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larisa [96]
2 years ago
8

(a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere’s velocity as a function of tim

e between , when the sphere is launched and , when the sphere hits the target. Label for the horizontal component of the sphere’s velocity and the vertical component of the sphere’s velocity.

Physics
1 answer:
jeka942 years ago
4 0

Answer:

Two identical spheres are released from a device at time t = 0 from the same ... Sphere A has no initial velocity and falls straight down. ... (b) On the axes below, sketch and label a graph of the horizontal component of the velocity of sphere A and of sphere B as a function of time. ... Which ball has the greater vertical velocity

Explanation:

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You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % of its init
Lady_Fox [76]

Answer:

T_{f} = 85.89 ° C

Explanation:

The linear thermal expansion process is given by

      ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

     ΔV = V β ΔT

Let's apply this equation to our case

     ΔV / V = ​​-0.507% = -0.507 10-2

     ΔT = (ΔV / V)  1 /β

     ΔT = -0.507 10⁻²  1 / 1.15 10⁻³

     ΔT = -4.409

     T_{f} –T₀ = 4,409

     T_{f} = T₀ - 4,409

     T_{f} = 90.3-4409

     T_{f} = 85.89 ° C

6 0
2 years ago
Read 2 more answers
A sample of water is heated at a constant pressure of one atmosphere. Initially, the sample is ice at 260 K, and at the end the
USPshnik [31]

In <u>370 K to 375 K </u>temperature intervals of 5 K, would be the greatest increase in the entropy of the sample.

Option: C

<u>Explanation</u>:

Because the largest difference in molar entropy occurs when a condensed phase (solid/liquid) transforms to the gas phase. Then change in entropy is equal to heat transfer divided by temperature: \Delta \mathrm{S}=\frac{\Delta Q}{\mathrm{m} T}.

According to given ice sample at 260 K, when this solid sample start converting into liquid sample it will gain positive temperature and steam will take place near 373 K (273 K ice temperature + 100^{\circ} \mathrm{C} temperature of boiling water). Therefore it’s very obvious that greatest increase in entropy will occur during 370 K – 375 K.

5 0
2 years ago
A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of
son4ous [18]

Answer:

<h2>35</h2>

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

7 0
2 years ago
A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa
Alekssandra [29.7K]

Answer:

(a) 16.777mi

(b)Yes, he was speeding

Explanation:

(a)

Let's do the proper operations in order to convert km to mi:

27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi

We can conclude that the trip length in miles was:

d=16.77706389mi

(b)

Let's calculate the speed of the man during the trip:

v=\frac{d}{t}

But first, let's do the proper operations in order to convert min to h:

16min*\frac{1h}{60min} =2.666666667h

Now, the speed is:

v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}

As we can see:

62.91398959\frac{mi}{h}>55\frac{mi}{h}

So, we can conclude that the driver was speeding

8 0
2 years ago
The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?
Klio2033 [76]
The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:
n= \frac{c}{v}
where the speed of light in vacuum is c=3 \cdot 10^8 m/s. The speed of light in benzene is v=2.00 \cdot 10^8 m/s, so we can use the previous relationship to find the refractive index of benzene:
n= \frac{3 \cdot 10^8 m/s}{2.00 \cdot 10^8 m/s}=1.5
7 0
2 years ago
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