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Akimi4 [234]
2 years ago
8

Assume that two cars have the same kinetic energy, but that the red car has twice the speed of the blue car. We then know that t

he red car has ____ mass as the blue car.
Physics
1 answer:
Anna71 [15]2 years ago
8 0

Answer:

equal

Explanation:

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Whose data did Kepler use to describe the motion of the planets?
Levart [38]

Answer:

Tycho Brahe

Explanation:

Tycho Brahe's accurate observations of planetary positions provided the data used by Johannes Kepler to derive his three fundamental laws of planetary motion.

7 0
2 years ago
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A pendulum has 711 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
Anika [276]
According to law of conservation of energy, 
<span>Energy can neither be constructed nor be destroyed but can be transformed from one form to another.
</span>
<span>At the highest point of the pendulum(point b), pendulum is associated with potential energy only and no kinetic energy.
</span><span>Therefore total energy at point b = potential energy = 711 J.... i
</span>
<span>At the bottom most point(point a), pendulum is associated only with kinetic energy and no potential energy.
</span>Therefore total energy at point a = kinetic energy ---- ii
<span>From i and ii,
</span>Kinetic energy = potential energy = 711 J.(Conserving energy)

Hence kinetic energy at the bottom most point is 711 J.
Hope this helps!!

7 0
2 years ago
-2 5/8 is bigger than -5/12
Neko [114]

Answer:

hey...................................

8 0
3 years ago
Can you help me on this fast just the second one
Mariulka [41]
4. E
5. D
6. F

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3 0
3 years ago
A textbook of mass 2.05 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d
blsea [12.9K]

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

a. Let T be the tension in the cord. For the textbook, T = ma since no other force acts on it and it is an horizontal force, and m = mass = 2.05 kg and a = acceleration. We find the acceleration from s = ut + 1/2at² where u = initial speed = 0 (since it starts from rest),  s = distance moved = 1.30 m and t = time = 0.850 s.

Substituting these values into s,

1.30 m = 0 × 0.850 + 1/2a × 0.850² = 0 + 0.36125a

1.30 = 0.36125a

a = 1.30/0.36125 = 3.6 m/s²

Substituting this into T, we have

T = ma = 2.05 kg × 3.6 m/s² = 7.38 N

b.  Let T be the tension in the cord attached to the book. The book has the only vertical forces acting on it as the tension, T(acting upwards) and its weight mg (acting downwards). So the net force acting on it is

T - mg = ma

T = m(a + g)

substituting a = 3.6 m/s² and g = 9.8 m/s² and m = 3.05 kg

T = 3.05(3.6 + 9.8) = 3.05 × 13.4 = 40.87 N

c. Since the tangential acceleration of the pulley is also the acceleration of the masses, the a = rα where r = radius of pulley = 0.200 m/2 = 0.100 m and α = angular acceleration of the pulley.

α = a/r = 3.6 m/s² ÷ 0.100 m = 36 rad/s²

Now, the torque on the pulley τ = Tr = Iα where I = moment of inertia of pulley about its rotational axis and T = tension in cord attached to book and r = radius of pulley = 0.200 m/2 = 0.100 m

From the equation above, I = Tr/α

Substituting the variables we have

I = 40.87 N × 0.100 m ÷ 36 rad/s² = 0.113 kg-m²

4 0
3 years ago
Read 2 more answers
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