What type of circuit is in the diagram?

During each heartbeat, about 80 g of blood is pumped into the aorta inapproximately 0.2 s. During this time, the blood is accele

taurus [48]

Answer:

Work is done by the heart on the blood during this time is 0.04 J

Explanation:

Given :

Mass of blood pumped, m = 80 g = 0.08 kg

Initial speed of the blood, u = 0 m/s

Final speed of the blood, v = 1 m/s

Initial kinetic energy of blood is determine by the relation:

$E_{1}=\frac{1}{2} m u^{2}$

Final kinetic energy of blood is determine by the relation:

$E_{2}=\frac{1}{2} m v^{2}$

Applying work-energy theorem,

Work done = Change in kinetic energy

W = E₂ - E₁

$W=\frac{1}{2} m (v^{2}-u^{2})$

Substitute the suitable values in the above equation.

$W=\frac{1}{2}\times0.08\times (1^{2}-0^{2})$

W = 0.04 J

6 0

2 years ago

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

7 0

2 years ago

you are building a race car. Your goal is to have a car that can go from 0 miles per hour to 80 miles per hour in 2 seconds. You

Vitek1552 [10]

<span>You want to focus on the car's acceleration.
(Hope this helps you! :) Have a nice day!)</span>

6 0

3 years ago

Read 2 more answers

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

2 years ago

Suppose multiple stations are connected in wired network where CSMA/CD implemented. Among them, station A wants to communicate w

anastassius [24]

Answer:

bro i think ur from the uni of arid agriculture rawalpindi and today is your networking paper

Explanation:

6 0

2 years ago