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Pani-rosa [81]
2 years ago
6

Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti

city of this collision? explain your answer.
Physics
1 answer:
Ket [755]2 years ago
4 0

Answer:

Perfectly inelastic collision

Explanation:

There are two types of collision.

1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.

2. Inelastic collision: When the momentum  the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

For a perfectly elastic collision, the two bodies stick together after collision.

Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.

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The number of significant figures on the measurement 0.050010 kg id
Alex17521 [72]
Its been some time so i might be wrong but i think the answer is 3 either or 2
6 0
2 years ago
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

y=\frac{m\lambda D}{d}

At m = 0

y_0=0

y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm

At m = 1

y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

3 0
2 years ago
A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
4 0
2 years ago
three neutral metal cans mounted on isulating stands are touching a negatively charge ballon is brought near can a can b is then
Nimfa-mama [501]
Charge on can A is positive. 
Charge on can C is negative.  
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon. 
 As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
7 0
2 years ago
"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f
ELEN [110]

Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c  = 25 ms = 25 × 10⁻³ s

The initial temperature T_1 = 273 K

The final temperature T_2 = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume =  0.4 mV/K × 473 K =  189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2  + (189.2 - 109.2)( 1 - \mathbf{e^{-t/c}})mV

y(t) = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram  attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 ×  189.2 mV

=  170.28 mV

170.28 mV = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

170.28 mV - 109.2 mV = 80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

61.08 mV =  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

0.7635  mV = ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

t = 1.44 × 25  × 10⁻³ s

t = 0.036 s

t = 36 ms

The error fraction = \dfrac{189.2-170.28  }{189.2}

The error fraction = 0.1

The error fraction = 10%

8 0
2 years ago
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