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4vir4ik [10]
3 years ago
7

Determine the stopping location of the prize wheel. At this moment it is centered on the number 11. It is spinning at a rate of

124.40 rpm. It is slowing at a rate of 1.400 rad/s/s.
Predict the time the wheel will remain spinning, the angular displacement it will go through, and then use this to predict the number on which the wheel will stop. To be safe, you will also get one number the right and to the left of the number you chose.
Physics
1 answer:
Goshia [24]3 years ago
5 0

Answer:

The wheel spins for 11.43s. The number cannot be determined.

Explanation:

We know that initial velocity is 124.4 rpm = 13.027137522 rad/s, acceleration is -1.14 rad/s^2, and final velocity is assumed to be 0 rad/s. We are asked to find t and displacement. We use the equation \omega=\omega_0+\alpha t where \omega=0 rad/s, \omega_0=13.027 rad/s, \alpha=-1.14rad/s^2, and t=?.

Rearrange the equation to obtain \frac{\omega-\omega_0}{\alpha}=t. Plug in the numbers and solve to obtain t=11.43s.

The number of the wheel cannot be determined as we do not know the placement of numbers on the wheel.

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A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘
hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of  mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution,T_f = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC

\Delta T_f=i\times K_f\times m

Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}

5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})

On solving we get:

\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of Ca(NO_3)_2 in the mixture:

\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%

The mass percentage of calcium nitrate is 31.23%.

5 0
3 years ago
The eiffel tower has a mass of 7.3 million kilograms and a height of 324 meters. its base is square with a side length of 125 me
uranmaximum [27]

Since the tower base is square with a side length of  125 m,

Therefore,

(125\ m)^2+ (125\ m)^2=31250 m^2

Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.

The volume of cylinder,

=\pi r^2h=3.14(88.38834765 m)^2\times 324 m =7948168.803\ m^3

Thus, the mass of the air in the cylinder,

=1.225\ kg/m^3 \times 7948168.803\ m^3=9736506.78\ kg

Hence, the mass of the air in the cylinder is this more  than the mass of the tower.


4 0
2 years ago
How do solutions differ from other mixtures?
zalisa [80]
The answer is D.
I hope this helps.
7 0
3 years ago
Read 2 more answers
A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz
uranmaximum [27]

Answer:

1) λ = 0.413 m , 2)v = 25,213 m / s , 3)  T = 0.216 N , 4) m = 22.04 10-3 kg

Explanation:

1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related

         λ = 2L / n               n = 1, 2, 3 ...

In this case L = 0.62 m and n = 3

Let's calculate

        λ = 2 0.62 / 3

        λ = 0.413 m

2) the velocity related to wavelength and frequency

      v =  λ f

      v = 0.413 61

      v = 25,213 m / s

3) let's use the equation

     v = √T /μ

     T = v² μ

     T = 25,213² 3.4 10⁻⁴

     T = 0.216 N

4) the rope tension is proportional to the hanging weight

      T-W = 0

     T = W

    W = m g

    m = W / g

    m = 0.216 / 9.8

    m = 22.04 10-3 kg

5) n = 2

     λ = 2 0.62 / 2

     λ = 0.62 m

6) v =  λ f

     v = 0.62 61

     v = 37.82 m / s

7) T = v² μ

   T = 37.82² 3.4 10⁻⁴

   T = 0.486 N

8) m = W / g

   m = 0.486 / 9.8

   m = 49.62 10⁻³ kg

9) n = 1

    λ = 2 0.62

    λ = 1.24 m

    v = 1.24 61

    v = 75.64 m / s

    T = v² miu

    T = 75.64² 3.4 10⁻⁴

    T = 2.572 10⁻² N

    m = 2.572 10⁻² / 9.8

    m = 262.4 10⁻³ kg

5 0
3 years ago
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
uranmaximum [27]

Answer:

fast. i have spoken

Explanation:

3 0
3 years ago
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