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schepotkina [342]
2 years ago
8

Why are joints important

Physics
1 answer:
zheka24 [161]2 years ago
5 0

Answer:

Because they help us move if we didn't have joints we would be like a wobbly balloon at a car dealership.

Explanation:

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Which change will always result in an increase in the gravitational force between two objects?
aleksklad [387]

Answer:

increasing the masses of the objects and increasing the distance between the objects

8 0
3 years ago
Two truckers are traveling directly away from each other at the same speed. If one trucker sounds her horn at a frequency of 221
8_murik_8 [283]

Answer:

v = 8.8 m /s

Explanation:

For listener and source going away from each other the formula of Doppler effect is as follows

\frac{f}{f_0} = \frac{V-v}{V+v}

V is velocity of sound , v is velocity of listner and source of sound

f₀ is apparent frequency and f is real frequency

V = 343 , v = ? ,f = 210 , f₀ = 221

Put these value in the relation above

[tex]\frac{210}{221} = \frac{343-v}{343+v}[/tex]

v = 8.8 m /s

6 0
3 years ago
What effect will a turning point have on an individuals life
ankoles [38]

Answer:

The turning points are those instants, moments or situations that happen in an absolutely unexpected way, as a result of which your life changes ... and nothing is the same as before.

4 0
3 years ago
Read 2 more answers
Three metal spheres are placed as shown in the image. Sphere A weighs 5 kg, B weighs 8 kg, and C weighs 3 kg.
Anni [7]

Answer:

B

Explanation:

7 0
2 years ago
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

F_{n} = 30.11N

Direction of the net force (β)

\beta =tan^{-1} (\frac{29.13}{7.62} )

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0
3 years ago
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