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sergij07 [2.7K]
3 years ago
15

If i have no swag and i dab what happens

Physics
2 answers:
vlabodo [156]3 years ago
7 0
You earn swag points.
Elena L [17]3 years ago
7 0

you earn swag points and from there you can buy certain items from the swag shop like air pods

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Ryan runs 50 m north and then 50 m east. What is his displacement?
san4es73 [151]

Answer:

Explanation:

Distance and direction of an object's change in position from a starting point. Displacement. 3. Jermaine runs exactly 2 laps around a 400 meter track. What is the displacement? 0 ... David walks 3 km north, and then turns east and walks 4 km. ... A person walks 50 meters directly north, stops, and then travels 32 meters

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3 years ago
What is the difference between free fall and weightlessness??
joja [24]
When you are in free fall, the force of gravity is stronger than your velocity perpendicular to where you're falling, and you move at a constant speed downwards.

Under feelings of weightlessness, you are still being pulled by gravity, but your perpendicular velocity and distance from the source can cancel each other out.
3 0
3 years ago
Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros
kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

v^2-u^2=2gh

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g=-9.8 m/s^2 is the acceleration of gravity

Solving for u,

u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s

The time needed to reach the maximum height can now be found by using the equation

v=u+gt

Solving for t,

t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

v'^2-u^2=2gh'

we find

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s

So the time to go from h' to h is

\Delta t = t-t'=0.52-0.32=0.20 s

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

\Delta t=2\cdot 0.20 = 0.40 s

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

v'^2-u^2=2gh'

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

\Delta t = 2\cdot 0.04 =0.08 s

8 0
3 years ago
Please I need the answer right now will mark you brainliest ​
Dmitrij [34]

Answer:

Explanation:

what do u need help with

3 0
2 years ago
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Creates an image that appears upside down behind the focal point
zheka24 [161]
An image that appears upside down behind the focal point is an image that is reflected on a concave mirror. Mirrors reflect different kinds of images based on the placement of an object that is reflected towards it. There are two kinds of mirrors, concave and a convex mirrors, the latter makes objects seem smaller and farther than where it is exactly.
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