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3 years ago

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What will happen to this current if a magnet is brought near the cord? A. It will exert a force on the voltage. B. The electric

current will stop flowing. C. The resistance of the wire will decrease. D. It will exert a force on the electric current.

1 answer:

Andreyy893 years ago

8 0

Answer:

The correct answer is D)

Explanation:

When an electric magnet is brought near a cord with an electric current, the cord will most likely deflect away from the magnet because electric fields flowing through a wire generates its own magnetic field.

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On his way to school, a student starts out walking quickly and then slows down as he gets closer to the school. Which graph best

Leni [432]

' A ' is the graph that shows it.

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3 years ago

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The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the

natulia [17]

Answer:

The value is $A = 2.80 *10^{-4} \ m^2$

Explanation:

From the question we are told that

The operating temperature is $T = 2450 \ K$

The emissivity is $e = 0.350$

The power rating is $P = 200 \ W$

Generally the area is mathematically represented as

$A = \frac{P}{ e * \sigma * T^2}$

Where $\sigma$ is the Stefan Boltzmann constant with value

$\sigma = 5.67 *10^{-8} \ W/m^2\cdot K^4$

So

$A = \frac{200}{0.350 * 5.67*10^{-8} * 2450^{4}}$

$A = 2.80 *10^{-4} \ m^2$

8 0

3 years ago

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh

S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

<u>V₀ = 5.47 m/s</u>

6 0

3 years ago

An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2

Varvara68 [4.7K]

Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

distance = 20 cm

$s_1 + s_2 = 10 + 4$

$s_1 + s_2 = 14\ kg$

taking moment about B

$s_1 \times 0.8 - 10 \times 0.4 - 4 \times 0.6 = 0$

$s_1 \times 0.8 = 6.4$

$s_1 = 8\ N$

$s_2 = 14 - s_1$

$s_2 = 14 - 8$

$s_2 = 6 N$

difference between two scale = 8 - 6

= 2 N

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3 years ago

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

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3 years ago