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sergey [27]
3 years ago
6

A genetics tool that uses letters to represent dominant and recessive alleles

Physics
1 answer:
Studentka2010 [4]3 years ago
7 0

its an punnent square

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A 30-gram bullet is fired and a 50-gram bullet is dropped simultaneously from the same height. Which will hit the ground first?
Marina86 [1]

Answer: the 30gram will hit the ground first

Explanation:the 30gram bullet will hit the ground first because it is fired

4 0
3 years ago
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A skydiver is falling at constant velocity. If a force of 600 N is pulling down on the skydiver, how much force must be acting u
skelet666 [1.2K]

Well, if the skydiver is at constant velocity, than there’s no acceleration, as stated by Newton’s first law. Thus the total net force would equate to 0. In order to make this statement true, the answer would have to be exactly 600 N.

7 0
3 years ago
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Vector vector b has x, y, and z components of 4.00, 4.00, and 2.00 units, respectively. calculate the magnitude of vector
Sav [38]
Good morning.

We see that \mathsf{\overset{\to}{b}} = \mathsf{(4.00, \ 4.00, \ 2.00)}

The magnitude(norm, to be precise) can be calculated the following way:

\star \ \boxed{\mathsf{\overset{\to}{a}=(x, y,z)\Rightarrow ||\overset{\to}{a}|| = \sqrt{x^2+y^2+z^2}}}


Now the calculus is trivial:

\mathsf{\|\overset{\to}{b} \| =\sqrt{4^2+4^2+2^2} =\sqrt{16+16+4}}\\ \\ \mathsf{\|\overset{\to}{b}\|=\sqrt{36}}\\ \\ \boxed{\mathsf{\|\overset{\to}{b}\| = 6.00 \ u}}
7 0
3 years ago
It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
Amiraneli [1.4K]

Answer:

354.72 m/s

Explanation:

m = mass of lead bullet

c = specific heat of lead = 128 J/(kg °C)

L = Latent heat of fusion of lead = 24500 J/kg

T_{i} = initial temperature = 27.4 °C

T_{f} = final temperature = melting point of lead =  327.5 °C

v = Speed of lead bullet

Using conservation of energy

Kinetic energy of bullet = Heat required for change of temperature + Heat of melting

(0.5) m v^{2} = m c (T_{f} - T_{i}) + m L\\(0.5) v^{2} = c (T_{f} - T_{i}) + L\\(0.5) v^{2} = (128) (327.5 - 27.4) + 24500\\(0.5) v^{2} = 62912.8\\v = 354.72 ms^{-1}

3 0
3 years ago
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
4 years ago
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