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2 years ago

What is conserved in a thermodynamically closed system?

Physics

1 answer:

kherson [118]2 years ago

3 0

Explanation:

C. Both energy and matter

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Chemical energy is the type of energy obtained from food. If you want more info look up information about the mitochondria and ATP. I hope this helps!

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2 years ago

In the inclined plane the objects that are thrown from a more inclined place go faster?

Mamont248 [21]

Assuming the objects roll down the inclined plane, yes.
If the object never touches the plane, then no.

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3 years ago

Read 2 more answers

The nucleus of a certain type of uranium atom contains

BaLLatris [955]

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=92+92

=184

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3 years ago

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

3 years ago

Light in a vacuum travels at a constant speed of 3x10^8 m/s. If the moons average distance from the earth is 38776106 km how lon

Karo-lina-s [1.5K]

Answer: 258.3 s

Explanation:

The speed $s$ is given by the following equation:

$s=\frac{D}{t}$

Where:

$s=3(10)^{8} m/s$ is the speed of light in vacuum

$D=2(38776106 km \frac{1000 m}{1 km})=7.75(10)^{10} m$ is the double of the distance between Earth and Moon, since the beam of light travels from Earth to the Moon and back to Earth again.

$t$ is the time it takes to the beam of light to travel the mentioned distance

Isolating $t$ and solving with the given information:

$t=\frac{D}{s}$

$t=\frac{7.75(10)^{10} m}{3(10)^{8} m/s}$

Finally:

$t=258.3 s \approx 258 s$

5 0

3 years ago