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3 years ago

18. A swimmer steps out of a swimming pool into a light breeze. Why does

1 answer:

OverLord2011 [107]3 years ago

6 0

When the swimmer gets out, the water on her skin begins to evaporate. This process requires takes a lot of energy and that energy comes from the heat of her body. This is similar to when someone sweats. Hope this helps!

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5 / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0

3 years ago

It is your parent's 25th Wedding Anniversary and you thought it would be special to throw them a surprise party. Your entire fam

KonstantinChe [14]

Answer:

120 miles per hour.

Explanation:

We need to find the time it takes my parents to drive home from the cottage. Since my father drives at 60 miles per hour, and the cottage is 240 miles from our home, and distance = speed × time. So, time = distance/speed = 240 mi/60 mi/h = 4 h.

So, it will take my father 4 hours to drive home from the cottage.

Since I have 2 hours to prepare for the party, the time left for me to drive to the cottage is 4 - 2 hrs = 2 hrs.

So, I'm supposed to drive to the cottage in at most 2 hours.

The speed at which I must drive in this time period is thus, speed = distance/time = 240 miles/2 hours = 120 miles per hour.

So, I must drive at a minimum speed of 120 miles per hour.

4 0

3 years ago

Appliances connected so that they form a single pathway for

Serga [27]

<h2>Question:</h2>

Appliances connected so that they form a single pathway for charges to flow are connected in a(n)

<h2>Answer:</h2>

<u>A</u><u>.</u><u> </u><u>Series</u><u> </u><u>circuit</u><u> </u>

<h2><u>#CARRYONLEARNING</u><u> </u></h2><h2><u>#STUDYWELL</u><u> </u></h2>

6 0

3 years ago

What is the force required to produce an acceleration of 46.4

Brilliant_brown [7]

That depends on the mass of the object, and the unit of the '46.4' .

If the '46.4' is ' meters per second² ' , then the force required is

(mass of the object in kilograms) x (46.4) newtons .

7 0

3 years ago