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Marina CMI [18]
3 years ago
10

Which does not contain a lens?

Physics
1 answer:
denis23 [38]3 years ago
8 0
Im pretty sure it’s A eye
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Hydropower _____.
zheka24 [161]
A, the energy of moving water
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3 years ago
In Part I of this experiment, the______was intentionally manipulated. This was the independent variable. The dependent variable
Masja [62]

Answer:

1. car hitting a barrier 2. velosity of the washer 3. force applied to the car 4. acceleration of the car.

Explanation: edge 2020

why because nice

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3 years ago
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Teresa drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Teresa drove
Ksenya-84 [330]

Question:

Teresa drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took  12 hours. When Teresa drove home, there was no traffic and the trip only took 8 hours. If her average rate was  20 miles per hour faster on the trip home, how far away does Teresa live from the mountains?

Do not do any rounding.

Answer:

480 miles

That is Teresa lives 480 miles  from the mountains

Explanation:

Here we have

Distance D = Speed S × Time t

On her way to the mountain we have

D = S × 12

On her way back we have

D = (S +20)×8

Therefore, by comparing the two above equations, we have

S × 12 = (S +20)×8

Therefore, 12·S = 8·S +160 or

4·S = 160

Hence S = 160/4 = 40 miles/hour

Therefore, from D = S × 12, we have

D = 40 × 12 = 480 miles.

7 0
3 years ago
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An electric vehicle starts from rest and accelerates at a rate a1 in a straight line until it reaches a speed of v. The vehicle
levacccp [35]

(a) t=\frac{v}{a_1}+\frac{v}{a_2}

In the first part of the motion, the car accelerates at rate a_1, so the final velocity after a time t is:

v = u +a_1t

Since it starts from rest,

u = 0

So the previous equation is

v= a_1 t

So the time taken for this part of the motion is

t_1=\frac{v}{a_1} (1)

In the second part of the motion, the car decelerates at rate a_2, until it reaches a final velocity of v2 = 0. The equation for the velocity is now

v_2 = v - a_2 t

where v is the final velocity of the first part of the motion.

Re-arranging the equation,

t_2=\frac{v}{a_2} (2)

So the total time taken for the trip is

t=\frac{v}{a_1}+\frac{v}{a_2}

(b) d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}

In the first part of the motion, the distance travelled by the car is

d_1 = u t_1 + \frac{1}{2}a_1 t_1^2

Substituting u = 0 and t_1=\frac{v}{a_1} (1), we find

d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}

In the second part of the motion, the distance travelled is

d_2 = v t_2 - \frac{1}{2}a_2 t_2^2

Substituting t_2=\frac{v}{a_2} (2), we find

d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}

So the total distance travelled is

d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}

7 0
3 years ago
I WILL GIVE BRAINLIEST
lubasha [3.4K]

Answer:

The answer is

A.Speed

and

D. Acceleration

7 0
3 years ago
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