Draw a Lewis dot structure for Hf

A 43-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d

lesya [120]

Answer:

$d_{2}=0.585m$

Explanation:

The torque is the force by the distance so to determinate that both torque are the same magnitude so

$T_{1}-T_{2}=0$

$T_{1}=T_{2}$

$F_{1}*d_{1}=F_{2}*d_{2}$

$43N*0.30m=21N*d_{2}$

Solve to d2

$d_{2}=\frac{41N*0.30m}{21N}$

$d_{2}=0.585m$

5 0

3 years ago

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

$P_2 = 44.15Lbf/in^2$

Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

5 0

3 years ago

Someone please help with this

SSSSS [86.1K]

Answer:

The new force is 2/3 of the original force

Explanation:

Coulomb's Law

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

$\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}$

Factoring out 2/3:

$\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}$

Substituting the original force:

$F'=\frac{2}{3}F$

The new force is 2/3 of the original force

7 0

2 years ago

It took Lightning McGreen 2.5 hours to travel

Law Incorporation [45]

Answer:

240km/h

Explanation:

speed=distance/time

speed=600/2.5

=240km/h

4 0

2 years ago

Read 2 more answers

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road