At time=0s, the object is at the 21.0-meter position along the roadway. Where is the object at time = 10 s?

An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the ce

Morgarella [4.7K]

Answer:

Angular velocity (w) = 8.86 rad/s

Explanation:

Angular velocity (w) = $\sqrt{} 4ghi/R^{2}$

g= 9.81 m/s

R= 0.5

hi (initial depth) = 0.5m

Hence= $\sqrt4* 9.81* 0.5/0.5^{2}$ = 8.86 rad/s

3 0

3 years ago

Claudia throws a baseball to her dog. Which free-body diagram shows the

chubhunter [2.5K]

Answer:

only the weight of the ball will act on the ball

Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown

6 0

3 years ago

A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. Suppose at f

Airida [17]

Answer:

c. The steady-state value of the current depends on the resistance of the resistor.

Explanation:

Since all the components are connected in series, when the switch is at first open, current will not flow round the circuit. As current needs to flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery.

But the moment the switch is closed, at the initial time t = 0, the current flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery. It then begins to increase at a rate that depends upon the value of the inductance of the inductor.

6 0

3 years ago

A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent

solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

$I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}$

The angular acceleration is given as

$a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}$

The angular velocity is given as

$w=at\\w=(0.905)(2.55)\\w=2.31rad/s$

So the Kinetic Energy is given as

$K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J$

3 0

3 years ago

Which is the second largest planet in the solar system

shutvik [7]

Answer:

Saturn is the answer of your question

8 0

2 years ago