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Vinil7 [7]
1 year ago
13

Seamus made an electromagnet from an iron nail, a piece of copper wire with three coils spread out across the nail, and a AAA-

Physics
1 answer:
never [62]1 year ago
7 0

Seamus can add batteries to increase the voltage, and he can decrease the space between the wire coils to accomplish this goal.

<h3>What is Attraction?</h3>

This is defined as any type of force that causes objects to come together and is possessed by objects such as a magnet.

In order to increase the power of the magnet it is best to increase the voltage and decrease the space of the coils so as to ensure it attracts at least four paper clips which makes option C the most appropriate choice.

Read more about Magnet here brainly.com/question/14997726

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Answer:

eksqijakojqnozjzbw.wlisjaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Explanation:

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A joule is an amount of energy, and a watt is a rate of using energy, defined as 1 W = 1 J / s. How many joules of energy are re
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How many joules of energy are required to run a 100 W light bulb for one day?

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4 0
2 years ago
Use the values provided to calculate the initial voltage
EleoNora [17]

Answer: V1= 360 V

Explanation:

7 0
3 years ago
Read 2 more answers
A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?
Kazeer [188]
-- She went up for 0.4 sec and down for 0.4 sec.

-- The vertical distance traveled in gravity during ' t ' seconds is

                   D  =  (1/2)  x  (g)  x  (t)²

                       = (1/2) (9.8 m/s²) (0.4 sec)²

                       =    (4.9 m/s²)  x  (0.16 s²)

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4 0
3 years ago
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A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
Pepsi [2]

Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
3 years ago
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