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3 years ago

2. A girl whose mass is 55kg stands on a spring weighing machine inside a lift. When the lift starts to ascend, its acceleration

Physics

1 answer:

Annette [7]3 years ago

5 0

Answer:

Explanation:

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Un acróbata de masa M, se impulsa hacia arriba con una velocidad v0 desde un

KatRina [158]

Answer:

La altura máxima que alcanzan el mono y el acróbata es $z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot v_{o}^{2}+\left(1-\frac{M\cdot g}{M+m} \right)\cdot h$ .

Explanation:

Asumamos que tanto el acróbata, el mono y el sistema acróbata-mono son conservativos y que el acróbata comienza su acción a una altura de cero. El estudio se divide en dos etapas: (i) <em>El acróbata se dirige al mono</em>, (ii) <em>El acróbata recoge al mono y alcanzan una altura máxima</em>.

Para resolver esta cuestión, nos valemos del Principio de Conservación de la Energía.

Parte I

La energía cinética traslacional inicial ( $K_{1,a}$ ) es igual a la suma de la energía cinética traslacional final ( $K_{2, a}$ ) y la energía potencial gravitacional final ( $U_{g,2,a}$ ).

$K_{1,a} = K_{2,a} + U_{g,2,a}$ (1)

$\frac{1}{2}\cdot M \cdot v_{o}^{2} = \frac{1}{2}\cdot M\cdot v_{1}^{2} + M\cdot g \cdot h$ (1b)

Donde:

$M$ - Masa del acróbata.

$g$ - Constante gravitacional.

$v_{o}$ - Rapidez inicial del acróbata.

$v_{1}$ - Rapidez del acróbata justo antes de recoger al mono.

$h$ - Altura inicial del mono.

Parte II

La suma de las energías iniciales cinética traslacional ( $K_{2, a}$ ) y potencial gravitacional de acróbata ( $U_{g,2,a}$ ) y la energía inicial potencial gravitacional del mono ( $U_{g,2,m}$ ) es igual a la suma de las energías potenciales gravitacionales iniciales del sistema acróbata-mono ( $U_{g,3,a+m}$ ), es decir:

$K_{2,a} + U_{g,2,a}+U_{g,2,m} = U_{g,3,a+m}$ (2)

$\frac{1}{2}\cdot M\cdot v_{1}^{2} + (M+m)\cdot g \cdot h = (M+m)\cdot g \cdot z$ (2b)

Donde:

$m$ - Masa del mono.

$z$ - Altura máxima del sistema acróbata-mono.

De (1b) tenemos que la rapidez del acróbata justo antes de recoger al mono es:

$\frac{1}{2}\cdot M \cdot v_{o}^{2} = \frac{1}{2}\cdot M\cdot v_{1}^{2} + M\cdot g \cdot h$

$v_{o}^{2} = v_{1}^{2}+2\cdot g\cdot h$

$v_{1} = \sqrt{v_{o}^{2}-2\cdot g\cdot h}$

Finalmente, la altura máxima alcanzada por el sistema acróbata-mono es:

$\frac{1}{2}\cdot M\cdot v_{1}^{2} + (M+m)\cdot g \cdot h = (M+m)\cdot g \cdot z$

$z = \frac{M\cdot v_{1}^{2}}{2\cdot (M+m)\cdot g}+h$

$z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot (v_{o}^{2}-2\cdot g\cdot h)+ h$

$z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot v_{o}^{2}+\left(1-\frac{M\cdot g}{M+m} \right)\cdot h$

La altura máxima que alcanzan el mono y el acróbata es $z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot v_{o}^{2}+\left(1-\frac{M\cdot g}{M+m} \right)\cdot h$ .

7 0

3 years ago

Calculate the change in entropy when 1.00 kg of water at 100 âˆ˜c is vaporized and converted to steam at 100 âˆ˜c. Assume that t

pickupchik [31]

Answer:6.04 kJ/K

Explanation:

Given

mass of water $m=1 kg$

Latent heat of vaporization is $L=2256\times 10^{3} J/kg$

entropy at constant temperature is given by

$\Delta S=\frac{Q}{T_0}$

$Q=m\times L$

$Q=1\times 2256\times 10^3$

$\Delta s=\frac{2256\times 10^3}{100+273}$

$\Delta s=6.04kJ/K$

5 0

4 years ago

The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex

lesya692 [45]

Answer:

A) 0.1477

B) 0.65388 mm

C) object is inverted

Explanation:

The formula for object - image relationships for spherical reflecting surface is given as;

n1/s + n2/s' = = (n2 - n1)/R

Where;

n1 & n2 are the Refractive index of both surfaces

s is the object distance from the vertex of the spherical surface

s' is the image distance from the vertex of the spherical surface

R is the radius of the spherical surface

We are given;

index of refraction of glass; n2 = 1.60

s = 24 cm = 0.24 m

R = 4 cm = 0.04 m

index of refraction of air has a standard value of 1. Thus; n1 = 1

a) So, making s' the subject from the initial equation, we have;

s' = n2/[((n2 - n1)/R) - n1/s]

Plugging in the relevant values, we have;

s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]

s' = 0.1477

b) The formula for lateral magnification of spherical reflecting surfaces is;

m = -(n1 × s')/(n2 × s) = y'/y

Where;

m is the magnification

n1, n2, s & s' remain as earlier explained

y is the height of the object

y' is the height of the image

Making y' the subject, we have;

y' = -(n1 × s' × y)/(n2 × s)

We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.

Thus;

y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)

y' = - 0.00065388021 m = -0.65388 mm

C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.

Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.

Thus, the object is inverted since m is negative.

6 0

3 years ago

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

$\terxt{time for compression}=\text{time for half revolution}$

$\terxt{time for compression}=\dfrac{1}{2}\times T$

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}$

Put the value into the formula

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60$

$\terxt{time for compression}=0.2\ sec$

We need to calculate the rate of internal energy

Using first law of thermodynamics

$U=Q-W$

$\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}$

Put the value into the formula

$\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)$

$\dfrac{\Delta U}{\Delta t}=6.8\ kW$

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

$\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}$

$\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}$

Put the value into the formula

$\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}$

$\Delta\theta=32.48^{\circ}$

Hence, The temperature change per compression stroke is 32.48°.

6 0

4 years ago

Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.

Vesnalui [34]

The angle of the planet is mathematically given as

dY= 704 degrees

<h3>What angle has planet Y rotated through during this time?</h3>

With Kepler's third rule, which states that a planet's orbit squared is a function of cubed radius, we can prove that this is the case.

Generally, the equation for the period is mathematically given as

(periodX / periodY)^2 = (radius X / radius Y)^3

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

\sqrt{(pX / pY )^2}= \sqrt{64}

(pX / pY=8

In conclusion, Because it takes 8 times longer to complete one orbit on planet X, planet Y travels 8 times farther than planet X does in the same time period...

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees