which a man is standing moving upward with a constant
speed of 10m/sec2. If a man drops a coin from a height of 2.5m. Find the time
taken by it to reach the oor of the elevation? g=9.8m/sec2

Answers

Answer 1

ANSWER

Lift frame :

Initial speed of the coin u=0 m/s

Acceleration a=9.8 m/s

2

Initial height of coin from the floor of elevator h=2.45 m

Time taken by coin to hit the floor T=

g

2H

⟹ T=

9.8

2×2.45

=

2

1

s

Answer 2
We are Given:

velocity of the lift (u₁) = 10 m/s

initial velocity of the coin (u₂) = 10 m/s  

(the coin is also moving with the Elevator)

acceleration of the coin (a₂) = -9.8 m/s²

acceleration of the Elevator (a₁) = 0 m/s

Distance covered by the coin (s)  = -2.5 m

__________________________________________________________

Relative Velocity and Acceleration:

Relative velocity:

We will find the velocity and acceleration of the coin with respect to the lift since we are monitoring the motion of the coin

velocity of the coin with respect to the elevator (u₂₁) = u₂ - u₁

u₂₁ = 10 - 10

u₂₁ = 0m/s

Relative Acceleration:

acceleration of the coin with respect to the elevator (a₂₁) = a₂ - a₁

a₂₁ = -9.8 - 0

a₂₁ = -9.8

__________________________________________________________

Solving for the Time taken:

From the second equation of motion, we know that:

s=  ut + 1/2at₂

we can rewrite this equation in terms of relative motion:

s = u₂₁(t) + 1/2(a₁₂)t²  

Notice that the time and the displacement are not relative, that's because displacement and time will remain the same no matter the frame of reference

replacing the known values in the equation:

-2.5 = (0)(t) + 1/2 (-9.8)(t²)

-2.5 = -4.9(t²)

dividing both sides by -4.9

t² = -2.5 / -4.9

t² = 25/49

t² = (5)² / (7)²

taking the square root of both the sides

t = 5/7   OR   0.71 seconds (approx)

Therefore, the coin will reach the floor of the Elevator in 0.71 seconds


Related Questions

match the following​

Answers

1 it would be 0
2 it would be B
3 would be A

What is your zodiac sign?
Capricorn: January 20th to February 16th.
Aquarius: February 16th to March 11th.
Pisces: March 11th to April 18th.
Aries: April 18th to May 13th.
Taurus: May 13th to June 21st.
Gemini: June 21st to July 20th.
CancerJuly 20th to August 10th.
Leo: August 10th to September 16th.
Virgo: September 16th to October 30th
Libra: October 30th to November 23rd
Scorpio: November 23rd to November 29th
Ophiuchus: November 29th to December 17th
Sagittarius: December 17th to January 20th

Answers

Answer: Gemini

Explanation: lol I just born that that sign ig

Estimate the radiation pressure due to a bulb that emits 25 W {\rm W} of EM radiation at a distance of 9.5cm \;cm from the center of the bulb. Assume that light is completely absorbed
P=? unit?

Answers

Answer:

[tex]7.3*10^{-7} \frac{N}{m^2}[/tex]

Explanation:

[tex]\frac{25}{(4*\pi*(0.095)^{2}*3*10^8 }[/tex]

Hope this is right and helps!

02
The ano ture of a converges
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The serie
atooaod
பி
5​

Answers

Im sorry i have no idea what that means ??? Auto correct

Planets A and B have the same mass, but planet A is half the size of planet B.
Which statement correctly explains the weight you would experience on each
planet?
O A. You would weigh the same on both planets because your mass
would be the same on both.
O B. You would weigh the same on both planets because the masses
of the planets are the same.
O c. You would weigh less on planet A because the distance between
you and the planet's center of gravity would be smaller.
O D. You would weigh more on planet A because the distance between
you and the planet's center of gravity would be smaller.

Answers

Answer:

c. You would weigh less on planet A because the distance between

you and the planet's center of gravity would be smaller.

Explanation:

The statement that best describes your weight on each planet is that you would weigh less on planet A because the distance between you and the planet's center of gravity would be smaller.

This is based on Newton's law of universal gravitation which states that "the force of gravity between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".

