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vesna_86 [32]
3 years ago
8

Which of the formulas represents anonpolar molecular?

Physics
1 answer:
makkiz [27]3 years ago
8 0
<span>CH4 I assume you're doing a test or something but I'm not good at explaining things so hopefully it's multiple choice! </span>
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22. The answer is 22.
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You are traveling in a spaceship at a speed of 0.70c away from Earth. You send a laser beam toward the Earth traveling at veloci
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The question is asking to describe and state and calculate what do the observer on the earth measure for the speed of the laser beam, and base on my research, the answer would be v = 1bc, I hope you are satisfied with my answer and feel free to ask for more 
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A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel
lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

\epsilon = - d\frac{\phi_B}{dt}

\epsilon = -NA\frac{dB}{dt}

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N = Number of loops

A = Cross sectional Area

B = Magnetic Field

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Therefore the correct answer is A.

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3 years ago
PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE PUT TRUE ANSWERS
SIZIF [17.4K]

Answer:

D

Explanation:

A is wrong bc best DNA will survive

B is wrong because organism adapt not only for food source but to be able to live in the environment as well

C is wrong because single celled organisms don't adapt

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In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
3 years ago
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