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Rzqust [24]
3 years ago
12

Please can anybody tell me what are these lab equipments ​

Physics
2 answers:
lana [24]3 years ago
8 0

Answer:

hii there

5 ) tripod

6 ) test tube holder

Explanation:

Hope it helps

Have a nice day : )

olga nikolaevna [1]3 years ago
4 0

Answer:

5 is the tripoid stand

Thanks have a bangtastic day

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A lizard accelerates from 2 m/s to 10 m/s in 4 seconds. what is the lizard average acceleration
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Acceleration = (change in speed) / (time for the change)

change in speed = (ending speed) - (starting speed)

change in speed = (10 m/s) - (2 m/s)  =  8 m/s

Acceleration = (8 m/s) / (4 sec)

Acceleration = (8/4) (m/s²)

<em>Acceleration = 2 m/s²</em>

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3 years ago
How does lithium react with cold water?
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Lithium metals reacts slowly with water to form a colourless solution of lithiums hydroxide (LiOH) and hydrogen gas (H2). The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic, but the reaction is slower than that of sodium (immediately below lithium in the periodic table)

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3 years ago
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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

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4 years ago
Because the moon rotates once for each revolution around Earth,
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Good Morning, how are you?

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Have a great day :)
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tan A = \frac{11.7^2}{50*9.8}

A = tan^{-1} (0.279367)

A = 15.608\°

Therefore the angle that should the curve be banked is 15.608°

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