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3 years ago

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An IGCSE student thinks it may be possible to identify different rocks (A, B and C) by measuring their

1 answer:

kherson [118]3 years ago

8 0

Answer:

See the answer below

Explanation:

a. The volume in the first measuring cylinder reads 70 $cm^3$ while that of the second reads 95 $cm^3$ . Hence;

V1 = 70 $cm^3$

V2 = 95 $cm^3$

b. <u>An object will always displace its own volume in a liquid</u>. Hence:

Volume V of the rock sample = V2 - V1

= 95 - 70 = 25 $cm^3$

c. Mass of A = 102 g

Volume of A = 25 $cm^3$

<em>Density = mass/volume</em>

Hence, density of A = 102/25 = 4.08 g/ $cm^3$

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Which event demonstrates electromagnetic waves transferring energy?

Citrus2011 [14]

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Sun heating a car sitting in a parking lot

Explanation:

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3 years ago

Read 2 more answers

Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

kvasek [131]

Answer:

a) $a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

b) $k = \frac{9.8}{9.74}=1.006$

Explanation:

Part a

For this case we can begin finding the period like this:

$T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s$

Then we know that the centripetal acceleration is given by:

$a= \frac{v^2}{r}$

And the velocity is given by:

$v=\frac{2\pi r}{T}$

If we replace this into the acceleration we got:

$a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}$

And we can replace the values and we got:

$a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

Part b

For this case we want to find a value of k such that:

$a= k 9.8$

Where a = 9.74, so then we can solve for k like this:

$k = \frac{9.8}{9.74}=1.006$

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3 years ago

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0

3 years ago