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2 years ago

What are alkaline earth metals used for?

Physics

2 answers:

Tomtit [17]2 years ago

6 0

Answer:

C firework

Explanation:

from Quizlet

Makovka662 [10]2 years ago

4 0

The answer is C. Fireworks

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An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

3 0

2 years ago

A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

Ivanshal [37]

Part 1)

here we know that supply took 10 s to reach the ground

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}\times 9.8 \times 10^2$

$y = \frac{1}{2}\times 9.8 \times 100$

$y = 490 m$

Part 2)

Here all the supply covered horizontal distance of 650 m in 10 s interval of time

so here we can say

$speed = \frac{distance}{time}$

$v = \frac{650}{10}$

$v = 65 m/s$

4 0

2 years ago

At sunset, red light travels horizontally through the doorway in the western wall of your beach cabin, and you observe the light

Nady [450]

Answer:

$9.8\cdot 10^{-6}m$

Explanation:

For light passing through a single slit, the position of the nth-minimum from the central bright fringe in the diffraction pattern is given by

$y=\frac{n \lambda D}{d}$

where

$\lambda$ is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have

$\lambda=700 nm = 7.00\cdot 10^{-7}m$ is the wavelength of the red light

D = 14 m is the distance of the screen from the doorway

d = 1.0 m is the width of the doorway

Substituting n=1 into the equation, we find the distance between the central bright fringe and the first-order dark fringe (the first minimum):

$y=\frac{(1)(7.00\cdot 10^{-7} m)(14 m)}{1.0 m}=9.8\cdot 10^{-6}m$

6 0

2 years ago

Help me out with this pe question plz :c

blagie [28]

Answer:

the first one is a group 1 and the second one is d all of the above

Explanation:

6 0

2 years ago

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A sprinter moving at 10 m/s slows down at a rate of 1.4 m/s2. how fast is the runner moving after 4 seconds?

AURORKA [14]

V=u+at
v=10+(-1.4 (4))
=10-5.6
=4.4m/s

3 0

2 years ago

Read 2 more answers