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kykrilka [37]
3 years ago
9

A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the spee

d and direction of the satellite after the crashes. Why would the crash affect the motion of the satellite, and which crash caused a greater change in motion for the satellite?
WILL GIVE BRAINLIEST

Physics
1 answer:
Komok [63]3 years ago
5 0

Answer:

The fastest satellite must change orbit

The most massive body (m₁) transfers more momentum to the satellite,

Explanation:

For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants

Most massive body (m1)

initial. Before the crash

      p₀₁ = M v + m₁ v₁

after the crash

      p_{f1} = M v´ + m₁ v₁´

how momentum is conserved

     p₀ = p_{f}

Lighter body (m2)

      p₀₂ = M v + m₂ v₂

       p_{f2} = M v´ + m₂ v₂´

           

Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.

We can see that  p₀₁> p₀₂

Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.

The fastest satellite must change orbit, thus rotating at a different distance from Earth

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Zepler [3.9K]

The magnitude and direction of the electric field in the wire are mathematically given as

L &=[(v / L) v / m] \hat{i}

<h3>What is the magnitude and direction of the electric field in the wire?</h3>

Generally, the equation for is  mathematically given as

A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides  

E d x=\int d v.

\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}

In conclusion, the magnitude and direction of the electric field in the wire are given as

L &=[(v / L) v / m]

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7 0
2 years ago
The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use
saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

M = \frac{Nl}{f_ef_0}

Where

N = Near point

l = distance between the object lens and eye lens

f_0= Focal length

f_e= Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

M = \frac{25*10}{3*5}

M = 16.67\approx 17\\

Therefore the correct answer is C.

3 0
3 years ago
A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits th
Gala2k [10]

Answer:

F_a_v_g=7093333.33N*s

Explanation:

The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:

F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}

Where:

m=mass\hspace{3}of\hspace{3}the\hspace{3}object

v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval

v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.

t_2=final\hspace{3}time

t_1=initial\hspace{3}time

Asumming v1=0 and t1=0:

F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s

8 0
3 years ago
A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface o
lidiya [134]
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
3 0
3 years ago
Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to
katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

OB=0.40 km

And the Max distance towards bookstore is,

OA=3.65 km

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

AB=\sqrt{OB^{2}+OA^{2}}

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km

Therefore the distance between there destination is 3.67 km.

6 0
3 years ago
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