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3 years ago

Gravitational pull is greatest

1 answer:

3 0

From that particular list of choices, the spot where you'll weigh the most is . . . . . at the beach. That's not because there's anything special about a beach. Mt. Everest and an airplane put you farther from the Earth's center, reducing the gravitational force. And the moon has much less mass than Earth has, resulting in much less gravitational force on its surface.

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If a machine exerts a force of 250 N on an object and no work is done, what must have occurred?

tresset_1 [31]

Answer:

The object also had a force of 250 going S (opposite way) towards the machines, hence nothing moved, or as the question mentions, nothing happened

Explanation:

7 0

3 years ago

Convert 141.2 kg to lbs and show all units and work

zavuch27 [327]

Your answer is 311.29271 lbs

5 0

3 years ago

What type of electromagnetic radiation is being used in the picture?

pashok25 [27]

Answer: b i think

Explanation:

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3 years ago

a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u

n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

$a_{c}$ = $\frac{v^{2} }{r}$

$a_{c}$ = 23.04/10.3

$a_{c}$ = 2.23 m/s²

It continues to fly but now with some tangential acceleration

$a_{t}$ = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

$a_{net}$ = $\sqrt{a_{c} ^{2} +a_{t} ^{2} }$

$a_{net}$ = $\sqrt{4.97 + 0.396}$

$a_{net}$ = 2.316 m/s²

So the magnitude of net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0

2 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

igomit [66]

Answer:

The maximum electric power output is $P_{max} =1.339*10^{9} \ W$

Explanation:

From the question we are told that

The capacity of the hydroelectric plant is $\frac{V}{t} = 690 \ m^3 /s$

The level at which water is been released is $h = 220 \ m$

The efficiency is $\eta =$ 0.90

The electric power output is mathematically represented as

$P = \frac{PE_l - PE _o}{t}$

Where $PE_l$ is the potential energy at level h which is mathematically evaluated as

$PE_l = mgh$

and $PE_o$ is the potential energy at ground level which is mathematically evaluated as

$PE_o = mg(0)$

$PE_o = 0$

$P = \frac{mgh}{t}$

here $m = V * \rho$

where V is volume and $\rho$ is density of water whose value is $\rho = 1000 kg/m^3$

$P = \frac{(\rho * V) * gh}{t}$

$P = \frac{V}{t} * gh \rho$

substituting values

$P =690 * 9.8 * 220 * 1000$

$P =1.488*10^{9} \ W$

The maximum possible electric power output is

$P_{max} = P * \eta$

substituting values

$P_{max} =1.488*10^{9} * 0.90$

$P_{max} =1.339*10^{9} \ W$

6 0

3 years ago