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Advocard [28]
3 years ago
10

Newton’s Second Law establishes the relationship between mass, net applied force, and acceleration given by F=ma. Consider a 4 k

g box of holiday candy on a horizontal surface such as a table. There is a 10N applied force to the right and a 2N force to the left. The box accelerates with a magnitude of _________?
Physics
1 answer:
netineya [11]3 years ago
7 0

Answer:

a= 2 m/s^2

Explanation:

take to the right as positive

let Ftot be the total forces acting on the box , m be the mass of the box and a be the acceleration of the box.

Ftot = 10 - 2 = 8 N

and,

Ftot = ma

   a = Ftot/m

      = 8/4

      = 2 m/s^2

therefore, the acceleration of the box is of magnitude of 2 m/s^2.

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Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

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Explanation:

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