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3 years ago

An image formed on a screen is always

Physics

1 answer:

Annette [7]3 years ago

6 0

Answer:

diminished and erect( upright)

Explanation:

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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

According to the law of reflection, the angle of incidence is equal to the angle of

Darina [25.2K]

Before going to answer this question first we have to understand reflection and laws of reflection.

Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .

Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.

Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]

The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]

There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.

There are two laws of reflection.

FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.

SECOND LAW- It states that that the angle of incidence is equal to the angle of reflection irrespective of the type of reflection.i.e i =r

Hence the correct answer will be angle of reflection.

3 0

3 years ago

Read 2 more answers

How many days does it take for a free to grow?

vovikov84 [41]

Idk what is growing but if it’s a free than c

7 0

3 years ago

Which of the following best describes a chemical equation that demonstrates the law of conservation of matter?

LiRa [457]

B, since it is the only one that actually conserves matter for certain. In each of the others, matter could still be imbalanced, since for A, for example, it could be 5 Carbons on the right and 5 Chlorines on the left, and that would not balance.

3 0

3 years ago

Here we will calculate the work per unit charge on an electron moving between two potentials. A 9.0 V battery is connected acros

Svet_ta [14]

Answer:

2000 V/m

1778021.69323 m/s

Explanation:

V = Voltage = 9 V = $\Delta V$

q = Charge of proton = $1.6\times 10^{-19}\ C$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

v = Velocity of electron

Electric field is given by

$E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{9}{4.5\times 10^{-3}}\\\Rightarrow E=2000\ V/m$

The electric field is 2000 V/m

Here, the energy of the system is conserved

$\dfrac{1}{2}mv^2=q\Delta V\\\Rightarrow v=\sqrt{\dfrac{2q\Delta V}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 9}{9.11\times 10^{-31}}}\\\Rightarrow v=1778021.69323\ m/s$

The velocity of the electron is 1778021.69323 m/s

4 0

3 years ago

An image formed on a screen is always​

An image formed on a screen is always