All categories

2 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

1 answer:

7 0

The question is asking to calculate the object's speed v1, v2, v3 at the certain time is the given of the problem, in my calculation, I would say that the speed would be 2m/s, 1.5m/s, 0.22m/s. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification

You might be interested in

During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.

Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

$Q = \frac{\Delta P \pi r^4}{8\eta l}$

Where:

$\eta_i =$ are the viscosities of the concrete before and after the increase

l = Length of the vessel

$r_1, R_2$ = Radio of the vessel before and after the increase

$\Delta P$ = Change in the pressure

$Q_{1,2} =$ The rates of flow before and after he increase

Our values are given as:

$Q_2 = 10Q_1 \rightarrow$ 10 times her resting rate

$\eta_2 = 0.95\eta_1$ 95% of its normal value

$\Delta P_2 = 1.5\Delta P_1$ Increase of 50%

Plugging known information to get

$Q_1 = \frac{\Delta P \pi r^4}{8\eta l}$

$Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4$

$r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}$

$r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}$

$r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}$

$r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}$

$r_2 = 1.586r_1$

Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

7 0

3 years ago

Temperature and kinetic energy have a_______relationship.

Alex73 [517]

The hotter an object is, the more kinetic energy it has, but i'm not sure what is the exact word missing??

8 0

2 years ago

Read 2 more answers

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

2 years ago

Part D

anygoal [31]

Answer: I didn't see a difference because the large ball's vertical displacement and velocity are the same as the small one's.

Explanation:

5 0

2 years ago

Early humans used rocks as tools to make other things as well as to construct buildings. Why was it better for them to use rocks

babymother [125]

Wood isn't as tough as rock. Wood also breaks down in weather and cracks under pressure. Plus rocks were more accessible.

6 0

2 years ago

Read 2 more answers