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dezoksy [38]
3 years ago
12

Caleb is swinging Rachel in a circle with a centripetal force of 533 N. If the radius of the circle is 0.75 m and Rachel has a m

ass of 16.4 kg, how fast is Rachel moving? Round to the nearest tenth. m/s
Physics
2 answers:
Maslowich3 years ago
7 0

Answer:

4.9

Explanation:

I just did it

Alika [10]3 years ago
6 0

Velocity = √((Centrifugal force *  radius)/mass)

Velocity = √(533 * 0.75 / 16.4)

Velocity = √(399.75 / 16.4)

Velocity = √24.375

Velocity = 4.9 m/s

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Can someone help me a bit on this? Will mark brainliest. ( no physical science option soooo)
Advocard [28]

Answer:

When a light wave goes through a slit, it is diffracted, which means the slit opening acts as a new source of waves. How much a light wave diffracts<em> (how much it fans out)</em> depends on the wavelength of the incident light. The wavelength must be larger than the width of the slit for the maximum diffraction. Thus, for a given slit, red light, because it has a longer wavelength, diffracts more than the blue light.

The corresponding relation for diffraction is

d sin(\theta) \approx \lambda,

where \lambda is the wavelength of light, d is the slit width, and \theta is the diffraction angle.

From this relation we clearly see that the diffraction angle \theta is directly proportional to the wavelength  \lambda of light—longer the wavelength larger the diffraction angle.

7 0
3 years ago
The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters
skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

<h2>ΔT = 17.11 °C</h2>

Hope this helps

5 0
2 years ago
If you know what a JW is LET ME KNOW and it is a type of ppl
lisabon 2012 [21]

Answer:

only thing I think of when I see that is 'Just Wondering'

Explanation:

3 0
2 years ago
Read 2 more answers
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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2 years ago
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How does the size of a planet affected the amount and type of gas in its atmosphere
dalvyx [7]
There are lots of variables that directly and indirectly contribute to the presence of gas on a surface
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