Approximately how far, in cm, does the blood move during one beat?

When evaluating data, why is it better to make a graph instead of just looking at the raw data in the table?(8th grade Science s

ziro4ka [17]

Answer:All professors planned and thought about data before graph construction. When reflecting on their graphs, professors and graduate students focused on the ... feel more inclusive in the learning process and gain better science process skills, ... We organized data in a table instead of a paragraph with numbers

Explanation:

hope that helps

3 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

The power is inversely proportional with time

trapecia [35]

Answer:

True

Explanation:

Given that Power whose unit is Watt equates to one joule of work per second. It implies that Power is directly proportional to the work done and inversely proportional to the time to do the work.

Therefore, in this case, the right answer to the question is that it is TRUE that the power is inversely proportional with time

8 0

3 years ago

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of

Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

$mg\ sin\ \theta = 120\times 10 \times sin\ 47^\circ{}=877.62 \ N$

Horizontal component of force(Normal reaction) :

$mg\ cos\ \theta = 120\times 10 \times cos\ 47^\circ{}=818.40 \ N$

Since, box is on the verge of slipping :

$mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07$

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0

3 years ago

While spinning down from 500 rpm to rest, a flywheel does 3.9 kj of work. this flywheel is in the shape of a solid uniform disk

Ksivusya [100]

The answer is 4.0 kg since the flywheel comes to rest the kinetic energy of the wheel in motion is spent doing the work. Using the formula KE = (1/2) I w².

Given the following:

I = the moment of inertia about the axis passing through the center of the wheel; w = angular velocity ; for the solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.

The radius = 1.2 m and KE = 3900 J

3900 J = (1/4) m (1.2)² (52.36)²

m = 3900 J / (0.25) (1.2)² (52.36)²

m = 3.95151 ≈ 4.00 kg

6 0

3 years ago