What's the answer, remember it says no friction

Skateboard A Force: 33 N Time: 0. 12 s Skateboard B Force: 33 N Time: 0. 57 s Skateboard C Force: 33 N Time: 0. 78 s Which skate

Snowcat [4.5K]

The skateboard that has the greatest impulse is Skateboard C (Impulse = 25.74 Ns)

From the question,

We are to determine which skateboard has the greatest impulse. To do this, we will calculate the impulse of each skate board.

Impulse can be calculated by using the formula

I = Ft

Where I is the impulse

F is the force

and t is the time

For Skateboard A

Force = 33 N

Time = 0. 12 s

∴ Impulse = 33 × 0.12

Impulse = 3.96 Ns

For Skateboard B

Force = 33 N

Time = 0. 57 s

∴ Impulse = 33 × 0.57

Impulse = 18.81 Ns

For Skateboard C

Force = 33 N

Time = 0. 78 s

∴ Impulse = 33 × 0.78

Impulse = 25.74 Ns

Hence, the skateboard that has the greatest impulse is Skateboard C (Impulse = 25.74 Ns)

Learn more here: brainly.com/question/21840495

7 0

3 years ago

PLEASE HELP ME VERY IMPORTANT

saul85 [17]

Answer: let's see, so it's a 1/4 chance of getting it right, hope your odds are good, have a wonderful day

Explanation:

:D

6 0

3 years ago

How much heat is required to convert 0.3 kg of ice at 0°C to water at the same temperature

madreJ [45]

Answer:

100,800 Jkg

The heat that is used to change the state of a mater is called latent heat.

In this case it is converting ice to water and it is called latent heat of fusion.

It is given by:

Heat = mc

where m is the mass of ice and l is the specific latent heat of fusion of ice.

l = 0.336 MJ

Heat = 0.3 × 0.336 MJ

= 0.3 × 0.336 × 10⁶

= 100,800 Jkg

6 0

4 years ago

Improvements in muscular strength will not affect muscular endurance.

olga2289 [7]

Answer:

the answer for your question is false ( F )

6 0

3 years ago

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

So

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

$\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

$\Delta \eta = 647 \pi$

$\Delta \eta = 2032.9$

4 0

4 years ago