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3 years ago

How does surface currents would be affected if eArth didn't rotate

1 answer:

6 0

Surface ocean currents are generally wind-driven. However, the rotation of the Earth affects the way the waters move through currents. Without rotation, currents may not exist.

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A centripetal force of 5.0 newtons is applied to a rubber stopper moving at a constant speed in a horizontal circle. If the same

mezya [45]

Answer:

Both the frequency f and velocity v will increase.

When the radius reduces, the circumference of the circular path becomes smaller which means that more number of revolutions can be made per unit time as long as the force is kept constant; this is an increase in frequency.

Explanation:

The centripetal force acting on a mass in circular motion is given by equation (1);

$F_c=\frac{mv^2}{r}....................(1)$

where m is the mass of the object and r is radius of the circle. From equation one we see that the centripetal force is directly proportional to the square of the velocity and inversely proportional to the radius of the circular path.

However, according to the problem, the force is constant while the radius and the velocity changes. Therefore we can write the following equation;

$\frac{mv_1^2}{r_1}=\frac{mv_2^2}{r_2}......................(2)$

Also recall that m is constant so it cancels out from both sides of equation (2). Therefore from equation we can write the following;

$v_2=\sqrt{\frac{v_1^2r_2}{r_1}} .................(2)$

By observing equation (2) carefully, the ratio $\frac{r_2}{r_1}$ will with the square root increase $v_1$ since $r_2$ is lesser than $r_1$ .

Hence by implication, the value of $v_2$ will be greater than $v_1$ .

As the radius changes from $r_1$ to $r_2$ , the velocity also changes from $v_1$ to $v_2$ .

3 0

3 years ago

Which of these circuit schematics has an ammeter? Α. Α B. B C. C D. D

madam [21]

Answer:

d is the correct answer for this question hope it helps

Explanation:

when you see a A in a circle the the ammeter

3 0

3 years ago

Read 2 more answers

What is the potential energy of the bowling ball as it sits on top of a building ?

tangare [24]

Answer:

The potential energy of the bowling ball will be mgh

Explanation:

Let the mass of bowling ball =m

The height of building on which bowling ball sits=h

So,

The potential energy of the bowling ball =P.E.= mgh

8 0

3 years ago

A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base

-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g= $426\times 10^{-3} kg$

1 kg=1000 g

Radius,r= $\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m$

1m=100 cm

Height,h=5m

$I=\frac{2}{2}mr^2$

a.By law of conservation of energy

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh$

$\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh$

$v=\omega r$

$gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2$

$\omega^2=\frac{6}{5r^2}gh$

$\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s$

Where $g=9.8m/s^2$

b.Rotational kinetic energy= $\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J$

Rotational kinetic energy=8.35 J

6 0

3 years ago

When light falls on objects, it interacts with them in different ways as shown in the images. What type of interaction is seen i

agasfer [191]

Answers with explanation

Transmission

If all the light passes through a medium without any absorption then transmittance is 100%.

Refraction

Refraction is the bending of a wave when it enters a medium where its speed is different and rays are again refracted when they leave that medium,

Reflection

Reflected rays don't pass through the medium instead rays bounces off an object at an angle.

7 0

3 years ago