All categories

2 years ago

Based on the data you found, about how many of the 100 aliens would become thin if the temperature were 35 C?

Physics

2 answers:

SOVA2 [1]2 years ago

7 0

Answer: d. more than 90

Explanation: edg 2020

saw5 [17]2 years ago

5 0

Answer:

Option D: More than 90

Explanation:

Thickness probably depends on temperature. The higher the temperature, the more aliens we expect to be thin.

You might be interested in

Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro

luda_lava [24]

Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$

8 0

3 years ago

Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER

lukranit [14]

Answer:

Explanation:

I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.

First row, KE = 0

Second row, KE = 220500 J

Third row, KE = 183750 J

Fourth row, KE = 205800 J

That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.

3 0

2 years ago

What horizontally-applied force will accelerate a crate of mass 400 kg at 1 meter per second per second across a factory floor a

son4ous [18]

Answer:

The horizontally applied force = 2360 N

Explanation:

<em>Force:</em> Force can be defined as the product of mass and acceleration. the S.I unit of force is Newton (N)

Fh = Fr + ma......... Equation 1

Where Fh = horizontally applied force, Fr = friction force, m = mass of the crate, a = acceleration of the crate.

<em>Given: m = 400 kg, a = 1 m/s²</em>

Fr = 1/2 W, W = mg ⇒W = 400×9.8 = 3920 N

∴Fr = 1/2(3920), Fr = 1960 N

Substituting these values into equation 1

Fh = 1960 + 400×1

Fh = 1960 + 400

Fh = 2360 N

Therefore the horizontally applied force = 2360 N

6 0

2 years ago

Choose all the answers that apply. The sun: -is the largest star in our solar system -is the largest star in our galaxy

Hatshy [7]

-is made mostly of hydrogen and helium.

-will eventually run out of fuel and die.

-creates energy through nuclear reactions

8 0

3 years ago

Read 2 more answers

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le

Nana76 [90]

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

6 0

3 years ago