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Charra [1.4K]
3 years ago
5

A toy car on a string is pulled across a table horizontally. The string is at an angle

Physics
1 answer:
ddd [48]3 years ago
6 0
Which way is it being pulled?
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A hockey puck is sliding across a frozen pond with an initial speed of 9.5 m/s. It comes to rest after sliding a distance of 37.
Bond [772]

Answer:

<h3>The coefficient of kinetic friction between the puck and the ice is \mu _{k} = 0.12</h3>

Explanation:

Given :

Initial speed  v_{o} = 9.5 \frac{m}{s}

Displacement x = 37.4 m

From the kinematics equation,

  v^{2} - v^{2} _{o}  = 2ax

Where v^{2}   = final velocity, in our example it is zero (v =0), a = acceleration.

   a =- \frac{90.25}{ 2 \times 37.4}

   a =- 1.21 \frac{m}{s^{2} }

From the formula of friction,

  F =- \mu _{k } N

Minus sign represent friction is oppose the motion

Where N = mg ( normal reaction force )

 ma = -\mu _{k}  m g                                                  ( ∵ g = 9.8 \frac{m}{s^{2} } )

So coefficient of friction,

 \mu_{k} = \frac{1.21}{9.8}

 \mu_{k} = 0.12

Therefore, the coefficient of kinetic friction between the puck and the ice is  \mu _{k} = 0.12 .

4 0
3 years ago
A star rotates with a period of 37 days about an axis through its center. The period is the time interval required for a point o
vlabodo [156]

Answer:

T = 1.1285 10⁻² day

Explanation:

For this exercise the forces in the premiere are internal, so the angular momentum is conserved

         L₀ = I₀ w₀

         L  = I w

         L₀ = L

         I₀ w₀ = I w

Angular velocity and period are related

         w₀ = 2π / T₀

         w = 2π  / T

         

The moment of inertia of a sphere is

       I₀ = 2/5 M R²

       I = 2/5 m r²

If we assume that the mass of the star does not change in the transformation

We substitute

         2/5 M R² 2π/T₀ = 2/5 M r² 2π/ T

          R² /T₀ = r² / T

          T = (r / R)² T₀

          T = (6.1 / 2.0 104) 37

          T = 1.1285 10⁻² day

5 0
3 years ago
Simple curiosity is not a legitimate reason to complete a scientific investigation
guapka [62]
Why not ? ? ?
If you can get the money, the laboratory to use, the experimental
equipment, and the people you need to work on it with you, then
curiosity is a fine, honorable, and perfectly good reason to go ahead,
do what you gotta do, follow your heart, satisfy your curiosity, and learn
new stuff. 
Neither the Large Hadron Collider, nor the Mars Rover, nor the Hubble
Space Telescope, nor the New Horizons probe, was built to cure cancer.
5 0
4 years ago
Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond.
pochemuha
It [the diamond] would act like a prism, and make a rainbow, or, the light would break up and disappear

Hope I helped!
~Mathlete12321

5 0
3 years ago
Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an
Lena [83]

Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

\frac{1}{f} = \frac{1}{265} + \frac{27}{10}

  = \frac{10+7155}{2650}

\frac{1}{f}  = \frac{7165}{2650}

⇒ f = \frac{2650}{7165}

      = 0.37

The focal length of the eye is 0.37 cm.

8 0
3 years ago
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