Please help have due very soon?thank you

How is it possible for one electrically neutral atom to exert an electrostatic force on other electrically neutral atom?

Free_Kalibri [48]

Answer:

Explained

Explanation:

Although Atom are electrically neutral. But atom atom is combination of nucleus and electrons. The nucleus of the atom is composed neutron and positively charged protons. On the outside of nucleus at some distance are the electrons which are negatively charged. So, there is difference in position of the two differently charge species. So, this way a electrically neutral atom can exert a electrostatic force on other electrically neutral atom

8 0

4 years ago

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th

kupik [55]

The equilibrium temperature is T13=3.12 ◦C

Explanation:

Given

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

1.When the first two liquids are mixed:

mC1(T1-T12)+mC2(T2-T12)=0
C1(6-13)=C2(23-13)=0
7C1=10C2
C1=1.42C2

2.When the second and third liquids are mixed:

mC2(T2-T23)+mC3(T3-T23)=0
C2(23-26.8)=C3(38-26.8)=0
3.8C2=12.8C3
C2=3.36C3

3.When the first and third liquids are mixed:

mC1(T1-T13)+mC3(T3-T13)=0
C1(6-T13)+C3(38-T13)=0
C1=1.42C2 C2=3.36C3
C1=1.42C2(3.36C3)
C1=4.77C3
C1(6-T13)+C3(38-T13)=0
4.77C3(6-T13)+C3(38-T13)=0
By solving the equation we get,
T13=3.12 ◦C
The equilibrium temperature is T13=3.12 ◦C

7 0

3 years ago

Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mas

timurjin [86]

Answer:

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

= 0.135 J + 0.11220 J

= 0.2472 J

≅ 0.247 J = 247 mJ

5 0

3 years ago

A ball rolls across the floor at a constant speed, what will happen to it if there is no friction?

Pepsi [2]

the ball will stay moving at the same rate of speed until something happens and stop it

7 0

3 years ago