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3 years ago

6

How many meters in 32 kilometers

2 answers:

Alisiya [41]3 years ago

7 0

Hello !

32 km = 32000 m

sleet_krkn [62]3 years ago

6 0

There are 32,000 meters in 32 kilometers.
You set up the problem: 32 kilometers=_____ meters
Then you times 32 by 1,000: 32 x 1,000 = 32,000
Going from kilometers to meters you times by a thousand and from meters to kilometers you divide by a thousand.
Hope this helps! :D

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Marina CMI [18]

convection

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3 years ago

3. Which one of these has the most inertia?

Arte-miy333 [17]

b. A Boeing 747 airplane

Explanation:

The Boeing 747 airplane will have the most inertia of all. Inertia is the tendency of an object to remain at rest or uniform motion.

Newton's first law of motion is also regarded as the law of inertia.
It states that "an object will remain at rest or uniform motion motion unless acted upon by an external force".
A body that has a large mass will be more resistant to any external force that acts on it.
The Boeing 747 has the most mass of all and will have the most inertia.

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3 years ago

SI unit for electrical current

Mashcka [7]

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5 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

What is the heat capacity of an object at 25.5∘C that absorbs 45 kJ of heat and is heated to 28.2∘C?

sergey [27]

Answer:

16.6 kJ/°C

Explanation:

given,

Amount of heat absorbed = 45 kJ

initial temperature, T₁ = 25.5°C

final temperature, T₂ = 28.2°C

change in temperature = T₂ - T₁

= 28.2 - 25.5 = 2.7° C

$Heat\ capacity = \dfrac{Heat\ absorbed}{\Delta T}$

$Heat\ capacity = \dfrac{45\ kJ}{2.7}$

$Heat\ capacity = 16.6\ kJ/^0C$

Heat capacity of the object is equal to 16.6 kJ/°C

4 0

3 years ago

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