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3 years ago

HELP PLEASE ............

Physics

1 answer:

Liono4ka [1.6K]3 years ago

7 0

1. solid
2. The molecules are tightly packed together so tha means there is less movement between the molecules.

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What is the mass of an object moving with 80N of force and an acceleration of 8 m/s2?

OverLord2011 [107]

Answer:

10 Kg

Explanation:

Force is equal to mass times acceleration

therefore mass is equal to force divided by acceleration

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2 years ago

What is the relationship between the spring constant and the period in a mass hanging on a spring oscillation and why?

anygoal [31]

Explanation:

Period of a mass on a spring is:

T = 2π√(m/k)

T is inversely proportional with the square root of k. So as the spring constant increases, the period decreases.

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3 years ago

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A rock is rolling down a hill. At position 1, its velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, its veloci

alukav5142 [94]

As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second

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3 years ago

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un astronauta cuya masa es 80kg permanece inmovil en el espacio exterior, al activar la unidad propulsora que lleva en la espald

beks73 [17]

Answer:

Explanation:

80(0) = 2(15) + (80-2)v

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2 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

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

We know that

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

In series

$R=R_1+R_2+R_3+R_4$

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

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

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago