Which energy source is one that is used to boil water to make steam in power stations

If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s

Sladkaya [172]

the orbital period is 5170 s

6 0

3 years ago

A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a

chubhunter [2.5K]

Answer:

The ballon will brust at

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Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

$R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}$

Rmax = 10.001 cm

Therefore, the max volume will be:

$V_{max} = \frac{4}{3} \pi R_{max}^3$

Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

$P_{max} = P_1 \frac{V_1T_2}{V_2T_1}$

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0

3 years ago

At an instant a traffic light turns green an automobile that has been waiting at an intersection of the road accelerates with 5m

joja [24]

1) The car overtakes the truck at a distance of 160 m far from the intersection.

2) The velocity of the car is 40 m/s

Explanation:

1)

The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by

$x_c(t) = \frac{1}{2}at^2$

where

$a=5 m/s^2$ is the acceleration

t is the time

On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by

$x_t(t) = vt$

where

v = 20 m/s is the velocity of the truck

t is the time

The car overtakes the truck when the two positions are the same, so when

$x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s$

So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is

$x_c(8) = \frac{1}{2}(5)(8)^2=160 m$

So, the car overtakes the truck 160 m far from the intersection.

2)

The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation

$v=u+at$

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

For the car in this problem, we have:

u = 0 (it starts from rest)

$a=5 m/s^2$

And we know that the car overtakes the truck when

t = 8 s

Substituting into the equation,

$v=0+(5)(8)=40 m/s$

Learn more about accelerated motion:

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#LearnwithBrainly

8 0

2 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

x = ( 0 - 15²) / 2 (-3.14)

x= 35.8 m

The shortest braking distance is 35.8 m

7 0

3 years ago

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{c}{f}$

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

7 0

3 years ago