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2 years ago

13

Using the graph, predict how many paper clips a 7.5 V battery would pick up for both the 25-coil electromagnet and the 50-coil e

lectromagnet. Explain your reasoning for your prediction.

1 answer:

Svet_ta [14]2 years ago

5 0

NOTE: The strength of an electromagnet is directly proportional to not only the current but the number of windings. Doubling the number of windings doubles the strength of the magnet.

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Which radiation has a higher frequency than visible light

Bogdan [553]

Infrared radiation<span> lies between the </span>visible<span> and microwave portions of the electromagnetic spectrum. Infrared waves have wavelengths longer </span>than visible<span> and shorter </span>than<span> microwaves, and have </span>frequencies<span> which are lower </span>than visible<span> and </span>higher than<span> microwaves.</span>

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3 years ago

Read 2 more answers

What are some indicators of energy transformations?

Annette [7]

Explanation:

Contact, vision, sound, flavor, and smell are all markers of energy transformations. The most basic example would be when we notice something has begun to pass through vision. Whenever an entity accelerates or slows down, energy is constantly transformed.

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3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

A space probe is coasting through space at a steady speed of 100p feet per second. the booster rocket fires for 1/2 seconds so t

aleksandr82 [10.1K]

Answer:

Acceleration: $9800 ft/s^2$

Explanation:

The acceleration of an object is equal to the rate of change of velocity:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the space probe in this problem, we have:

u = 100 ft/s (initial velocity)

v = 5000 ft/s (final velocity)

t = 0.5 s (time taken)

Therefore, the acceleration is

$a=\frac{5000-100}{0.5}=9800 ft/s^2$

6 0

3 years ago

A container of rocks is brought back from the Moon's surface where the acceleration due to gravity is 162 meters per second if t

spin [16.1K]

Answer:

6318 N

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Mass (m) of container = 650 kg

Weight (W) of container on the earth =.?

Next, we shall determine the acceleration due to gravity of the earth. This can be obtained as follow:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Acceleration due to gravity of the earth (gₑ) =.?

gₘ = 1/6 × gₑ

1.62 = 1/6 × gₑ

1.62 = gₑ /6

Cross multiply

gₑ = 1.62 × 6

gₑ = 9.72 m/s²

Finally, we shall determine the weight of the container on the earth as follow:

Mass (m) of container = 650 kg

Acceleration due to gravity of the earth (gₑ) = 9.72 m/s²

Weight (W) of container on the earth =.?

W = m × gₑ

W = 650 × 9.72

W = 6318 N

Therefore, the weight of the container on earth is 6318 N

6 0

3 years ago