Please help me with 15,14qqqq

Why might an electromagnet be used to pick up old cars in junkyards?

siniylev [52]

B) Hope it helps ,Have a nice day :)

7 0

4 years ago

What kind of image is formed when an object is placed at the focal point of a convex lens? Explain WHY this is the image that oc

mariarad [96]

Answer:

Since light does not actually pass through this point, the image is referred to as a virtual image. Observe that when the object in located in front of the focal point of the converging lens, its image is an upright and enlarged image that is located on the object's side of the lens.

Explanation:

4 0

3 years ago

Efficiency of electrons causes negative charge why?

Paraphin [41]

Answer:

Electron is negatively charged because you don't get work or in simple words charge out of it.

Explanation:

4 0

3 years ago

The rear window in a car is approximately a rectangle, 1.3 m wide and 0.30 m high. The inside rear-view mirror is 0.50 m from th

maria [59]

Answer:

Height of mirror 0.075 m

width of mirror 0.325 m

Explanation:

given data

wide = 1.3 m

high = 0.30 m

driver’s eyes = 0.50 m

rear window = 1.50 m

solution

we take here height / width of the mirror is

height / width = $\frac{h}{w}$ .................1

and

height /width of the window is

height /width = $\frac{h_w}{w_w}$ .................2

and

distance of eye / window by the mirror is

distance of eye / window = $\frac{x_e}{x_w}$ .................3

so here

θ = θi = θr ....................4

and tanθ for vertical is

tanθ = $\frac{h}{x_e}$

tanθ = $\frac{h_w}{(x_e + x_w)}$ ....................5

so

h = $h_w \times \frac{x_e}{(x_e + x_e)}$ ....................6

put here value and we get

$h = 0.30 \times \frac{0.50}{(0.50 + 1.50)}$

h = 0.075 m

and

when we take here tanθ for horizontal than it will be

tanθ = $\frac{w}{x_e}$

tanθ = $\frac{w_w}{(x_e + x_w)}$ .......................7

so

$w = w_w \times \frac{x_e}{(x_e + x_w)}$ ....................8

put here value and we get

$w = 1.3 \times \frac{0.50}{(0.50 + 1.50)}$

w = 0.325 m

7 0

3 years ago

Consider a two-dimensional projectile motion problem involving a ball starting and ending at the same height. the projectile lea

mariarad [96]

The range of the projectile is maximum at angle 45°, the range is 10.2m, the time to reach the maximum height is 0.72 s and the maximum height is 2.6 m

<h3>What is a Projectile ?</h3>

The stone, or object or anything projected in a trajectory path is known as a projectile.

In the given two-dimensional projectile motion problem involving a ball starting and ending at the same height. the projectile leaves the ground with an initial velocity of 10 m/s at some angle Ф with the horizontal,

(a) The value of the angle Ф (in degrees) which will makes the range of the projectile maximum is angle 45° because Range = u²sin2Ф ÷ g

At Ф = 45

Range = u²sin90° ÷ g

where sin 90 = 1

So, Range = u² / g which is the maximum range.

(b) The range at this angle will be

Range = 10² ÷ 9.8

Range = 100 / 9.8

Range = 10.2 m

(c) To calculate how long it takes the ball to reach maximum height, we will use the formula

t = usinФ ÷ g

t = 10 sin 45 ÷ 9.8

t = 0.72s

(d) The maximum height will be

H = u²sin²Ф ÷ 2g

H = 10²(sin45)² ÷ 2 × 9.8

H = 100 × 0.5 ÷ 19.6

H = 50 ÷ 19.6

H = 2.55 m

Therefore, the range of the projectile is maximum at angle 45°, the range is 10.2m, the time to reach the maximum height is 0.72 s and the maximum height is 2.6 m

Learn more about Projectile here: brainly.com/question/12870645

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6 0

2 years ago