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2 years ago

Hypothermia is an excessive decline in a body temperature. true/ false

Physics

1 answer:

Fantom [35]2 years ago

5 0

It's True that Hypothermia is an excessive decline in a body temperature.

<h3>What is Hypothermia?</h3>

Hypothermia is condition that occurs when the body’s temperature drops below 95° F (35° C).The normal body temperature is 98.6° F (37° C). Hypothermia is medical emergency.

When a person’s body temperature is dangerously low, the brain and the body cannot function properly. Left untreated, hypothermia can lead to the cardiac arrest (heart stops beating) and death.

<h3>What are the treatments of Hypothermia?</h3>

The first steps when treating hypothermia are:

Move to warm, dry location. Remove the wet clothing and replace with the dry clothing. Cover up with the jacket, hat, and blanket

When hypothermia is the more severe, doctors may also need to:

Insert IV into your vein and pump warm fluids into your body. Give you warm oxygen through a mask or a breathing tube. Use machine that warms your blood and pumps it back into your body

To know more about oxygen visit: brainly.com/question/13905823

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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers

A uranium nucleus 238U may stay in one piece for billions of years, but sooner or later it decays into an α particle of mass 6.6

ale4655 [162]

Answer: The recoil speed is - 8.9604. $10^{4}$ m/s.

Explanation: According to the Third Law of Newton, every action has an oppsite and equal reaction, and the Second Law of Newton, Force=mass·acceleration. Acceleration is a variation in velocity by any given time, so Force = mass·velocity·time.

Combining the two laws, there is : m1·v1 = - m2·v2. This is the Law of Conservation of Momentum.

Substituting and calculating:

v2 = - ( $\frac{m1}{m2}$ ) · v1

v2 = - $\frac{6.64.10^{-27} }{3.88.10^{-25} }$ · 5.24. $10^{6}$

v2 = - 8.9604. $10^{4}$

The recoil speed of the thorium nucleus is - 8.9604. $10^{4}$ m/s.

8 0

4 years ago

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

4 years ago

A new electric car makes use of storage batteries as it source of energy. Its mass is 1300kg and it is powered 26 batteries, eac

Lorico [155]

Answer:

a) Horse power required = 3.575 hp

b) The battery must be charged after approximately 225 km

Explanation:

a) Frictional force, F = 240 N

Average speed of the car = 40 km/h = 40 * (1000/3600) = 11.11 m/s

Horse power required = Power dissipated by the frictional force

Power = Force * Velocity

Power = 240 * 11.11

Power = 2666.67 Watt

746 Watt = 1 hp

2666.67 Watt = 2666.67/746

Horse power required = 3.575 hp

The charge of 1 battery, Q = 52 Ah = 52 * 3600 = 187,200 C

Q = 187,200 C

For 24 batteries, Q = 24 * 187,200

Q = 4492800 C

Power consumed by the batteries, P = QV/t

P = 2666.67 W

2666.67 = (4492800 * 12)/t

t = (4492800 * 12)/2666.67

t = 20217.575 s

Velocity = distance/time

11.11 = distance/20217.575

distance = 20217.575 * 11.11

distance = 224617.26 m

distance = 224.62 km

5 0

3 years ago

1. What would be the final temperature of a mixture of 50g of water at 200C temperature and

ad-work [718]

It would be the average I guess..300c temperature.Heat flows from higher temp. to lower temp. till the temp. of both the liquids are same.

8 0

3 years ago