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m_a_m_a [10]
3 years ago
14

In Example Suppose the rod stays on the wall in the same position, but the hinge is moved to be at the center of mass position o

f the rod. The rod is then held out horizontally and released. How do the energiesof the system change after the rod is released? Check all that apply.
a. The change in the system’s gravitational potential energy would be zero
b. The change in the system’s gravitational potential energy is positive
c. The change in the system’s gravitational potential energy is negative
d. The change in the system’s kinetic energy would zeroThe change in the system’s kinetic energy is positive
e. The change in the system’s Kinetic energy is negative
Physics
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

A

Explanation:

The change in the system’s gravitational potential energy would be zero

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Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
6–23 an automobile engine consumes fuel at a rate of 22 l/h and delivers 55 kw of power to the wheels. if the fuel has a heating
Anna007 [38]

Explanation & answer:

Given:

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P  =  55 kW

heating value, H = 44,000 kJ/kg

Solution:

Calculate energy intake

E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

= 215111.1 j/s

Calculate output power

P = 55 kW

= 55000 j/s

Efficiency

= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6% to 1 decimal place.

8 0
3 years ago
Which of the following is known as the following cycle because it's reservoir is a rock.
inessss [21]

Letra a obrigadoaijahahavqvqgqgqg

3 0
3 years ago
Imagine a 15 kg block moving with a velocity of 20 m/s to the left. Calculate the kinetic Energy of this block.
ivann1987 [24]

Answer:

3000 J

Explanation:

Kinetic energy is:

KE = ½ mv²

If m = 15 kg and v = -20 m/s:

KE = ½ (15 kg) (-20 m/s)²

KE = 3000 J

3 0
3 years ago
Which type of electromagnetic wave has the greatest wavelength?
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Radio waves are the longest
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