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3 years ago

HELLLPPPPPP!!!!!!

Physics

1 answer:

Bond [772]3 years ago

3 0

DIVIDE THEN MUTPLY THEN YOU AEMPLY SUCK

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A massless, rigid board is placed across two bathroom scales that are separated by a distance of 1.71 m. A person lies on the bo

Nata [24]

Answer:

Let the weight of the person be W and be located at a distance 'a' from the left scale as shown in the figure

Since the body is in equilibrium we can use equations of statics to analyse the problem.

Taking Sum of Moments about A we have

$316\times 1.71-W\times a=0\\\\$

Taking Sum of Moments about B we have

$475\times 1.71-W\times (1.71-a)=0\\\\$

Solving the above 2 equations for W and 'a' we get

$316\times 1.71=W\times a\\\\475\times 1.71-W\times 1.71=-Wa\\\\\therefore W=\frac{316+475}{1}=791N\\\\\therefore a=\frac{316\times 1.71}{791}=0.683m$

8 0

3 years ago

Which of the following statements about energy is false?

Artemon [7]

Answer:

Explanation:

5 0

3 years ago

A 0.145 kg baseball is thrown with a velocity of 25.0 m/s. How much work was done on the baseball to bring it from rest to 25.0

sergey [27]

Answer:

45.31 J

Explanation:

We are given that

Mass of baseball , m=0.145 kg

Initial velocity, u=0

Final velocity, v=25 m/s

We have to find the work done on the baseball to bring it from rest to 25 m/s

We know that

Work done = Change in kinetic energy

Work done, W= $\frac{1}{2}m(v^2-u^2)$

Using the formula

Work done, W $=\frac{1}{2}(0.145)((25)^2-0)$

Work done= $\frac{1}{2}(0.145)(625)$

Work done, W=45.31 J

Hence, the work done on the baseball to bring it from rest to 25 m/s

=45.31 J

3 0

3 years ago

Does time effect the work done?

ahrayia [7]

Sometimes if you're a fast worker then more job can be done. if you're slow it will take a lot more time. So I guess depends on situation.

4 0

4 years ago

The temperature of a plastic cube is monitored while the cube is pushed 3.4 m across a floor at constant speed by a horizontal f

Nesterboy [21]

Answer:

$\Delta E_{floor} = 51 J$

Explanation:

The work (W) done on the cube to be pushed across the floor is equal to the total thermal energy (ΔE) of the system:

$W = \Delta E_{T} = \Delta E_{cube} + \Delta E_{floor}$ (1)

Also, the work done on the cube by the horizontal force is giving by:

$W = F \cdot d$ (2)

<em>where F: force applied to the cube , d: displacement of the cube </em>

<em>By equaling the equations (1) and (2)</em>, we can find the thermal energy of the floor:

$\Delta E_{cube} + \Delta E_{floor} = F \cdot d$

$\Delta E_{floor} = F \cdot d - \Delta E_{cube}$

$\Delta E_{floor} = 20 N \cdot 3.4 m - 17 J$

$\Delta E_{floor} = 51 J$

So, the increase in the thermal energy of the floor is 51 J.

Have a nice day!

7 0

3 years ago