Which wave has the highest frequency when traveling through the indicated metal?

Question 1 of 10

Neko [114]

Answer:

D: A mathematical model

Hope its right if not so sorry :)

3 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago

What are the charged particles that form when atoms gain or lose electrons?

beks73 [17]

Answer:

Ions

Explanation:

6 0

3 years ago

13. Direct instruction takes place when

jekas [21]

A skill set is explicitly taught or an activity is completed

3 0

4 years ago

A projectile is thrown with an initial speed vo = 25.0 m/s at an initial angle with the horizontal ao = 45.0. a . Find the time

Natali [406]

Answer:

See the explanation below

Explanation:

To solve this problem we must decompose the initial speeds into x & y.

$v_{o}_{x}=25*cos(45)=17.67[m/s]\\v_{o}_{y}=25*sin(45)=17.67[m/s]\\$

The acceleration of gravity is equal to g = 9.81[m/s^2] downward.

The maximum height is when the velocity of the projectile is zero in the component y, that is, it will not be able to go higher, by means of the following kinematic equation we can find that time, for that specific condition.

a)

$v_{y}=(v_{y})_{0}+a*t\\0 = 17.67 - 9.81*t\\17.67 = 9.81*t\\t=1.8 [s]$

Note: Acceleration is taken as negative as it is directed downwards.

b)

The position in the x component can be found using the following kinematic equation

$x=(v_{x})_{o}*t\\x=17.67*1.8\\x=31.82[m]$

The position in the y component can be found using the following kinematic equation

$y =(v_{y})_{o}*t+\frac{1}{2} *g*t^{2} \\y=17.67*1.8-0.5*9.81*(1.8)^{2}\\ y=15.91[m]$

c)

Since the motion on the X-axis is at constant speed, there is no acceleration, so the only acceleration that exists is due to gravity

d)

In the attached image we can see, the projectile with the vectors of acceleration and velocity.

7 0

3 years ago