Stored energy due to the interactions between objects

A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon

Anettt [7]

First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed $v_x=30 m/s$ , and an accelerated motion on the y-axis, with initial speed $v_y=20 m/s$ and acceleration $g=9.81 m/s^2$ :
$S_x(t)=v_xt$
$S_y(t)=v_y t- \frac{1}{2} gt^2$
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
$S_y(t)=0$
Therefore:
$v_y t - \frac{1}{2}gt^2=0$
which has two solutions:
$t=0$ is the time of the beginning of the motion,
$t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s$ is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
$S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m$

4 0

3 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

mrs_skeptik [129]

Answer:

surface charge density on each sphere is $440 \times 10^{-9}$ C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = $(\frac{1}{4\pi\epsilon })\times (\frac{Q_{1} }{R^{2} } )$ .................1

put here value

660000 = $9 \times 10^9 \times \frac{Q1}{0.12^2}$

Q1 = 1056 × $10^{-9}$

and

here field inside a conductor is zero so that electric potential ( V ) is constant

$\frac{Q{1} }{R} = \frac{Q{2} }{r}$ ..................2

so Q2 will be

Q2 = $\frac{5}{12} \times 1056 \times 10^{-9}$

Q2 = $440 \times 10^{-9}$ C

6 0

3 years ago

Trial 1: Get a textbook and put a sheet of paper on top of it. Fold the paper as needed to keep the paper from sticking over the

siniylev [52]

Answer:

1) the two objects reach the floor at the same time.

2)the book reaches the floor much earlier than the foil

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more

Explanation:

This interesting experiment has the following results

1) first case. Sheet on top of book

In this case the two objects reach the floor at the same time.

This shows that the acceleration in the two objects is the same and we call it the acceleration of gravity.

The speed of the body increases as it goes down linearly.

This occurs because the book that receives air resistance is much heavier, so the resistance has almost no effect on its movement, the sheet does not have the air resistance because it goes down next to the book.

2) second case. Book and sheet next to each other.

In this case the book reaches the floor much earlier than the foil.

This is because the resisting force of the air has almost no effect on the book and its movement is little affected by this force.

In the case of the blade, it has very little weight, therefore as its speed increases, the resistance force of the air rapidly equals the weight of the blade.

W_sheet - fr = 0

so after this, since the acceleration is zero, it goes down at constant speed, this speed is called the terminal velocity.

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more.

3 0

3 years ago

find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed

Cloud [144]

Answer:

$U=5*10^7ft.Ib$