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

Analogy: atom is to element as ____ is to compound

Masja [62]

Molecule is the answer to what you are looking for

5 0

4 years ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

6 0

3 years ago

Thing 1 and Thing 2 push with a force of 2000 N to move a rather small 500 kg elephant 30 m across the kitchen floor. How much w

Amanda [17]

Answer:

60000 J

Explanation:

Assuming the force is applied parallel to the displacement of the elephant, the work done to move it across the floor is

$W=Fd$

where

F = 2000 N is the force applied

d = 30 m is the displacement of the elephant

Substituting the numbers into the formula, we find

$W=(2000 N)(30 m)=60,000 J$

3 0

3 years ago

The latent heat of fusion of alcohol is 25 kcal/kg and its melting point is -114 o C. It has a specific heat of 0.60 in its liqu

UNO [17]

Answer:

=170kcal

Explanation:

We first calculate the amount of energy required to melt the alcohol using the formula: MLf, where Lf is the latent heat of fussion

We then calculate amount of heat required to raise the temperature of liquid alcohol to -14° C using MC∅. We then add the two.

Thus ΔH=MLf+MC∅

ΔH=2kg×25kcal/kg+ 2kg×(0.6kcal/kg.K×(-14-⁻114)

=50kcal+120kcal

=170kcal

3 0

3 years ago

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .

3 0

3 years ago