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3 years ago

True or false: Decreased output from the vasomotor center allows arterioles and veins to constrict.

Physics

1 answer:

wolverine [178]3 years ago

3 0

<h3><u>Answer;</u></h3>

The above statement is False

<h3><u>Explanation;</u></h3>

Decreased output from the vasomotor center allows arterioles and veins to dilate.
The vasomotor center controls vessel tone or contraction of the smooth muscle in the tunica media.It is responsible for central regulation of cardiac electrical activity, myocardial performance, and peripheral vascular tone.
Changes in diameter affect peripheral resistance, pressure, and flow, which in turn affect cardiac output.

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A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad

sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

$\theta = \frac{S}{r}$

where, S = length of the rod

r = radius

Putting the given values into the above formula as follows.

$\theta = \frac{S}{r}$

= $\frac{4}{2}$

= $2 radians (\frac{180^{o}}{\pi})$

= $114.64^{o}$

Now, we will calculate the charge density as follows.

$\lambda = \frac{Q}{L}$

= $\frac{18 \times 10^{-9} C}{4 m}$

= $4.5 \times 10^{-9} C/m$

Now, at the center of arc we will calculate the electric field as follows.

E = $\frac{2k \lambda Sin (\frac{\theta}{2})}{r}$

= $\frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}$

= 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

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3 years ago

In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas

Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

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zzz [600]

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3 years ago

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