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ki77a [65]
3 years ago
12

A ball connected to a 1.1 m string and is swing in circular fashion. It’s tangential velocity is 15 m/s. What is its centripetal

acceleration?
Physics
1 answer:
musickatia [10]3 years ago
6 0

Answer:

ac = 204 [m/s²]

Explanation:

To solve this problem we must use the following equation that relates the tangential velocity to the radius of rotation.

ac = v²/r

where:

v = tangential velocity = 15 [m/s]

r = radius = 1.1 [m]

Now replacing we have:

ac = (15)²/1.1

ac = 204 [m/s²]

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A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,
Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem: 

s=1.80m when t=0.245s 

<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation): 

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2 

v=8.549m/s 

<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t) 

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like 

0=(v)+(a)(t) 

Substituting and solving for t: 

0=(8.549m/s)+(-9.81m/s^2)(t) 

t=0.8714s 

<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

3.725m – 1.8m=1.925m</span>

 

Answer:

<span>1.925m</span>

3 0
3 years ago
determine the loudness (in decibels) of the sound at a rock concert if the intensity of the sound is 1 x 10–1 w/m2. remember, th
EleoNora [17]

The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is  110 dB.

Here we are dealing with loudness which is the perception of the Intensity of the sound.

The formula  to refer to in order  to  find the value of the loudness of a sound is ,

  db= 10log(I/I₀)

As we are provided with the current intensity which is  1 x 10⁻¹ Wm⁻². and the initial intensity which is  1 x 10⁻¹² Wm⁻².

So, by substituting the required values in the formula we get

db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)

 = 10 * 11 log(10)

 = 110

So, the result is 110 dB.

To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.

To know more about questions related to loudness refer to the link brainly.com/question/21094511?referrer=searchResults.

#SPJ4

4 0
2 years ago
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
A stone has a volume of 0.50 cm³ and a mass of
Dafna1 [17]

Answer:

ρ = M / V = 2 g / .5 cm^3 = 4 g/cm^3

8 0
2 years ago
It took Lilly 7 seconds to ride her bike 35 meters to the end of the block.What was her speed?
Zolol [24]
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4 0
3 years ago
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