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user100 [1]
3 years ago
8

Identifying the guilty party was mainly based on eyewitness accounts during what time period?

Physics
2 answers:
kondaur [170]3 years ago
8 0

The time period for guilty party was between 1900-1988.

laiz [17]3 years ago
5 0

Answer:

1900-1988

Explanation:

This was the time period during which identifying the guilty party was mainly based on eyewitness accounts. The use of eyewitness accounts became common as witnesses often provided information that no one else could provide. Moreover, witnesses were believed to be unbiased. However, this is not the main method used to identify the guilty party anymore.

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A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric fiel
AveGali [126]

Answer:

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

j=\frac {I}{A} where I is current and A is area and A=\pi r^{2}

Substituting this into the first equation then E=P\times \frac {I}{\pi r^{2}}

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

E=1.72\times 10^{-8}\times \frac {30}{\pi \times 0.001295^{2}}=9.79\times 10^{-2} N/c=0.0979 N/c

4 0
3 years ago
Why does the lens need to be thicker for viewing nearby objects?
Maurinko [17]

Answer: To focus on a near object – the lens becomes thicker, this allows the light rays to refract (bend) more strongly. To focus on a distant object – the lens is pulled thin, this allows the light rays to refract slightly.

Explanation:

5 0
3 years ago
A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?
vesna_86 [32]

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

v^2=u^2+2as

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

v=u+at

where symbols have the usual meaning

Applying the given values we get

t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds

4 0
3 years ago
if you push a box a distance of 2000 meters with a force of 1 newton, how many calories have you used
kap26 [50]
Note that
1 J = 0.239 cal

By definition,
Work = Force x Distance

Therefore work done is
W = (1 N)*(2000 m) = 2000 J

In calories,
W = (2000 J)*(0.239 cal/J) = 478 cal

Answer: 478 calories

4 0
3 years ago
A disgruntled autoworker pushes a small foreign import off
Andrews [41]

Answer:

v = a/√(2h/g) m/s

Explanation:

Lets say the distance away from the cliff is a.

then, a = v t

where v is velocity with which it was thrown and t is time taken to fall.

Using equations of motion, we can also say that

h=1/2gt^2

where h is the height of the cliff

Thus, t^2 = 2h/g and t = √(2h/g)

Thus, v = a/√(2h/g).

the vehicle was pushed off  the cliff with the velocity , v = a/√(2h/g). m/s

5 0
2 years ago
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