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3 years ago

How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?

1 answer:

5 0

Power is defined as the rate at which the body is doing work:
$P=\frac{W}{t}$
Work is defined as displacement done by the force times that displacement:
$W=F\cdot h$
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
$W=62N\cdot10m=620J$
of work.
Now we just divide that by 5s to get how much power is required:
$P=\frac{620J}{5s}=124W$

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A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.

babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

$p=\frac{F}{A}$

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

$p\propto \frac{1}{A}$

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

8 0

3 years ago

I need help on all of them

oksano4ka [1.4K]

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6 0

2 years ago

Considerando que los coeficientes de dilatación de los siguientes metales son: hierro 11.7 x 10-6; plomo 27.3 x 10-6; cobre 16.7

Rasek [7]

Answer:

el plomo será el más largo

Explanation:

Dado que;

longitud inicial (l1) = 4m

Longitud final l2

aumento de temperatura (θ) = 10 ° C

Coeficiente de expansión lineal α

Ahora para el hierro;

α = 11,7 x 10-6

Desde;

l2-l / l1θ = α

l2 = α l1θ + l1

l2 = l1 (αθ + 1)

l2 = 4 ((11,7 x 10-6 * 10) + 1)

l2 = 4.00044 m

Para el plomo

l2 = 4 ((27,3 x 10-6 * 10) + 1)

l2 = 4,00109 m

Para cobre

l2 = 4 ((16,7 x 10-6 * 10) + 1)

l2 = 4.000668 m

Por lo tanto, el plomo será el más largo

7 0

3 years ago

A circular sector has an area of 64 in. squared. The radius of the circle is 8 in. What is the arc length of the sector?

Rom4ik [11]

Answer:

L = 16 [in]

Explanation:

In order to find the arch we must first find the angle that forms the circular sector, using the following formula:

$A = \frac{\alpha*r^{2} }{2} \\where:\\A= area = 64[in^{2}]\\r = radius = 8[in]\\\\\alpha = angle[rad]therefore\\\alpha =\frac{2*A}{r^{2} } \\\alpha =\frac{2*64}{8^{2} } \\\alpha =2[rad]$

Now using the following equation we can calculate the arc length

$L=\alpha *r\\L=2*8\\L=16[in]$

5 0

3 years ago

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI

OverLord2011 [107]

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

<u>Explanation: </u>

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } \times \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} \times 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^{-2}$

So, the work done can be expressed in $k g m s^{-2}$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2}$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^{2} / \mathrm{s}^{2}$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2} \times \frac{1000 \mathrm{g}}{1 \mathrm{kg}} \times \frac{100 * 100 \mathrm{cm}^{2}}{m^{2}}=3200 \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2}$

Thus, work done in ergs is 3200 ergs.

6 0

3 years ago