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Vilka [71]
3 years ago
15

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60

° = 0.87; cos 60° = 0.50. Ignore friction and the weights of the pulleys.)
Physics
1 answer:
miskamm [114]3 years ago
5 0

Answer:

Answer:196 Joules

Explanation:

Hello

Note:  I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work  is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules

Answer:196 Joules

I hope it helps

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The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-
just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}

6.5\text{ Light-years}=1.9929065\text{ Parsecs}

6.5\text{ Light-years}\approx 1.99\text{ Parsecs}

Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.

5 0
3 years ago
A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr
Zinaida [17]

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

m'=9.00 \ g

or,

    =0.009 \ kg

mass of wooden block,

m=1.20 \ kg

speed,

s=0.390 \ m

Coefficient of kinetic friction,

\mu=0.20

As we know,

The Kinematic equation is:

⇒  v^2=u^2+2as

then,

Initial velocity will be:

⇒  u=v^2-2as

        =v^2-2 \mu gs

On substituting the given values, we get

⇒  u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}

       =\sqrt{-1.5288}

       =1.23 \ m/s

As we know,

The conservation of momentum is:

⇒  mu=m'u'

or,

⇒ Initial speed, u'=\frac{mu}{m'}

On substituting the values, we get

⇒                            =\frac{1.20\times 1.23}{0.009}

⇒                            =\frac{1.476}{0.009}

⇒                            =164 \ m/s                              

3 0
2 years ago
A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t
Anastaziya [24]

Answer:

The gauge pressure is  P_g  =  2058 \ P_a

Explanation:

From the question we are told that

       The height of the water contained is  h_w  =  30 \ cm  =  0.3 \ m

        The height of liquid in the cylinder is  h_t  =  40 \ cm  = 0.4 \ m

       

At the bottom of the cylinder the gauge pressure is  mathematically represented as

        P_g  =  P_w + P_o

Where  P_w is the pressure of water which is mathematically represented as

      P_w  =  \rho_w  *  g * h_w

Now  \rho_w is the density of water with a constant values of  \rho_w  = 1000 \ kg /m^3

   substituting values

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     P_w  =  2940 \  Pa

While P_o is the pressure of oil which is mathematically represented as

          P_o  =  \rho_o *  g *  (h_t -h_w )

Where \rho _o is the density of oil with a constant value

         \rho _o  = 900 \ kg / m^3

substituting values

       P_o  =  900 *  9.8 * (0.4 - 0.3)

       P_o  =  882 \ Pa

Therefore

      P_g  =  2940 - 882

      P_g  =  2058 \ P_a

6 0
3 years ago
A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass
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Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope (F_{t}), vertical gravitational force (F_{g}) and vertical normal force (F_{n}), due to the surface. Since there is no vertical movement, F_{g} and F_{n} cancels it out. So, for this block, net force is horizontal due to the rope F_{t}.

The block of mass m is hanging from the pulley, so there is the force of the rope (F_{t}) and the gravitational force (F_{g}). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

F_{r}=m.a

a=\frac{F_{r}}{m}

a=\frac{18.8}{3.6}

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

F_{r}=m.a

F_{t}-F_{g}=m.a

F_{t}-m.g=m.a

18.8-9.8m=5.22m

15.02m=18.8

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: F_{g}=m.g.

The weight for the hanging block is

F_{g}=1.25*9.8

F_{g}= 12.25 N

Comparing tension and weight:

\frac{12.25}{18.8} ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

4 0
3 years ago
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