Pls,help ASAP .the question is in attachment

During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h

QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

3 0

3 years ago

Atoms that have a positive charge will be attracted to atoms with a

kap26 [50]

Answer:

negative

Explanation:

positive charges attract negative charges and vice versa. and are possible to nullify

4 0

2 years ago

If you were standing at the center of curvature in front of a concave mirror, what image would be projected?

matrenka [14]

"<span>The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.</span>

6 0

2 years ago

Read 2 more answers

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$