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3 years ago

A monkey hangs motionless by one arm from a tree branch. Draw the free-body diagram.

Physics

1 answer:

Alexeev081 [22]3 years ago

7 0

Find the free-body diagram in attachment.

Explanation:

The free-body diagram of an object is a diagram representing all the forces acting on a body. Every force is represented using an arrow, where:

- The length of the arrow is proportional to the magnitude of the force

- The direction of the arrow shows the direction of the force

For the monkey in this problem, we have two forces acting on it:

- The weight of the monkey, labelled as (mg), acting downward, where

m is the mass of the monkey

g is the acceleration of gravity

- The tension in the tree branch, labelled as T, which acts upward

The monkey is motionless: this means that its acceleration is zero, so the net force actin on it is zero - this means that the forces acting on the monkey, the weight and the tension, are balanced, therefore they are shown as two arrows of same length, but opposite direction.

Learn more about forces and weight here:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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What total mass must be converted into energy

Eduardwww [97]

This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
E = m c²

   1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

   Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

   = (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

   = 1.144 x 10⁻³⁰ kg . (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:

The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

   = (2,242 x 10⁶ x 86,400 x 365) joules

   = 7.0704 x 10¹⁶ joules .

How much converted mass is that ?

   E = m c²

Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s

    Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

      = 0.786 kilogram ! ! !

THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

</span>

7 0

3 years ago

Read 2 more answers

A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.

wolverine [178]

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = $m=2.70\times 10^3kg$

We know that kinetic energy is given by

$KE=\frac{1}{2}mv^2$

So $v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec$

(B) We know that work done is given by

W = Fd

So $F=\frac{W}{d}=\frac{4780}{25.5}=187.45N$

4 0

3 years ago

What is your mass in kilograms if you weigh 120 pounds

schepotkina [342]

It depends where you are.

-- If you weigh 120 pounds on the Moon,
then your mass is 329.1 kilograms.

-- If you weigh 120 pounds on Mars,
then your mass is 143.8 kilograms.

-- If you weigh 120 pounds on the Earth,
then your mass is 54.4 kilograms.

6 0

3 years ago

It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f

Elena-2011 [213]

Answer:

B) the change in momentum.

Explanation:

The impulse is defined as the product between the force applied on an object (F) and the duration of the collision ( $\Delta t$ ):

$J=F \Delta t$ (1)

We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

$F=ma$

So, (1) becomes

$J=ma \Delta t$

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time: $a=\frac{\Delta v}{\Delta t}$ . If we substitute into the previous equation, we find

$J=m\frac{\Delta v}{\Delta t}\Delta t=m\Delta v$

And the quantity $m\Delta v$ is equivalent to the change in momentum, $\Delta p$ .

6 0

3 years ago

1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it

spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0

3 years ago