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2 years ago

13

What's the melting point of aluminum

1 answer:

PSYCHO15rus [73]2 years ago

8 0

660.3 degrees centigrade/celsius

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A 2.0-kg ball rolls to the right at 3.0 m/s. A 4.0-kg ball rolls to the left at 2.0 m/s . What is the momentum of the system aft

charle [14.2K]

Answer:

Final momentum after a head on collision is -2kgm/s

Explanation:

One ball moves to the right and the other moves opposite and momentum is a vector quantity so that considering the direction

Initial momenta are P₁=2x3=6kgm/s P₂=4x(-2)=-8kgm/s

Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s

4 0

2 years ago

Pls answer quick will mark brainlest and 5 stars

QveST [7]

Answer:A

Explanation:

It’s bigger I am not sure

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2 years ago

Read 2 more answers

How is life determined on math

olga nikolaevna [1]

Answer:

you can use math as a banker, a doctor, a scientist, the president probably uses math, you use math to see how much less juice you gave your sibling, and you can use math to help in collage! (sorry if its wrong tell me if it is)

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3 years ago

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If velocity is positive, which woul

shusha [124]

Answer:

C. An inital volocity that is faster than the final volocity

Explanation:

.

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2 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago