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Oksana_A [137]
2 years ago
9

Bob wanted to investigate the effects of plant fertilizer. He bought three identical plants and gave plant 1 fertilizer every mo

rning; plant 2 fertilizer once a week, and he didn't use any fertilizer on plant 3. Bob watered each plant every morning and set all three plants on the windowsill so they would all have the same amount of sunlight. He measured the height of each plant once a week . He plotted plant height (y-axis)vs time (x-axis) for each plant. What is the independent variable * 1 point
Physics
1 answer:
olga_2 [115]2 years ago
8 0

Answer:

TIME he applied the fertilizer to each plant

Explanation:

Independent variable in an experiment is the variable that is subject to change or manipulation by the experimenter. In this experiment, Bob wanted to investigate the effects of plant fertilizer. Bob sets up the experiment by applying the fertilizer to each plant at DIFFERENT TIMES i.e. plant 1-every morning, plant 2-once a week, plant 3-never.

Based on this, it is obvious that the independent or manipulated variable is the TIME at which he applied the fertilizer. On the other hand, the dependent or measured variable is the height of the plants.

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The bird that migrates the farthest is the Arctic tern. Each year, the Arctic tern travels
Tanzania [10]

Answer: 131.14km per day

Explanation: since the second half of the terns migration takes 122 days we can assume that the full migration would take 244 days. using this we can divide the total distance by the total amount of days it takes (because speed = distance/time) which is 32,000/244, which would be 131.14

8 0
2 years ago
Please help really easy worth 20
vovangra [49]

Answer:

Stationary

20N

Explanation:

From the graph, we see that the body traveling is on a fixed position. Therefore, it is a stationary body.

The graph given is a position - time curve.

 This curve depict a body changing position with given time.

Since the line of the curve is on a single position, the body is not changing position with the passage of time therefore, it is a stationary object.

B. 20N

From Newton's third law of motion we understand that "action and reaction force are equal but oppositely directed".

 Since the person is exerting a force of 20N on the balance.

So, the reaction force by the balance is 20N upward.

4 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

(1)(2.5)/(0•50)=P₂

P₂=5

Pressure is now 5.0 atm

8 0
3 years ago
A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the
Shalnov [3]

Heat required to raise the temperature of a given system is

Q = ms\Delta T

here we know that

m = mass

s = specific heat capacity

\Delta T = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2

Q = 5 * 1700 * 25 + 2 * 900 * 25

Q = 212500 + 45000

Q = 257500 J

So heat required to raise the temperature of the system is 257500 J

4 0
3 years ago
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