Since weight is dependent on the force of gravity and mass, the planet with more gravitational pull will have masses on them weighing more.

Since the distance between the person and the center of the planet is smaller, therefore, the weight will be lesser.

Answer:it’s D for ape x

Explanation:          

what is the speed od sail boat that us traveling 100 meters in 120 seconds?​

Answers

The sailboat is about 0.83m/sec (rounded to the nearest tenth).

To find speed, you would calculate distance over time.

100 meters/120 seconds = 0.83 meters per second

A ball of mass 2 kg is moving with a speed of +6 m/s directly towards
another ball of mass 1 kg moving with a speed of -4 m/s. If they collide
inelastically, what will be the speed of the resulting 3 kg ball
immediately after the collision?

Answers

Answer: 2.7

Explanation:

In a football game the running back is running up the field. He starts from rest and runs 4 seconds with an acceleration of 1.3m/s^2.
What is the magnitude of his displacement

Answers

Answer:

3.38m

Explanation:

Given parameters:

Time  = 4s

Acceleration  = 1.3m/s²

Unknown:

Magnitude of the displacement = ?

Solution:

The body starts at rest and the initial velocity is 0m/s. To solve this problem, we have to use the expression below;

    S   = Ut  + [tex]\frac{1}{2}[/tex]at²

 S  = displacement

t is the time

  a is the acceleration

  U is the initial velocity

  V is the final velocity

Insert the parameters and solve;

   S = (0 x 4)  +  [tex]\frac{1}{2}[/tex] x 1.3² x 4  = 3.38m

3. A coin is tossed vertically upward.
a. What happens to its velocity while it is in the air?
b. Does its acceleration increase, decrease, or remain constant while it is in the air?

Answers

Answer:

a. The initial velocity will decrease until it reaches its final velocity/maximum (which is 0).

b. The acceleration always remain constant.

Newton’s Law of Cooling. Newton’s law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the medium M(t) and the temperature of the body. That is, dT/dt = K[M(t) - T(t)] , where K is a constant. Let K = 0.04 (min)-1 and the temperature of the medium be constant, M(t) = 293 kelvins. If the body is initially at 360 kelvins, use Euler’s method with h = 3.0 min to approximate the temperature of the body after

(a) 30 minutes.

(b) 60 minutes.

Answers

Answer:

After 30 minutes, the temperature of the body is: T₁₀ = 311.60 K

After 60 minutes, the temperature of the body is: T₂₀ = 298.18 K

Explanation:

Given that:

[tex]\dfrac{dT}{dt}= K \bigg [ M(t) -T(t)\bigg][/tex]

where;

K = 0.04

M(t) = 293

Then;

[tex]\dfrac{dT}{dt}= 0.04 \bigg [ 293 -T\bigg][/tex]

[tex]\dfrac{dT}{dt}= 11.72 -0.04 \ T[/tex]

Using Euler's Formula;

[tex]T_{n+1} = T_n + hf( t_n, T_n)[/tex]

where;

[tex]f(t_n,T_n) = 11.72 - 0.04 T_n[/tex]

Then;

[tex]T_{n+1} = T_n + 3.0 (11.72-0.04 \ T_n)[/tex]

[tex]T_{n+1} = 0.88T_n + 35.16 \ ---(1)[/tex]

At initial state [tex]t_0[/tex]  (0);  [tex]T_0[/tex] = 360

At t₁ = 3.0 when T₀ = 360

[tex]T_1= 0.88 T_o + 35.16[/tex]

[tex]T_1= 0.88 (360) + 35.16[/tex]

[tex]T_1 = 351.96 \ K[/tex]

