Ask question

Ask question

All categories

3 years ago

8

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward.

If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

1 answer:

fgiga [73]3 years ago

4 0

complete question:

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answer:

F = 1776 N

Explanation:

mass of ball = 60 g = 0.06 kg

velocity of downward direction = 22 m/s = v1

velocity of upward direction = 15 m/s = v2

Δt = 1/800 = 0.00125 s

Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

p = mv

When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .

I = pf − pi = ∆p

F = ∆p/∆t = I/∆t

let the upward velocity be the positive

Δp = mv2 - m(-v1)

Δp = mv2 - m(-v1)

Δp = m (v2 + v1)

Δp = 0.06( 15 + 22)

Δp = 0.06(37)

Δp = 2.22 kg m/s

∆t = 0.00125

F = ∆p/∆t

F = 2.22/0.00125

F = 1776 N

You might be interested in

A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w

myrzilka [38]

The centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, $a_c = 15.625 m/s^2$ and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
$v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s$

8 0

3 years ago

A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p

frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

$a(t)=(5-10t)\ m/s^2$

(a) Since, $a=\dfrac{dv}{dt}$

v = velocity

$dv=a.dt$

$v=\int(a.dt)$

$v=\int(5-10t)(dt)$

$v=5t-\dfrac{10t^2}{2}=5t-5t^2$

(b) $v=\dfrac{dx}{dt}$

x = position

$x=\int v.dt$

$x=\int (5t-5t^2)dt$

$x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3$

(c) Velocity function is given by :

$v=5t-5t^2$

$5t-5t^2=0$

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0

3 years ago

Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force

xeze [42]

Coulomb's law:

Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

= (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C) / (1.0 m)²

= (8.99×10⁹ x 1×10⁻¹² / 1.0) N

=    8.99×10⁻³ N

=    0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses'; don't need it.
   Mass has no effect on the electric force between them.

-- 'frictionless table'; don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>

7 0

3 years ago

Justin depends 600watts if power pushing a car 10m in 5sec how much force did he exert

Natasha_Volkova [10]

Force exerted by Justin=300 N

Here power= 600 W

distance traveled=10 m

time=5 s

power is given by P= Work done/ time

600=work/5

so work= 60x5=3000J

now work done= force* distance

3000=F *10

F= 3000/10

F=300 N

5 0

4 years ago

An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har

sineoko [7]

Answer:

Acceleration will be equal to $2m/sec^2$

Explanation:

We have given mass of the object m = 0.4 kg

Spring constant k = 8 N/m

Maximum displacement of the spring is given x = 0.1 m

From newton's law force is equal to $F=ma$ .....eqn 1

By hook's law spring force is equal to $F=kx$ .....eqn 2

From equation 1 and equation 2

$ma=kx$

$0.4\times a=8\times 0.1$

$a=2m/sec^2$

So acceleration will be equal to $2m/sec^2$

8 0

3 years ago