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likoan [24]
3 years ago
12

4. A 75 kg bobsled is pushed along a horizontal surface by two athletes. After the

Physics
1 answer:
vlada-n [284]3 years ago
8 0

The net force on the sled is 300 N

Explanation:

First of all, we start by finding the acceleration of the bobsled, by using the suvat equation:

v^2-u^2=2as

where:

v = 6.0 m/s is the final velocity of the sled

u = 0 is the initial velocity

a is the acceleration

s = 4.5 m is the displacement of the sled

Solving for a, we find

a=\frac{v^2-u^2}{2s}=\frac{6.0^2-0}{2(4.5)}=4 m/s^2

Now we can find the net force on the sled by using Newton's second law:

F = ma

where

F is the net force

m = 75 kg is the mass of the sled

a=4 m/s^2 is the acceleration

Solving the equation, we find the net force:

F=(75)(4)=300 N

Learn more about acceleration and Newton laws here:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

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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
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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

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

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