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hram777 [196]
2 years ago
6

Find the magnitude: || 5-3i || ...?

Physics
1 answer:
____ [38]2 years ago
5 0
The magnitude would be :

\sqrt{5^2 + 3^2}

= √34

Hope this helps
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If two items are equal in size, the denser one will
Talja [164]
Be heavier

density=mass÷volume

if two items have the same size they have the same volume so the heavier one will be the denser one
6 0
3 years ago
A force off 700 newtons is applied to a 600 kg bowling ball. What is the acceleration of the bowling ball once the force is appl
Readme [11.4K]

Answer:

<h2>1.17 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{700}{600}  =  \frac{7}{6}  \\  = 1.1666666...

We have the final answer as

<h3>1.17 m/s²</h3>

Hope this helps you

6 0
3 years ago
Awnser these pls they r for middle schoolers lol
soldi70 [24.7K]

Answer:

Find the set of value of x Find the set of value of x which satisfy the inequality 2r2- 5x 2 18which satisfy the inequality 2r2- Find the set of value of x which satisfy the inequality 2r2- 5x 2 185x 2 18

8 0
2 years ago
If you live on the path of a jet stream, will the temperature get hotter or colder?
Evgen [1.6K]
Jet streams are tied to global warming so I’m guessing hotter
8 0
3 years ago
Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re
leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

F_ed_e=F_ld_l

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

F_l=\frac{F_ed_e}{d_l}

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

F_l=\frac{25*60}{33}

F_l=45.5 lbs

7 0
3 years ago
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