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3 years ago

13

What is the difference between heat exhaustion and heat stroke?

2 answers:

balu736 [363]3 years ago

6 0

I believe the answer is D, Heat exhaustion involves a lack of sweating, while heat stroke involves extreme sweating. Also just to add the on if heat exhaustion is left untreated then it could turn into a heat stroke.

laila [671]3 years ago

5 0

Answer: the answer is D, Heat exhaustion involves a lack of sweating, while heat stroke involves extreme sweating. Also just to add the on if heat exhaustion is left untreated then it could turn into a heat stroke.

Explanation:

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A car has a mass of 1000 kg. What is the acceleration produced by a force of 2000 N?

EastWind [94]

F=ma
a=F/m
a=2000/1000
a=2 m/s^2

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2 years ago

In an automobile collision, a 44-kilogram passenger moving at 15 meters per second is brought to rest by an air bag during a 0.1

const2013 [10]

Answer:

6,600N

Explanation:

According to second law of motion, Force = mass × acceleration

If acceleration = change in velocity/time = 15/0.10

Acceleration = 150m/s²

Given mass = 44kg

Force = 44× 150

Force = 6,600N

Magnitude of the average force exerted on the passenger during this time is 6,600N

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3 years ago

The star Betelgeuse is about 600 light-years away. If it explodes tonight,

Thepotemich [5.8K]

Answer:

C

Explanation:

It has to travel 600 light years before we would be able to observe the explosion.

4 0

4 years ago

Read 2 more answers

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

B)

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago

A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf

makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A

Resposta: 6 m/s^2 e 12N

Bate com o gabarito, man? Ou eu tô viajando aqui?

Abç!

6 0

3 years ago