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3 years ago

How much energy will it take to increase the temperature of 200 milliliters if water by 12 degrees Celsius?

1 answer:

7 0

Answer: The specific heat capacity of water is 4,200 Joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C. Lead will warm up and cool down fastest because it doesn't take much energy to change its temperature.

Explanation:

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A branch falls from a tree how fast is the branch moving after drug .28 seconds

alexdok [17]

Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s

We have to find final velocity.

The equation we use is

Final velocity=initial velocity+acceleration x time

Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s

We would round this to:

Vf (final velocity)=2.7m/s

6 0

3 years ago

A train passes a stationary observer. Which of

Lyrx [107]

Explanation:

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4 0

2 years ago

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

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5 0

2 years ago

The counter is 1.5 m high, and the bowl has a mass of 0.5 kg. How much gravitational energy is stored in the bowl-earth system?

alina1380 [7]

Answer:

7.35 J

Im assuming, upon answering the question, that the gravity in this scenario is 9.8? As 9.8 is the gravitational force upon the earth.

5 0

2 years ago

Explain imprinting in relation to Lorenz's geese. What would most likely have happened if the first thing the goslings saw was a

iragen [17]

If the first thing that the goslings saw was a dog, they would have followed the dog as a mother.

Imprinting refers to the process of training an animal to bond with anything it sees after birth even if it is not its real mother. Lorenz first achieved imprinting in 1935 using geese which followed him as their mother shortly after they were born.

If the geese were exposed to a dog, they could also have seen the dog as their mother and followed it accordingly shortly after birth.

Learn more: brainly.com/question/11401513

7 0

1 year ago