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nordsb [41]
3 years ago
5

A space shuttle is approaching the surface of the moon to land. It approaches the moons surface with an initial velocity of 28m/

s and is slowing at a rate of -2m/s^2. Show all work and units for problems. a) Draw velocity vs time graph
b) When will shuttles velocity equal zero?
c) How far above the surface of the moon was the shuttle when it began to slow down?
Physics
1 answer:
saul85 [17]3 years ago
3 0

Answer:

Explanation:

a ) Velocity vs time graph will be a straight line with negative slope , making a slope of 2 because slope of v-t graph gives the value of acceleration.

b ) v = 0 , u = 28 m/s

t = ?

v = u - at

0 = 28 - 2 t

t = 14 s

shuttle velocity will become zero after 14 sec.

c ) v ² = u² - 2 a s

0 = 28² - 2 x 2 x s

s = 28² / 4

= 196 m .

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7 0
3 years ago
Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i
leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are \dfrac{\lambda}{2}

Thus; for both speakers; the wavelength of the sound is:

\dfrac{\lambda}{2} = (10+30) cm

\dfrac{\lambda}{2} = (40) cm

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{8}+ \Delta \phi_o

\pi \ rad  = \dfrac{ \pi}{4}+ \Delta \phi_o

\Delta \phi_o  =  \pi -\dfrac{ \pi}{4}

\Delta \phi_o  = \dfrac{ 4\pi - \pi}{4}

\Delta \phi_o  = \dfrac{ 3\pi}{4} \ rad

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad

\Delta \phi = \dfrac{3 \pi}{4} \ rad

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)

A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)

A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)

A = 0.765a

7 0
2 years ago
Que fuerza se obtendrá en el émbolo mayor de una prensa hidraúlica cuya área es de 167 cm2, cuando el émbolo menor de área igual
stiv31 [10]

Responder:

<h2>5.368N, </h2>

Explicación:

Según el principio pascal, establece que la presión aplicada en un punto sobre un líquido en un recipiente cerrado es igual a igual a la presión en cualquier otro punto del líquido.

Matemáticamente Presión ejercida por el pistón más pequeño = Presión ejercida por el pistón más grande.

La presión es la relación entre la fuerza y ​​su área de sección transversal.

P = Fuerza / Área de sección transversal

Sea P1 la presión sobre el pistón más pequeño y P2 la presión ejercida por el pistón más grande.

Como P1 = P2 entonces;

F1 / A1 = F2 / A2

Dado F1 = 450N, A1 = 14 cm², A2 = 167 cm² y F2 =?

Sustituyendo el valor conocido en la fórmula para obtener el requerido, tenemos;

\frac{450}{14} = \frac{F2}{167}\\  \\Cross\ multiplying\\\\14*F_2 = 450*167\\\\14F_2 = 75,150\\\\F_2 = \frac{75,150}{14} \\\\F_2 = 5,367.9N

F_2 \approx 5, 368N

Por lo tanto, la fuerza que se obtendrá en el pistón más grande de una prensa hidráulica cuya área es de 167 cm² es aproximadamente 5,368N,

4 0
3 years ago
Why is the level of glycerin initially decreases when heated ?​
Mila [183]
Because glucerin is hotter so it decreases faster
4 0
2 years ago
Answers are - <br><br>A 235 N<br><br>B 376 N<br><br>C 271 N<br><br>D 188 N<br><br>E 470 N
amm1812

Answer:

C

Explanation:

8 0
3 years ago
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