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ehidna [41]
3 years ago
7

1. A car with a mass of 1,500 kilograms is moving around a circular curve at a uniform velocity of 16 m/s. The curve has a radiu

s of 100 meters. What's the centripetal force on the car?
2. A car with a mass of 1,200 kg is moving around a circular curve at a uniform velocity of 20 m/s. The centripetal force on the car is 6,000 N. What's the radius of the curve?


Write your answers using 1–2 paragraphs.

3. Describe the universal gravitational law.

4. Evaluate the techniques and models used by Ptolemy and Copernicus to explain the setup of the universe.
Physics
2 answers:
Crank3 years ago
6 0

Discovered by Sir Isaac Newton, this law states that every object in the universe that has mass attracts every other object in the universe that has mass. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers. When applying this to a situation with two objects, the object with the smaller mass will do most of the moving because the other object has too much inertia to move any noticeable amount.


4. Without advanced technology like we have today, Ptolemy and Copernicus tried to best explain the model of the universe through observation. Ptolemy’s model came first and placed a stationary earth at the center of the model. Everything else moved in respect to earth. This was widely accepted since it seemed like earth was not moving. Ptolemy stated that the planetary bodies moved around the earth in circular paths. However, this wasn’t always witnessed through observation. He adjusted his model to state that some planets must be moving in secondary orbits.


Copernicus put a rotating earth in a sun-centered model. The rotation of earth was able to account for the rising and setting of stars. The orbital motion of the earth and moon also accounted for the motion of the sun and moon with respect to the stars. This was easier to understand but encountered scrutiny due to its differences from religious teachings.


One big difference between the approaches in the two is that Copernicus didn’t try to adjust his model to match what was going on; he used observations to develop the model. In addition, one common trend in science is that the simplest explanation is usually most accurate or closer to accurate. Copernicus’ model was more straightforward; Ptolemy’s was more complex.

svet-max [94.6K]3 years ago
5 0

<u>Answers</u>

1)  3,840 N

2) 80 m


<u>Explanations</u>

Centripetal force is the force that maintenance objects that are moving in a circular motion.

F = mv²/r

  = (1500 × 16²)/100

   = (1500×256)/100

     = 384000/100

     = 3,840 N

2) F = (mv²)/r

       6000 N = (1200 × 20²)/r

r = (1200 × 400)/6000

   = 480,000/6000

     = 80 m

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Answer:

Weight on Jupiter will be equal to 2940 N

Explanation:

We have given given acceleration due to gravity on Jupiter is 3 times of acceleration due to gravity on earth

Acceleration due to gravity on earth g=9.8m/sec^2

So acceleration due to gravity on Jupiter = g'=3\times 9.8=29.4m/sec^2

Mass is given m = 100 kg

We have to find the weight

Weight is equal to W = mg, here m is mass and a is acceleration

So weight W=100\times 29.4=2940N

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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

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Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

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3 years ago
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