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3 years ago

A(n) has charge but negligible mass, whereas a(n) has mass but no charge.

Physics

2 answers:

Dahasolnce [82]3 years ago

7 0

Answer:

An electron has charge but negligible mass, whereas a neutron has mass but no charge.

Explanation:

Atoms are made of equal number of protons and electrons. Almost all the mass of an atom is in the nucleus of the atom. The nucleus is made of protons and neutrons (except hydrogen, which has no neutron)

The charge of an electron is $-1.6 \times 10^{-19}$ C

The charge of a proton is $+1.6 \times 10^{-19}$ C

The charge of a neutron is 0 C

The mass of a proton is $1.6726 \times 10^{-27}$ kg or 1.007276 u

The mass of a neutron is $1.6749 \times 10^{-27}$ kg or 1.008664 u

The mass of an electron is $9.1 \times 10^{-31}$ kg or 0.00054858 u

As we can see the proton is 1,838 times heavier than an electron and the neutron is 1,840 times heavier than an electron.

So the electron has a negligible mass and a negative charge but the neutron has no charge but it is heavier than electron and proton

vfiekz [6]3 years ago

6 0

Electron;Neutron is the correct answer.

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grandymaker [24]

Answer:

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(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

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Explanation:

Given that,

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Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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