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mario62 [17]
3 years ago
10

Mechanical energy is a term that is used to describe A. kinetic energy only. B. both potential and kinetic energy. C. potential

energy only. D. neither potential nor kinetic energy.
Physics
2 answers:
gayaneshka [121]3 years ago
7 0
B. both potential and kinetic energy

myrzilka [38]3 years ago
5 0

Answer: B. both potential and kinetic energy.

Explanation:

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What is the acceleration of a 1000kg car subject to a 550N net force?
igomit [66]

Answer:

a=550÷1000

a=0.55m/s²

5 0
2 years ago
A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce
Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

x=x_0+v_0t+at^2

Where

x_0 = Initial position

v_0 = Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

v_f = v_0 + at

Where,

v_f = Final velocity

For our case we have that there is neither initial position nor initial velocity, then

x= at^2

With our values we have x = 401.4m, t=4.945s, rearranging to find a,

a=\frac{x}{t^2}

a = \frac{ 401.4}{4.945^2}

a = 16.41m/s^2

Therefore the final velocity would be

v_f = v_0 + at

v_f = 0 + (16.41)(4.945)

v_f = 81.14m/s

Therefore the final velocity is 81.14m/s

8 0
3 years ago
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball. As ir shrinks in size, the cl
Trava [24]

Answer:

rotates faster

Explanation:

A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball As it shrinks in size, the cloud rotates faster. Because Angular momentum is conserved, so when it shrinks the moment of inertia decreases, then angular speed must increase. So it rotates fast.

4 0
3 years ago
True or false. The planet's speed never changes. That is why we have a stable orbit.
Svetradugi [14.3K]
This answer is true the earth always stays at one speed
8 0
1 year ago
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