At t₂ = 6.0 when T₂ = 0.88T₁ + 35.16

T₂ = 0.88(351.96) + 35.16

T₂ = 344.89 K

At t₃ = 9.0 when T₃ = 0.88T₂ + 35.16

T₃ = 0.88(344.89) + 35.16

T₃ =338.66 K

At t₄ = 12.0 when T₄ = 0.88T₃ + 35.16

T₄ = 0.88(338.66) + 35.16

T₄ = 333.18 K

At  t₅ = 15.0 when T₅ = 0.88T₄ + 35.16

T₅ = 0.88(333.18) + 35.16

T₅ = 328.36 K

At t₆ = 18.0 when T₆ =  0.88T₅ + 35.16

T₆ = 0.88(328.36) + 35.16

T₆ = 324.12 K  

At t₇ = 21.0 when T₇ = 0.88T₆ + 35.16

T₇ = 0.88(324.12) + 35.16

T₇ = 320.39 K

At t₈ = 24.0 when T₈ = 0.88T₇ + 35.16

T₈ = 0.88(320.29) + 35.16

T₈ = 317.02 K

At t₉ = 27.0 when T₉ = 0.88T₈ + 35.16

T₉ = 0.88(317.02) + 35.16

T₉ = 314.14 K

At t₁₀ = 30 when T₁₀ = 0.88T₉ + 35.16

T₁₀ = 0.88(314.14) + 35.16

T₁₀ = 311.60 K

At t₁₁ = 33.0 when T₁₁ = 0.88T₁₀ + 35.16

T₁₁ = 0.88(311.60) + 35.16

T₁₁ = 309.37 K

At t₁₂ = 36.0  when T₁₂ = 0.88T₁₁ + 35.16

T₁₂ = 0.88(309.37)+ 35.16

T₁₂ = 307.41 K

At t₁₃ = 39.0  when T₁₃ = 0.88T₁₂ + 35.16

T₁₃ = 0.88( 307.41) + 35.16

T₁₃ = 305.68 K

At t₁₄ = 42.0  when T₁₄ = 0.88T₁₃ + 35.16

T₁₄ = 0.88(305.68) + 35.16

T₁₄ = 304.16 K

At t₁₅ = 45.0  when T₁₅ = 0.88T₁₄ + 35.16

T₁₅ = 0.88(304.16) + 35.16

T₁₅ = 302.82 K

At t₁₆ = 48.0  when T₁₆ = 0.88T₁₅ + 35.16

T₁₆ = 0.88(302.82) + 35.16

T₁₆ = 301.64 K

At t₁₇ = 51.0  when T₁₇ = 0.88T₁₆ + 35.16

T₁₇ = 0.88(301.64) + 35.16

T₁₇ = 300.60 K

At t₁₈ = 54.0  when T₁₈ = 0.88T₁₇ + 35.16

T₁₈ = 0.88(300.60) + 35.16

T₁₈ = 299.69 K

At t₁₉ = 57.0  when T₁₉ = 0.88T₁₈ + 35.16

T₁₉ = 0.88(299.69) + 35.16

T₁₉ = 298.89 K

At t₂₀ = 60  when T₂₀ = 0.88T₁₉ + 35.16

T₂₀ = 0.88(298.89) + 35.16

T₂₀ = 298.18 K

Bugs Bunny, having a mass of 75 kg, is pulled by Babs Bunny and Buster Bunny. Babs
Bunny tugs Bugs with a 50 N force to the north, towards a carrot cake stand. Buster pulls
Bugs with a 120 N force to the south, towards a basket ball court. There is a force of
friction of 15 N. What will be Bug's acceleration?

Answers

Answer:

a = 0.73 m/s²

Explanation:

Considering South direction to be positive and North to be negative. The unbalanced force on bugs bunny will be given as:

Unbalanced Force = Force of Buster - Force of Babs - Friction

Unbalanced Force = 120 N - 50 N - 15 N

Unbalanced Force = 55 N

But, from Newton's Second Law:

Unbalanced Force = ma

55 N = ma

where,

m = mass of Bugs Bunny = 75 kg

a = acceleration of Bugs Bunny = ?

Therefore,

55 N = (75 kg)a

a = 55 N/75 kg

a = 0.73 m/s²

Answer:

a = 0.73 m/s²

Explanation:

-_- im tired -_-

An airplane accelerates with a constant rate of 3.0 m/s2

starting at a velocity of 21 m/s. If the final velocity is 60 m/s,

what is the displacement during this period?

Answers

Answer:

526.5m

Explanation:

Given parameters:

Acceleration = 3m/s²

Initial velocity  = 21m/s

Final velocity  = 60m/s

Unknown:

Displacement = ?

