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hoa [83]
4 years ago
15

How much does a 725 kg elevator weigh? F = [?]N

Physics
2 answers:
olga2289 [7]4 years ago
8 0
725 Kg equals 725x9.8= 7105 N
Assoli18 [71]4 years ago
7 0

Answer:

7105N

Explanation:

725 kg = 725.9.8=7105N

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At what point on a line through the ends of the meter stick is the electric field equal to zero?
Gnesinka [82]

The point on a line through the ends of the meter stick is the electric field equal to zero is 4.7 m from the 0 cm mark.

<h3>What is electric field?</h3>

An electric field is the physical field that surrounds the electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to physical field for a system of charged particles. Electric fields originate from the electric charges and time-varying electric currents. Electric fields and the magnetic fields are both manifestations of the electromagnetic field, one of the four fundamental interactions (also called forces) of nature.

Electric fields are important in many areas of the physics, and are exploited in electrical technology. In atomic physics and chemistry, for the instance, the electric field is the attractive force holding the atomic nucleus and electrons together in atoms. It is also force responsible for chemical bonding between atoms that result in molecules.

The electric field is defined as a vector field that associates to each point in the space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point. The derived SI unit for electric field is the volt per meter (V/m), which is equal to the newton per coulomb (N/C)

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The correct question will be:

"A +5.00-μC point charge is placed at 0.0 cm mark of meter stick and a -4.00-μC point charge is placed at 50.0 cm mark. At what point on a line through the ends of meter stick is electric field equal to zero?"

4 0
2 years ago
An object with mass 80 kg moved in outer space. When it was at location &lt;7, -34, -7&gt; its speed was 14.0 m/s. A single cons
Sergio [31]

Answer:

W = -2080 J

Explanation:

initial position vector of the object is given as

r_i = 7\hat i - 34\hat j - 7 \hat k

similarly final position vector is given as

r_f = 12\hat i - 42\hat j - 11\hat k

now the displacement of the object is given as

\vec d = \vec r_f - \vec r_i

now we will have

\vec d = (12\hat i - 42\hat j - 11\hat k) - (7\hat i - 34\hat j - 7 \hat k)

\vec d = 5\hat i - 8\hat j- 4\hat k

now the force on the object is given as

\vec F = (200\hat i + 460 \hat j - 150 \hat k)

so here in order to find the work done

W = \vec F . \vec d

W = (200\hat i + 460 \hat j - 150 \hat k). (5\hat i - 8\hat j- 4\hat k)

W = 1000 - 3680 + 600 = -2080 J

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3 years ago
A 5.0 kg book is lying on a 0.25 meter high table.
Wittaler [7]

Answer:

Below in the picture:-

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