In an elastic collision, _______ energy is conserved.

Sandy is whirling a ball attached to a string in a horizontal circle over his head. If Sandy doubles the speed of the ball, what

jeka57 [31]

The tension in the string B) It quadruples.

Explanation:

The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:

$T= m\frac{v^2}{r}$

where:

T is the tension in the string

m is the mass of the ball

v is the speed of the ball

r is the radius of the circle (the lenght of the string)

In this problem, we are told that the speed of the ball is doubled, so

v' = 2v

Substituting into the previous equation, we find the new tension in the string:

$T' = m \frac{(2v)^2}{r}=4(m\frac{v^2}{r})=4T$

Therefore, the tension in the string will quadruple.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

6 0

3 years ago

What type of electromagnetic wave does a light bulb and a radio antenna release?

andrew-mc [135]

I believe that a light bulb releases visible light and a radio antenna releases a radio waves

4 0

3 years ago

The normal force which the path exerts on a particle is always perpendicular to the _________________ tangent to the path. trans

Marianna [84]

Tangent to the pathh.

7 0

3 years ago

What Converts sound to a signal in a telephone

Irina-Kira [14]

The part you talk into, that converts the sound of your voice
into an electrical signal, is a tiny microphone.

-- The sound waves from your voice are ripples in the air.

-- In most microphones, there's a tiny coil of wire hanging
between the ends of a tiny magnet.

-- When the ripples in the air hit the little coil of wire, they
make it vibrate (wiggle) slightly.

-- When a coil of wire wiggles in the field of a magnet,
a current flows in the wire.

There's your electrical signal !

6 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