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STatiana [176]
3 years ago
10

When a person's temperature reaches 104 degrees, the chances of survival are decreased dramatically. Group of answer choices Tru

e False
Physics
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

True

Explanation:

With the increase in temperature hypothalamus fails and heatstroke occurs due to this failure. Hypothalamus is the region of our brain that act as a thermostat. It co-ordinates our physiological response to excessive heat. When the person’s temperature reaches to 104 degrees then it causes heatstroke. This heatstroke is very sudden and can kill person. Hence, we can conclude that when person’s temperature reaches to 104 degrees chances of survival decreases dramatically.

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Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.
Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

F=\frac{\Delta p}{\Delta t}

where

F is the force applied

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be written as

\Delta p=m(v-u)

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

F=\frac{m(v-u)}{\Delta t}

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s

4 0
3 years ago
A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a
PSYCHO15rus [73]

Answer:

1.28 x 10^4 N

Explanation:

m = 1500 kg, h = 450 km, radius of earth, R = 6400 km

Let the acceleration due to gravity at this height is g'

g' / g = {R / (R + h)}^2

g' / g = {6400 /  (6850)}^2

g' = 8.55 m/s^2

The force between the spacecraft and teh earth is teh weight of teh spacecraft

W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N

8 0
3 years ago
1. why is<br> experimentation so important to science
Alekssandra [29.7K]

Answer: to provide evidence to a theory

Explanation: Experimentation allows for multiple trials to provide evidence to a scientific theory. Without experimentation there would be no data to back up your hypothesis.

5 0
3 years ago
Find the acceleration of a hamster when it increases its velocity from rest to 8.0 m/s in 3.0 s .
Eva8 [605]
<span>Acceleration is the change in velocity divided by time.
We can find acceleration a, by the following formula

a=v-u/t

where,
v is the final velocity (in this question v=8.0 m/s)
u is the initial velocity (since the hamster starts from rest, u=0)
t is the time taken (i,e 3.0 second)
now by applying the formula we have,

a = 8.0 - 0 / 3
= 8 / 3
= 2.65 m/s</span>²<span> 

The acceleration is 2.65 meters per second squared</span>
6 0
3 years ago
A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10r, where a and fare in m/s^2
xz_007 [3.2K]

Answer:

Mistake in question

The correct question

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

Explanation:

Given the function

a = 50 —10t

The car started from rest u = 0

And it accelerates to a speed of 125m/s

Then, let find the time in this stage

Acceleration can be modeled by

a = dv/dt

Then, dv/dt = 50—10t

Using variable separation to solve the differentiation equation

dv = (50—10t)dt

Integrating both sides

∫ dv = ∫ (50—10t)dt

Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'

∫ dv = ∫ (50—10t)dt

v = 50t —10t²/2. Equation 1

[v] 0<v<125 = 50t —10t²/2 0<t<t'

125—0 = 50t — 5t² 0<t<t'

125 = 50t' — 5t'²

Divide through by 5

25 = 10t' — t'²

t'² —10t' + 25 = 0

Solving the quadratic equation

t'² —5t' —5t' + 25 = 0

t'(t' —5) —5(t' + 5) = 0

(t' —5)(t' —5) = 0

Then, (t' —5) = 0 twice

Then, t' = 5 seconds twice

So, the car spent 5 seconds to get to 125m/s.

The second stage when the parachute was deployed

We want to the time parachute reduce the speed from 125m/s to 10m/s,

So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''

The function of deceleration is give as

a = - 0.02v²

We know that, a = dv/dt

Then, dv/dt = - 0.02v²

Using variable separation

(1/0.02v²) dv = - dt

(50/v²) dv = - dt

50v^-2 dv = - dt

Integrate Both sides

∫ 50v^-2 dv = -∫dt

(50v^-2+1) / (-2+1)= -t

50v^-1 / -1 = -t

- 50v^-1 = -t

- 50/v = - t

Divide both sides by -1

50/v = t. Equation 2

Then, v ranges from 125 to 10 and t ranges from 0 to t''

[ 50/10 - 50/125 ] = t''

5 - 0.4 = t''

t'' = 4.6 seconds

Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.

So the total time is

t = t' + t''

t = 5 + 4.6

t = 9.6 seconds

b. Total distanctraveleded.

First case again,

We want to find the distance travelled from t=0 to t = 5seconds

a = 50—10t

We already got v, check equation 1

v = 50t —10t²/2 + C

v = 50t — 5t² + C

We add a constant because it is not a definite integral

Now, at t= 0 v=0

So, 0 = 0 - 0 + C

Then, C=0

So, v = 50t — 5t²

Also, we know that v=dx/dt

Therefore, dx/dt = 50t — 5t²

Using variable separation

dx = (50t —5t²)dt

Integrate both sides.

∫dx = ∫(50t —5t²)dt

x = 50t²/2 — 5 t³/3 from t=0 to t=5

x' = [25t² — 5t³/3 ]. 0<t<5

x' = 25×5² — 5×5³/3 —0

x' = 625 — 208.333

x' = 416.667m

Stage 2

The distance moved from

t=0 to t =4.6seconds

a = -0.002v²

We already derived v(t) from the function above, check equation 2

50/v = t + C.

When, t = 0 v = 125

50/125 = 0 + C

0.4 = C

Then, the function becomes

50/v = t + 0.4

50v^-1 = t + 0.4

Now, v= dx/dt

50(dx/dt)^-1 = t +0.4

50dt/dx = t + 0.4

Using variable separation

50/(t+0.4) dt = dx

Integrate both sides

∫50/(t+0.4) dt = ∫ dx

50 In(t+0.4) = x

t ranges from 0 to 4.6seconds

50In(4.6+0.4)—50In(4.6-0.4) = x''

x'' = 50In(5) —50In(4.2)

x'' = 8.72m

Then, total distance is

x = x' + x''

x = 416.67+8.72

x = 425.39m

The total distance travelled in both cases is 425.39m

5 0
3 years ago
Read 2 more answers
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