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3 years ago

As a rocket rises, its kinetic energy changes

Physics

2 answers:

djverab [1.8K]3 years ago

5 0

The answer is True
hope i helped

Bond [772]3 years ago

3 0

<span>Transformed into potential energy</span>

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Which best describes the energy of a sound wave as it travels through a medium

Solnce55 [7]

C. It depends on the medium

3 0

3 years ago

Yung's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a sc

Lady bird [3.3K]

Answer:

1.082 mm

Explanation:

From the question, we can see that we were given The following

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1.1 m

We know that Y(20) = 10.2 mm = 10.2*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1.1 * 502*10^-9)/10.2*10^-3

d = 1.1*10^-5 / 10.2*10^-3

d = 1.082 mm

Therefore, we can say that the distance of separation between the two slits is 1.082 mm

4 0

3 years ago

PLEASE HELP ASAP WILL REWARD BRAINLIEST:))

laiz [17]

Answer:

120s^-1

Explanation:

v=12v

I=10A

and since rate is with time, therefore rate=energy/time.

H=IV

10×12=120/s

therefore the rate is 120s^-1

6 0

1 year ago

A wire loop with 70 turns is formed into a square with sides of length ???? . The loop is in the presence of a 1.20 T uniform ma

Oksi-84 [34.3K]

Answer:

l= 3.002 cm

Explanation:

Given that

n= 70 turns

B= 1.2 T

θ= 15°

I= 1.5 A

τ = 0.0294 N⋅m

Lets take length of sides is l.

We know that

τ = n I A B sin θ

Area of square ,A= l²

Now by putting the value

τ = n I A B sin θ

0.0294 = 70 x 1.5 x l² x 1.2 x sin 15°

l² = 0.000901 m²

l² = 9.01 cm²

l= 3.002 cm

3 0

3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

1 year ago