Solution:

To solve this problem, we apply the right motion equation which is shown below;

     V²  = U²  + 2aS

  V is the final velocity

  U is the initial velocity

  a is the acceleration

 S is the displacement

 Insert the parameters and solve;

    60²  = 21² + 2(3)S

    3600 = 441 + 6S

    3600 - 441  = 6S

    3159  = 6S

       S = 526.5m

HELP!!!!!! Need Help !!!

Answers

Answer:

its y

Explanation:

A large, open at the top, upright cylindrical tank contains fuel oil with a density of 0.890 ✕ 103 kg/m3.
(a) If the air pressure is 101.3 kPa, determine the absolute pressure (in Pa) in the fluid at a depth of 28.0 m.

(b) Determine the force (in N) exerted by only the fluid on the window of an instrument probe at this depth if the window is circular and has a diameter of 3.20 cm.

Answers

Answer:

a) Pabs = 345.7 [kPa]

b) F = 278.08 [N]

Explanation:

Manometric pressure in liquids is defined as the product of density by gravitational acceleration by the height of the liquid.

[tex]P_{g} =Ro*g*h[/tex]

where:

Pg = manometric pressure [Pa]

Ro = density = 0.890*10³[kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = liquid height = 28 [m]

Now the absolute pressure is defined as the sum of the atmospheric pressure plus the manometric pressure

a)

[tex]P_{abs}=P_{man}+P_{gau}\\P_{abs}= 101300 + (890*9.81*28)\\P_{abs} = 345765.2 [Pa] = 345.7 [kPa][/tex]

b) To determine the force at that point

The pressure is now defined as the relationship of a force over the area.

[tex]P=F/A\\F = P*A\\[/tex]

But the area of a circle can be calculated as follows:

[tex]A=\frac{\pi }{4} *(0.032)^{2}\\A = 0.0008042[m^{2}] \\F = 345765.2*0.0008042\\F = 278.08 [N][/tex]

When forces in a system are in balance there is

Answers

Answer:

Equilibrium

Explanation:

When forces in a system are in balance, there is equilibrium.

A balance of forces around a system causes equilibrium of a system.

A body will remain its state of rest or continue in uniform motion unless acted upon by an external force. This balance of forces on a body keeps a body at rest. Even in motion, the body is at rest unless an external force acts on it.

Pls help me progress report just came in pls help me

Answers

Answer:

3rd. outer

Explanation:

they are electropositive have a tendency to give electrons

It is said that Galileo dropped objects off of the Leaning Tower of Pisa to determine whether heavy or light objects fall faster. If Galileo had dropped a 5 kg cannonball from a height of 12 meters,

Answers

Answer:

ΔP.E = - 588 J

Explanation:

The complete question is as follows:

It is said that Galileo dropped objects off the Leaning Tower of Pisa to determine whether heavy or light objects fall faster. If Galileo had dropped a 5.0-kg cannon ball to the ground from a height of 12 m, what would have been the change in PE of the cannon ball?

Answer: So, in order to calculate the change in Potential Energy of the cannon ball, we will use the general formula of Potential Energy as follows:

ΔP.E = mgΔh

where,

ΔP.E = Change in Potential Energy = ?

m = mass of cannon ball = 5 kg

g = acceleration due to gravity = 9.8 m/s²

Δh = Change in Height of Cannon Ball = Final Height - Initial Height

Δh = 0 m - 12 m = - 12 m

Therefore,

ΔP.E = (5 kg)(9.8 m/s²)(- 12 m)

ΔP.E = - 588 J

Negative sign shows decrease in Potential Energy

Please any one can help solving this question

Answers

Which part are you stuck on? Or is it all of these questions?

A skateboard rolls off a horizontal ledge that is 1.12 m high, and lands
1.48 m from the base of the ledge.
How much time was he in the air?
(Unit = s)

Answers

Answer:

He was 0.4781 s in air.

Explanation:

This question is on an object launched horizontally with zero angle and at a height H.

The formula to apply is ;

[tex]T=\sqrt{\frac{2H}{g} }[/tex]

where ;

T=time of flight = ?

H= height of the object was launched = 1.12 m

T= √2*1.12/9.8

T=√0.22857

T= 0.4781 seconds

During a long drought, many deer become weak because a basic need is not being met.

What requirements for life are the deer most likely lacking?

