All categories

3 years ago

If you wanted to produce more power,what could you do to maximize power

2 answers:

6 0

In physics, power is defined as energy per unit time. You will also hear it described as work per unit time. The standard unit of measure for power is the watt, where a watt is defined as joules (energy) per second (time). This is expressed as a fraction as J/s. If you wanted to increase the power in any operation, you can either increase the energy (more joules) or reduce the time (fewer seconds).

Vikentia [17]3 years ago

6 0

Same work in less time, or more work in same time

You might be interested in

A 20 kg object is floating in space. What is its mass?

trapecia [35]

It's still 20 kg. Mass doesn't change according to gravity, only weight changes.

5 0

3 years ago

A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is

AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to $W=mg$ (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to $B=m_w g$ ( $m_w$ =mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

$W'=W-B$

which can be rewritten as

$m'g = mg-m_w g$

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

$m_w = m-m'=540 g-342 g=198 g$

(b) $1.98\cdot 10^{-4} m^3$

If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.

The volume of water displaced is given by

$V_w = \frac{m_w}{\rho_w}$

where

$m_w = 198 g = 0.198 kg$ is the mass of the water displaced

$\rho_w = 1000 kg/m^3$ is the density of the water

Substituting,

$V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3$

And so this is also the volume of the rock.

(c) $2727 kg/m^3$

The average density of the rock is given by

$\rho = \frac{m}{V}$

where

m = 540 g = 0.540 kg is the mass of the rock

$V=1.98\cdot 10^{-4} m^3$ is its volume

Substituting into the equation, we find

$\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3$

3 0

3 years ago

What type of tectonic plate boundary exists along the edge of the North American plate near the coast of Northern California, Or

kobusy [5.1K]

Answer:

-transform plate boundary

- false

4 0

3 years ago

"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front

cluponka [151]

If i wouldve know i would tell you sorry

4 0

3 years ago

The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses

Orlov [11]

Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

8 0

3 years ago