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tigry1 [53]
3 years ago
6

A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s. What is th

e average speed of the object?
Physics
1 answer:
kumpel [21]3 years ago
3 0
The average speed of the object is given by the ratio between the total distance covered and the total time needed for the motion:
v= \frac{S}{t}

The total distance covered is:
S=2m + 80 m = 82 m
while the total time of the motion is
t=2s+40 s=42 s

Therefore, the average speed of the object is
v= \frac{S}{t}=  \frac{82 m}{42 m}=1.95 m/s
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Explanation:

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A 1502.7 kg car is traveling at 33.1 m/s when
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By definition of average acceleration,

<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

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3 years ago
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How does the ratio of tin to copper affect the properties of the alloy bronze?
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7 0
2 years ago
3
Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

3 0
2 years ago
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