A) enough space
B) enough water
C) adequate shelter
D) ability to find mates

Answers

B) enough water
because it’s a drought

Answer:

B) enough water I agree with what the person under/over me said

Explanation:

Bobby runs with an average velocity of 7.5 m/s over a straight-line distance of 90 meters. Calculate the time Bobby ran?

Answers

Answer:

12s

Explanation:

Given parameters:

Velocity  = 7.5m/s

Distance  = 90m

Unknown:

Time taken  = ?

Solution:

Speed is the rate of distance traveled with time. It is mathematically expressed as;

      Speed  = [tex]\frac{distance}{time}[/tex]

Since the unknown is time;

        Time  = [tex]\frac{distance}{speed}[/tex]  

      Insert the parameters and solve;

  Time  = [tex]\frac{90}{7.5}[/tex]   = 12s

1. Synthesize Information You push your
younger sister on a swing in a park. Then you
give her a harder push. Explain what happens
in each case, in terms of the second and third
laws of motion

Answers

Answer:separate

Explanation:

What is the half-life of a radioisotope?

half of the time needed for a sample of radioisotope to become stable
the time needed for half of a sample of radioisotope to decay
the time needed before the first half of a radioisotope begins to decay
half of the time needed for a radioisotope to leave the body

Answers

Answer: The amount of time it takes for half of the material to decay

B. The time needed for half of a sample of radioisotope to decay is the half-life of a radioisotope.

What is radioactive decay?

Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive.

The most common types of decay are alpha decay, beta decay, and gamma decay, all of which involve emitting one or more particles.

The radioactive half-life means, the time required for half the atoms of a particular radioisotope to decay into another isotope.

The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay.

The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
Therefore,

The half-life of a radioisotope is the time needed for half of a sample of radioisotope to decay.

The correct option is B

Learn more about radioactive decay here:

https://brainly.com/question/1770619

#SPJ5

HELP ME ASAP!!!!
Select the correct answer.
According to Erik Erikson's psychosocial theory of personality, which stage occurs during adolescence?
A. identity versus identity confusion
B.
initiative versus guilt
C. generativity versus stagnation
D.
intimacy versus isolation
Reset
Next

Answers

Answer:

Identity versus confusion is the fifth stage of ego according to psychologist Erik Erikson's theory of psychosocial development.

Explanation:This stage occurs during adolescence between the ages of approximately 12 and 18. During this stage, adolescents explore their independence and develop a sense of self.

Answer:

D) autonomy versus shame and doubt

Explanation:

Please Help Asap :( if you help may god bless you

Answers

Chlorine Cl: 17 protons, 17 electrons, 7 Valence electrons, Nonmetal, Give 3, 14 new electrons, +1, Cl-

Magnesium Mg: 12 protons, 12 electrons, 3 Valence electrons, Metal, Give 2, 10 new electrons, 2-, Mg ^2+

Oxygen O: 8 protons, 8 electrons, 6 Valence electrons, Nonmetal, Give 6, 2 new electrons, -2, 0 ^2-

Nitrogen N: 7 protons, 7 electrons, 5 Valance electrons, Nonmetal, Give 4, 3 new electrons, -3, N ^3-

I really hope this helps :)

A plane flies for 2 hours at a speed of 430 km/h. How far did the plane travel?

Answers

Answer:

860 km

Explanation:

430*2=860

Is Nuclear Energy renewable? Why or why not? Use in your own words.

Answers

Explanation:

Nuclear energy is argued to be renewable because it is found in the atoms that make up the earth, does not emit carbon and its use can be made unlimited by the use of technologies.

Answer:

Nuclear energy is a non-renewable energy source that comes from the nucleus of atoms.

A 75 kg student walks up three flights of stairs, a vertical height of about 50 ft. If the magnitude of the average rate at which the gravity force does work on the student equals 500 W, how long would it take the student to travel up the three flights of stairs?

Answers

Answer:

22.4 s

Explanation:

The following data were obtained from the question:

Mass (m) = 75 Kg

Height (h) = 50 ft

Power (P) = 500 W

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

Next, we shall convert 50 ft to metres (m). This can be obtained as follow:

1 ft = 0.3048 m

Therefore,

50 ft = 50 ft × 0.3048 m / 1 ft

50 ft = 15.24 m

Thus, 50 ft is equivalent 15.24 m.

