5 points

What is/are the external cause code(s) for a passenger involved in a mva that lost control on the highway and hits a guardrail?

Sholpan [36]

Here is the answer. The external cause code for a passenger that is involved in a MVA that lost control on the highway and hits a guardrail is E815. Your fifth digit is a 1 for a motor vehicle passenger. You would use an additional E code when a place of occurrence (example, home or parking lot) is documented. In this case, the location is documented as the highway. Hope this helps.

6 0

4 years ago

Please help asap. I'm terrible with atoms

11Alexandr11 [23.1K]

<span>Correct answer: more, smaller</span>

6 0

4 years ago

Read 2 more answers

Which description properly describes the plant structures involved in photosynthesis?

raketka [301]

Answer:

D

Explanation:

Stomata take in carbon dioxide (i think they realease oxygen too), Phloem transports the glucose throughout the plant, Xylem transports the water and minerals from the roots. If you want me to explain more I can

3 0

3 years ago

Read 2 more answers

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

4 years ago

In which situation would a space probe most likely experience centripetal acceleration?

tester [92]

Question: In which situation would a space probe most likely experience centripetal acceleration?

as it revolves around a planet

as it flies straight past a moon

as it is pulled in a line toward the Sun

as it lifts off from Earth

Answer:

When "space probe revolves around a planet" most likely to experience centripetal acceleration

Explanation:

Centripetal acceleration defined as the rate in change of tangential velocity. Also, as per Newton's second law, any kind of force will be directly proportional to the acceleration attained by the object. So, for centripetal acceleration, the force will be the centripetal force. The centripetal force will be acting on an object rotating in a circular motion with its direction of force towards the center. Thus, centripetal acceleration will be experienced by an object or a space probe when it is in a circular motion. It means the space probe is revolving around a planet.

8 0

3 years ago