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Nadusha1986 [10]

2 years ago

7

I need help ASAP!!!! Here’s the question. Stephanie is going to buy an ice cream cone. She can pick a sugar cone or a waffle con

e. She can have vanilla, chocolate, or mint ice cream. How many possible combinations of ice cream cones can she have?

1 answer:

PolarNik [594]2 years ago

7 0

Answer:

Six

Explanation:

Assuming she can only have ONE flavor in her cone

2 cone types * 3 flavors = 6 combos

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What is the acceleration of a Ford Mustang GT that can go from 0.00 to 27.8<br> m/s in 5.15 seconds?

bixtya [17]

Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)

Finial speed= 27.8s
Initial speed= 0s
Time taken= 5.15

So..

27.8-0/5.15= 5.40m/s (rounded to two decimal places)

4 0

2 years ago

A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive i

klio [65]

Answer:

1.08

Explanation:

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer , so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. = λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08

5 0

3 years ago

How is velocity ratio of wheel and axle calculatedaad<br>

iogann1982 [59]

Answer:

VR = $\frac{Radius of wheel}{Radius of axle}$

Explanation:

A machine is a device that can be used to overcome a load by the application of an effort through a pivot. Examples are: pulleys, wedge, screw jack, wheel and axle etc.

The wheel and axle is a simple device that can be used to lift a load through a height. Its velocity ratio (VR) can be determined by:

VR = $\frac{Radius of wheel}{Radius of axle}$

Note that for a practical wheel and axle, the radius of the wheel is greater than the radius of the axle.

3 0

3 years ago

Three charges are enclosed inside a spherical closed surface. The net flux through the surface is −216 N · m2/C. If two of the c

sladkih [1.3K]

Answer:

q₃ = -4.81 nC

Explanation:

We can use the Gauss Law here:

∅ = q/∈₀

where,

∅ = Net Flux = - 216 N.m²/C

q = total charge enclosed inside sphere = ?

∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²

Therefore,

- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²

q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)

q = - 1.91 nC

So, the total charge will be sum of all three charges:

q = q₁ + q₂ + q₃

- 1.91 nC = 1.74 nC + 1.16 nC + q₃

q₃ = - 1.91 nC - 1.74 nC - 1.16 nC

<u>q₃ = -4.81 nC</u>

5 0

3 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

<em>∵mass is constant at speeds v << c (speed of light)</em>

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0

3 years ago