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melomori [17]
2 years ago
14

A car left skid marks 78 m long on a road as it slid to a stop. If the car's acceleration was -3.9 m/s2, what was its velocity b

efore it began to skid, to the nearest m/s?
Physics
1 answer:
spayn [35]2 years ago
7 0
It is 25 I’m sure of it
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A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef
inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0
2 years ago
Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa
rewona [7]

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

\theta =1.22\cdot \frac{\lambda }{D}

Applying the given values we get

\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads

thus the linear separation equals L=\theta \times Distance

Applying the given values we get

L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes

6 0
3 years ago
The two loudspeakers in the drawing are producing identical sound waves. The waves spread out and overlap at the point P. What i
Damm [24]

Answer:

5/2π

Explanation:

According to quizlet the answer is 5/2π

5 0
3 years ago
Two small insulating spheres with radius 3.50×10−2 m are separated by a large center-to-center distance of 0.555 m. One sphere i
xz_007 [3.2K]

Answer:

E = 7.83 \times 10^5 N/C

Explanation:

Since we know that two sphere is oppositely charged so net electric field at the mid point of two balls will be sum of the electric field due to each ball at the mid point

So we know that

E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}

here we know that

q_1 = 4.20 \mu C

q_2 = 2.50 \mu C

r = \frac{0.555}{2}

so we have

E = \frac{(9\times 10^9)(4.20 + 2.50) \times 10^{-6}}{0.2775^2}

E = 7.83 \times 10^5 N/C

4 0
2 years ago
All sound waves need a(n) _______________________________________ to travel through. 4. _____________________ and ______________
Paha777 [63]

Answer:

Material medium

compressions and rarefactions

Explanation:

A sound wave is an example of a mechanical wave. All mechanical waves require a material medium for propagation. The medium for the propagation of sound is air. This is the reason why, if you cover your mouth, it will be difficult for another person to hear whatever you are saying.

Sound is also a longitudinal wave. Longitudinal waves are described in terms of compressions and rarefactions. Compressions refer to areas where air molecules crowd together while rarefactions refer to areas where the air molecules spread out.

4 0
3 years ago
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