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3 years ago

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As a rocket is accelerating vertically upward at 92 m/s^2 near the Earth's surface, it releases a projectile. Immediately after

release the acceleration (in m/s^2) of the projectile is: A) 9.8 down. B) 9.8 up C) 9.2 down D) 92 up E) none of the above

1 answer:

valkas [14]3 years ago

6 0

Answer:

A)

Explanation:

Once released, the projectile is subject only to the influence of gravity, which is always downward, and produces on the object an acceleration of 9.8 m/s².

The vertical acceleration of the rocket only supplies an initial velocity to the projectile, in the vertical direction, upwards.

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Answer:

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b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance

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= 98,000 J

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b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

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KE = 10,000 J

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If the out come of a 1.0-kg ball moving at 3.0 m/s is 4.5 kinetic energy what would the kinetic energy be if the mass and speed

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If a non-human civilization were to develop on Saturn's largest moon, Titan, its scientists might well develop a temperature sca

klasskru [66]

Answer: $-330.47 \°M$

Explanation:

Let's begin by explaining that the relation between the Celsius scale and Kelvin scale is:

$\°C +273.15=K$

This means the absolute zero point of the Kelvin scale is $0 K=-273.15 \°C$

Keeping this in mind, we have the freezing point and boling point of Methane as:

Freezing point: $0 \°M=-182.6 \°C=90.55 K$

Boiling point: $100 \°M=-155.2 \°C=117.95 K$

According to this, there is a linear relation between the methane scale ( $\°M$ ) and the Kelvin scale in the form:

$T=a+bt_{M}$ (1)

Where:

$T$ is the temperature in Kelvin

$t_{M}$ is the temperature in degrees Methane

Firstly, we need to find the value of $a$ and $b$ with the two given points ( $90.55 K$ and $117.95 K$ ):

When $T=90.55$ and $t_{M}=0$ :

$90.55=a+b(0)$

$a=90.55$ (2)

When $T=117.95$ and $t_{M}=100$ :

$117.95=90.55+b(100)$

$b=0.274$ (3)

Now we have the linear equation:

$T=90.55+0.274t_{M}$ (4)

Isolating $t_{M}$ :

$t_{M}=\frac{T-90.55}{0.274}$ (5)

Evaluating for $T=0 K$ :

$t_{M}=\frac{0-90.55}{0.274}$ (6)

Finally:

$t_{M}=-330.47 \°M$

This means $0 K=-330.47 \°M$

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