Ask question

Ask question

All categories

sergiy2304 [10]

3 years ago

7

Two types of energy can be either kinetic or potential energy. Identify those two types. For each type, explain why it can be ei

ther kinetic or potential. -Help fast i have my test soon!

2 answers:

love history [14]3 years ago

8 0

Answer:

All forms of energy are either kinetic or potential. The energy associated with motion is called kinetic energy . The energy associated with position is called potential energy . Potential energy is not "stored energy".

Explanation:

Leno4ka [110]3 years ago

3 0

All forms are mostly kinetic!!

You might be interested in

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is

NikAS [45]

<span>4.0 m/s2

it's 9 squared divided by 6</span>

8 0

3 years ago

Read 2 more answers

What is the speed of an object that travels 60 metres in 4 seconds

Zina [86]

Answer:

$s = \frac{d}{t} = \frac{60}{4} \\ \boxed{speed = 15m. {sec}^{ - 1} }$

4 0

3 years ago

Read 2 more answers

1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic

Korvikt [17]

51.448 g is the required answer!

8 0

3 years ago

A mountaintop is a height y above the level ground. A woman measures the angle of elevation of the

Kamila [148]

Refer to the diagram shown below.
We want to find y in terms of d, φ and θ.

By definition,
$tan (\theta) = \frac{y}{x} \\\\ tan( \phi) = \frac{y}{x-d}$

Therefore
y = x tan(θ) (1)
y = (x - d) tan(φ) (2)

Equate (1) and (2).
$(x - d) \, tan(\phi) = x \, tan(\theta) \\ x[tan(\phi) - tan(\theta)] = d \, tan(\phi) \\ x= \frac{d tan(\phi)}{tan(\phi)-tan(\theta)}$

From (1), obtain the required expression for y.

Answer:
$y= \frac{d \, tan(\phi) \, tan(\theta)}{tan(\phi)-tan(\theta)}$

7 0

3 years ago