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2 years ago

I need help with this question, could you help me? :))) ASAP

Physics

1 answer:

Pepsi [2]2 years ago

4 0

Oh ya the correct answer for this is B I think because structure B

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When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in pa, must its hull be able to withstand? how many times l

nordsb [41]

<span>Depth = 5.0 Ă— 10^2 m
Density of sea water = 1.025 x 10^3
Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
Since the normal pressure is retained in the hull, no need to bother about Po Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
So it is 49.56 times larger.</span>

5 0

3 years ago

Read 2 more answers

Please Answer these quick!!!! I WILL GIVE BRAINLIEST<br> Do as many as you can!

shepuryov [24]

Answer:

said

1 a

2 b

3 a

\idonotcare.net

Explanation:

6 0

3 years ago

the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is

cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0

3 years ago

You and your little cousin sit on a see-saw. You sit 0.5 m from the fulcrum, and your cousin sits 1.5 m from the fulcrum. You we

svetoff [14.1K]

Answer:

200N

Explanation:

0.5/1.5=x/600N

1/3=x/600

x=200N

7 0

3 years ago

Using Mirror equation A, Calculate The Frequency Of The Long Stand And The Shortest Wave Length Of That An Object Is Placed Of A

Leya [2.2K]

The Image distance and Magnification of The Image will be 30 cm and 3.

<h3>What is focal length?</h3>

The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.

Given data;

Focal length,f=?

Image distance,v=?

Object distance,u= 10 cm

Magnification,m= 2.85

The focal length is half of the radius;

f=R/2

f=30 Cm/2

f= 15 Cm

The mirror equation is found as;

$\rm \frac{1}{f} =\frac{1}{v} +\frac{1}{u} \\\\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{10} } \\\\\ v= -30 \ cm$

The magnification of the lens is found as;

$\rm m=\frac{30}{10}\\\\ m=3$

Hence, the image distance and magnification of The image will be 30 cm and 3.

To learn more about the focal length refer;

brainly.com/question/16188698

#SPJ1

6 0

2 years ago