All categories

3 years ago

If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?

Physics

2 answers:

Alekssandra [29.7K]3 years ago

8 0

Answer:

2 meters per second²

Explanation:

PilotLPTM [1.2K]3 years ago

7 0

Answer:2m/s^2

Explanation:

mass=0.25kg

Force =0.5N

Acceleration=force/ mass

Acceleration=0.5/0.25

Acceleration=2

Acceleration =2m/s^2

You might be interested in

What is in between the nucleus and the electrons in an atom?

hoa [83]

Answer:

D. Empty Space

Explanation:

4 0

3 years ago

A campus bird spots a member of an opposing football team in an amusement park. The football player is on a ride where he goes a

kenny6666 [7]

Answer:

No..

Explanation:

As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.

As the person is on circular arc constantly moving it will never meet that drop.

6 0

3 years ago

Blood is tissue .how?

Levart [38]

Yes, blood is a tissue. It is a tissue because it is a group of similar cells that have functions.

3 0

2 years ago

Read 2 more answers

A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

$v^2=u^2+2as$

$0^2=14.5^2+2\times a\times 18.25$

$a=-5.76m/sec^2$

We know that acceleration is given by

$a=\mu g$

$5.76=\mu\times 9.81$

$\mu =0.587$

So coefficient of friction will be 0.587

6 0

3 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

slega [8]

We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
$dI =r^2*dm$
where $dm =M(b*dr)/(ab)$
Now we find the moment of inertia by integrating from $-a/2$ to $a/2$
The moment of inertia is
$I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3)$ (from (-a/2) to $I=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12$ (a/2))

4 0

3 years ago