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pogonyaev
3 years ago
13

Jeff is investigating factors that affect the growth rate of potted bean plants. Which of the following experimental variables w

ould not be relevant to his investigation?
A. the type of soil used
B. the amount of water given
C. the color of the pot
D. the amount of light exposure
Physics
2 answers:
Kipish [7]3 years ago
8 0
It would be C. the color of the pot. its pretty obvious that i would not effect the project.

aksik [14]3 years ago
5 0

Answer:

C. the color of the pot

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A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule
saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

m_1v_1+m_2v_2=0

170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0

v_2=-0.17\ m/s

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0
3 years ago
Sound travels through water at a speed of 1500 m/s. If the frequency of a sound is 1000 Hz, what is the wavelength?
ratelena [41]

Answer:

1.5m

Explanation:

Velocity=1500m/s

Frequency=1000hz

Wavelength =velocity ➗ frequency

wavelength =1500 ➗ 1000

Wavelength=1.5m

3 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
Who live in a pineapple under the sea?
saw5 [17]

Answer:

SpongeBob SquarePants

5 0
3 years ago
0. 85 kg of lead, specific heat 128 J/kgC, is heated from 50 C to 95 C. How much heat energy did the sample absorb?
Zanzabum
4896

0.85 x 45 x 128 = 4896

Change in energy = specific heat capacity x mass x change in temperature
5 0
2 years ago
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