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mars1129 [50]
3 years ago
7

Application

Physics
1 answer:
natulia [17]3 years ago
3 0

Answer:

\[0.900\hat{i}+0.600\hat{j}\,\text{N}\]

Explanation:

https://opentextbc.ca/universityphysicsv1openstax/chapter/5-7-drawing-free-body-diagrams/

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Define physical quantity​
NemiM [27]

Answer:

The quantity which can be measured is called physical quantity.

7 0
3 years ago
Read 2 more answers
A stone is thrown straight up from the ground with an initial speed of 15.2 m/s. At the same instant, a stone is dropped from a
Harman [31]

Answer:

47 m

Explanation:

Time of flight of stone thrown up = time of fall of stone dropped down.

Time of flight = 2u /g

where u is initial velocity and g is acceleration due to gravity.

= 2 x 15.2 / 9.8

=3.10 s.

For fall of second stone

time of fall t = 3.1 s

Height h = ?

h = ut + .5 g t²

= 0 + .5 x 9.8 x 3.1 x 3.1

= 47 m

5 0
3 years ago
Using the force table, components of a vector can be found experimentally by suspending masses from 2 orthogonal strings which o
SCORPION-xisa [38]
Answer: 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).

Explanation:

1) Since the mass of <span>155 g is suspended at 210 degrees, you need to find the components of its weight on the orthogonal coordinate system (0° and 90°).
</span>

<span>2) You do that using the trignometric ratios sine and cosine.
</span>

<span>Weight is mass × g.
</span>
<span>Weight of the object = 155g × g
</span>
<span>Angle, α = 210°
</span>

<span>Horizontal component (0°)
</span>
<span>cosα = horizontal / hypotenuse ⇒ horizontal = hypotenuse × cosα
</span>
⇒ horizontal = 155g × g × cos(210°) = - 134.23g  × g

Vertical component
sinα = vertical / hypotenuse ⇒ vertical = hypotenuse × sinα
⇒ vertical = 155g × g × sin(210°) = -77.5g × g

3) Conclusion:

Therefore, the masses that must be suspended to balance the forces of the 155g mass are 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).




8 0
3 years ago
An object travels at constant velocity of 3 m/s for a time period of 7.15 s. What is its displacement over this time?
son4ous [18]

Hi there!

We can use the following equation for constant velocity:

\large\boxed{d = vt}

d = displacement (m)

v = velocity (m/s)

t = time (s)

Plug in the givens:

d = 3 * 7.15 = \boxed{21.45 m}

5 0
3 years ago
When the velocity of a moving object stays the same, it has a what speed?
Vera_Pavlovna [14]
If velocity is constant, then the object is moving
at constant speed in a straight line.
7 0
3 years ago
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