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2 years ago

7

How many times more gravitational potential energy would the same lander have on top of a 350-m hill on Earth? It would have tim

es more gravitational potential energy

1 answer:

algol132 years ago

5 0

Answer:

6

Explanation:

Just did it.

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At present most of the world's energy needs are supplied by what kind of energy

antiseptic1488 [7]

Fossil fuels . . . coal, oil, natural gas

Among primitive cultures, wood is an important source.

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3 years ago

What makes a model not useful?

lina2011 [118]

Answer is b hope this helps

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2 years ago

calculate the distance a plane flies on a 7.95 hour flight from Chicago to London. Assume a constant speed of 800.0 km/h

soldier1979 [14.2K]

Answer:

6360 km

Explanation:

Use the kinematics equation $x=v_ot+\frac{1}{2}at^2$ . We are given t = 7.95 hours and a = 0 m/s^2 (constant speed means there is no acceleration). Solve for x.

$x=(800)(7.95)+\frac{1}{2}(0)(7.95^2)\\x=6360 \ km$

4 0

3 years ago

What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?

igomit [66]

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

4 0

2 years ago

Read 2 more answers

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

2 years ago