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jolli1 [7]
3 years ago
10

A car, initially traveling 28.0ft/s, steadily speeds up to 44.0ft/s in 4.50s. Determine all unknowns and answer the following qu

estion.
How far did the car travel during this time?
Physics
1 answer:
dexar [7]3 years ago
5 0

Explanation:

hi do u mind helping me with something lmak

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When water in a lake freezes, the ice that forms floats on top of any water that is still liquid. Why does the ice float?
ddd [48]

Answer:

it floats because the lake is cold at that moment so when part of the lake freezes it still remains solid and floats because of the lake and the surrounding of the lake is still cold

5 0
3 years ago
If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these
insens350 [35]

To develop this problem it is necessary to apply the concepts related to Sound Intensity.

By definition the intensity is given by the equation

\beta = 10Log(\frac{I}{I_0})

Where,

I = Intensity of Sound

I_0= Intensity of Reference

At this case we have that 15 engines produces 15 times the reference intensity, that is

I= 15I_0

And the total mutual intensity is 100 dB, so we should

\beta = 100-10*log(\frac{15I_0}{I_0})

\beta = 100-10*log(15)

\beta = 100-11.76

\beta = 88.23dB

Therefore each one of these engines produce D. 88dB.

5 0
3 years ago
If a car is traveling at 35 mph how far will the car travel in 2.5 hours
Vanyuwa [196]

Answer:

87.5 miles

Explanation:

d = s * t

d = 35 * 2.5

d =87.5

4 0
3 years ago
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

4 0
3 years ago
Many people love roller coasters. Some people call them scream machines. They agree that
Alex73 [517]

Answer:

There is some sort of "rollers" on the trail part that is used to make the roller coaster move when it is going up. Then because of that, the speed gets quicker when it drops down, so then it could move for a longer time until it reaches the next arch.

Explanation:

hope this helps :)

4 0
2 years ago
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