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yanalaym [24]
2 years ago
9

Body Systems Directly Involved (Give at Least 2)

Physics
1 answer:
Natali [406]2 years ago
6 0

Answer:

Nervous system and immune system

Explanation:

The nervous system is involved because you immediately jumped away, that ur brain and nerves initiating the fight or flight response. The immune system is involved because the tack most likely pierced your skin, so your immune system needs to work to make sure the wound does not get infected.

You might be interested in
What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static
JulijaS [17]

Answer:

<h2>          v= 21.47m/s      </h2>

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

U*m*g=\frac{mv^2}{r}

U*g=\frac{v^2}{r}

substituting our given data in to expression we can solve for the speed V

0.5*9.81=\frac{v^2}{94}

making v the subject of formula we have

0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47

v= 21.47m/s

<em><u>hence the maximum velocity of the car is 21.47m/s</u></em>

7 0
3 years ago
A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
4 0
3 years ago
Eats plants and meats
VMariaS [17]
Omnivores e<span>at both plants and meats. </span>
8 0
3 years ago
Read 2 more answers
ASAP ANSWER PLEASE WILL MARK BRAINLIEST
lisov135 [29]

Yes

Explanation:

It is a reasonable result obtained.

Error = true value - measured value

true value = 24.5

measured value = 24.2

Error =  24.5 - 24.2 = 0.3g

 The error reported in the reading is 0.3g

The reason why we had a disparity in the figures obtained from this measurement is primarily due to some erroneous scale.

The mixture at the end of the day is a solution.

We are expected to have the same mass but due to experimental or some form of random error introduced, we noticed a difference.

The value obtained is quite logical as we only had a deviation of 0.3g.

learn more:

Error brainly.com/question/2764830

#learnwithBrainly

6 0
3 years ago
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
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