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Semenov [28]
3 years ago
13

9.

Physics
1 answer:
zepelin [54]3 years ago
8 0
A I believe it would be true
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The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co
motikmotik

Explanation:

The position vector r:

\overrightarrow{r(t)}=lcos\theta\hat{i}+lsin\theta\hat{j}

The velocity vector v:

\overrightarrow{v(t)}=\overrightarrow{\frac{dr}{dt}}=\dot{l}cos\theta-lsin\theta\dot{\theta}\hat{i}+\dot{l}sin\theta+lcos\theta\dot{\theta}\hat{j}

The acceleration vector a:

\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}

\overrightarrow{v(t)}=0.13\hat{i}+0.18\hat{j}

\overrightarrow{a(t)}}=-0.3\hat{i}-0.1\hat{j}

5 0
3 years ago
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe
RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}

where,

P = Friction Loss per foot of pipe = \frac{0.4}{1000\ ft} = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}

<u>Q = 913.9 gpm</u>

5 0
3 years ago
Help with science for A P E X
eimsori [14]

Answer:

inertia

Explanation:

7 0
3 years ago
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