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3 years ago

During the new moon phase, why is the Moon not visible in the sky?

Physics

1 answer:

Gennadij [26K]3 years ago

3 0

Answer:

Explanation:

The moon gets the light from the sun. When the moon lies between the sun and the earth, only the back portion of the moon gets the light from the sun. So the side facing the sun does not get any light and appears to be dark or does not appear at all.

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Why does a light bike have more kinetic energy

FrozenT [24]

Because it doesn't use energy it uses mechanical and kinetic

3 0

3 years ago

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final:

$A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]$

$A_{f} = 5102.752\,cm^{2}$

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0

3 years ago

HELP ME PLEASE

Stels [109]

The answer is B. the germinal stage

7 0

3 years ago

Read 2 more answers

The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s

just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c) 78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0

3 years ago

A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s

ankoles [38]

Answer:

Explanation:

Radius of the ball is $R=11cm=0.11m$

Initial speed of the ball is $v_{com0}=6.0m/s$

Initial angular speed of the ball is $\omega = 0$

Coefficient of kinetic friction between the ball and the lane

is $\mu =0.35$

Due to the presence of frictional force, ball moves with

decreasing velocity.

(a)

velocity $v_{com0}$ in terms of $\omega$ is

$V_{com0} = -R\omega\\\\=-(0.11m)\omega\\\\= (-0.11\omega)m/s$

(b)

Ball's linear acceleration is given by

$a=-\mu g\\\\=-(0.35) (9.8 m/s^2)\\\\= -3.43m/s^2$

(c)

During sliding, ball's angular acceleration is calculated as

$\alpha=-\frac{\tau}{I}\\\\-\frac{\mu mgR}{(\frac{2}{5}mR^2)}\\\\-\frac{2}{5}\frac{\mu g}{R}\\\\-\frac{2}{5}\frac{(0.35)(9.8)}{0.11}\\\\-77.95rad/s^2$

(d)

The time for which the ball slides is calculated from the

equation of motion is

$V_{cm}= V_{cm0} + at\\\\V_{cm} = V_{cm0} + (-\mu g )t\\\\-0.11\omega=6.0m/s -(3.43m/s^2 )t\\\\-(0.11) (\alpha t) =6.0 m/s - (3.43 m/s^2)\\\\- (0.11)(-77.95 rad/s^2)t = 6.0m/s - (3.43 m/s^2 )t\\\\8.5745t + 3.43t= 6.0\\\\12.0045t = 6.0\\\\t= 0.4998s$

(e)

Distance traveled by the ball is

$X= V_{com,0}+ \frac{1}{2}at^2\\\\= (6.0m/s)(0.4998 s)+ 0.5(-3.43m/s^2) (0.4998 s)^2\\\\=2.57m$

(for)

The speed of the ball when smooth rolling begins is

$V_{cm} = V_{com, 0}+ at\\\\=6.0 m/s +(-3.43m/s^2 )(0.4998 s)\\\\= 4.29m/s$

5 0

3 years ago

Read 2 more answers