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sergij07 [2.7K]
3 years ago
5

In concave lenses a Distance object appears ​

Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

it appears to be farther away than it actually is, and therefore smaller then the object itself.

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SELECT ALL THE CORRECT ANSWERS
Aleonysh [2.5K]
I think the answer is x
3 0
3 years ago
A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. Whe
Allisa [31]

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N  weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

8 0
3 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
3 years ago
Why you do scientists collect multiple trials?
sergiy2304 [10]

"When we do experiments it's a good idea to do multiple trials, that is, do the same experiment lots of times. When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time."

6 0
3 years ago
How long would it take to go 11 miles at 22 mph?
harkovskaia [24]
It would tack about 3.2 h

5 0
3 years ago
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