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Rzqust [24]
3 years ago
9

Which of the following is considered evidence of a chemical change?

Physics
1 answer:
shusha [124]3 years ago
7 0

When it has lost a main function of the previous object. When a piece of paper is burned, it is no longer paper. Paper can be wrote on, but ash cannot.

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What is the most advanced degree commonly held by Business, Management, and Administration workers? associate degree bachelor’s
s2008m [1.1K]

Answer: Master Degree- it shows that your our more professional in what you wen tto get your degree in

Explanation:

4 0
4 years ago
Read 2 more answers
A 3.0 kg ball is lifted two meters upward in two seconds, and a 6.0 kg ball is lifted one meter upward in one second? Which requ
Sav [38]

Answer:

1. They both uses same energy

2. The 6 kg ball requires more power than 3kg ball

Explanation:

Sample 1

m = 3kg

g= 10m/s^2

h = 2m

t = 2secs

W = mgh = 3 x 10 x 2 = 60J

P= w/t = 60/2 = 30watts

Sample 2

m = 6kg

g= 10m/s^2

h = 1m

t = 1sec

W = mgh = 6 x 10 x 1 = 60J

P= w/t = 60/1 = 60watts

They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball

3 0
4 years ago
Two automobiles of equal mass approach an intersection. One vehicle is traveling with speed 13.0 m/s toward the east, and the ot
puteri [66]

Answer:

Explanation:

We shall apply law of conservation of momentum to know the Speed of northward moving vehicle before collision to check the veracity of driver's statement .

Let v be the velocity of composite mass after collision

Applying law of conservation of momentum in north direction

m v₂ = 2m v sin55.08

Applying law of conservation of momentum in east  direction

m x 13 = 2m v cos55.08

Dividing these two equations

v₂ / 13 = tan55.08

v₂ = 13  tan55.08

= 18.62 m/s

= (18.62 x60 x 60) / 1000

= 67 km/h

= 67 x 5/8 mi/h

= 42 mi/h

So he is lieing.

7 0
4 years ago
The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the
Ivanshal [37]

Answer:

(a) \alpha = - 1.32\ rev/m^{2}

(b) \theta = 13674\ rev

(c) \alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}

(d) a = 22.458\ m/s^{2}

Solution:

As per the question:

Angular velocity, \omega = 190\ rev/min

Time taken by the wheel to stop, t = 2.4 h = 2.4\times 60 = 144\ min

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

\omega' = \omega + \alpha t

omega' = final angular velocity

\omega = initial angular velocity

\alpha = angular acceleration

Now,

0 = 190 + \alpha \times 144

\alpha = - 1.32\ rev/m^{2}

Now,

(b) The no. of revolutions is given by:

\omega'^{2} = \omega^{2} + 2\alpha \theta

0 = 190^{2} + 2\times (- 1.32) \theta

\theta = 13674\ rev

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}

(d) The radial acceleration is given by:

\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s

Linear acceleration is given by:

a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}

a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}

5 0
3 years ago
If you see lightening bolt and count for 4 seconds before you hear the thunder how far away was the lightening strike ? It was n
rjkz [21]

well, you divide 4 by 5, so .8

.8 of a mile is 4224 feet.

i really hope this helps :)

4 0
4 years ago
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