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alexdok [17]
2 years ago
5

A 66.0 kg diver is 3.10 m above the water, falling at speed of 6.10 m/s. Calculate her kinetic energy as she hits the water. (Ne

glect air friction)
Physics
1 answer:
castortr0y [4]2 years ago
3 0

Kinetic energy as she hits the water is 3300 joule.

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to determine the final velocity?</h3>
  • The final velocity is determined as

V²=U²+2aS

  • V= final velocity, U= initial velocity, a= acceleration and S= distance
<h3>What's the final velocity of the driver falling from 3.10m with initial velocity of 6.10m/s?</h3>
  • Here, a= 9.8m/s², U= 6.10m/s and S= 3.10m
  • So, V²= 6.1²+2×9.8×3.10= 98
  • V= √98= 10m/s
<h3>What's the kinetic energy of the driver when touches the water?</h3>

Kinetic energy= 1/2×mass×velocity²

= 1/2 × 66 × 10²

= 3300J

Thus, we can conclude that the kinetic energy of the driver is 3300 Joule.

Learn more about the kinetic energy here:

brainly.com/question/25959744

#SPJ4

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A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is
Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d=\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m

Wavelength of light,\lambda=510 nm=510\times 10^{-9} m

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

sin\theta=\frac{m\lambda}{d}

For first bright fringe, =1

sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305

\theta=sin^{-1}(0.305)=17.76^{\circ}

The distance between two m=1 fringes

x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

sin\theta=1

sin\theta=\frac{m\lambda}{d}

Substitute the values

1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}

m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3

Maximum number of bright fringes on the scree=2m+1=2(3)+1=7

8 0
4 years ago
The following objects have velocity and mass as given: Object A: m=5 kg, ⃗v=−11 ^j Object B: m=6 kg, ⃗v=5^ i +8.7 ^ j Object C:
Fofino [41]

Answer:

For object A

m = 5 kg ,   v= -11 j

For B object

m = 6 kg  ,  v= 5 i +8.7 j

For object C

m = 10 kg   ,   v= -10 i

We know that

Linear momentum   P= m v   kg.m/s

a) A and C

Momentum in y direction

Py=- 5 x 11 j= - 55 j  kg.m/s

Momentum in x direction

Px=- 10 x 10 j= - 100 i    kg.m/s

b) B and C

Momentum in y direction

Py=6 x 8.7 j= 52.2 j    kg.m/s

Momentum in x direction

Px=( 6 x 5 - 10 x 10 ) i = - 70 i   kg.m/s

c) A ,B and C

By using data of a and b

Momentum in y direction

Py= 6 x 8.7 - 5 x 11 j= -2.8 j    kg.m/s

Momentum in x direction

Px=  6 x 5 -10 x 10 i = -70 i   kg.m/s

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Answer:

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Explanation:

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50 because read step by step explanation
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3. Plumb line → A. Line of C.O.G

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