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PilotLPTM [1.2K]
3 years ago
8

If we drop different weights from same height which will fall first to ground? Both of them hit the ground at the same time righ

t?
Physics
2 answers:
STALIN [3.7K]3 years ago
7 0
Gravity acceleration is the same so yes, b<span>oth of them hit the ground at the same time.</span>
crimeas [40]3 years ago
5 0
Yes tthey would both hit the ground
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A paper airplane used to study wing designs is an example of a ?
kow [346]
Model I'm guessing. Coz that's using an object to explain
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3 years ago
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A car on a roller coaster starts at zero speed at an elevation above the ground of 26 m. It coasts down a slope, and then climbs
nirvana33 [79]

The speed of the car at the top of the hill is 14m/s

<u>Explanation:</u>

given that

Initial velocity u of the car=0 m/s

The distance can be determined by finding out the difference between the elevation of the first slope and second slope.

elevation of the first slope=26 m

elevation of second slope=16m

distance s=26-16=10 m

acceleration due to gravity g=9.8 m/s2

speed of the car at the top of the hill can be determined by using the equation

v^2=u^2+2as\\v^2=0^2+2\times 9.8\times 10\\=196\\v=\sqrt{196} =14m/s

speed of the car at the top of the hill is 14m/s

5 0
3 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
iren2701 [21]
1) The electric potential at a distance r from a single point charge is given by
V(r) = k  \frac{q}{r}
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
q=2.2 \mu C =2.2 \cdot 10^{-6} C
So the potential at distance r=6.3 m is
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2})  \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2})  \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V
3 0
3 years ago
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A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of
nlexa [21]

Answer:

a_c=1.44\ m/s^2

Explanation:

<u>Centripetal Acceleration</u>

It's the acceleration that an object has when traveling on a circular path to take into consideration the constant change of velocity it must have in order to keep going in the circular path.

Being v the tangent speed, and r the radius of curvature of the circle, then the centripetal acceleration is given by

\displaystyle a_c=\frac{v^2}{r}

We can compute the value of v by using the distance and the time taken to travel:

\displaystyle v=\frac{x}{t}=\frac{200\ m}{26.4\ s}

v=7.58\ m/s

Now we calculate the centripetal acceleration

\displaystyle a_c=\frac{7.58^2}{40}=1.44\ m/s^2

\boxed{a_c=1.44\ m/s^2}

4 0
3 years ago
Physics question i appreciate your help please
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There are two numbers i.e (c) and (d) with 4 significant figures, since we start counting the first non-zero digit from the left, therefore, A. Has 1 significant figure, B has 5 significant figurea and C,D have 4 significant figures.
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