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2 years ago

Wave energy can only be transmitted through a material medium. wave energy can only be transmitted through a material medium.

1 answer:

3 0

The correct answer is False.
In fact, electromagnetic waves (which carry energy), do not need a material medium to travel, because they can travel in vacuum as well. So, wave energy can be transmitted also through vacuum.

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What is the gravitational force that two objects would feel if they are 3.5 meters apart? Object 1 has a mass of 10x10^5 kg and

polet [3.4K]

Answer:

1.63 N

Explanation:

F = GMm/r^2

= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2

= 1.63 N ( 3 sig. fig.)

6 0

3 years ago

Which is true of oxidation? PLS HELP FAST and thank you!

elena-14-01-66 [18.8K]

Answer:The Answer Is D

Explanation:

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2 years ago

A bird is flying north with a mass of 2.5kg has a momentum of 17.5kg m/s at what velocity is it flying

klasskru [66]

Momentum = mass • velocity
v= 17.5/2.5
= 7 m/s

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3 years ago

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What is the mass of a cylinder of lead that is 1.80 in in diameter, and 4.12 in long. the density of lead is 11.4 g/ml?

Sedbober [7]

1950 g This is the answer due to the kilograms of lead being distributed

5 0

3 years ago

Read 2 more answers

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

3 years ago