True or false: The therapeutic alliance is defined as the level of commitment a patient has to therapy.

What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these

max2010maxim [7]

Answer:

E. all of these

Explanation:

The designation of a point in space all the points that necessary

- reference point

- a direction

- fundamental units

- a direction

- motion

all are necessary to designate a point in space. Hence option E is correct.

For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.

8 0

3 years ago

Bob runs 1800 seconds at an average speed of 1.5 m/sec. How far did he go? 25 Points!!!!

german

Distance= Time×Speed

= 1800×1.5

= 2700 m

I am not sure it's right. the question itself is confusing.

4 0

2 years ago

) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

$P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s$ or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0

3 years ago

What is the combined resistance for two resistors with resistances of 10 Ohms and 23.7 Ohms if they are connected in series

Agata [3.3K]

Answer: 33.7Ω

Explanation:

Since there are two resistors connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i.e Rtotal = R1 + R2

R1 = 10Ω

R2 = 23.7Ω

Hence, Rtotal = 10Ω + 23.7Ω

Rtotal = 33.7Ω

Thus, the combined resistance for two resistors is 33.7Ω

3 0

3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

2 years ago