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jarptica [38.1K]
3 years ago
8

About what fraction of earth's freshwater is in the form of ice

Physics
1 answer:
Bingel [31]3 years ago
3 0
68% So 68/100 of freshwater is found in ice
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How does radioactive decay work
Lesechka [4]

Answer:

Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.

Explanation:

6 0
3 years ago
The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m
IgorC [24]

Answer:

0.0360531138247 V/m

Explanation:

\rho = Resistivity of gold = 2.44\times 10^{-8}\ \Omega .m (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = \dfrac{\pi}{4}d^2

E = Electric field

Resistivity is given by

\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m

The  electric field in the wire is 0.0360531138247 V/m

6 0
3 years ago
Read 2 more answers
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?
miv72 [106K]

2m/s²

Explanation:

Given parameters:

Radius = 50m

Speed = 10m/s

Unknown:

Acceleration of the cyclist

Solution:

The acceleration of the cyclist is directed inside of the corner because his motion is inward. This is a form of centripetal acceleration;

Centripetal acceleration is given by;

     a = \frac{v^{2} }{r}

v is the velocity

r is the radius

   a =  \frac{10^{2} }{50} = 2m/s²

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

3 0
3 years ago
A 6.5-l sample of nitrogen at 25°c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin
Katarina [22]
The formula that we will going to use in this question is simply P1V1/T1 = P2V2/T2 where T is the constant.
P1V1 = P2V2 
P2 = P1V1/V2 = (V1 / V2) x P2 = (13.0 L / 6.5 L) x 0.76 atm = 1.5 atm 
The answer in this question is 1.5 atm
5 0
3 years ago
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