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-Dominant- [34]
4 years ago
12

In July 2005, NASA's "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the s

urface at 37000 km/h relative to the surface. The original speed of the comet at that time was about 40000 km/h, and its mass was estimated to be in the range (0.10−2.5)×1014kg. Use the smallest value of the estimated mass.
Physics
1 answer:
eimsori [14]4 years ago
5 0

Question:

(a) What change in the comet’s velocity did this collision produce? Would this change be noticeable? (b) Suppose this comet were to hit the earth and fuse with it. By how much would it change our planet’s velocity? Would this change be noticeable? (The mass of the earth is 5.97×1024kg.)

Answer:

The answers to the question are;

(a) The change in the comet’s velocity produced by the collision is 2.86 × 10⁻⁶km/h or 7.944 × 10⁻⁷ m/s

(b) It would change our planet’s velocity by  6.70× 10⁻⁸ km/h or 1.86× 10⁻⁸ m/s

Change is too small to be noticeable

Explanation:

We not that the question is about conservation of liner momentum

Therefore we have, by listing out the known parameters

m₁ = Mass of "Deep Impact" = 372 kg

m₂ = Mass of Tempel 1 comet  = (0.1 to 2.5) × 10¹⁴ kg,

v₁ = Vaelocity of "Deep Impact" = 37000 km/h

v₂ = Velocity of Tempel 1 comet = 40000 km/h

From the principle of linear momentum, we have, for both bodies moving in opposite direction;

m₁×v₁ + m₂×v₂ = m₁×v₃ + m₂×v₃ since it was a crash, it is assumed that they both have the same final velocity

This gives

372 kg ×37000 km/h  - 0.1 × 10¹⁴ kg × 40000 km/h = (372 kg + 0.1 × 10¹⁴ kg )×v₃

13764000 kg·km/h - 4.0 × 10¹⁷  kg·km/h = 10000000000372×v₃

v₃ = (-399999999986236000 kg·km/h)/10000000000372 = -39999.999997 km/h ≈ - 40000  km/h in the direction of Deep Impact

Change in comet velocity

= 40000  km/h - 39999.999997 km/h

= 2.86 × 10⁻⁶km/h = 7.944 × 10⁻⁷ m/s

(b) If the colission is with earth, we have

m₃ = Mass of earth

From the principle of conservation of linear momentum, we have

m₂v₂+m₃ v₃ = (m₂ + m₃) v₄

v₃ = Initial velocity of Earth = 0 km/h

m₃ = Mass of Earth = 5.97 × 10²⁴ kg

Therefore, pluggin in the vaalues gives

0.1 × 10¹⁴ kg × 40000 km/h + 5.97 × 10²⁴ kg × 0 km/h = (0.1 × 10¹⁴ kg + 5.97 × 10²⁴ kg) × v₄

Therefore,

v₄ = (4.0 × 10¹⁷  kg·km/h + 0 kg·km/h)/ (5970000000010000000000000 kg)

= 6.70× 10⁻⁸ km/h = 1.86× 10⁻⁸ m/s

Change is too small

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Explanation:

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6 0
3 years ago
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Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
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Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

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x=-1-0.4365=-1.4365\ m

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The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?
anastassius [24]
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
\rho =  \frac{AR}{L}
where
\rho is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
A=\pi r^2
where r is the radius of the wire. Substituting in the previous equation ,we find
\rho =  \frac{\pi r^2 R}{L}

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
\rho' =  \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4  \frac{\pi r^2 R}{L}   = 4 \rho
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5 0
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A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

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at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

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2gh = u²

h = u²/2g

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at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

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