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Alenkasestr [34]

3 years ago

7

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 N when the cab is stand

ing still. What is the reading when the cab is moving upward (a) with a constant speed of 7.6 m/s and (b) with a speed of 7.6 m/s while decelerating at a rate of 2.4 m/s2

1 answer:

lana66690 [7]3 years ago

7 0

Answer:

(A) Reading will be 65 N

(B) Net force on the elevator will be 49.076 N

Explanation:

We have given the balance force = 65 N

Acceleration due to gravity $g=9.8m/sec^2$

We know that W=mg

So $65=m\times 9.8$

m = 6.632 kg

(a) In first case as the as the speed is constant so the force on the elevator will be 65 N

(B) In second case as the elevator is decelerating at a rate of $2.4m/sec^2$

So net acceleration = 9.8-2.4= $7.4m/sec^2$

So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N

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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for <em>M</em>, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

5 0

3 years ago

Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°

lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía ( $S$ ), medida en joules por Kelvin, tenemos la siguiente expresión:

$dS = \frac{\delta Q}{T}$ (1)

Donde:

$Q$ - Ganancia de calor, en joules.

$T$ - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

$dS = \frac{L_{v}}{T}\cdot dm$ (1b)

Donde:

$m$ - Masa del sistema, en kilogramos.

$L_{v}$ - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

$\Delta S = \frac{\Delta m \cdot L_{v}}{T}$

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que $m = 0.50\,kg$ , $L_{v} = 2.7\times 10^{5}\,\frac{J}{kg}$ and $T = 630.15\,K$ , entonces el cambio de entropía es:

$\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}$

$\Delta S = 214.235 \,\frac{J}{K}$

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0

3 years ago

Which of the following expressions for power or dimensionally correct?

Slav-nsk [51]

Power=F.V

dimension: ML^2T^-2

8 0

3 years ago

PLEASE HELP ME IM ON A TIMER

garri49 [273]

Answer:

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).

Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).

And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)

And the displacement will be defined by the folliwing vector operation:

$A (0,0) = oi + 0j\\F (3,6) = 3i + 6 j\\Displacement vector = (3-0)i + (6-0)j = 3i + 6j$

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

$Displacement = \sqrt{(3)^2+ (6)^2} \\Displacement = 6.70$

And the angle will be defined by:

tan(beta)=3/6

beta = tan^-1(6/3)

beta = 63.43°

5 0

3 years ago

Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi

yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron= $v=6.1\times 10^7m/s$

Charge on electron, $q=e=1.6\times 10^{-19} C$

Mass of electron, $m_e=9.1\times 10^{-31} kg$

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

$V=\frac{v^2m_e}{2e}$

Where V=Potential difference

$m_e=$ Mass of electron

v=Velocity of electron

Using the formula

$V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}$

$V=10581.59 V$

Hence, the potential difference=10581.59 V

8 0

3 years ago