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alina1380 [7]
3 years ago
6

If a car is moving backward and has negative acceleration, what can be said about the speed of the car?

Physics
2 answers:
Alona [7]3 years ago
7 0
C? I really don’t know -.-
KATRIN_1 [288]3 years ago
4 0
I think its A let me know its wrong or not
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Im pretty sure its neptune. Its mot pluto because its not a planet. Good luck.
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wide tube that extends down from the bag of solution, which hangs from a pole so that the fluid level is 90.0 cm above the needl
Bingel [31]

Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa

6 0
3 years ago
Difference between uniform circular motion and non uniform circular motion
Goryan [66]
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6 0
4 years ago
What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J
Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)

and on the surface the current is

I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A

(b)

For Jb = J0(1 - r/R)

I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )

and on the surface

I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi  \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A

and in a simmilar way for Jb

I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi   J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0
3 years ago
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