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MA_775_DIABLO [31]
3 years ago
10

Describe the motion represented by a horizontal line on a distance time graph

Physics
2 answers:
AfilCa [17]3 years ago
8 0

A horizontal line on a distance time graph is saying that the distance doesn't change as time goes on.  This tells us that the object's speed is zero, and its velocity is zero, which pretty much says that the object is stationary ("at rest").

MissTica3 years ago
6 0
A horizontal line on a distance-time graph indicates that an object is stationary.
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Can you explain this a bit more I don’t quite understand
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3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame
sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0
3 years ago
An open system starts with 52 J of mechanical energy. The energy changes
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5 0
2 years ago
A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision
Nimfa-mama [501]

Answer:

Δ KE =  249158.6 kJ  

Explanation:

given data

Truck mass  M =  1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum  that is express as

Mu = (M+m) V     .......................1

put here  value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy  will be here

Δ KE =   \frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2         .................2

put here value and we get

 Δ KE = \frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2  

solve it we get

Δ KE =  249158.6 kJ  

6 0
3 years ago
Some machines do not multiply the force that is applied to them
Katyanochek1 [597]

Answer:

Machine Efficiency

Explanation:

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5 0
3 years ago
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