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hram777 [196]
3 years ago
6

Mr. Smith challenged his students to create an electromagnet that will pick up more paper clips than anyone else in the class. T

immy is very excited and begins to collect all of the necessary materials. He finds a D-cell battery, an iron nail, and a large roll of copper wire. What could Timmy do, using just these materials, to increase the strength of his electromagnet to pick up the maximum amount of paper clips?
Physics
2 answers:
kaheart [24]3 years ago
6 0

Answer:

c) increase the number of coils of copper he puts around the nail

this is the answer

Explanation:

aliya0001 [1]3 years ago
5 0
Wrap the wire very tight around the iron nail.
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An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels fo
stepan [7]

Answer:

Explanation:

For circular path in magnetic field

mv² / R = Bqv ,

m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .

a )

R = mv / Bq

If v is changed  to 2v , keeping other factors unchanged , R will be doubled

b )

magnitude of acceleration inside field

= v² / R

= Bqv / m

As v is doubled , acceleration will also be doubled

c )

If T be the time inside the magnetic field

T = π R / v

=  π  / v x  mv / Bq

= π m / Bq

As is does not contain v that means T  remains unchanged .

d )

Net force acting on electron

= m v² / R = Bqv

Net force = Bqv

As v becomes twice force too becomes twice .

So a . b , d are correct answer.

4 0
3 years ago
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

5 0
4 years ago
Anybody know the answer??
omeli [17]

Answer:

Centimeter

Explanation:

We know the measurement is defined as:

1 meter= 100cm

1 cm= 1/100 meter

1 centimeter= 0.01 meter

Therefore, 0.01 meter is equal to 1 cm or 1 centimeter.

6 0
3 years ago
Read 2 more answers
How does the mass of a bowling ball that has been rolled down the lane affect the kinetic energy?
solniwko [45]
The heavier it is the more frictions it creates sliding on the floor. this make more kinetic energy.
3 0
3 years ago
Read 2 more answers
A playground ride consists of a disk of mass M = 49 kg and radius R = 1.7 m mounted on a low-friction axle. A child of mass m =
HACTEHA [7]

Answer:

The angular speed is 0.83 rad/s.

Explanation:

Given that,

Mass of disk M=49 kg

Radius = 1.7 m

Mass of child m= 29 kg

Speed = 2.6 m/s

Suppose if the disk was initially at rest , now how fast is it rotating

We need to calculate the angular speed

Using conservation of momentum

m\omega_{i}=(mr^2+\dfrac{Mr^2}{2})\omega_{f}

mvR=(mr^2+\dfrac{Mr^2}{2})\omega

Put the value into the formula

29\times2.6\times1.7=(29\times1.7^2+\dfrac{49\times1.7^2}{2})\omega_{f}

\omega_{f}=\dfrac{29\times2.6\times1.7}{(29\times1.7^2+\dfrac{49\times1.7^2}{2})}

\omega_{f}=0.83\ rad/s

Hence, The angular speed is 0.83 rad/s.

4 0
3 years ago
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