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Ann [662]
3 years ago
9

if the net force of an object is in a positive direction what will the direction of the resulting acceleration be

Physics
1 answer:
myrzilka [38]3 years ago
3 0
Resulting acceleration would be in Positive direction, 'cause their direction doesn't change & they both have same direction.

Hope this helps!
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An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a
vladimir1956 [14]
I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s
8 0
3 years ago
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Two states of matter are described below. State A: Cannot be compressed and retains its shape State B: Highly compressible Which
telo118 [61]

Answer:

State A = piece of metal; State B = air

Explanation:

For the three main states of matter here's how it breaks down.

Solid - Cannot be compressed and retains its shape

Liquid - Cannot be compressed and does not retain its shape

Gas - Compressible and does not retain its shape.

Knowing this State A has to be solid.  Only one of the options has A as a solid, so that's the answer.   Worth knowing state B is a gas though, only one compressible, just like solid is the only one that retains its shape.

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3 years ago
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A father places his daughter in a swing that is 0.60\,\text{m}0.60m0, point, 60, start text, m, end text above ground. Then he r
schepotkina [342]

This question involves the concepts of the law of conservation of energy and kinetic energy.

The girl's fastest speed is "3.7 m/s".

According to the law of conservation of energy, the girl will have the fastest speed at mean position, which will be calculated as follows:

Loss in Potential Energy = Gain in Kinetic Energy

mg\Delta h=\frac{1}{2}mv^2\\\\v=\sqrt{2g\Delta h}

where,

v = maximum speed = ?

g = acceleration due to gravity = 9.81 m/s²

Δh = change in height = 1.3 m - 0.6 m = 0.7 m

Therefore,

v=\sqrt{2(9.81\ m/s^2)(0.7\ m)}

<u>v = 3.7 m/s</u>

<u></u>

Learn more about the Law of Conservation of Energy here:

brainly.com/question/381281?referrer=searchResults

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2 years ago
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Answer:

3 electron hai bro of puch mujhe sab aata h

6 0
3 years ago
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A point charge is placed 3m from a 4uc charge what is the strength of the electric field on the point charge at this distance ro
Vlada [557]

The strength of the electric field on the point charge at this distance will be 4000 V/m.

<h3>What is the strength of the electric field?</h3>

The strength of the electric field is the ratio of electric force per unit charge.

The given data in the problem is;

Qis the unit charge = 4.0 × 10⁻⁶ C

E is the strength of the electric field

R is the distance from point charge = 3 m

The strength of the electric field is;

\rm E = \frac{KQ}{R^2} \\\\ \rm E = \frac{9 \times 10^9 \times 4 \times 10^{-6} \ C}{3^2} \\\\ E= 4000 V/m

Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.

To learn more about the strength of the electric field refer to the link;

brainly.com/question/15170044

#SPJ1

7 0
1 year ago
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