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Free_Kalibri [48]
2 years ago
9

(b) Figure 4 shows a car travelling on a motorway.

Physics
1 answer:
Alik [6]2 years ago
5 0

Answer:

To calculate anything - speed, acceleration, all that - we need <em>data</em>. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to <em>measure </em>it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring <em>time </em>and <em>distance</em>, and a method for measuring each.

Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:

  1. Press start on the stopwatch right as you pass a lamp post
  2. Each time you pass another lamp post, press the lap button on the stopwatch
  3. Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
  4. Record your data
  5. Calculate the time intervals for passing each lamp post using the lap data
  6. Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed

Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.

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Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot, isothermal metal substrate at Th = 300°C. The proper
Ad libitum [116K]

Answer:

74.52s

Explanation:

The solution is shown in the picture below

7 0
2 years ago
A 60.0-kg mass person wishes to push a 120-kg mass box across a level floor. The coefficient of static friction between the pers
Aneli [31]

Answer:

μ = 0.350

Explanation:

For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:

F_{p} = F_{b}

In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):

m_{p}*g*\mu_{p}=m_{b}*g*\mu_{b}\\m_{p}*\mu_{p}=m_{b}*\mu_{b}\\60*0.7=120*\mu_{b}\\\mu_{b}= 0.35

Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.

8 0
2 years ago
A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of
Flauer [41]

Answer:

v = 4.375\,\frac{m}{s}

Explanation:

The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}

The initial speed of Ryan is:

v = 4.375\,\frac{m}{s}

5 0
2 years ago
Read 2 more answers
A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq
Snowcat [4.5K]

Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

Work done by block on the spring is equal to change in elastic potential energy

i.e. Work done by block W=\frac{1}{2}kx^2

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block W=-\frac{1}{2}kx^2

Thus option d is correct

6 0
2 years ago
Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.
Stels [109]
The hoop is attached.

Consider that the friction force is given by:
F = μ·N
   = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ

Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
   = </span>(1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
   = tan⁻¹(2·0.9)
   = 61°

Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>

5 0
2 years ago
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