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3 years ago

1. How much energy is needed just to melt 0.56kg of ice at 0◦ C?

Physics

1 answer:

ivann1987 [24]3 years ago

5 0

Answer:

This will require 266.9 of heat energy.

Explanation:

To calculate the energy required to raise the temperature of any given substance, here's what you require:The mass of the material, m The temperature change that occurs, ΔT The specific heat capacity of the material,

(which you can look up). This is the amount of heat required to raise 1 gram of that substance by 1°C.

Here is a source of values of

c for different substances:

Once you have all that, this is the equation:

Q=m×c×ΔT(Q is usually used to symbolize that heat required in a case like this.)For water, the value of c is 4.186g°C So, Q=750×4.186×85=266=858=266.858

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What adaptation of a cheetah is shown in the image?

pychu [463]

The adaptation of a cheetah is shown in the image which shows it running is its long and strong legs to run fast which is denoted as option B.

<h3>What is Adaptation?</h3>

These are the physical and behavioral features which ensures that they survive in their changing environment.

Cheetahs are known to have long and strong legs which is the reason why it runs at a very high speed.

Read more about Adaptation here brainly.com/question/29594

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2 years ago

Canada geese migrate essentially along a north–south direction for well over a thousand kilo-meters in some cases, traveling at

kotykmax [81]

Answer:

a) 66.4 relative to the west in the south-west direction

b) 5.455 hours

Explanation:

a)If the wind is blowing east-ward at a speed of 40km/h, then the west component of the geese velocity must be 40km/h in order to counter balance it. Geese should be flying south-west at an angle of

$cos(\alpha) = 40 / 100 = 0.4$

$\alpha = cos^{-1}(0.4) = 1.16 rad = 180\frac{1.16}{\pi} = 66.4^0$ relative to the West

b) The south-component of the geese velocity is

$100sin(\alpha) = 100sin(66.4^0) = 91.65 km/h$

The time it would take for the geese to cover 500km at this rate is

t = 500 / 91.65 = 5.455 hours

7 0

3 years ago

How can damage to living things occur from exposure to X-rays or gamma

OLEGan [10]

Answer:C

Explanation:

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3 years ago

Read 2 more answers

A population has a birthrate of 10/1000 , a death rate of 9/1000, an immigration rate of 3/1000, and an emigration rate of 7/100

Charra [1.4K]

Birth rate 10/1000 = +population

Death rate 9/1000 = -population

Immigration rate 3/1000 = +population

Emigration rate 7/1000 = -population

10/1000 - 9/1000 + 3/1000 - 7/1000= -3/1000

Growth rate = (-3/1000)(100%)= -0.3%

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3 years ago

For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red

barxatty [35]

Answer:

Explanation:

To calculate the red shift you use the following formula:

$z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1$

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

$z=\sqrt{\frac{c+v}{c-v}}-1$

- A red dwarf moving away from Earth at 39.1 km/s :

$z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}$

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

$z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}$

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

$z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}$

- A blue dwarf moving away from Earth at 25.9 km/s $z=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}$

- A red dwarf moving transversely at 14.1 km/s

$z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}$

6 0

4 years ago