A ball is dropped from a window sill and hits the ground in 4.1s. What is the final velocity?

Why weigh is not constant planet to planet?

ahrayia [7]

Gravitational pull changes between each planet while its the "MASS" which doesnt change

3 0

3 years ago

What is the force of the drag for a 65 kg bicyclist, initially at rest at the top of a hill coasts down the hill, reaching a spe

Fofino [41]

Who knows, not me, not me and not me again

4 0

2 years ago

Urgent!!! 50 points to who answers this question in a clear and simple explanation :

lina2011 [118]

You only need to use the right two simple formulas:

Work = (force) x (distance)

-- A mass that weighs 500N is being pulled down by gravity with a force of 500N. (That's what "weighs" means.) If you want to lift it straight up against gravity, you have to lift with a force of 500N.

The WORK you do on the mass is (500N) x (height you raise it to).

Power = (work) / (time)

-- 4,000 watts = (500N x Height) / 20 seconds

From here, the rest is just algebra ... pulling the height out of this equation:

Multiply each side by (20 sec):

(500N x height) = (4,000 watts x 20 sec)

Divide each side by (500N):

Height = (4,000 watts x 20 sec) / (500N)

Height = (4,000 x 20 / 500) (watt x second / Newton)

(Remember that "watt-second" = "Joule",

and "Joule" = "Newton-meter".)

<em>Height = 160 meters</em>

4 0

3 years ago

How much work do you do when you lift a child with 50N of force .5m?

Anettt [7]

Answer:

25 joules

Explanation:

Work Done = Force × Distance

Force = 50N

Distance = 0.5m

$50 \times 0.5 \\ = 25$

Therefore, the amount of work done is 25Nm which is 25 joules

7 0

3 years ago

A 95 kg fullback is running at 3.0 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west. C

STatiana [176]

(a) Original momentum of the fullback: 285 kg m/s

The original momentum of the fullback is equal to the product between his velocity and its mass:

$p_i=mv$

where:

m = 95 kg is the mass

v = 3.0 m/s is the velocity

Substituting the numbers into the formula, we find

$p_i=(95 kg)(3.0 m/s)=285 kg m/s$

(b) Impulse exerted on the fullback: -285 kg m/s

The impulse exerted on the fullback is equal to his variation of momentum:

$I=\Delta p=p_f -p_i$

where:

$p_f = 0$ is the final momentum of the fullback (zero because he comes to a stop)

$p_i = 285 kg m/s$ is the initial momentum

Substituting,

$I=0-(285 kg m/s)=-285 kg m/s$

(c) Impulse exerted on the tackler: 285 kg m/s

The total momentum of the fullback and the tackler must be conserved:

$p_i + P_i = p_f + P_f$

where $p_i$ and $p_f$ are the initial and final momentum of the fullback, while $P_i, P_f$ are the initial and final momentum of the tackler.

We can re-arrange the equation as follows:

$P_f - P_i = p_i -p_f\\\Delta p_{tackler} = -\Delta p_{fullback}\\I_{tackler} = -I_{fullback}$

which means that the impulse exerted on the tackler is the negative of the impulse exerted on the fullback, so:

$I=-(-285 kg m/s)=285 kg m/s$

(d) Average force exerted on the tackler: 335.3 N

The impulse on the tackler is equal to the product between the average force and the time of the collision:

$I=F \Delta t$

Since $\Delta t=0.85 s$ , we can find the average force:

$F=\frac{I}{\Delta t}=\frac{285 kg m/s}{0.85 s}=335.3 N$

5 0

4 years ago