Next, we shall determine the energy used by the student to walk up three flights of stairs. This can be obtained as follow:

Mass (m) = 75 Kg

Height (h) = 15.24 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 75 × 9.8 × 15.24

E = 11201.4 J

Finally, we shall determine the time taken for the student to travel up the three flights of stairs. This can be obtained as follow:

Power (P) = 500 W

Energy (E) = 11201.4 J

Time (t)

Power (P) = Energy (E) / Time (t)

500 = 11201.4 /t

Cross multiply

500 × t = 11201.4

Divide both side by 500

t = 11201.4 /500

t = 22.4 s

Therefore, it will take the student 22.4 s to travel up the three flights of stairs.

The time taken for the student to travel up the three stairs is 22.4 s.

The given parameters;

mass of the student, m = 75 kgheight of the stairs, h = 50 ft = 15.24 mwork done on the student by gravity, P = 500 W

The gravitational potential energy of the student due to vertical position of the three stairs is calculated as;

[tex]P.E = mgh\\\\P.E = 75 \times 9.8 \times 15.24 \\\\P.E =11,201.4 \ J[/tex]

The time taken for the student to travel up the three stairs is calculated as follows;

[tex]E = Pt\\\\t = \frac{E}{P} \\\\t = \frac{11,201.4}{500} \\\\t = 22.4 \ s[/tex]

Thus, the time taken for the student to travel up the three stairs is 22.4 s.

Learn more here:https://brainly.com/question/25510584

A source of sound produces a note of 512 Hz in air at 17 degree celsius with wavelength 66.5 cm. Find the ratio of molar heat capacities at constant pressure to constant volume at NTP. Densities of air and mercury at NTP are 1.293 kg/m^3 and 13600 kg/m^3 respectively.​

Answers

Answer:

a. 0.32 b. 1448 m/s

Explanation:

We know v ∝ √T where v = velocity of sound and T = absolute temperature.

Let v₁ = velocity of sound at 17°, v₁ = fλ where f = frequency of sound = 512 Hz and λ = 66.5 cm = 0.665 m

So, v₁ = fλ = 512 Hz × 0.665 m = 340.48 m/s

T₁ = 17 + 273 = 290 K

Let v₂ = velocity of sound in air at NTP = unknown and T₂ = temperature at NTP = 0°C + 273 = 273 K

Now v₁/v₂ = √T₁/√T₂

So, v₂ = (√T₂/√T₁)v₁

= [√(T₂/T₁)]v₁

substituting the values of the variables, we have

v₂ =  [√(273 K/290 K)]340.48 m/s

v₂ =  [√0.9413]340.48 m/s

v₂ = (0.9702)340.48 m/s

v₂ = 330.35 m/s

Also v = √(γP/ρ) where v = velocity of sound in air at NTP  = 330.35 m/s, γ = ratio of molar heat capacities, P = pressure at NTP = 1.013 × 10⁵ Pa and ρ = density of air = 1.293 kg/m³

Since,  v = √(γP/ρ)

making γ subject of the formula, we have

γ = v²ρ/P

substituting the values of the variables, we have

γ = (330.35 m/s)² × 1.293 kg/m³/1.013 × 10⁵ Pa

= 31975.36 kg/m²s² ÷ 1.013 × 10⁵ Pa

= 0.32

b. Speed of sound in mercury v₃ = √(B/ρ) where B = Bulk modulus of mercury = 28.5 × 10⁹ Pa and ρ = density of mercury = 13600 kg/m³

v₃ = √(B/ρ)

= √(28.5 × 10⁹ Pa/13600 kg/m³)

= √(28.5 × 10⁹ Pa/13.6 × 10³ kg/m³)

= √(2.096 × 10⁶) m/s

= 1.448 × 10³ m/s

= 1448 m/s

Under what conditions does moving electric charge produce a magnetic force
in a copper wire?
A. Under any conditions
B. Only when the wire is connected to the magnet
C. Only when the wire is wrapped around an iron bar
D. Only when the wire is coiled

Answers

Answer:

I'm pretty sure that the answer is C. Hope this helps.

Explanation:

Answer:

Answer is A. Under any conditions

Explanation:

Just got it correct on A p e x

Other Questions
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