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3 years ago

13

How to the magnitude of the applied force of a suspended string which applies a horizontal force to make an angle to the vertica

l

1 answer:

nataly862011 [7]3 years ago

8 0

Answer:

Vector have magnitude and direction

Explanation:

Use the component X and component y to find the magnitude and direction

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A person weighing 50 newtons climbs a 2-meter ladder. How many joules of work is done? joules.

Mariulka [41]

50N x 2m =100J
Work= force x distance

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3 years ago

When a person floating in swimming pool, they experience an upthrust. what causes this effect?

velikii [3]

When your head sticks out of the water the upthrust reduces as your head is no longer displacing water, and there is 0 resultant force so you float at that level. hope this helps, can i get brainliest please

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4 years ago

Calcular la distancia recorrida por un móvil que tiene una rapidez de 35 m/s inicia un moviendo uniformemente acelerado con una

kumpel [21]

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What are the areas of the SAARC cooperation? Make their list.

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3 years ago

A single force acts on a 2.0 kg particle-like object in such a way that the position of the object as a function of time is give

kaheart [24]

Answer:

The work done by the force is 1435 Joule.

Explanation:

You have to apply the work-energy theorem, which associates the work and the mechanical energy

W of all non conservative forces = ΔEm = Emb - Ema

where Emb is the mechanical energy in point b and Ema is the mechanical energy in point a

Is also known that mechanical energy is the sum of potential energy and kinetic energy.

In this case, there is only kinetic energy because the potential energy is given by conservative forces like the weight but the weight is perpendicular to the displacement and because of that it doesn't produce work.

So, the work is given by:

W= Ekb - Eka, where Ek is the kinetic energy

By definition the kinetic energy is given by:

$Ek=\frac{1}{2}mV^{2}$ where m is the mass and V is the speed.

Replacing the definition of Ek in the equation of work:

$W=\frac{1}{2}mVb^{2} - \frac{1}{2}mVa^{2}$

So, you have to calculate the speed in point a and in point b

You can calculate the velocity of the object as a function of time derivating the equation of the position:

$V(t) = \frac{dX(t)}{dt} = \frac{d(3.0t-4.0t^{2}+1.0t^{3}) }{dt}$

$V(t)=3.0-8.0t+3.0t^{2}$

Replacing t=0 s to obtain the speed in the initial point (point a)

Va=3.0-8(0)+3.0(0)(0)

Va=3.0 m/s

Replacing t=5.0 to obtaing the speed in final point (point b)

Vb= 3-8(5)+3(5)(5) = 38 m/s

Therefore the work is:

$W = \frac{1}{2}(2.0)(38)^{2} - \frac{1}{2}(2.0)(3^{2}) =1444 - 9 = 1435 [J]$

<u>Where J is Joules, the unit for work.</u>

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3 years ago

when an incandescent light bulb is turned on, its thin wire filament heats up quickly. as the temperature of this wire filament

Leno4ka [110]

Well, we usually assume that the resistance of a circuit component
is constant and doesn't change. But the truth is that for anything
that conducts current, its resistance always increases somewhat
when it warms up.

For things like light bulbs, electric toasters, space heaters, electric
stove burners, the heat coils in a blow-dryer ... anything that's
designed to be really hot when it's doing its job ... the resistance
of those things increases significantly when they come up to their
operating temperatures.

6 0

3 years ago