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2 years ago

Which image represents a solenoid?

Physics

2 answers:

KengaRu [80]2 years ago

5 0

Answer:

the first one

Explanation: The 2nd one is almost the same

oksano4ka [1.4K]2 years ago

5 0

Answer:

Explanation:

Just did this on edgen, B and C are magnets, not solenoids

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A spearfisher stands in shallow water and sees a fish a few feet in front of her. She throws her spear directly toward the posit

DENIUS [597]

Answer:

the spear will end up above the fish relative to the actual position of the fish.

Explanation:

due to refraction of light coming from the fish the fish will appear slightly above from its real position

So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.

So here we can say that the spear will end up above the fish relative to the actual position of the fish

5 0

3 years ago

Lance arrives early at the airport (with flowers and balloons in hand) to welcome a friend. Her plane is delayed. While waiting,

Tju [1.3M]

Answer:

Explanation:

Lance was never a bright young fella so he rolled down a hill and lost his left boot

4 0

2 years ago

Nick parks his car at Target and walks 150 ft. north to get to the store. On his way back to the car, he walks 50 ft. before pau

mr Goodwill [35]

75 steps is what he’s missing

7 0

2 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

ANTONII [103]

Answer:

$11.3 \Omega$

Explanation:

We can find the resistance of the wire by using Ohm's law:

$V=RI$

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

$R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega$

6 0

3 years ago