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3 years ago

6

Which of the following is NOT true about sweat? Sweat helps cool the body. Sweat glands are located all over the body. Sweat hel

ps regulate body temperature. Sweat is used to sense pain or pressure changes.

2 answers:

erma4kov [3.2K]3 years ago

8 0

The answer is c!!!!!!!!!!!!!!!

Klio2033 [76]3 years ago

5 0

Sweat is used to sense pain or pressure change.
I eliminated all the ones i wouldnt choose

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At what time were the racers at the same position?<br> O 15<br> O 3s<br> O 5s<br> 0 4s

Jlenok [28]

The Answer is:

O 3s

Hope you got it right.

5 0

3 years ago

Read 2 more answers

A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo

guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}$

If we consider than the direction of the electric field is $\vec{E}=E_0\hat{x}$ , we can solve the problem differentiating the integral for each face of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6$

E₀ is a constant and each surface is equal to each other, so: $S_1=S_2=S_i=S$

Therefore:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c$

3 0

3 years ago

1. A boy walk 140 m due north, 85 m due east, 35 m due southeast, 38 m due west and then 19 m due northwest. Calculate the displ

Dima020 [189]

Answer:

141 m at 65.6° N of E

Explanation:

Let E be along the positive x axis of a unit circle

N = 90°

E = 0°

SE = -45°

W = 180°

NW = 135°

east displacement

x = 140cos90 + 85cos0 + 35cos-45 + 38cos180 + 19cos135 = 58.313708... m

north displacement

y = 140sin90 + 85sin0 + 35sin-45 + 38sin180 + 19sin135 = 128.6862915... m

d = √(128.6862915² + 58.313708²) = 141.28216525... m

tanθ = 128.6862915 / 58.313708

θ = 65.622521...

7 0

3 years ago

A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: $E_k=\frac{1}{2}mv^2$

2. gravitational potential energy of mass m, grav. acc. g and height h: $E_g=mgh$

3. potential energy in a spring with spring constant k and displacement from equilibrium x: $E_s=\frac{1}{2}kx^2$

Calculating x:

$\frac{1}{2}mv_a^2=\frac{1}{2}kx^2$

$x=\sqrt{\frac{m}{k}}v_a$

Calculating the speed:

$\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}$

$h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R$

$\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R$

Solving for $v_b$ :

$v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}$

7 0

3 years ago

A machine puts out 6,000 J of work. To produce that much work the machine

topjm [15]

Answer:

Efficiency of the machine = 75%

Explanation:

Given:

Input work = 8,000 J

Output work = 6,000 J

Find:

Efficiency of the machine

Computation:

Efficiency of the machine = [Output work / Input work]100

Efficiency of the machine = [6,000 / 8,000]100

Efficiency of the machine = 75%

5 0

2 years ago