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3 years ago

A chandelier has sockets for 12 light bulbs. Suppose there are 75 of these

Physics

1 answer:

barxatty [35]3 years ago

3 0

-- First, we need to know how many total sockets there are in the auditorium.

(75 chandeliers) x (12 sockets/chandelier) = 900 sockets

-- Now, in order to fill 900 sockets, we'll need 900 bulbs.

(900 bulbs) / (4 bulbs/box) = 225 boxes

If all the bulbs we buy are good ones, and we don't drop any while we're installing them, then we can just exactly do the job with <em>225 boxes</em>.

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Fed [463]

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3 years ago

ingrid kicks a football with an initial velocity of 12 m/s at an angle 45 relative to the ground. what is the horizontal compone

nydimaria [60]

Let's assume that the coordinate system is defined as positive in the first quadrant.

The system has its origin in the place where Ingrid kicks the football ball.

We have then that the horizontal component is given by:

$x = 12cos (45)$

Rewriting we have:

$x = 12 ( \frac{\sqrt{2}}{2})\\x = 8.485281374$

Rounding the obtained result we have:

$x = 8.5 \frac{m}{s}$

Answer:

The horizontal component of the initial velocity to the nearest tenth is:

$x = 8.5 \frac{m}{s}$

4 0

3 years ago

Read 2 more answers

The resultant of two forces is 250 N and the same are inclined at 30° and 45° with resultant one on either side calculate the ma

Varvara68 [4.7K]

Answer:

The two forces are;

1) Force 1 with magnitude of approximately 183.013 N, acting 30° to the left of the resultant force

2) Force 2 with magnitude of approximately 129.41 N acting at an inclination of 45° to the right of the resultant force

Explanation:

The given parameters are;

The (magnitude) of the resultant of two forces = 250 N

The angle of inclination of the two forces to the resultant = 30° and 45°

Let, F₁ and F₂ represent the two forces, we have;

F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force

The components of F₁ are $\underset{F_1}{\rightarrow}$ = -F₁ × sin(30°)·i + F₁ × cos(30°)·j

The components of F₂ are $\underset{F_2}{\rightarrow}$ = F₂ × sin(45°)·i + F₂ × cos(45°)·j

The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j

The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out

We have;

F₂ × sin(45°)·i = F₁ × sin(30°)·i

F₂ ·√2/2 = F₁/2

∴ F₁ = F₂ ·√2

∴ F₂ × cos(45°)·j + F₁ × cos(30°)·j = 250·j

Which gives;

F₂ × cos(45°)·j + F₂ ·√2 × cos(30°)·j = 250·j

F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N

F₂ ≈ 129.41 N

F₁ = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N

F₁ ≈ 183.013 N

The two forces are;

A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.