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mrs_skeptik [129]
3 years ago
13

A snake slithered across the sidewalk in between two sign posts. At the first sign post his speed was

Physics
1 answer:
masha68 [24]3 years ago
5 0

This is an example of Uniform Speed

EXPLANATION

Uniform Speed => no change in speed

Uniform Accelerations Speed => There is a change in speed, but the acceleration remains the same.

#LearnWithEXO

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We have seen that when you generate a wave on a stretched spring, as long as the medium doesn’t change (that is, the tension and
TiliK225 [7]

Answer:

No, the acceleration is not always zero.

Explanation:

It does not mean that the acceleration of the particle is zero.

The velocity of wave is different from the velocity of particle.

The acceleration of wave is different from the acceleration of particle.

the acceleration of the particle is given by

a =- w^2 y

where, w is the angular frequency and y is the displacement from the mean position.

So, the acceleration is zero at mean position only and it varies as the position changes.  

6 0
3 years ago
The police department is excited to have some new motorcycle units. One officer said that these motorcycles can go from 0 miles
vaieri [72.5K]
The officer is describing the motorcycle's acceleration

That is, he's describing how the motorcycle's velocity changes when compared to the time. That's the formula for calculating the average acceleration of any body.
3 0
3 years ago
Tony drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Tony drove hom
polet [3.4K]

Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

For the second journey going home

Distance =y×(a+b)

Distance Mountain= distance home

ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

Distance=speed ×time

Distance= a×x=ax

Now, applying the questions

So from the questions

x=6hours, y=4hours

Also, b=22miles/hour

Then,

a=yb/(x-y)

a=4×22/(6-4)

a=88/2

a=44miles/hour

Then, the house distance from the mountain is

Distance=ax

Distance =44×6

Distance =264miles

4 0
4 years ago
Read 2 more answers
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
Which of the following has the most biomass?
Crazy boy [7]
Where are the questions so that I can deliver a more accurate answer. 
8 0
3 years ago
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