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3 years ago

5

What’s is a object that moves through the air space acted on only by gravity

2 answers:

inessss [21]3 years ago

7 0

This describes projectile motion while neglecting any outside forces other than gravity

Arada [10]3 years ago

4 0

I believe the answer is projectile
As the object reach its maximum height after it thrown, its speed will falls into 0. But due to the gravity force, the speed of the object will start to accelerate again after it came down until it hit another medium.
Example of projectiles: a thrown baseball, a catapult shot.

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There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

3 years ago

Which form of energy does a plant store when light is transformed during photosynthesis

Stolb23 [73]

The form of energy a plant stores when light is transformed during photosynthesis is chemical energy. Hope this helps!

8 0

3 years ago

What are non-contact forces please include example in your answer

Alina [70]

Answer:

Gravitational force. Magnetic force. Electrostatics. Nuclear force.

Explanation:

Apple falling from a tree

raindrops falling from the sky

4 0

3 years ago

Read 2 more answers

If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gr

Len [333]

its a i just have to type more but its a

5 0

3 years ago

Read 2 more answers

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

Lesechka [4]

Answer:

$\Delta V = 0.053 A$

Explanation:

Electric field in a given region is given by equation

$E = \frac{A}{r^4}$

as we know the relation between electric field and potential difference is given as

$\Delta V = -\int E. dr$

so here we have

$\Delta V = - \int (\frac{A}{r^4}) .dr$

$\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}$

here we know that

$r_1 = 1.71 m$ and $r_2 = 2.89 m$

so we will have

$\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})$

so we will have

$\Delta V = 0.053 A$

8 0

3 years ago