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2 years ago

8

What is the density of a piece of wood with a mass of 25 kg and a volume of 0.0385 m³?

1 answer:

Orlov [11]2 years ago

7 0

Answer:

649kg/m^3

Explanation:

Let p be the density of this particular object.

Formula for density:

$p = \frac{mass \: (in \: kg)}{volume \: (in \: {m}^{3}) }$

We can substitute the givenmass and volume to find density of the object.

$p = \frac{25kg}{0.0385 {m}^{3} } \\ = 649kg \: per \: {m}^{3}$

Therefore the density of this object is 649kg/m^3.

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A comet is cruising through the solar system at a speed of 50,000 kilometers per hour for 4 hours time. What is the total distan

dimulka [17.4K]

Answer:

A comet is cruising through the solar system at a speed of 50,000 kilometers per hour for 4 hours time. What is the total distance traveled by the comet during this time?

✓ Distance = Speed * Time Distance = 50,000 * 4 Distance = 200,000 km. Hope this helps

Explanation:

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2 years ago

5. A man has a weight of 100 Newtons. How much work is done if he climbs 4 meters up a ladder? Plug numbers under the equation.

Virty [35]

Answer:

<h2>400 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 100 × 4

We have the final answer as

<h3>400 J</h3>

Hope this helps you

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3 years ago

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

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3 years ago

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Katena32 [7]

Answer:

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3 years ago

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on

leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

$q_1=50\ nC at y=6\ m$

$q_2=-80\ nC at x=-4\ m$

$q_3=70\ nC at y=-6\ m$

Electric Potential is given by

$V=\frac{kQ}{r}$

Distance of $q_1$ from $x=8\ m$

$d_1=\sqrt{6^2+8^2}$

$d_1=\sqrt{36+64}$

$d_1=10\ m$

similarly $d_2=8-(-4)$

$d_2=12\ m$

$d_3=\sqrt{(-6)^2+8^2}$

$d_3=\sqrt{36+64}$

$d_3=10\ m$

Potential at $x=8\ m$ is

$V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}$

$V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]$

$V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}$

$V_{net}=9\times 5.33$

$V_{net}=47.97\approx 48\ V$

5 0

3 years ago