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sweet-ann [11.9K]
4 years ago
9

Force has _____. magnitude direction gravity weight

Physics
2 answers:
Contact [7]4 years ago
7 0

Answer:

magnitude

Explanation:

Aleksandr-060686 [28]4 years ago
6 0

Answer:

A and B

Explanation:

Force has both magnitude and direction since it is a vector quantity

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How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=m*g*h)​
frez [133]

Answer:

<h2>98 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 5 × 9.8 × 2

We have the final answer as

<h3>98 J</h3>

Hope this helps you

5 0
3 years ago
Kilogram and second are called fundamental units why​
Digiron [165]

Answer:

It is because they are physical quantities

Explanation:

I think so..

6 0
3 years ago
Read 2 more answers
How many outer-orbital electrons are found in an atom of:a) Na?b) P?c) Br?d) I?e) Te?f) Sr?
lys-0071 [83]

Answer:

  1. Na=1
  2. P=5
  3. Br=7
  4. I=7
  5. Te=6
  6. Sr=2

6 0
2 years ago
Please answer ASAP!!
ch4aika [34]

Answer:

O So you can ask questions about steps you do not understand.

Explanation:

One of the reasons why it is encouraged that the laboratory procedure must be read at the beginning of any type of experiment is familiarize the student with the experimental procedure. Also, questions that can help make the smooth running of the experiment to proceed can be cleared.

  • It is to this regard that understanding laboratory procedures is very vital.
  • Students must grasp the details of each steps of the experiment before proceeding with their investigation.
  • Other critical information needed to set up the experiment can also be brought into the lime light.
3 0
3 years ago
A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w
steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

h = L - L\cos{31°} = L(1 - \cos{31°})

so its initial gravitational potential energy U_0 is

U = mgh = mgL(1 - \cos{31°})

\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})

\:\:\:\:\:=0.23\:\text{J}

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U

We know that the initial kinetic energy K_0, as well as its final gravitational potential energy U are zero so we can write the conservation law as

mgL(1 - \cos{31°}) = \frac{1}{2}mv^2

Note that the mass gets cancelled out and then we solve for the velocity v as

v = \sqrt{2gL(1 - \cos{31°})}

\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}

\:\:\:\:\:= 1.3\:\text{m/s}

5 0
3 years ago
Read 2 more answers
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