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ZanzabumX [31]
3 years ago
11

If you had a mass of 44 g and a volume of 40.5ml what is the density

Physics
1 answer:
klio [65]3 years ago
4 0
Here, We know, Density = Mass / Volume
Here, mass = 44 g
volume = 40.5 ml = 40.5 cm³

Substitute their values, 
d = 44 / 40.5
d = 1.086 g/cm³

In short, Your Answer would be  1.086 g/cm³

Hope this helps!
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The product of 2 x 104 cm and 4 x 10–12 cm, expressed in scientific notation is ____.
sdas [7]
Answer:    8 * 10⁻⁸  cm²  .
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Explanation:
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(2 * 10⁴ cm) * (4 * 10⁻¹² cm)  =

2 *4 * 10⁴ * 10⁻¹² = 8 * 10⁽⁴⁺⁽⁻¹²⁾⁾ =  8 * 10⁻⁸  cm² .
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Note the follow property of exponents:
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xᵃ * xᵇ = x⁽ᵃ ⁺ ᵇ⁾  ;  as such:  " 10⁴ * 10⁻¹² = 10⁽⁴⁺⁽⁻¹²⁾⁾ = 10⁻⁸ " .
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6 0
3 years ago
How much force (in Newtons) acts on a 1700 kg car going around a
Daniel [21]

Answer:

b) 4781 N

Explanation:

Because there is a redius do this question is talking about the acceleration force which= mv^2/r

so a=15^2/80=2.8125 m^2/s

so the force will be = m.a

F =1700×2.8125=4781.25 N

5 0
3 years ago
do bowling balls float? ( I know the answer i just want to see how many other people know. also answer as in depth as you can)
crimeas [40]

yes they would float cause of the weight of the ball

8 0
4 years ago
"List, from beginning to end, the stages in the weather as a cold front goes by: (a) warm air is forced upward where it cools; (
RSB [31]

Answer:

(a) warm air is forced upward where it cools;

(c) cumulonimbus or nimbo-stratus clouds form

(d) thunderstorms with heavy showers and gusty winds occur"

(b) air cools and sinks, pressure rises, rain stops;

Explanation:

  • A cold for front is a denser air , as that forms under the warmer and lighter air mass and this causes a low pressure along the  cold form and thus causes the formation of a thunderstorm where enough moisture is present and  drop in temperature occurs and in the northern hemisphere the cold front shifts the winds to from the southwest to northwest clockwise.
8 0
3 years ago
A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and
trapecia [35]

Answer:

a) f'=544.66 \textup{Hz}

b) f'=584.75 \textup{Hz}

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, v_s = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

f'=f\frac{v\pm v_d}{v\pm v_s}

where,

v_d = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

f'=f\frac{v}{v+v_s}

on substituting the values, we get

f'=564\times\frac{343}{343+12.1752}

or

f'=544.66 \textup{Hz}

b) for maximum frequency, we have the formula as:

f'=f\frac{v}{v-v_s}

on substituting the values, we get

f'=564\times\frac{343}{343-12.1752}

or

f'=584.75 \textup{Hz}

3 0
3 years ago
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