Which term is another name for fraunhofer diffraction?

A 108 kg clock initially at rest on a horizontal floor requires a 639 N horizontal force to set it in motion. After the clock is

labwork [276]

Answer:

$\mu_s=0.60$

Explanation:

It is given that,

Mass of the clock, m = 108 kg

Force acting on it when it is in motion, $F=639\ N$

After the clock is in motion, a horizontal force of 521 N keeps it moving with a constant velocity, F' = 521 N

It is assumed to find the coefficient of between the clock and the floor. The force of friction is given by :

$F=\mu_smg$

$\mu_s=\dfrac{F}{mg}$

$\mu_s=\dfrac{639\ N}{108\ kg\times 9.8\ m/s^2}$

$\mu_s=0.60$

So, the coefficient of static friction between the clock and the floor is 0.6. Hence, this is the required solution.

5 0

3 years ago

If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot

Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=I_{1}+I_{2}$

$I=(m+m_{1})\times r^2+(m+m_{2})\times r^2$

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

$I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2$

$I=117.45\ kg-m^2$

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0

4 years ago

You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are pres

Dominik [7]

Answer:

<em>The required constant friction force for the last 20 m is 6,862.8 N</em>

Explanation:

<u>Energy Conversion</u>

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

$E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J$

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

$E_1=K_1+W_1$

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

$W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J$

This means that the kinetic energy that remains when the car reaches ground level is

$K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J$

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

$W_2=F_{r2}\cdot x_2=137,256$

Solving for Fr2

$\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N$

The required constant friction force for the last 20 m is 6,862.8 N

3 0

3 years ago

The weld bead in SMAW is formed by the?

SVETLANKA909090 [29]

The answer is C I just took this quiz.

7 0

3 years ago

Whic two types of cells are required during the process of osteogenesis

Mkey [24]

Answer:

Bone homeostasis involves multiple but coordinated cellular and molecular events. Two main types of cells are responsible for bone metabolism: osteoblasts (which secrete new bone), and osteoclasts (which break bone down).

Explanation:

3 0

3 years ago