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3 years ago

Based on its chemical makeup, where do you think amazonite is found? Please help me!

1 answer:

5 0

It can be found in the U.S. and other parts around the world! It doesn't come from the Amazon River, but it's been found in Brazil before. Also, Amazonite is a green variety of "Microline Feldspar".

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You serve a volleyball with a mass of 0.77 kg. The ball leaves your hand with a velocity of

dimaraw [331]

<h2>Given :</h2>

mass (m) = 0.77 kg

velocity (v) = 2.33 m/s

<h2>Solution :</h2>

$\boxed{ \mathrm{Kinetic \: Energy = \frac{1}{2} m {v}^{2} }}$

$\dfrac{1}{2} \times 0.77 \times (2.33) {}^{2}$

$\dfrac{418.0253}{2}$

$209.01265 \: \: joules$

$\mathrm{209.01 \: \: joules} \: \: \: (approx)$

_____________________________

$\mathrm{ \#TeeNForeveR}$

7 0

3 years ago

A stone of weight 10N falls from the top of a 250m high cliff. a) Calculate how much work is done by the force of gravity in pul

tiny-mole [99]

Answer:

work done = ( force × displacement)

(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.

work done by gravity = (250 × 10) = 2500 joule

(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.

4 0

2 years ago

The amount of usable energy from anaerobic respiration is less than the amount of useable energy from aerobic respiration becaus

Ann [662]

B. The waste products of anaerobic respiration contain a lot of unprocessed energy.

Hope this helps!

5 0

3 years ago

Read 2 more answers

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center

Bess [88]

Answer:

10.93 rad/s

Explanation:

If we treat the student as a point mass, her moment of inertia at the rim is

$I_r = mr^2 = 67.8*3.7^2 = 928.182 kgm^2$

So the system moment of inertia when she's at the rim is:

$I_1 = I_d + I_r = 274 + 928.182 = 1202.182 kgm^2$

Similarly, we can calculate the system moment of inertia when she's at 0.456 m from the center

$I_2 = I_d + 67.8*0.456^2 = 274 + 14.1 = 288.1 kgm^2$

We can apply the law of angular momentum conservation to calculate the post angular speed when she's 0.456m from the center:

$I_1\omega_1 = I_2\omega_2$

$\omega_2 = \omega_1\frac{I_1}{I_2} = 2.62*\frac{1202.182}{288.1} = 10.93 rad/s$

8 0

3 years ago

4. What is the momentum of a 70 kg object traveling at 20 m/s?

Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

6 0

3 years ago