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3 years ago

7

Galilee said that if you rolled a ball along a level surface it would be

1 answer:

lana [24]3 years ago

3 0

The answer would be stay because the surface is flat so it will stay!

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In the attachment there is a density column where there is colour

Anettt [7]

Answer:

For your kind information, read the question thoroughly!

The bottom is the container part and not the liquid.

I hope it will be useful.

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3 years ago

Read 2 more answers

A 7.0-kg rock is subject to a variable force given by the equation f(x)=6.0n−(2.0n/m)x+(6.0n/m2)x2 if the rock initially is at r

stealth61 [152]

For Newton's second law, the force is equal to the product between the mass and the acceleration of the rocket:
$F=ma$
From which we can rewrite the acceleration as
$a(x)= \frac{F}{m} = \frac{1}{m} (6-2x+6x^2)$
where m=7.0 kg.

The velocity of the rocket is the derivative of the acceleration:
$v(x) = \frac{1}{m} (-2+12 x)$
and if we substitute x=9.0 m, we find the rocket velocity after 9.0 m:
$v(9)= \frac{1}{7}(-2+12\cdot 9)=15.1 m/s$

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3 years ago

Which idea would most likely be dangerous for a student to think while entering a lab?

aleksley [76]

Your answer should be a i can figure out how the tools work as i go along

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2 years ago

Which process binds together sediment to form new rock?

vekshin1

Answer:

its cementation i got it correct

Explanation:

6 0

3 years ago

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

$Mv+mu=Mv_1+mv_2$

(900 x 47) + (200 x -30) = (900 x $v_1$ ) + (200 x $v_2$ )

36300 = (900 x $v_1$ ) + (200 x $v_2$ )

$9v_1 + 2v_2 = 363$ ..............(i)

$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

= 19

Thus, only impulse of importance is used to find final velocity.

8 0

3 years ago