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GrogVix [38]
3 years ago
11

5. Peter had an inflated balloon that he released and flies across the room. The balloon slows down and then stops on top of the

dinning table, As the balloon slows downs the force becomes
A.balanced
B.frictional
C.restricted
D.unbalanced​
Physics
2 answers:
Fiesta28 [93]3 years ago
7 0
Balanced as it reaches terminal velocity
Sergio039 [100]3 years ago
5 0
I think it is c because it was slowing down I don’t know for sure
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Electrons are continuously being knocked out of air molecules by cosmic ray particles from space. Once the electrons are free, t
nata0808 [166]

(a) 1.25\cdot 10^{-14} J

The change in potential energy of the electron is given by:

\Delta U=q E d

where

q=1.6\cdot 10^{-19}C is the magnitude of the electron's charge

E=150 N/C is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

\Delta U=(1.6\cdot 10^{-19}C)(150 N/C)(520 m)=1.25\cdot 10^{-14} J

(b) 78 kV

The potential difference the electron has moved through is given by

\Delta V=Ed

where

E=150 N/C is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

\Delta V=Ed=(150 N/C)(520 m)=78,000 V=78 kV

4 0
3 years ago
12. A cross-country race car driver sets out on a 1.00 hour, 100.0 km race. At the halfway marker
Akimi4 [234]

Answer:

A.) Time = 0.625 hrs = 37.5 minutes

B.) Speed required = 133.33km/hr

Explanation:

Given the following :

Total race distance = 100km

Total time required = 1 hour = 60 minutes

Average speed after traveling 50km = 80km/hr

80km/hr : This means it will take one hour to cover a distance of 80km

Therefore, time taken to cove first 50km at that average speed equals :

1 hour = 80 km

t hours = 50km

80t = 50

t = 50/80

t = 5/8 hours

t = 0.625hours

t = 0.625 * 60 = 37.5minutes

B)

Average speed required to complete the race in 1 hour = 100km/hr

Time used to complete first 50km = 0.625 hour

Time remaining: (1 - 0.625) hour = 0.375

Speed required = Distance left / time left

Speed required = 50 / 0.375

Speed required = 133.33km/hr

8 0
3 years ago
a 10 kg solid disk of radius 0.5 m is rotated about an axis through its center.the disk accelerates from rest to angular speed o
mezya [45]

Answer:

τ ≈ 0.90 N•m

F =

Explanation:

I = ½mR² = ½(10)0.5² = 1.25 kg•m²

α = ω²/2θ = 3.0² / 4π = 0.716... rad/s²

τ = Iα = 1.25(0.716) = 0.8952... ≈ 0.90 N•m

τ = FR

Now we have the unanswered question of reference frame.

80° from what?

If it's 80° from the radial

F = τ/Rsinθ = 0.90/0.5sin80 = 1.818...  ≈ 1.8 N

If it's If it's 80° from the tangential

F = τ/Rcosθ = 0.90/0.5cos80 = 10.311...  ≈ 10 N

There are an infinite number of other potential solutions

7 0
3 years ago
Image caught<br>on Screen is called​
bixtya [17]

Answer:

<h2>Virtual image</h2>

Explanation:

<h3><em>Virtual</em><em> </em><em>image</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>caught</em><em> </em><em>on</em><em> </em><em>a</em><em> </em><em>screen</em></h3>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em>

<em>will</em><em> </em><em>give</em><em> </em><em>the</em><em> </em><em>brainliest</em><em>!</em>

<em>follow</em><em> </em><em>~</em><em>H</em><em>i</em><em>1</em><em>3</em><em>1</em><em>5</em><em>~</em>

4 0
3 years ago
Read 2 more answers
So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

in this case we'll use the average radius of earch which is 6,371 km, so we get:

r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

a=24.07 mm/s^{2}

5 0
3 years ago
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