Answer:
<em>The end of the ramp is 38.416 m high</em>
Explanation:
<u>Horizontal Motion
</u>
When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.
The maximum horizontal distance traveled by the object can be calculated as follows:
If the maximum horizontal distance is known, we can solve the above equation for h:
The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:
h= 38.416 m
The end of the ramp is 38.416 m high
Answer:
The planets in order from the sun are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and finally the dwarf planet Pluto. Most people have at least heard about our solar system and the planets in it.
Explanation:
<h3><u><em>
please mark me brainliest</em></u></h3>
206Pb = 1.342 x10^22 atoms
<span>To find the number of atoms, you must first find the number of moles. If 238U is 238.029g/mol, and we have 1.75 grams, how many moles is that? 1.75 divided by 238.029 = 0.007352045 moles. To find the number of atoms in 0.007352045 moles, you multiply by a mole: </span>
<span>0.007352045 x 6.02 x 10^23 = 4.426 x10^21 atoms. </span>
<span>Same procedure for 206Pb: </span>
<span>4.59 divided by 205.97446 = 0.022284316 moles </span>
<span>0.022284316 x 6.02 x 10^23 = 1.342 x10^22 atoms. </span>
<span>Hope that helps you!
https://answers.yahoo.com/question/index?qid=20100331153014AAoMXcu
</span>
Answer:
4.96×10¯¹⁰ N
Explanation:
The following data were obtained from the question:
Mass 1 (M1) = 300 Kg
Mass 2 (M2) = 300 Kg
Separating distance (r) = 110 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Gravitational force (F) =?
The gravitational force between the two goal posts can be obtained as follow:
F = GM1M2 / r²
F = 6.67×10¯¹¹ × 300 × 300 / 110²
F = 6.003×10¯⁶ / 12100
F = 4.96×10¯¹⁰ N
Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N
Answer:
Please find the answer in the explanation.
Explanation:
Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?
We can make the weak one to carry less load through two different ways or means.
First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.
Second, the strong porter can carry the long timbre wood almost to the fulcrum and allow the weak one to support at the other end. By doing this, the weak one will only carry light portion of the load.
