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Aleks04 [339]
3 years ago
8

1). An Owl and bat share the same kingdom and phylum; an owl and a cardinal share the same kingdom, phylum, and class. The owl a

nd __________________ have more characteristics in common.
2). The ______________ kingdom is the only kingdom of eukaryotes that contains both autotrophs and heterotrophs and both unicellular and multicellular organisms.

Trying to finish my homework -_- Halppp!! and I have more questions on the way......so yeah sorry for being a bother.
If you answer this it gives you 24pts.
Physics
2 answers:
strojnjashka [21]3 years ago
8 0
The first is the Cardinal because the Owl shares more levels of classification with it.

The second is Protista. The only way I see to enforce the idea is just the knowledge that it is a fact.

I hope this Helps!
kolbaska11 [484]3 years ago
7 0
1. cardinal

2.  Protista

Glad to help :)
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If the mass of a material is 42 grams and the volume of the material is 15 cm^3, what would the density of the material be?
Fantom [35]

Density = mass / volume

Density = 42g / 15cm^3

Density = 2.8g/cm^3

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What is the maximum number of primary partitions that gpt supports?
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Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb
GuDViN [60]

Answer:

Q = 12540  J

Explanation:

It is given that,

Mass of water, m = 50 mL = 50 g

It is heated from 0 degrees Celsius to 60 degrees Celsius.

We need to find the energy required to heat the water. The formula use to find it as follows :

Q=mc\Delta T

Where c is the specific heat of water, c = 4.18 J/g°C

Put all the values,

Q=50\times 4.18\times (60-0)\\Q=12540\ J

So, 12540 J of energy is used to heat the water.

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If a car is moving backward and has negative acceleration, what can be said about the speed of the car?
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Read 2 more answers
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
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