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3 years ago

A 42 kg object is 2.5 m from a 55kg object. How much is the force of attraction between them?

Physics

1 answer:

melisa1 [442]3 years ago

8 0

Answer:

Force of attraction is 2,46*10^-8N $F= G*(m1*m2/r^2)\\where \\\\G=6,67*10^-11 (N*m^2/kg^2)\\m1=42kg\\m2=55kg\\r=2,5m\\therefore\\F=6,67*10^-11*(42*55/2.5^2)\\ F=2,46*10^-8 N$

Explanation:

Using the Law of Universal Gravitation, proposed by Newton.

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Mark and Paul are in a race. Mark is 20 meters from the finish line and running at a constant 3.5 m/s. Paul is 5 meters behind h

USPshnik [31]

Answer:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

Explanation:

For this case we have an illustration for the problem on the figure attached.

And we can solve this problem analyzing each one of the runner. Let's begin with Mark

Mark

For this case we know that $V_M = 3.5 m/s$ and th velocity is constant. The distance from Mark and the finish line is $D_M = 20 m$

Since the velocity is constant we can create the following relation:

$D_M = V_M t_M$

And solving for $t_M$ we got:

$t_m = \frac{D_M}{V_M}= \frac{20m}{3.5 m/s}= 5.71 s$

So then Mark will nd the race after 5.71 seconds

Paul

We know that the initial velocity for Paul is given $V_{iP}= 2.7 m/s$ we also know that the total distance between Paul and the finish line is 25 m and we want to find the acceleration that Paul needs to apply in order to tie the race, and Paul have 5.71 sconds in order to reach the finish line.

We can use this formula in order to find the acceleration (because we assume that the acceleration is constant) that he needs to apply:

$x_f = x_i + v_i t + \frac{1}{2} a t^2$

And since $\Delta x = x_f - x_i$ we have this:

$\Delta x= v_i t + \frac{1}{2} a t^2$

And if we replace we have this:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

And the final velocity for Paul using this acceleration would be:

$V_{fP}= V_{iP}+ a_P t = 2.7m/s + 0.588 m/s^2 (5.71s)= 6.057 m/s$

3 0

4 years ago

. Is there net work done on an object at rest or moving at a constant velocity? <br> WHICH ONE ???

Grace [21]

If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).

8 0

3 years ago

3. A soccer player applies an average force of 180 N during a kick. The kick accelerates a .45 kg soccer ball from rest to a spe

liq [111]

Answer:

8.1Ns

Explanation:

Impulse is expressed as the product of Force and time

Impulse = Ft

F is the force = 180N

Get the time using the equation of motion

v = u+at

18 = 0+F/m(t)

18 = 180/0.45(t)

18 = 400t

t = 18/400 secs

Recall that;

Impulse = Ft

Impulse = 180×18/400

Impulse = 8.1Ns

Hence the impulse imparted to the ball is 8.1Ns

4 0

3 years ago

if the bar and block are too heavy, one of the strings may break. assuming that the two strings are identical, which one will br

DerKrebs [107]

The string that will break first depends on the weight of the block or bar subjected to such string.

According to Hook's law, the force applied to an elastic material is directly proportional to the extension of the material.

F= kx

where;

<em>F is the force applied to the object = weight of the object</em>

For given two identical strings, the string that will break first depends on the mass of the bar and the block.

If the bar is heavier than the block, then the string subjected to the bar will break first.
On the other hand, if the block is heavier than the bar, then the string subjected to the block will break first.

Thus, the string that will break first depends on the weight of the block or bar subjected to such string.

Learn more here:brainly.com/question/4404276

7 0

3 years ago

a car travels at 50 m/s for first 20 seconds , 30 m/s for next 30 seconds and finally 20 m/s for next 10 seconds . find its aver

kaheart [24]

The average speed traveled by car is $35m/s$ .