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rosijanka [135]

2 years ago

15

A 1000 kg elevator is rising and its speed is increasing with an acceleration of 3 m/s^2. What is the resulting tension in the v

ertical cable that is lifting the elevator?

1 answer:

Alchen [17]2 years ago

7 0

Answer:

12800 N

Explanation:

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What is this feature?

Viktor [21]

Answer: c Delta

Explanation:

A delta is a landform created by deposition of sediment that is carried by a river as the flow leaves its mouth and enters slower-moving or stagnant water. This occurs where a river enters an ocean, sea, estuary, lake, reservoir, or (more rarely) another river that cannot carry away the supplied sediment.

Hope this helps!!

3 0

2 years ago

Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,

jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

Therefore: $E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}$

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$

7 0

3 years ago

A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T

Mamont248 [21]

Let $y_0$ be the height of the building and thus the initial height of the ball. The ball's altitude at time $t$ is given by

$y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity.

The ball reaches the ground when $y=0$ after $t=2.9\,\mathrm s$ . Solve for $y_0$ :

$0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2$

$\implies y_0\approx16.258\,\mathrm m$

so the building is about 16 m tall (keeping track of significant digits).

3 0

2 years ago

Hi gir_ls join nkd-mbja-nuj

GREYUIT [131]

Answer:

never lol

studying is your work

but why all are doing I don't know=_=

3 0

2 years ago

Which interaction of nature depends on the distance through which it acts and is involved in beta decay? a. strong c. electromag

USPshnik [31]

The answer is weak. The interaction of nature that will depend on the distance through the way it acts and involved in beta decay is the weak interaction or the weak force. This interaction is the responsible for radioactive decay which also plays a significant role in nuclear fission.

5 0

2 years ago