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2 years ago

PLEASE HELP ASAP

Physics

2 answers:

Sholpan [36]2 years ago

4 0

D, excess body heat evaporates which cause sweat.

Dmitrij [34]2 years ago

3 0

The answer should be B

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Svetllana [295]

Answer:

B.) Visit a healthcare professional

Explanation:

Their judgement is much better than a friend's or your own.

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2 years ago

An actor who memorizes his lines word-for-word has demonstrated _____ learning.

Nana76 [90]

<span>Mechanical association learning used by an actor to memorize his lines</span>

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2 years ago

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What is the mass of a stone moving at a speed of 15 meters/second and having a momentum of 7.5 kilogram - meters / seconds

Elanso [62]

Answer:

0.5 kg

Explanation:

The momentum of an object is defined as

p = mv

where

m is the mass

v is the velocity

In this problem we have,

v = 15 m/s is the velocity of the stone

p = 7.5 kg m/s is the momentum

Solving for m, we can find the mass of the stone:

$m=\frac{p}{v}=\frac{7.5 kg m/s}{15 m/s}=0.5 kg$

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2 years ago

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A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

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2 years ago

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respecitvely; $a$ is the acceleration; and $\Delta x$ is the change in position.

So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0

2 years ago