*15 points*

What is role of skin in balancing body temperature during strenous excervise

trapecia [35]

Answer:

The skin releases sweat

Explanation:

The body temperature is necessary to maintain homeostasis. Skin controls body temperature by sweating when it is too heated.

3 0

3 years ago

Can I have answers of all these questions?<br> Urgent plz!

Leona [35]

Q3. (a) 0m/s, as they are asking for initial velocit.
(b)(i) The paper has a large surface area or weighs less than the coin,thus,falls smoothly.
(ii)The coin has more mass than the paper.
(c) They fall at the same acceleration and hit the bottom at the same time.
Their mass doesn't matter in the vacuum.

7 0

2 years ago

The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit

Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V= 30 m/s ( + x direction )

d) λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

As we know that standard form of wave equation given as

$y=A sin(\phi -\omega t)$

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s ( + x direction )

V = f λ

30 = 95.49 x λ

λ = 0.31 m

We know that speed is the rate of displacement

$U=\dfrac{dy}{dt}$

$U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

$U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s$

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

3 years ago

What does the person in back of the ambulance experience? A) a lower frequency of the siren B) a lower amplitude of the siren C)

hoa [83]

The answer to your question is "A. a lower frequency of the siren.

Because the person in back of the ambulance will hear a lower frequency of the siren. This is because the waves are stretched out. A longer wavelength results in a lower frequency.

5 0

3 years ago

Read 2 more answers