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galben [10]
3 years ago
14

Which landform represents the boundary between the land and an ocean or a lake?

Physics
2 answers:
aleksklad [387]3 years ago
8 0

Answer:

coastline

Explanation:

its the coastline

Likurg_2 [28]3 years ago
3 0
Coastline or seashore
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A scientist measures the number of flies in a room over time. The results are
snow_lady [41]
A. The number of files was constant.
8 0
3 years ago
Which statement describes the difference between renewable and nonrenewable energy sources
castortr0y [4]
Renewable souces are sources that replenish naturally, they can never be depleted. Nonrenewable sources have a limited amount of and cannot be replenished forever...
3 0
4 years ago
Please help me with physics ks3!! Sound waves and hearding (pic incl)
Nina [5.8K]
Your answers for a), b), and c) are correct.  Good work !

d). The high noise levels in Technology and PE are more of
a concern to the teachers than to the students because the
students are only in there for 1 or 2 periods a day, but the
Technology and PE teachers are in there ALL day.

e).  Mr. Jones can't hear as high frequency as Jenny can because
he is much older than Jenny is.  Sadly, even without damage due to
loud noise, the ability to hear high frequencies does decrease with age.

f).  The wooden surfaces in the gym cause the gym to be louder
than it would be if it had carpet on the floor.  Carpet ... and soft
walls and ceilings ... absorb a lot of the sound that hits them. 
But hard surfaces don't absorb much of the sound that hits them,
so it just keeps bouncing around until it finally fades away. 

You can see this easily ... just go into the gym at your school, clap
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7 0
3 years ago
Read 2 more answers
A 55 meter rope is lying on the floor and has has a weight of 4040 N in total. How much work is required to lift up one end of t
Kazeer [188]

Answer:

Explanation:

Given

Length of rope L=55\ m

Weight of rope W=4040\ N

weight density \lambda =\frac{4040}{55}=73.45\ N/m

Work done to lift rope 33 m

W=\int_{0}^{33}\lambda hdh

W=\int_{0}^{33}73.45hdh

W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0

W=39.993\ kJ      

6 0
3 years ago
At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar
swat32

Answer:

C=1,25\cdot 10^{5} kJ/^{\circ}C

Explanation:

First of all let's define the specific molar heat capacity.

C = \frac{-Q}{n\cdot \Delta T} (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system  

Now, we can find n with the molar mass (M) the mass of the compound (m).

n=\frac{m}{M}=6.95\cdot 10^{-3} moles      

Using (1) we have:

C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}

C=1,25\cdot 10^{5} kJ/^{\circ}C

I hope it helps!

6 0
3 years ago
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