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2 years ago

WHAT WOULD HAPPEN IF A HUMAN'S BODY TEMPERATURE WAS 98.6 DEGREES CELSIUS??

Physics

2 answers:

Delvig [45]2 years ago

4 0

Answer:

You would die you would be 209.48 Fahrenheit and being higher than 107 Fahrenheit is very bad so imagine what would happen to you :(

Explanation:

qwelly [4]2 years ago

3 0

You would die you would be 209.48 Fahrenheit and being higher than 107 Fahrenheit is very bad so imagine what would happen to you :(

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You have been training and can run a mile in 6.5 minutes. How long would it take to run a lap around the outside track, which is

Paha777 [63]

Answer:

1 minute 36.85 seconds

Explanation:

First we need to convert the miles into meters, as the demanded result should be in meters.

1 mile = 1,609.34 meters

Also, 6.5 minutes should be converted into seconds.

1 minute = 60 seconds

6.5 x 60 = 390 seconds

Now we need to divide the miles with the seconds to see how much meters have been run in a second.

1,609.34 / 390 = 4.13 meters

The suggested meters now should be divided with the distance run in one second.

400 / 4.13 = 96.85 seconds

So we get a result of 96.85 seconds, or 1 minute 36.85 seconds.

5 0

2 years ago

Cognitive,emotional, and physical reasons why adolescents engage in riskier behavior

taurus [48]

Cognitive:
This is because teens tend to excrete more hormones when they are in their adolescent ages causing them to act in a riskier behavior.
Emotions:
Because of their feeling that they want to have freedom.
Physical:
Their hormones are affecting their actions.

8 0

2 years ago

What are the products of the double-replacement reaction between potassium bromide and silver nitrate? Name the resulting compou

timofeeve [1]

Explanation:

A double replacement reaction is a reaction in which two different compounds are mixed together and both their cations and anions get exchanged with each other respectively.

When potassium bromide reacts with silver nitrate then it results in the formation of potassium nitrate and silver bromide.

The chemical reaction equation is as follows.

$KBr + AgNO_{3} \rightarrow KNO_{3} + AgBr$

6 0

2 years ago

Read 2 more answers

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

2 years ago

As a pole of a 2nd-order discrete-time system moves away from the origin in the z-plane, while its phase remains constant, the d

Rudik [331]

Answer:

False

Explanation:

When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.

As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.

ωd = ω₀√(1 - ζ)

Where ζ is called damping ratio.

For small value of ζ

ωd ≈ ω₀

4 0

2 years ago