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Fofino [41]
3 years ago
8

What is Gravity?PLEASE ANSWER ​

Physics
2 answers:
JulsSmile [24]3 years ago
8 0

Answer:

<u>GRAVITY</u>:

In physics, <u>gravity</u> is the force that attracts a body towards the centre of the earth, or towards any other physical body having mass.

jolli1 [7]3 years ago
7 0

Answer:

Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun.

<em><u>Please mark as brainliest</u></em>

Have a great day, be safe and healthy

Thank u  

XD

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I need the ans for this question QUICK PLEASE!!!
Basile [38]

Explanation:

A) 1.05

B) 1.33

C) 1.16

D) 0.62

All units in cm

3 0
2 years ago
Which term best describes the basic unit that makes up all matter?
neonofarm [45]

Answer:

C. Atoms

Explanation:

The basic unit of matter and the smallest, indivisible unit of a chemical element. It comprises a nucleus (neutrons + protons) that is surrounded by a cloud of electrons.

6 0
2 years ago
3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
andreev551 [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0
2 years ago
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
2 years ago
In middle ages people believed that abnormal behavior was cause by ?
Svetach [21]
C, they didn't know any better
8 0
3 years ago
Read 2 more answers
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