Ask question

Ask question

All categories

Svetradugi [14.3K]

3 years ago

6

Which of these helps you understand what people are saying to you? A. Interrupting them B. Looking away while they're speaking C

. Thinking about how you'll respond D. Thinking about what they're saying

2 answers:

Vaselesa [24]3 years ago

7 0

D ...................................

densk [106]3 years ago

3 0

I believe the answer is <span> D. Thinking about what they're saying
Interrupting them will make them unable to finish their full message/thought to you. Looking away while they're speaking will make you reduce your attention and unable to process the information properly. Thinking about how you'll response also limit your attention to them and could make you cloud your opinion and misinterpret what's they're saying.</span>

You might be interested in

A bird flew 16 km west in 5 hours , then flew 20 km east in 6 hours . What was the birds velocity?

LenaWriter [7]

During that period of time, the bird's displacement was 4 km east. So its velocity was (4km east)/(11hrs). That's 0.36 km/hour east. (rounded)

5 0

2 years ago

Ben rushin is waiting at a stoplight. when it finally turns green, ben accelerated from rest at a rate of a 6.00 m/s2 for a time

jasenka [17]

In the 4.10 seconds that elapsed, Ben reaches a velocity of

$v_f=v_0+at\implies v_f=0\,\dfrac{\mathrm m}{\mathrm s}+\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.10\,\mathrm s)$

$\implies v_f=24.6\,\dfrac{\mathrm m}{\mathrm s}$

In this time, his displacement $\Delta x$ satisfies

${v_f}^2-{v_0}^2=2a\Delta x\implies\left(24.6\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=50.4\,\mathrm m$

4 0

3 years ago

How does cancer change the normal life cycle of a cell ?

melamori03 [73]

In normal cells, the cell cycle is controlled by a complex series of signaling pathways by which a cell grows, replicates its DNA and divides.

In cancer, as a result of genetic mutations, this regulatory process malfunctions, resulting in uncontrolled cell proliferation.

4 0

3 years ago

A hockey puck slides across the ice at a constant velocity. Is the sliding puck in equilibrium? Why or why not?

geniusboy [140]

Answer:

Since the hockey puck is moving at constant velocity

So here we will have

$F_{net} = ma$

where we have

a = acceleration of the object

m = mass of object

so here since we know that acceleration is defined as rate of change in velocity

so here we will say that

$a = \frac{dv}{dt}$

$a = 0$

so we have

$F_{net} = 0$

so we will say that hockey puck is in equilibrium as there is no net force of it

3 0

3 years ago

As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile

Ber [7]

Answer:

a) $\Delta T= 100^{\circ}C$

b) $\bigtriangledown T=1^{\circ}C.mile^{-1}$

c) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

d) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

$T=(d+30) ^{\circ}C$ ...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

$\Delta T= T_f-T_i$ ..............................(2)

where:

$T_f$ = final temperature

$T_i$ = initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

$\Delta T= (100+30)-(0+30)$

$\Delta T= 100^{\circ}C$

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

$\bigtriangledown T=\frac{\Delta T}{\Delta d}$ ......................(3)

where:

$\Delta d$ = change in distance

$\bigtriangledown T=$ change in temperature with respect to distance

putting the respective values in eq. (3)

$\bigtriangledown T=\frac{100}{100}$

$\bigtriangledown T=1^{\circ}C.mile^{-1}$

(c)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

(d)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

4 0

3 years ago