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viktelen [127]
4 years ago
8

Express 0.00000000062 kg in scientific notation

Physics
2 answers:
Dima020 [189]4 years ago
8 0

Answer:

6.2 x 10^-10

Explanation:

move the decimal place.

Sonbull [250]4 years ago
3 0

Explanation:

In this question, we need to express 0.00000000062 in scientific notation.

Any number can be written in the form of significant figures as :

N=a\times 10^b

Where

a is any non zero number

b is any integer

Here, there are 9 zeroes before 6. It can be expressed in the form of significant figures as :

0.00000000062=6.2\times 10^{-10}

Hence, the given number is expressed in the form of significant figures.

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How is work calculated?
BartSMP [9]

Work is force times distance. If there's no distance, there's no work being done.

6 0
4 years ago
PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS<br>​
butalik [34]

Answer:

g=GM/R^2

Universal Gravutation Constant:

f=GM×m/R^2

Force can be also expressed as

f=m×g

so,

mg=GMxm/R^2

The m gets cancelled so

g=GM/R^2

8 0
4 years ago
the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t
givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

a_{t} = v = 0.8 m/s^{2}

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

         = 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

          = 0.8

at x= 5 Km

dy/dx = 0.8(5)

         = 4

p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} }   }{\frac{d^{2y} }{dx^{2} } }

now insert the values,

p = \frac{[1+(4)^{2}]^{\frac{3}{2} }  }{0.8}  = 87.62 km

→ Now the normal component of acceleration is given by

a_{n} = \frac{v^{2} }{p}

    = (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}

a = [(0.8)^{2} + (0.457)^{2}]^{0.5}

a = 0.921 m/s²

4 0
3 years ago
Can anyone help me with this question please​
JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

  1. v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

5 0
3 years ago
If you were to stand in the exact center of a rotating disc, you would only have what kind of
Dmitriy789 [7]

Answer:

Tangential speed or Rotational speed

3 0
3 years ago
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