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Stolb23 [73]
2 years ago
8

Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s

ignal your interest to the speaker?
Physics
1 answer:
BARSIC [14]2 years ago
4 0
This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others. 
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Ram has power of 550 watt. What does it mean?
WARRIOR [948]
It means you can do 550 Newton Meters of work every second. Power is the rate of doing work, I hope this helps
4 0
2 years ago
An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init
elena-14-01-66 [18.8K]

Answer:

Approximately 71\%.

Explanation:

The formula for the kinetic energy \rm KE of an object is:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2,

where

  • m is the mass of that object, and
  • v is the speed of that object.

Important: Joule (\rm J) is the standard unit for energy. The formula for \rm KE requires two inputs: mass and speed. The standard unit of mass is \rm kg while the standard unit for speed is \rm m \cdot s^{-1}. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

m = 63\; \rm g = 0.063\; \rm kg.

Initial \rm KE of this arrow:

\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}.

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%.

8 0
2 years ago
A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must
mezya [45]
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction. 
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end) 
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
4 0
2 years ago
Read 2 more answers
The hammer throw was one of the earliest Olympic events. In this event, a heavy ball attached to a chain is swung several times
Aleonysh [2.5K]

Answer:

Given that

T= 0.43 s

Radius of the ball path's , r=2.1 m

a)

We know that

f= 1/T

Here f= frequency

      T= Time period

Now by putting the values

f= 1/T

T= 0.43 s

f= 1/0.43

f=2.32 Hz

b)

We know that

V= ω r

ω = 2 π f

ω=Angular speed

V= Linear speed

ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s

V= ω r= 14.60 x 2.1 = 30.66 m/s

c)

Acceleration ,a

a =ω ² r

a= 14.6 ² x 2.1 = 447.63 m/s²

We know that g = 10 m/s²

So

a= a/g= 447.63/10 = 44.7 g m/s²

a= 44.7 g m/s²

7 0
2 years ago
A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h
Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0
2 years ago
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