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Rudik [331]
3 years ago
11

Who ever helps can you help me with some other ones?

Physics
1 answer:
Ierofanga [76]3 years ago
8 0

Answer:

2 seconds

Explanation:

It only went up by 2

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Which lobes of the brain receive the input that enables you to feel someone scratching your back? (2 points)
andrew-mc [135]

Answer: Parietal

Explanation: The parietal lobe is where the primary somatosensory cortex is located. This cortex is where all tactile stimulation is processed in the brain and allows to you detect/feel someone scratching your back.

3 0
3 years ago
What is the difference between the law of conservation of energy and conserving energy
babymother [125]
<span>law of conservation of </span>energy<span> is </span><span><span>states that energy of the universe remains constant cant be created nor destroyed and conserving energy is not using as much power as you was like trying to make power bill lower while law of conservation is constant </span> </span>


6 0
3 years ago
In an electron cloud, an electron farther east away from the nucleus has?
vladimir2022 [97]

An electron that is far away from the nucleus have higher energy than an electron near the nucleus. Nucleus are positively charged and those electrons near it get attracted; those electrons gain kinetic energy hence reducing their internal energy. The electrons far from nucleus have low kinetic energy hence more internal energy.

8 0
3 years ago
what is the pressure exerted, what is the pressure exerted by 50kg girl as she places her weight on one shoe if the heels area i
lana [24]

Answer:

The pressure exerted by the girl is 245,000 N/m²

Explanation:

Given;

mass of the girl, m = 50 kg

area of the girl's shoe, A = 0.002 m²

The pressure exerted by the girl is calculated as follows;

P = \frac{F}{A} \\\\Where;\\F \ is \ the \ force \ exerted \ by \ girl's \ weight\\\\P = \frac{F}{A} = \frac{mg}{A} = \frac{50 \times 9.8}{0.002} = 245,000 \ N/m^2

Therefore, the pressure exerted by the girl is 245,000 N/m²

6 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
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