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Ymorist [56]
3 years ago
13

What is the new kinetic energy of the 1900 kg ship on the right moving at 4 m/s?

Physics
2 answers:
Kay [80]3 years ago
8 0

Answer:

15,200J

Explanation:

The equation is 1/2 x m x v², where m= mass and v= velocity.

Elis [28]3 years ago
7 0

Answer:

soi nuevo

Explanation:me  regalan puntos

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Hydrogen is the second most abundant gas in the atmosphere? True or false?
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False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.
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What are the order of the 7 prefixes you are required to know in order from largest to smallest
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10^9 giga, 10^6 mega, 10^3 kilo, 10^-3 milli, 10^-6 micro, 10^-9 nano, 10^-12 pico
Potentially they might want centi which is 10^-2
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The convection currents in the asthenosphere cause the movement of the
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Convection currents generated within the asthenosphere push magma upward through volcanic vents and spreading centres to create new crust. Convection currents also stress the lithosphere above, and the cracking that often results manifests as earthquakes.
7 0
3 years ago
A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above
Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed v_{1}= 2\ m/s

Speed v_{2}=7\ m/s

Pressure in main pipe P_{1}=2\times10^{5}\ Pa

(I). We need to calculate the diameter

Using equation of continuity

Av_{1}=Av_{2}

\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}

(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7

d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}

d_{2}=5.345\ cm

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})

P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)

P_{2}=1.28500\times10^{5}\ Pa

(III).  If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}

P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})

Here, h_{3}=0

Put the value in the equation

P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17

P_{3}=3.5400\times10^{5}\ Pa

Hence, This is required solution.

7 0
3 years ago
A solid nonconducting sphere of radiusRcarries a chargeQdistributed uniformly throughout itsvolume. At a certain distancer1(r1&l
lara [203]

Answer:

E' = \frac{1}{8} E

Explanation:

Given data:

first case

Distance of electric field from center of sphere is r_1 <R

Electric field at r_1< R

E = \frac{kQr_1}{R^3}

second case

Distance of electric field from centre of sphere is r_1 < 2R

Electric field at r_1< 2R

E' = \frac{kQr_1}{8R^3}

so, we have

E' = \frac{1}{8} E

3 0
3 years ago
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