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2 years ago

Answer this question fast please i will mark you brainliest

Physics

2 answers:

Leya [2.2K]2 years ago

6 0

Answer:

b) 72 F°

Explanation:

In Fahrenheit, the difference between freezing and boiling is 180 F° (that’s 212 °F minus 32 °F). In Celsius, the difference is 100 C° (by definition of 0 °C and 100 °C for those two points). So 100 C° = 180 F°; dividing by 100 gives us 1 C° = 1.8 F°.

So a difference of 40 C° is equal to 40 times (1.8 F°) = 72 F°.

Notice I am using °F and °C for actual temperatures, and F° and C° for temperature DIFFERENCES.

barxatty [35]2 years ago

5 0

Answer:

respuesta d)

Explanation:

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2 years ago

Which of the following statements is TRUE?

KIM [24]

Hello there!

<span>Which of the following statements is TRUE?

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3 years ago

Read 2 more answers

1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational

enyata [817]

Answer:

d = <23, 33, 0> m , F_W = <0, -9.8, 0> , W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

x = vox t

Y Axis

y = $v_{oy}$ t - ½ g t²

It's displacement is

d = x i ^ + y j ^ + z k ^

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d = <23, 33, 0> m

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In vector notation, in general the upward direction is positive

W = (0 i ^ - 9.8 j ^ + 0K ^) N

W = <0, -9.8, 0>

Work is defined

W = F. dy

W = F dy cos θ

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6 0

2 years ago

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

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3 years ago

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Answer:

If an object has a high density then the molicules making up that object are closly packed togeather. Because of this, objects with a higher density will have more mass than objects of the same size that have a lesser density.

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3 years ago

Answer this question fast please i will mark you brainliest​

Answer this question fast please i will mark you brainliest