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3 years ago

What do you need to make 100 minutes charges neutral

1 answer:

4 0

Neutrons are neutral and do not have any charge at all. Protons carry a positive charge, and electrons carry the negative charge.

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If 100 kg person weighed 400 N on the planet Zorg,

Brut [27]

Answer:

4 m/s²

Explanation:

Formula

Weight = Mass x Gravity due to acceleration

Solving :

400 N = 100 kg x g
g = 400/100
g = 4 m/s²

5 0

2 years ago

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a waves amplitude is 0.5 meters. If the amplitude is increased to 1 metro, how does its energy change

marysya [2.9K]

Answer:

The energy becomes 4 times greater.

Explanation:

We know that the energy of a wave is proportional to the square of its amplitude

E ∝ Amplitude^2

Since the original amplitude = 0.5 m

and the new amplitude becomes = 1 m

We are doubling the amplitude. This means that the new energy will be affected by a factor of 4

E_new ∝ (2*Amplitude)^2 =

E_new ∝ 4*(Amplitude)^2

E_new = 4*E

5 0

3 years ago

What is shearing? is it means taking out the hair of the sheep

ICE Princess25 [194]

Answer:

Thats a greate quiestionm look it up

Explanation:

5 0

4 years ago

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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

Question about generators and electrical currents. Please, no absurd answers! Thank you

Kamila [148]

The first answer should be correct if not then the second one

3 0

3 years ago

Read 2 more answers