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Ostrovityanka [42]

3 years ago

10

A person walks 12 meters to the west and then 16 meters back to the east. what is the distance covered and the direction of disp

lacement

1 answer:

devlian [24]3 years ago

6 0

The distance is 28 meters and the direction of displacement is East I think

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Where are the nucleus and the electrons located in this atom

levacccp [35]

In the middle the nucleus is thr brain of it and the electrons spin around the whole atom

5 0

3 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force f is applied to the other end of th

Vlada [557]

For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

$F\times L = mg \times \frac{L}{2}$

here weight will act at mid point of door so its distance is half of the total distance where force is applied

here we know that

$mg = 145 N$

now we will have

$F = \frac{mg}{2}$

$F = \frac{145}{2} = 72.5 N$

so our applied force is 72.5 N

7 0

4 years ago

A man travels 15 km east before turning around and heading 5 km west. What is his displacement?

navik [9.2K]

Answer:

10 km East

Displacement is the shortest path between two points.

6 0

3 years ago

Read 2 more answers

An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result

mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

$s=ut+\frac{1}{2}gt^{2}$

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

$s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}$

So the distance covered by the acorn is 3.136 m.

8 0

3 years ago

In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of t

a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot = Pgas

= 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵ =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

ΔP = ρgh

6.46 × 10⁵ = 1000 x 9.81 x h

h = 65.85 meters

7 0

3 years ago