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EleoNora [17]
3 years ago
8

Do you have quantitative data (numerical results or calculations)? Do you have qualitative data (written observations and descri

ptions)? How can you organize this date for your report?
--------------------------------------------
Edge 2020, 6th grade, The lab is on motion and about cars on a slope.
--------------------------------------------
Physics
1 answer:
marin [14]3 years ago
6 0
Organizers can help your writing what comes first second or third or miss like a addy
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If the period of a certain wave (wavelength = 4.5 m) is 2 seconds, what is the speed of the wave?
taurus [48]

Answer:

2.25m/s

Explanation:

wavelength=4.5m

period=2sec

frequency= 1/period

=1/2

speed= wavelength×wave frequency

=4.5×1/2

=2.25 m/s

so, none of the options are correct

5 0
3 years ago
A heavy copper ball of mass 2 kg is dropped from a fiftieth-floor apartment window. Another one with mass 1 kg is dropped immedi
STALIN [3.7K]

Answer:

C

Explanation:

Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!

5 0
3 years ago
How much energy (in joules) is required to evaporate 0.0005 kg of liquid ammonia to vapor at the same temperature?
Lana71 [14]

Answer:

685.6 J

Explanation:

The latent heat of vaporization of ammonia is

L = 1371.2 kJ/kg

mass of ammonia, m = 0.0005 Kg

Heat = mass x latent heat of vaporization

H = 0.0005 x 1371.2

H = 0.686 kJ

H = 685.6 J

Thus, the amount of heat required to vaporize the ammonia is 685.6 J.

6 0
3 years ago
Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area
Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

Explanation:

Given that,

Magnetic field = 42 μT

Area A=3.71\times10^{5}\ km^2

A=3.71\times10^{11}\ m^2

We need to calculate the flux per unit area

flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}

flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2

We need to calculate the total earth's surface area

A'=4\pi r^2

A'=4\times\pi\times(6.3781\times10^{6})^2

A'=5.1120\times10^{14}\ m^2

We need to calculate the rest of earth's area

A''=A-A'

Put the value into the formula

A''=5.1120\times10^{14}-3.71\times10^{11}

A''=5.10829\times10^{14}\ m^2

We need to calculate the net magnetic flux through the rest of Earth's surface

B'=5.10829\times10^{14}\times1.132\times10^{-16}

B'=5.782\times10^{-2}\ T

Hence, The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

3 0
4 years ago
What unit do we use for pressure
-Dominant- [34]

Answer:

Pascal

Explanation:

5 0
3 years ago
Read 2 more answers
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