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monitta
3 years ago
10

Braddy connected the loose wire to the battery and created an electromagnet. He picked up 45 thumb tacks with his electromagnet,

though his goal was to pick up 50 thumb tacks. What could Braddy do to increase the strength of his electromagnet and pick up more thumbtacks? A) Use fewer coils of wire. B) Use a screw instead of a nail. C) Use two batteries instead of one. D) Replace the nail with a piece of steel.
Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:

C) Use two batteries instead of one.

Explanation:

-Strength of an electromagnet depends on the electrical current which flows through the wires.

-Two batteries have a higher current than one and thus higher strength in the electromagnet.

You might be interested in
4.- Una vagoneta de 1000 kg de peso parte del reposo en el punto 1 y desciende, sin rozamiento, por la vía indicada en la figura
Akimi4 [234]

Answer:

A) 49,050 N

B) 16 m

Explanation:

Question:

El dibujo de la pregunta se obtiene de un documento titulado "TRABAJO DIVERSO Y ENERGÍA" que se encuentra en línea y se presenta aquí.

La masa dada del vagón, m = 1,000 kg

La altura del punto en el que descansa el vagón, punto 1, h₁ = 12 m

A) El radio en el punto 2, el punto más bajo, R = 6 m

La fuerza, 'N', que la vía ejerce sobre el vagón en el punto 1 viene dada por la siguiente relación;

N = El peso del vagón + La fuerza de movimiento del vagón

∴ N = m × g + m × a

Dónde;

g = La aceleración debida a la gravedad ≈ 9,81 m / s²

a = La aceleración del vagón

Observamos que para el movimiento circular, la fuerza de movimiento del vagón, m × a = La fuerza centrípeta que actúa sobre el vagón = m × v² / R

∴ m × a = m × v² / R

Dónde;

v² = La velocidad del vagón en el punto 2 = 2 · g · h₁

Por lo tanto;

N = m × g + m × a = m × g + m × v² / R = m × g + m × 2 · g · h₁ / R

∴ N = 1000 × 9,81 + 1000 × 2 × 9,81 × 12/6 = 49,050

La fuerza que ejerce el vagón en el punto 2, N = 49,050 N

B) En el punto 3, tenemos;

N = m · g - m · a₃

La fuerza centrípeta en el punto 3, m · a₃ = m · v₃² / R₃

∴ La altura en el punto 3, h₃ = 4 m

El cuadrado de la velocidad en el punto 3, v₃² = 2 · g · (h₁ - h₃)

Para que el vagón esté seguro en el punto 3, la fuerza de la vía sobre el vagón, N = 0 para que el vagón permanezca en la vía actuando

Por lo tanto;

N = m · g - m · a₃ = 0

m · g = m · a₃ = m · v₃² / R₃ = m · (2 ​​· g · (h₁ - h₃)) / R₃

∴ R₃ = (2 · g · (h₁ - h₃)) / g = (2 · (h₁ - h₃)) = 2 × (12 - 4) = 16

El radio de curvatura en el punto 3 para que el punto sea seguro es R₃ = 16 m.

5 0
3 years ago
A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit
Alecsey [184]

Answer:

  F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

          I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

        v = d / t

        t = d / v

Reduce SI system

          m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

          d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

         t = 50 10⁻³ / 600

         t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

        F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

       F = m (v-vo) / t

       F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

       F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

5 0
3 years ago
The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad
Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

                               A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

                                 A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

                                T = 2*p / w

                                T = 2*p / (2*p / 92.8)

Hence,                     T = 92.8 min

7 0
3 years ago
Hiii please help i’ll give brainliest if you give a correct answer please thanks!
lakkis [162]

Answer: the first one

Explanation: good luck!

8 0
3 years ago
Read 2 more answers
How much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate​
Nuetrik [128]

Answer:

<h3><em>2</em><em>4</em><em>7</em><em>9</em><em> </em><em>Newton</em></h3>

<em>Sol</em><em>ution</em><em>,</em>

<em>Mass</em><em>=</em><em>1</em><em>0</em><em>0</em><em> </em><em>kg</em>

<em>Accele</em><em>ration</em><em> </em><em>due</em><em> </em><em>to</em><em> </em><em>gravity</em><em>(</em><em>g</em><em>)</em><em>=</em><em>2</em><em>4</em><em>.</em><em>7</em><em>9</em><em> </em><em>m</em><em>/</em><em>s^</em><em>2</em>

<em>Now</em><em>,</em><em>.</em>

<em>weight = m \times g \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 100  \times 24.79 \\  \:  \:  \:  \:  \:  \:  = 2479 \: newton</em>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em>

5 0
3 years ago
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