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Gennadij [26K]
3 years ago
14

Consider two soap bubbles having radii r1 and r2 (r1 < r2) connected via a valve. What happens if we open the valve?

Physics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

Explanation:

The radius of the smaller bubble, r1 will decrease and that of the bigger bubble, r2 will increase.

The pressure that is present in the smaller bubble usually is greater than the pressure that exists inside that of the bigger bubble. This then makes air to flow from r1 to r2 thereby making the radius of the smaller bubble r1, to decrease while keeping that of the bigger bubble r2 higher.

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Compute the dot product of the vectors u and v​, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and
OLga [1]

Answer:

\theta = 106.3 degree

Explanation:

As we know that

\vec w = -\hat i + 7\hat j

\vec v = 7\hat i - \hat j

also we know that

\vec v. \vec w = -14

it is given as

\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)

\vec v. \vec w = - 7 - 7 = -14

also we can find the magnitude of two vectors as

|v| = \sqrt{(-1)^2 + (7)^2}

|v| = \sqrt{50}

similarly we have

|w| = \sqrt{(7^2) + (-1)^2}

|w| = \sqrt{50}

now we know the formula of dot product as

\vec v. \vec w = |v||w| cos\theta

-14 = (\sqrt{50})^2cos\theta

\theta = cos^{-1}(\frac{-14}{50})

\theta = 106.3 degree

3 0
3 years ago
What is it called the number that tells how much of something there is
Dvinal [7]
A cardinal númber is a number that says how many of something there are such a one , two , three , four , five
4 0
3 years ago
Define SI unit of force. A force of 8N acting on a body changes its velocity uniformly from 2 m/s to 5    m/s in 20s. Calculate
laila [671]

Answer:

si unit of force is Newton or dyne

6 0
1 year ago
A mass m at the end of a spring vibrates with a frequency
Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

6 0
3 years ago
A ball is launched straight up with initial speed of 30.0 m/s. What is the ball's velocity when it comes back to its original po
Zanzabum

Answer:

We could get the time taken by the ball to return back to earth, using the formula:

s = u t + ½ a t², where

s = displacement of the body moving with initial velocity u, acceleration 'a' in time t.

In the present case s=0 (as the ball returns back to starting time)

u= 30 m/s; a = -10 m/s² ( negative sign as a is in opposite direction to u); t=?

0 = 30 t - ½ ×10 ×t²; ==> 5 t = 30, t= 6 second.

So ball will return back after 6 second after being thrown up.

Explanation:

I looked it up

Hope this helps

3 0
3 years ago
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