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nordsb [41]
3 years ago
10

For each of the following statements determine whether it is correct or incorrect.

Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

a) T; b) F; c) F; d) T; e) T; f) T

Explanation:

a) True, since the equation of work done on a spring is \frac{kx^2}{2}

b) False, it will be only mgh.

c) False, since it is mv^2, no direction involved.

d) True, since the measured speed of objects depends from the reference frame of the observer.

e) True, since its formula is mv^2 and both terms are positive.

f) True, since the centripetal force the Earth feels, which is the gravitational atraction the Sun exerts, is perpendicular to its circular trajectory.

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The average speed is 8 miles
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When objects are heated they tend to expand because
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4 years ago
___ acceleration occurs when an object speeds up
AleksandrR [38]

Explanation:

Acceleration is defined as the change in velocity over time.

When there is an increment or increase in the magnitude of velocity of a moving body then it is known as positive acceleration.

Whereas when there is a decrease in magnitude of velocity of a moving body then it is known as negative acceleration.

Thus, we can conclude that positive acceleration occurs when an object speeds up.

3 0
3 years ago
Read 2 more answers
Consider the following three objects, each of the same mass and radius: (1) a solid sphere (2) a solid disk (3) a hoop All three
Vinil7 [7]

Answer:

The correct answer is

a) 1, 2, 3

Explanation:

In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus

PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²

The transformation equation fom potential to kinetic energy is =

m×g×h = \frac{1}{2} mv^{2} + \frac{1}{2} (\frac{2}{5} mr^{2} )(\frac{v}{r}) ^{2}

v_{Sphere} = \sqrt{\frac{10}{7} gh}

v_{Hoop} = \sqrt{gh}

v_{Disc}=\sqrt{\frac{4}{3} gh}

Therefore the order is with increasing rotational kinetic energy hence

the first is the sphere 1 followed by the disc 2 then the hoop 3

the correct order is a, 1, 2, 3

8 0
4 years ago
A car enters a horizontal, curved roadbed of radius 50 m. the coefficient of static friction between the tires and the roadbed i
UkoKoshka [18]
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
  v = √(kgr)
  v = √(0.20·9.8 m/s²·50 m)
  = 7√2 m/s ≈ 9.90 m/s
6 0
3 years ago
Read 2 more answers
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