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4 years ago

Does anyone know what Base jump is?

Physics

1 answer:

Sever21 [200]4 years ago

4 0

Answer:

BASE jumping is the recreational sport of jumping from fixed objects, using a parachute to descend safely to the ground. "BASE" is an acronym that stands for four categories of fixed objects from which one can jump: building, antenna, span, and earth (cliff).[1][2] Participants exit from a fixed object such as a cliff, and after an optional freefall delay, deploy a parachute to slow their descent and land. A popular form of BASE jumping is wingsuit BASE jumping.

Explanation:

In contrast to other forms of parachuting, such as skydiving from airplanes, BASE jumps are performed from fixed objects which are generally much lower altitude, and BASE jumpers typically carry only one parachute. BASE jumping is significantly more hazardous than other forms of parachuting, and is widely considered to be one of the most dangerous extreme sports

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kifflom [539]

Answer: $219.65\ \text{km}$

Explanation:

Given

Car drives 215 km east and then 45 km North

Displacement is East direction is

$\Rightarrow \vec{r_1}=215\hat{i}$

Now, the displacement from that to 45 km North is given by

$\Rightarrow \vec{r_{21}}=45\hat{j}$

Net displacement is $\vec{r_2}$

$\Rightarrow \vec{r_2}=\vec{r_1}+\vec{r_{21}}\\\Rightarrow \vec{r_2}=215\hat{i}+45\hat{j}$

Magnitude of the displacement is

$\Rightarrow \left | r_{2} \right |=\sqrt{\left ( 215 \right )^2+\left ( 45 \right )^2}\\\Rightarrow \left | r_{2} \right |=\sqrt{48250}\\\Rightarrow \left | r_{2} \right |=219.65\ \text{km}$

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Answer:

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A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounc

erma4kov [3.2K]

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

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