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Angelina_Jolie [31]
3 years ago
11

Which formula is used to find an object's acceleration?

Physics
1 answer:
andreyandreev [35.5K]3 years ago
7 0

Answer:

c

Explanation:

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A car has a mass of 1200 kg. A man can comfortably push down with a force of 400 N. If the small piston in a hydraulic jack has
kotegsom [21]
A) F1/A1=F2/A2
F2 = 12000 Newtons, while F1 = 400 Newtons and area A1= 6 cm²
400/6 = 12000/A2
A2 = (6 × 12000)/400
     =  180 cm²
 
b) How long does this lift the car
 d2 =F1/F2×d1
      = (400/12000)× 20
       = (1/30)×20
       = 20/30
       = 0.67 cm
This lifts the car a distance of 0.67 cm
        

4 0
3 years ago
A pure substance can be a
valina [46]
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4 years ago
How can we DESCRIBE force between two charged bodies
disa [49]

Answer:

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7 0
3 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
Q3. If each tape represents the distance travelled by the object for 1
xxMikexx [17]
No clue brother this one diff kit
5 0
3 years ago
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