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joja [24]
3 years ago
12

Which is not a good not a function of proteins

Physics
1 answer:
Yuki888 [10]3 years ago
3 0

Answer:

options please???

Explanation:

You might be interested in
if heat energy is transferred from direct contact between a warm and a cold object it has been transferred by what ?
Katarina [22]

When heat energy is transferred from direct contact between a warm and a cold object , it is known as heat transfer by conduction.

In conduction, the heat transfer takes place within an object or between two objects in contact until the temperature becomes uniform. this kind of heat transfer continues until the temperature at two ends between which the heat transfer take place , becomes equal.  Heat transfer takes place from point at high temperature to point at lower temperature.

5 0
3 years ago
A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
a scientist is studying the effects of sunlight on rose bushes. which statement is a hypothesis that he can test be preforming a
DaniilM [7]

Answer:

If a rose bush is placed in direct sunlight then it will grow taller than a rose bush placed in indirect sunlight.

Explanation:

8 0
3 years ago
A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
jeka57 [31]

Answer:

a. 3.1 m/s^2

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

mg sin \theta - \mu_k R = ma

where :

m = 6.0 kg is the mass  of the box

g = 9.8 m/s^2 the acceleration of gravity

\theta=39^{\circ}  is the angle of the slope

\mu_k = 0.40 is the coefficient of friction

R is the normal reaction  

a is the acceleration

- Along the perpendicular direction:

R-mg cos \theta =0

From the 2nd equation, we get an expression for the reaction force:

R=mg cos \theta

And substituting into the 1st equation, we can find the acceleration:

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2

6 0
3 years ago
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