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Zielflug [23.3K]
3 years ago
14

What part of the crane is used to lift and lower the load?

Physics
2 answers:
Flura [38]3 years ago
7 0
I think the answer is b.boom
Cerrena [4.2K]3 years ago
3 0

Answer:

its the hoist

Explanation:

i did my osha 10

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Which of the following would increase​
bazaltina [42]

Answer:

1 and 3

Explanation:

<u>1 and 3  </u>

Increasing coils increases strength

   COOLING the wire would increase current flow and strength of magnet

Adding an iron core will definitely increase the strength of the electromagnet

7 0
2 years ago
Read 2 more answers
A fishbowl has a circular opening with a diameter of 13 cm. The fishbowl sits upright on a table in a magnetic field of 0.00110
azamat

Answer:

Did you ever get the answer?

Explanation:

5 0
3 years ago
If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?
aleksklad [387]
Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
6 0
3 years ago
Why is a protective apron or lab coat important to use when working with acids?
Kipish [7]

Answer:

Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.

  • An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
  • A strong acid ionizes completely in solution.
  • When they come in contact with a fabric, they break them down violently.
  • So, if they come in contact with the skin, it causes a violent break down of body tissues.
  • The apron acts a protective layer.
4 0
3 years ago
Read 2 more answers
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
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