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Natali [406]
3 years ago
15

a 3.00 kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2 find the resultant force acting on the object an

d the magnitude of the resultant force
Physics
2 answers:
il63 [147K]3 years ago
6 0

F = m.a

F = 3(2 I + 5 j)

F = 6i + 15j is resultant force

Magnitude force = √6+15 = √21

katen-ka-za [31]3 years ago
6 0

Answer:

(6 i + 15 j) N ,  16.16 N

Explanation:

m = 3 kg, a = ( 2 i + 5 j) m/s^2

Force = mass x acceleration

F = 3 x ( 2i + 5j) = 6 i + 15 j

The resultant force, F = (6 i + 15 j) N

Magnitude of net force is given by

F = \sqrt{6^{2}+15^{2}}

F = 16.16 N

You might be interested in
In a solution,a Is the substance being dissolved
Elena-2011 [213]
In a solution, a SOLUTE is the substance being dissolved.

In a solution, a SOLVENT is the substance in which something is dissolved.

For example in salt water, the salt is the solute. And the water is the solvent. Usually, which substance has a larger amount in the solution can be called a solvent. It doesn't matter if it's a gas, solid, or liquid. They can all be solute or solvents.
4 0
4 years ago
Consider a large spring, hanging vertically, with spring constant k = 3220 N/m. If the spring is stretched 25.0 cm from equilibr
Olegator [25]

Answer:

m = 82.1 kg

Explanation:

For this exercise we must use Hook's law that states that the force exerted by a spring is proportional to the displacement

        Fe = -k x

Let's use Newton's second law to establish equilibrium, the elastic force up and the body weight down

       Fe - W = 0

       Fe = W = mg

       k x = m g

       m = k x / g

Let's reduce the distance to SI units

       x = 25 cm (1 m / 100cm) = 0.250 m

      m = 3220 0.250 /9.8

      m = 82.1 kg

8 0
3 years ago
If the moon orbited twice as far from earth how far would it "fall" each second?
Kamila [148]

I can't imagine that this is going to do you much good, but
I'm sure going to enjoy solving it.
-------------------------
Skip this whole first section.
It was an attempt to master a bunch of trees, while
the forest was right there in front of me all the time.
Drop down below the double line.
-------------------------

Kepler's 3rd law says:

       (square of the orbital period) / (cube of the orbital radius) = constant

           T₀² = K R₀³

I put the zero subscripts in there, because you doubled 'R'
and I need to know how that affected 'T'.

           new-T² = K(2 R₀)³

           new-T² = 8 K (R₀)³  =  8 old-T₀²

          <u> new-T = √8  old-T</u>     <=== that's what I was after 

I just teased out the Moon's new orbital period if it's distance were doubled.
Instead of 1 month, it's now  √8  months.

To put a somewhat sharper point on it, the moon's period of revolution
changes from  27.322 days to 27.322√8  =  77.278 days (rounded) .

Using 385,000 km for the moon's current average distance, the current orbital speed is
             (2π x 385,000 km) / (27.322 days) = 1,024.7  m/s
(One online source says 1.023 km, so we're not doing too badly so far.)

================================================

I'm such a dummy.  I don't need to go through all of that.

If the moon were twice as far from Earth as it really is, then it would
average 770,000 km instead of the present 385,000 km.

That's 120.86 times the Earth's radius of  6,371 km.

So the acceleration of gravity out there would be

     (1 / 120.86)² of the (9.807 m/s²) that it is here on the surface.

     new-G  =  0.000671  m/s²


  Distance a dropped object falls = 1/2 g t²

                 In the first second, that's  1/2 g (1)² = 1/2 g

For an orbiting object, every second is the "first"second, because ...
as we often explain orbital motion qualitatively ... the Earth "falls away"
just as fast as the curved orbit falls.

Distance an object falls in the 1st second =

             1/2 G  =   0.000336 m/s  =  <em>0.336 millimeter per second</em>

I estimate the probability of a mistake somewhere during this process
at approx 99.99% .  But I don't have anything better right now, and I've
wasted too much time on it already, so I'll stick with it.


5 0
4 years ago
A rod is pivoted about its center and oriented horizontally. A 5.0-N force directed upward is applied 4.0 m to the left of the p
Usimov [2.4K]

Answer:

The total torque is 27.5 Nm

Explanation:

Given;

5.0-N force directed upward is applied 4.0 m to the left of the pivot,

5.0-N force directed upward is applied 1.5 m to the right of the pivot,

Taking the moment about the pivot, the total torque is given by;

τ = Fr

where;

F is the appllied force

r is the radius of the force arm

τ = (5 N x 4 m) + (5 N x 1.5 m)

τ = 27.5 Nm

Therefore, the total torque is 27.5 Nm

5 0
3 years ago
If you comb your hair and the comb becomes positively charged what does your hair become
elena-s [515]

If you comb your hair and the comb becomes positively charged, then your hair becomes negatively charged.

7 0
3 years ago
Read 2 more answers
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