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3 years ago

Glassy and metallic are terms that describe what mineral property? A.luster B.color C.shape D.hardness

Physics

2 answers:

lana66690 [7]3 years ago

6 0

The answer is A.luster

Otrada [13]3 years ago

5 0

The correct answer is option A. Luster

Hope this helps! :)

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The slope of the line on a speed-time graph tells the speed , true or false ?

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An example of a folkway violation would be driving 70 mph in a 50-mph zone. Group of answer choices True False

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Explanation:

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2 years ago

What is the speed of sound in air when the temperature is 20°C?

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3 years ago

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John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and

EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

7 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

2 years ago