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stealth61 [152]

3 years ago

15

Where would you expect to have more touch receptors: on the palm of your hand or on the back of your hand? Explain your reasonin

g
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1 answer:

anygoal [31]3 years ago

6 0

Answer:

ive answered this

Explanation:

please check your previose question

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A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

3 years ago

The trough of the sine curve used to represent a sound wave corresponds to

iren [92.7K]

Answer:

The correct answer is a rarefaction.

Explanation:

Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.

There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.

Have a nice day!

4 0

3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

Angelina_Jolie [31]

Answer:

a). $\frac{\dot{W}}{m}= 311$ kJ/kg

b). $\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

$\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0$

$\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})$

$\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})$

$\frac{\dot{W}}{m}= -30+1.1(980-670)$

$\frac{\dot{W}}{m}= 311$ kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

$\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}$

$\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978$

$\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

5 0

3 years ago

An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel

Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

S = ut + ½ gt²

∴ g = 2S/t²

Substituting the given values,

g = 2 x 144 /6²

= 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

7 0

3 years ago

Knowing that the board is 4 meters long, one child weighs 300 n and the other 250 n, where should one of the children sit so tha

Neko [114]

Answer:v nxfgdjngdnmgndjfnncnfndngndsnbxzmnfn

Explanation:

8 0

3 years ago