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3 years ago

10

Where can you find the neutrons of an atom? in the nucleus with the protons orbiting the nucleus In the nucleus with the electro

ns this is not part of an atom
ANSWER PLEASE ITS A EXAM 10 POINTS PLEASE FASTTTTTTTTTTT!!!!!!!1

1 answer:

cricket20 [7]3 years ago

3 0

Answer:

Nuclease is the answer I know

I hope this is the answer

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A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:

AfilCa [17]

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

F = mv/t

Given the following parameters

Force F = 8.0N

mass m = 16kg

time t = 4.0s

Required

speed v

Substitute the given parameters into the formula

v = Ft/m

v = 8 * 6/16

v = 48/16

v = 3m/s

Hence the change in speed of this object is 3m/s

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2 years ago

What is matter explain verifly

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A 78 kg skydiver can be modeled as a rectangular "box" with dimensions 24 cm × 35 cm × 170 cm . If he falls feet first, his drag

ANTONII [103]

Answer:

The terminal speed of his is 137.68 m/s.

Explanation:

Given that,

Mass of skydiver = 78 kg

Area of box $A =24\times35=840\ cm$

Drag coefficient = 0.80

Density of air $\rho= 1.2\times kg/m^3$

We need to calculate the terminal velocity

Using formula of drag force

$F_{d} = \dfrac{1}{2}\rho v^2Ac$

Where,

$\rho$ = density of air

A = area

C= coefficient of drag

Put the value into the formula

$78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80$

$v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}$

$v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}$

$v=137.68\ m/s$

Hence, The terminal speed of his is 137.68 m/s.

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3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

$A_1=1.6 cm^2=1.6\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

$A_2=1.2 cm^2=1.2\times 10^{-4} m^2$

$A_3=4.4 cm^2=4.4\times 10^{-4} m^2$

$A_4=7 cm^2=7\times 10^{-4} m^2$

Potential difference,V=140 V

We know that

$R=\frac{\rho l}{A}$

According to question

$l_1=l_2=l_3=l_4=l$

In series

$R=R_1+R_2+R_3+R_4$

$R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})$

$R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})$

$R=\rho l(18284.6)$

$I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}$

Potential across 1.2 square cm= $V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V$

Hence, the voltage across the 1.2 square cm wire=63.8 V

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3 years ago