Marx argued that what happens when a worker is separated

A spring stretches by 0.0177 m when a 2.82-kg object is suspended from its end. How much mass should be attached to this spring

Artemon [7]

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

$f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}$ (1)

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

$F=kx$

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

$mg=kx$

and using $g=9.8 m/s^2$ and $x=0.0177 m$ , we find

$k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m$

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

$m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg$

#LearnwithBrainly

6 0

3 years ago

Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 AM, each heading for Wildwood. One car’s average speed is 10

olganol [36]

Answer:

The average speed of the cars is 50mph

Explanation:

Option C

4 0

3 years ago

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

3 0

3 years ago

Please help im failing!!!

natta225 [31]

1- Kinetic , Mass , Speed
2- Speed
3- Speed, Mass
4- Mass, More
5- Transferred, collide
6- Kinetic, electricity
7- Transferred, Destroyed

:)

5 0

3 years ago

3.

Alexxx [7]

Answer:

120°

Explanation:

Draw a free body diagram. There are three forces acting on the traffic light. Two tension forces acting along the cables, and weight.

The tension forces have an angle θ between them. That means each tension force forms an angle of θ/2 with respect to the vertical. So the y component of each tension force is:

Ty = T cos (θ/2)

Sum of the forces in the y direction:

∑F = ma

Ty + Ty − W = 0

2 Ty = W

Substituting:

2 T cos (θ/2) = W

If W = T, then:

2 W cos (θ/2) = W

2 cos (θ/2) = 1

cos (θ/2) = 1/2

θ/2 = 60°

θ = 120°

6 0

4 years ago