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Sever21 [200]
4 years ago
10

Which are long chains of subunits formed by the process of polymerization? Check all that apply.

Physics
2 answers:
Leya [2.2K]4 years ago
8 0
B,D,F just took the test
vampirchik [111]4 years ago
6 0

The long chains of subunits formed by the process of polymerization are:

B. Macromolecules

D. Polymers

F. DNA

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saul85 [17]

Explanation:

Newton thus the S I unit of force is newton

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3 years ago
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Mass of mercury = 57 g Volume of mercury = 4.2 mL
kari74 [83]
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7 0
3 years ago
Two cars leave towns 800 kilometers apart at the same time and travel toward each other. One car's rate is 18 kilometers per hou
vivado [14]

Answer:

speed of slower car is 71 km/h

Explanation:

The distance between two cars is given as

d = 800 km

time taken by two cars to meet is given as

t = 5 hours

now the relative speed of two cars is given as

v_{rel} = \frac{d}{t}

v_1 + v_2 = \frac{800}{5} = 160 km/h

also it is given that the difference of speed of two cars is

v_1 - v_2 = 18 km/h

now from above two equations we have

v_1 = 89 km/h

v_2 = 71 km/h

7 0
3 years ago
A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the
levacccp [35]

Answer:

M = 6.32 x 10⁻⁶ H

Explanation:

given,

Length of solenoid = 10 cm = 0.1 m

diameter = 0.40 cm

radius = 0.2 cm = 0.002

number of turns, N₁ = 800

                            N₂ = 50

mutual inductance will be equal to  

     M = \dfrac{\mu_0N_1N_2A}{l}

     M = \dfrac{4\pi \times 10^{-7}\times 800\times 50 \times \pi \times (0.002)^2}{0.01}

           M = 6.32 x 10⁻⁶ H

hence, mutual inductance of the combination of two coil is equal to  M = 6.32 x 10⁻⁶ H

7 0
4 years ago
A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant
Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}\ \ =-B\frac{A_2-A_1}{t_2-t_1}   (1)

ФB: magnetic flux =  BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm

To find the areas A1 and A2 you calculate the radius:

r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2

Finally, the emf induced, by using the equation (1), is:

emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0
3 years ago
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