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3 years ago

The temperature -273°C is —.

Physics

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

-273°C = Absolute zero

Explanation:

Average Kinetic energy of particles in matter refers to the temperature of matter. This value is not arbitrary zero temperature. It explains the lowest kinetic energy of the particles in matter.

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Name 3 muscles of the leg

xxTIMURxx [149]

Answer:

calf

Explanation:

jdjdjddjdjdhdhdj

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3 years ago

What is the acceleration of a 30.0-kg go kart if the thrust of its engine is 300.0 N?

VikaD [51]

Answer : $a=10\ m/s^2$

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

$F = m\times a$

a is the acceleration of the engine.

$a=\dfrac{F}{m}$

$a=\dfrac{300\ N}{30\ Kg}$

$a=10\ m/s^2$

So, the acceleration of the engine is $10\ m/s^2$ .

Hence, this is the required solution.

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3 years ago

Read 2 more answers

A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What vol

Gwar [14]

Answer:

The volume of water evaporated is 199mL

Explanation:

Concentration is calculated with the following formula

$C=\frac{n}{V}$

where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.

So we isolate the variable n to know the amount of moles, using the volume given in liters

$230mL=0.23L$

$n=C*V=0.240 M*0.23L=0.055 mol$

Now, we isolate the variable V to know the new volume with the new concentration given.

$V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL$

Finally, the volume of water evaporated is the difference between initial and final volume.

$V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL$

4 0

3 years ago

If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?

cricket20 [7]

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

3 0

3 years ago

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

Setler [38]

Answer:

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R = $R_{m}$ + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s

3 0

3 years ago