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3 years ago

8

A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,

what is the new volume of the balloon in the hot room?

2 answers:

grin007 [14]3 years ago

8 0

I think the answer is 85 degrees

Anestetic [448]3 years ago

6 0

this would be equal to 3.68 L

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An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?

Luden [163]

Answer:

$a=2.8\ m/s^2$

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2$

So, the acceleration of the object is $2.8\ m/s^2$ .

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3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for <em>M</em>, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

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3 years ago

What are the first organisims to live within a newly created patch of land

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4 years ago

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3 years ago

1. What is the kinetic energy of a 1.75 kg ball travelling at a speed of 54 m/s?

Over [174]

Answer:

We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

Explanation:

Given

Mass m = 1.75 kg

Velocity v = 54 m/s

To determine

Kinetic Energy (K.E) = ?

We know that a body can possess energy due to its movement — Kinetic Energy.

Kinetic Energy (K.E) can be determined using the formula

$K.E=\frac{1}{2}mv^2$

where

m is the mass (kg)

v is the velocity (m/s)

K.E is the Kinetic Energy (J)

now substituting m = 1.75, and v = 54 in the formula

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\left(1.75\right)\left(54\right)^2$

$K.E=1458\times 1.75$

$K.E=2551.5$ J

Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

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3 years ago