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3 years ago

12

Se realiza un experimento en el que se mide el alargamiento de un muelle debido a la acción de una pesa; se trata de comprobar l

a ley de Hooke. Los datos que se han obtenido son: Medición 1 2 3 4 5 Alargamiento 42.0 48.4 51.3 56.3 58.6 Masa 2 4 6 8 10 El coeficiente de determinación es:

1 answer:

mel-nik [20]3 years ago

4 0

Answer:

m = 1,975 m / kg , b = 38.05 m

Explanation:

In this experiment, the elongation is plotted against the applied mass

getting a straight line

y = m x + b

where b would be the initial length of spring let's calculate the slope for which we use two well separated points

m = (56.3 -48.4) / (8 - 4)

m = 1,975 m / kg

the equation remains

y = 1,975 x + b

for x = 2 kg y = 42.0 m

we substitute in the equation

42 = 1,975 2 + b

b = 42 - 3.95

b = 38.05 m

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A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side. (a) H

Debora [2.8K]

Answer:

Explanation:

Given

balloon is ascending at the rate of $u=12\ m/s$

Balloon is at a height of $s=80\ m$

As the package is dropped from the balloon it must possess the same initial velocity as the balloon

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2=12^2+2\times 9.8\times 80$

$v^2=1712$

$v=41.37\ m/s$

time taken is given by

$v=u+at$

substituting values

$41.37=-12+9.8\times t$

as we consider downward direction as positive

$t=\frac{53.37}{9.8}$

$t=5.44\ s$

therefore time taken to reach the bottom is t=5.44 s

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where would the spaceprobe experience the strongest net (or total) gravitional force exerted on it by Earth and Mars

Arte-miy333 [17]

Answer:

r = 41.1 10⁹ m

Explanation:

For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal

∑ F = 0

F (Earth- probe) - F (Mars- probe) = 0

F (Earth- probe) = F (Mars- probe)

Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system

the distance from Earth to the probe is R (Earth-probe) = r

the distance from Mars to the probe is R (Mars -probe) = D - r

where D is the distance between Earth and Mars

$G \ \frac{m \ M_{Earth}}{r^2} = G \ \frac{m \ M_{Mars}}{(D-r)^2}$

M_earth (D-r)² = M_Mars r²

(D-r) = $\sqrt{ \frac{M_{Mars}}{ M_{Earth}} }$ r

r ( $1 + \sqrt{ \frac{M_{Mars}}{M_{Earth}} }$ ) = D

r = $\frac{D}{ 1+ \sqrt{\frac{M_{Mars}}{ M_{Earth}} } }$

We look for the values in tables

D = 54.6 10⁹ m (minimum)

M_earth = 5.98 10²⁴ kg

M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg

let's calculate

r = 54.6 10⁹ / (1 + √(0.642/5.98) )

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An electron moving with a velocity = 5.0 × 10 7 m/s enters a region of space where perpendicular electric and a magnetic fields

inna [77]

Answer:

The magnetic field is $2 \times 10^{-4}$ T

Explanation:

Given:

Velocity of electron $v = 5 \times 10^{7} \frac{m}{s}$

Electric field $E = 10^{4} \frac{V}{m}$

The force on electron in magnetic field is given by,

$F = qvB \sin \theta$ ......(1)

The force on electron in electric field is given by,

$F = qE$ ......(2)

Compare both equation,

$qE = qvB \sin \theta$

Here $\sin \theta = 1$

$E = vB$

$B= \frac{E}{v}$

$B = \frac{10^{4} }{5 \times 10^{7} }$

$B = 2 \times 10^{-4}$ T

Therefore, the magnetic field is $2 \times 10^{-4}$ T

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3 years ago