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NemiM [27]
3 years ago
10

Pls help.No random links pls.

Physics
2 answers:
Elden [556K]3 years ago
8 0

Answer:

latitude my guy

Explanation:

north south

jek_recluse [69]3 years ago
6 0

Answer:

I believe its the second oneee

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What of the following is the mathematical form of boyles law?
viva [34]

Answer:

C

Explanation:

V=1/p

By means of cross multiplication so by that we will have pv=1 which also implies p1v1=p2v2 coz boyles law states that the volume of a given mass of gas is inversely proportional to pressure provided that the temperature in kelvin remains constant

7 0
3 years ago
A loaded 500 kg sled is traveling on smooth horizontal snow at 5 m/s when it suddenly comes to a rough region. The region is 10
sergejj [24]

Answer:

400 N

Explanation:

Change of Kinetic Energy to Friction Wok

∆KE = W

½ x m x (v(5)² - v(3)²) = f x d

½ x 500 x (5² - 3²) = f x 10

250 x (25 - 9) = f x 10

25 x 16 = f

f = 400 N

6 0
3 years ago
The law that states that the farther away a galaxy is, the faster it is moving away from us is called ______.
lorasvet [3.4K]

Answer: Hubble's Law

Explanation: Hubble's law that states that the farther away a galaxy is, the faster it is moving away from us

5 0
1 year 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
A 6.00-kg crate slides down a ramp from rest. The ramp is 1.00 m in length and inclined 30.0° above the horizontal. The crate ex
Maslowich

Answer: a) V = 9.81 m/a

b) S = 3.905m

c) V2 = 8.29m/s

d) Yes. The speed reduces.

Explanation:

Please find the attached files for the solution

4 0
3 years ago
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