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Alex_Xolod [135]
3 years ago
14

A skateboarder is traveling at a velocity of 8 m/s. She slows down and comes to a stop in 4 seconds. Calculate the skater’s acce

leration.
Physics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

Go to this link it has many examples - https://staff.retsd.mb.ca/public/or4xe53elrzxg5ldmjuwe3i/Lists/SharedDocuments/Physics/14.2.PDF

Explanation:

Link ^

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Why are experiments often performed in laboratories?
Marrrta [24]

Answer:

B

Explanation:

Laboratories are usually controlled: however, in public places, there are way too many factors that could change an experiment

6 0
3 years ago
A device for measuring atmospheric pressure is a _____. thermometer manometer barometer seismometer
vodomira [7]

A device used to measure atmospheric pressure is a barometer.

8 0
3 years ago
Read 2 more answers
A potter’s wheel of radius 17 cm starts from rest and rotates with constant angular acceleration until at the end of 32 s it is
dedylja [7]

Answer:

α=0.625rad/s^2

v=340m/s

w=10rad/s

θ=320rad

Explanation:

Constant angular acceleration = ∆w/∆t

angular acceleration = 20/32

α=0.625rad/s^2

Linear velocity v=wr

v = 20×17= 340m/s

Average angular velocity

w0+w1/2

w= 0+20/2

w= 20/2

w=10rad/s

What angle did it rotate with

θ=wt

θ= 10×32

=320rad

4 0
3 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
How would you define compound
aliina [53]

Answer:

as a way to put two things together and yeeah

Explanation:

i know because this is what my mom told me

5 0
3 years ago
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