Lightning heats the surrounding air to

How far can your little brother get if he can travel at 2.5 m/s and in 5?

KatRina [158]

d=? v=2.5 u=0 and t=5 therefore the formula to be used to find the distance my brother covered is d=1/2(v-u)t

d=1/2(2.5-0)5

=6.15m

4 0

2 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

$r'=\frac{1}{4}r$

So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0

3 years ago

Explain how water circulates from deep oceans to the surface.

Alexxandr [17]

"These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation. In the Earth's polar regions ocean watergets very cold, forming sea ice." this is what I found...

4 0

3 years ago

A car with mass m traveling at speed v has kinetic energy k. what is the kinetic energy of a second car that has the same mass m

vampirchik [111]

Kinetic energy, KE, is modeled by the formula $KE = \frac{1}{2}mv^2$ , where m is the mass in kg and v is the velocity in m/s.

In this scenario, mass and one-half are constant but the velocity changes.

You can see that by squaring twice the velocity, that is equal to four times the original KE. Therefore, the answer is 4k.

7 0

3 years ago

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago