What happens when the arms of the milky move away from the center of the galaxy

Which of the following placards allows a driver to park in a handicapped parking space.

malfutka [58]

All of the above ......

4 0

3 years ago

Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig

marin [14]

Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

5 0

3 years ago

Answer?...................

torisob [31]

Answer:

The correct option is;

c. Leaving the chuck key in the drill chuck

Explanation:

A Common safety issues with a drill press leaving the chuck key in the drill chuck

It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.

It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable

Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.

5 0

2 years ago

P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si

klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V 0.184 W

2.499 mV 125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

$V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\$ *5

= 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:

V_o = A_voc*V_i*(R_o/R_l+R_o)

= 4.545*(100/100+50)

= 3.03 V

Now we can determine delivered power:

P_L = V_o^2/R_L

= 0.184 W

Apply voltage-divider principle to the circuit:

V_o = (R_o/R_o+R_s)*V_s

= 50/50+100*10^3*5

= 2.499 mV

Now we can determine delivered power:

P_l = V_o^2/R_l

= 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.

4 0

3 years ago

A body whose velocity is constant has a. positive acceleration b. negative acceleration g. zero acceleration d. all of the above

adoni [48]

Answer:

option (c) is the correct answer which is zero acceleration.

Explanation:

It is given in the question that the velocity is constant.

Now,

the options are provided in relation to the acceleration.

We know,

acceleration is rate of change of velocity per unit time i.e

acceleration = $\frac{dV}{dt}$

since, the change in velocity is given to be zero,

thus, dV/dt = 0

hence,

acceleration = 0

therefore, option (c) is the correct answer which is zero acceleration.

4 0

3 years ago