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3 years ago

Earl _______ go to the game today.* cantcan'tcan 'ntcant'

Physics

2 answers:

Helen [10]3 years ago

7 0

Can't. It is the combining of the two words to short form them, it is a common thing in the English language

mariarad [96]3 years ago

4 0

Can't

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A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em

Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

$P = \ro g h = 1000 * 10 * 20 = 200000 Pa$

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

$P_1V_1 = P_2V_2$

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

$V_1 = Ah$

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

$P_1Ah_1 = P_2Ah_2$

$h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m$

b) If the temperatures changes, we can still reuse the ideal gas equation above:

$\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}$

$h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m$

3 0

3 years ago

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance s between the spokes along the rim is

$s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}$

The 20-cm arrow, traveling at 6 m/s, will travel its length in

$t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}$

The fastest speed that the wheel can spin without clipping the arrow is

$v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}$

The angular velocity $\omega$ of the wheel is given by

$\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}$

In terms of rev/s, we can convert the answer above as follows:

$23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}$

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

3 0

3 years ago

What is the state of y when y = k/x, x is halved.

tangare [24]

Answer:

y becomes doubled.

Explanation:

If;

y = $\frac{k}{x}$

what is the state of y when x is halved;

the given expression is an inverse relationship. When y increases, x is supposed to decrease and vice versa.

if x is halved; x = $\frac{x}{2}$

$\frac{k}{\frac{x}{2} }$ = $\frac{2k}{x}$

Now compare :

$\frac{k}{x}$ : $\frac{2k}{x}$

we see that y becomes doubled

8 0

3 years ago

What type of plate movement are formed between the Eurasian and Indian plates

stira [4]

A convergent plate movement happened, this movement also cause the Himalayan mountains, where Mt. Everest is located.

Hope this helps!

4 0

3 years ago

4. While traveling on a dirt road, the bottom of a car hits a sharp rock and a small hole develops at the bottom of its gas tank

Ede4ka [16]

Answer:

Initial velocity at hole = 2.426 m/s

Upon closure of lid, Velocity will change to: v2 = √[(2gz1) - 2(P2 - P1)/ρ]

Explanation:

This is a Bernoulli equation problem.

Now, let's assume that the outer and inner pressure of the gas tank are the same. Also, that the velocity of the fluid inside the tank v1 = 0 and the outside height z2 = 0 which will be the reference height.

The Bernoulli equation is;

P1 + ρ(v1)²/2 + ρgz1 = P2 + ρ(v2)²/2 + ρgz2

Putting v1 and z2 as 0, and P1 and P2 will cancel out as they are the same value, we have;

ρgz1 = ρ(v2)²/2

ρ will cancel out to give;

g•z1 = (v2)²/2

v2 = √2gz1

In the question, we are told that the height of the gasoline in the tank is 30 cm. So, z1 = 30cm = 0.3m

So,

v2 = √2 × 9.81 × 0.3

v2 = √5.886

v2 = 2.426 m/s

From the equation of v2 gotten earlier, we can see that the exit velocity depends on the height of the gas in the tank and that as the gas leaks out, the speed at which it is leaking out will decrease.

Now, if we close the lid of the tank, the inner and outer pressure would be not be the same as earlier assumed and so the velocity equation gotten earlier would change to;

v2 = √[(2gz1) - 2(P2 - P1)/ρ]

3 0

3 years ago