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liraira [26]
3 years ago
10

You walk 6 m North and then 4 m south what distance did you travel? What is your displacement?

Physics
1 answer:
mestny [16]3 years ago
6 0
<span>Distance is the actual path covered and displacement is the shortest distance from the object to the point of origin.

Distance is a total of 10m because you walked 6m initially and then another 4m. 6m + 4m = 10m

Displacement includes the starting point. You walked 6m North. Then you turned around and walked 4m South. Your total displacement is 2m because that is your current distance from the starting point.

Please mark as brainliest if satisfied with answer
</span>
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You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g&#10;            y- 0 = 10.0²/2 9.8&#10;            y - 0 = 5.10 m&#10;            &#10;The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system&#10;             y₂ = 5.1 + 44&#10;             y₂ = 49.1 m&#10;Let's use the other equation to find the time&#10;              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
Describe how the student should measure the time taken for the toy parachute to
PolarNik [594]

The time must be measured with respect to gravity. As it falls, it has free fall that is the force acting on it will be the gravity.With the distance in account, d = 1/2 gt²

t = √(2d/g)

4 0
3 years ago
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
NARA [144]

Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = 6\times 10^6\ J

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s

The time taken to lift the load is 141.26626 seconds

5 0
3 years ago
The escape speed from the moon is much smaller than from earth. True or False
Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

v_{e} = \sqrt{\frac{2GM}{r}}  (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is 5.972x10^{24}kg and its radius is 6371000m

Then, replacing those values in equation 1 it is gotten.

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}  

v_{e} = 11.18m/s

For the case of the Moon:

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}  

v_{e} = 2.38m/s

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

4 0
3 years ago
PLZ HELP!!!!
FrozenT [24]
I would think the answer is c.
5 0
3 years ago
Read 2 more answers
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