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scoray [572]
2 years ago
8

Help please!!! Physics circular motion

Physics
1 answer:
svp [43]2 years ago
6 0

Answer:

2. 3.1415 m/s

3. 0.63m

4. 0.006 m/s^2

Explanation:

2. v=(2*pi*r))/T, put in values and solve.

3. Circumference=2*pi*radius, radius is 0.1m, plug in and solve.

4. a=(v^2)/r, (don't worry about converting to meters, since units are the same in kilometers, they will cancel out) plug in values and solve.

That's it!

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On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera
pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

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8 0
3 years ago
A motorcycle and its driver has a mass of 145 kg. Answer the following questions about it, using correct units.
inysia [295]

a) KE=0.5*mv^2==0.5*145*25^2=45312.5 J

b) PE=mgh=145*9.8*3.5=4973.5 J

c) ME=KE+PE=m(0.5v^2+gh)=62524 J

4 0
3 years ago
Read 2 more answers
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
Rank the objects below in order from smallest to largest, where 1 is the smallest and 5 is the largest. local group of galaxies
sasho [114]
1. earth
2. solar system
3. milky way galaxy
4. local group of galaxies
5. universe
4 0
3 years ago
A 35kg cannonball sits at rest on a flat level surface. What is the normal force exerted on the cannon ball. Use -9.8 m/s2 for a
Tom [10]

Answer:

a) N = 343 [N]

Explanation:

We must remember Newton's third law, which tells us that the force acting on a body is equal to the normal reaction force of equal magnitude but acting in the opposite direction.

N=m*g

where:

m = mass = 35 [kg]

g = gravity acceleration = 9.81 [m/s²]

N = 35*9.8\\N= 343 [N]

8 0
2 years ago
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