1. How does RNA differ from DNA?

Pls pls help me out AHH, what is not true about MEIOSIS?

agasfer [191]

Answer:

B.

Explanation:

This is not true as the number of chromosomes in the daughter cells are half the number in the parent's cells

4 0

3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0

3 years ago

Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par

kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so: