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3 years ago

2. In contrast to Venus, the coldest temperature yet measured on the surface of

Physics

1 answer:

garik1379 [7]3 years ago

4 0

Answer:

Explanation:

Temperature Scales

There are three temperature scales in the modern sciences: Fahrenheit, Celsius, and Kelvin. Fahrenheit temperature scale assigns the value 32 for the freezing point of water and 212 for the boiling point of water and divides that interval into 180 parts. Celsius scale has a similar reference, giving 0 to the freezing point of water and 100 for the boiling point of water. The conversion between them is as follows

$\displaystyle F=\frac{9}{5}C+32$

$K=C+273.15$

The coldest temperature yet measured on the surface of any body in the solar system is -235°C. Converting to Fahrenheit

$\displaystyle F=\frac{9}{5}(-235)+32=-391^oF$

$K=-235+273.15=38.15^oK$

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Help! PROJECTILE PROBLEM: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25

dsp73

Here,

height at failure, h1 = 525 m,
upward acceleration, a = 2.25 m/s^2,
velocity = v m/s,

SO, 

v^2 = 2*a*h = 2*2.25*525 = 2362.5 
Now, acceleration, g = 9.8 m/s^2,

SO, 

heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters 
Hence,

a) 
Total height = 525+120.54 = 645.54 meters

b)
time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec

---------------------------------

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4 0

4 years ago

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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.55 times a second. A tack is stuck in the tire a

earnstyle [38]

Answer:

Tangential speed = 5.72 m/s

Centripetal acceleration = $91.6\text{ m/s}{}^2$

Explanation:

The tangential speed, V, is given by

$v=\omega r$

where $\omega$ is the angular speed and is given by $2\pi f$ (f is the angular frequency or frequency of rotation)

Thus,

$v=2\pi f r = 2\times3.14\times2.55\times0.357 = 5.72\text{ m/s}$

The centripetal acceleration,a, is given by

$a=\dfrac{v^2}{r}$

$a=\dfrac{5.72^2}{0.357} = 91.6\text{ m/s}{}^2$

7 0

4 years ago

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If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

4 years ago

A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm

Sunny_sXe [5.5K]

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

$t=5\cdot 10^{-5}s$

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

$c=3.0\cdot 10^8 m/s$

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

$d=vt$ (1)

where

$v=c=3.0\cdot 10^8 m/s$

$t=5\cdot 10^{-5}s$

$d=2L$ , where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

$L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m$

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4 years ago

Which option is an example of chemical energy being transformed into kinetic

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D I know this because I know lol.

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3 years ago

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