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3 years ago

What is the difference between density and specific gravity

1 answer:

4 0

Density is defined as mass per unit volume. It has the SI unit kg m-3 or Kg/m-3 and is an absolute quantity. Specific gravity is the ratio of a material's density with that of water at 4 °C (where it is most dense and is taken to have the value 999.974 kg·m-3). It is therefore a relative quantity with no units.

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When is the next total solar eclipse in illinois

atroni [7]

April 8, 2024

Eclipse will be total over the southern part of the state. Roughly everything south of Decatur and Springfield.

7 0

2 years ago

A man pushes on piano with mass 170 kg; it slides at constant velocity down a ramp that is inclined at 20.0 ∘ above the horizont

nikdorinn [45]

Answer

given,

mass of the piano = 170 kg

angle of the inclination = 20°

moves with constant velocity hence acceleration = 0 m/s²

neglecting friction

so, force required to pull the piano

F = m g sin θ

F = 170 × 9.81 × sin 20°

F = 570.39 N

so, force required by the man to push the piano is F = 570.39 N

4 0

2 years ago

A man takes 20 seconds to climb 5m up a ladder. He weighs 720N. Calculate the power he must deliver to do this.

Inessa05 [86]

Answer:

180 W

Explanation:

The work done by the man against gravity is equal to its gain in gravitational potential energy:

$W=mg\Delta h$

where

(mg) = 720 N is the weight of the man

$\Delta h= 5 m$ is the change in height

Substituting,

$W=(720)(5)=3600 J$

The power he must deliver is given by

$P=\frac{W}{t}$

where

W = 3600 J

t = 20 s is the time taken

Substituting,

$P=\frac{3600}{20}=180 W$

3 0

3 years ago

Study the image of the moving car.

gayaneshka [121]

Answer:

While traveling downhill, the car’s potential is <u>increasing</u> and kinetic energy is <u>decreasing</u>

Explanation:

hope this helps!

6 0

2 years ago

Read 2 more answers

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

2 years ago