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Serggg [28]
3 years ago
9

A flea jumps by exerting a force of 1.07 10-5 N straight down on the ground. A breeze blowing on the flea parallel to the ground

exerts a force of 1.14 10-6 N on the flea. Find the direction and magnitude (in m/s2) of the acceleration of the flea if its mass is 6.0 10-7 kg. (Let us assume that).
Physics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Answer:

a) 15.77 m/sec2

b) 13.3 deg

Explanation:

we are given;

Flea force = F1=1.07×10⁻5 N j

Breeze force = F2 = 1.14× 10⁻6 N (-j

mass of flea =6.0 ×10⁻7 kg

So net force on the flea=F1+F2+weight of flea=1.07×10⁻5 j +1.14× 10⁻6 i + 6.0 ×10⁻7 (-j) ×9.8= ma

==> ma = 1.07×10⁻5 j - 0.588×10⁻5 j + 0.114×10⁻5 i

==> ma= 0.482 ×10⁻5 j +0.114×10⁻5 i

==> ma = 0.114×10⁻5 i +0.482 ×10⁻5 j

== a = (0.114×10⁻5 i +0.482 ×10⁻5 j) / 6.0 ×10⁻7

==> a =

==>a= (1.9 j+8.03 i ) m/sec2

mag of a \sqrt{1.9^{2}  +8.03^{2} = 15.77 m/sec2

direction angle = tan⁻1(1.9/8.03)=13.3°

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At 15°C air is transmitted <br>at 340 m/s. Express this speed<br>in Kilometers per hour.​
EleoNora [17]

Answer:

1224km/hr

Explanation:

To convert from m/s to km/hr

1000m = 1km

Divide both sides by 1000

1m = 1/1000 km................. (1)

60×60 seconds = 1 hr

3600s = 1hr

Divide both sides by 3600

1s = 1/3600 .............(2)

Divide (2) by (1)

1m/s =  1/1000 ÷ 1/3600 km/hr

1m/s = 1/1000 × 3600/1  km/hr

1m/s = 3600/1000  km/hr

1m/s = 3.6 km/hr .............(3)

To convert 340m/s to km/hr

Multiply (3) by 340

1× 340m/s = 3.6 × 340 km/hr

340m/s = 1224km/hr

I hope this was helpful, please mark as brainliest

7 0
3 years ago
Read 2 more answers
Determine the net work a hiker must do on a 3.35-kg backpack to carry it up a
Sphinxa [80]

Answer:

410.4J

Explanation:

Step one:

given

mass= 3.35kg

weight= 3.35*9.81= 32.86N

h=12.49m

Required

The net work done

Step two:

the work done is given  as

WD= force* distance

WD= 32.86*12.49

WD= 410.4J

8 0
3 years ago
At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately
Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

I = 2 kJ/s*m^2

area  on which photons are received is given as

A = 4.80 cm^2 = 4.80 * 10^-4 m^2

now we can find the power received due to sunlight

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so here we can say

P = N\frac{hc}{\lambda}

given here that

\lambda = 510 nm

0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}

0.96 = N * 3.88 * 10^{-19}

N = \frac{0.96}{3.88*10^{-19}}

N = 2.47 * 10^{18}

so it will strike 2.47 * 10^18 photons on given area per second

3 0
4 years ago
In which direction does centripetal force act on an object
Gwar [14]
In a circle or circular motion
5 0
4 years ago
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During an early phase of the planet's existence, earth's atmosphere was similar in composition to the atmospheres of mars and ve
KengaRu [80]

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8 0
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