The topic is finding volume. help on number 2?

Physics

1 answer:

arsen [322]2 years ago

6 0

50 times 2.5 and them subract the product from 2500
I THINK THAT IS THE ANSWER
HOPE THIS HELPS!!!!!!! ;-)

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Because of the high sugar and caffeine content in energy drinks, disregarding the unpredictable risks, what are two other increa

Wittaler [7]

Answer:

It increases heart rate and blood pressure, heart problems, dehydration, seizures, and dangerously high blood pressure. (From overconsumption of caffine.)

Explanation:

6 0

2 years ago

A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?

Korvikt [17]

$\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}$

Speed of the mobile = 250 m/s
It starts decelerating at a rate of 3 m/s²
Time travelled = 45s

$\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}$

Velocity of mobile after 45 seconds

$\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}$

We can solve the above question using the three equations of motion which are:-

v = u + at
s = ut + 1/2 at²
v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

$\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}$

We are provided with,

u = 250 m/s
a = -3 m/s²
t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ Final velocity of mobile = 115 m/s

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4 0

3 years ago

One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.

Zepler [3.9K]

Answer:

The answer is " $\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}$ "

Explanation:

Given:

$\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\$

calculation:

$Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978$