Which of the following choices is an example of an allele?

Consider a pool shaped like the bottom half of a sphere, that is being filled. The radius of the pool is 10 ft. If the height is

Stells [14]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

As electric current moves through a wire, heat generated by resistance is conducted through a layer of insulation and then conve

mestny [16]

Answer:

Hello your question is incomplete attached below is the complete question

answer : ri ( thickness of wire ) = 14.167 m

Explanation:

attached below is a detailed solution

using the given data to determine the thickness of wire

5 0

2 years ago

A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f

scoray [572]

Answer:

BA

0

$\dfrac{NBA}{\Delta t}$

Explanation:

B = Magnetic field

A = Area

$\theta$ = Angle

t = Time taken

Before rotation the magnetic flux is given by

$\phi_i=BAsin\theta\\\Rightarrow \phi_i=BAsin90\\\Rightarrow \phi_i=BA$

Magnetic flux is BA

After rotation the magnetic flux is given by

$\phi_i=BAsin\theta\\\Rightarrow \phi_i=BAsin0\\\Rightarrow \phi_i=0$

The magnetic flux is 0

Magnitude of emf is given by

$\epsilon=NA\dfrac{d\phi}{dt}\\\Rightarrow \epsilon=\dfrac{NBA}{\Delta t}$

The magnitude of the average emf induced in the entire coil is $\dfrac{NBA}{\Delta t}$

4 0

3 years ago

A small metal sphere weighs 0.34 N in air and has a volume of 13 cm3 . What is the acceleration of the sphere as it falls throug

Xelga [282]

To solve this problem we will apply the concepts related to the balance of forces. Said balance will be given between buoyancy force and weight, both described as derived from Newton's second law, are given as

Buoyancy force

$F_B = V\rho g$

Here,

V = Volume

$\rho$ =Density of air

g = Acceleration due to gravity

Weight

$F_W = mg$

m = mass

g = Gravity

Our values are given as,

$\text{Weight of the sphere} = W = 0.34 N$

$\text{Volume} = V = 13 cm^3 = 13*10^{-6}m^3$

$\text{density of air} =\rho =1.29kg/m^3$

$\text{gravity}= g = 9.8 m/s^2$

Then,

$F = V\rho g$

Replacing,

$F = (13*10^{-6}m^3 )(1.29Kg/m^3)( 9.8 m/s^2) = 1.6434* 10^{-4} N$

Now net force is ,

$F_{net} = mg - F$

Mass of the sphere is

$m = \frac{W}{g} = \frac{0.34N}{9.8m/s^2} = 0.03469 kg$

Now acceleration of the sphere is

$a = \frac{F_{net}}{m}$

$a = \frac{( 0.34 N)- (1.6434* 10^{-4} N)}{0.03469 kg}$

$a = 9.822m/s^2$

Therefore the acceleration of the sphere as it falls through water is $9.822m/s^2$

5 0

3 years ago

A first-order reaction has a rate constant of 0.241/min. if the initial concentration of a is 0.859 m, what is the concentration

nordsb [41]

Answer: 0.077 M

Explanation:

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant = $0.241minute^{-1}$

t = time taken for decay process = 10 minutes

a = initial amount of the reactant= 0.859 M

a - x = amount left after decay process =?

Putting values in above equation, we get:

$0.241 minutes^{-1}=\frac{2.303}{10.0}\log\frac{0.859}{a-x}$

$(a-x)=0.077M$

Thus the concentration of a after 10.0 minutes is 0.077 M.

5 0

3 years ago

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