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ahrayia [7]
2 years ago
13

Alan and Monica tested how fertilizer X affected the growth of a plant. Alan and Monica put 100 grams of fertilizer X in pot 1.

They put 0 grams of fertilizer X in pot 2.
What is pot 2 called in science?
A.
a test group
B.
a data group
C.
an observational group
D.
a control group
Physics
1 answer:
BARSIC [14]2 years ago
7 0

Answer:

D.

a control group

Explanation:

In a scientific experiment such as the one above, there is an experimental group and a control group. The experimental group is the group that receives the treatment while the control group does not receive any treatment. The control group helps the researcher to observe if the treatment had any significant effect.

In this case, it will help Alan and Monica to determine if fertilizer X actually had an effect on the plant. Therefore, the pot with o grams of fertilizer in it is the control group.

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The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o
blsea [12.9K]

Answer:

F_f=840N

Explanation:

From the question we are told that

Weight of fireman W_f= 87.5kg

Pole distance D=4.10m

Final speed is V_f 1.75m/s

Generally the equation for velocity is mathematically represented as

v^2 = v_0^2 + 2 a d

Therefore Acceleration a

a'= v^2 / 2 d

a'= 0.21m/s^2

Generally the equation for Frictional force F_f is mathematically given as

F_f=m*a

F_f=m*(g-0.21)

F_f=87.5*(9.81-0.21)

Therefore

F_f=840N

6 0
3 years ago
For phyics again , pls !!!
CaHeK987 [17]

Answer: same

Explanation: They both weigh a kilogram and there is no friction

3 0
3 years ago
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le
BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7  ×  10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5  × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) ×  5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0
3 years ago
A moving particle encounters an external electric field that decreases its kinetic energy from 9970 eV to 6340 eV as the particl
marusya05 [52]

Answer:

q = -1.61x10⁻¹⁷ C

Explanation:

The charge of the particle can be found using the definition of the work done by electric force:  

W = -q\Delta V         (1)

<u>Where</u>:

q: is the charge

ΔV: is the difference in electric potential

The work is also equal to:

W = E_{p_{A}} - E_{p_{B}}    (2)

<u>Where</u>:

E_{p_{A}} and E_{p_{B}} are the electric potential energy of the points A and B, respectively.

Now, by conservation of energy we have:

K_{A} + E_{p_{A}} = K_{B} + E_{p_{B}}     (3)

<u>Where</u>:  

K_{A} and K_{B} are the kinetic energy of the points A and B, respectively.

Rearranging equation (3):  

K_{B} - K_{A} = E_{p_{A}} - E_{p_{B}}      

K_{B} - K_{A} = W

K_{B} - K_{A} = -q\Delta V

Solving the above equation for q:

q = -\frac{K_{B} - K_{A}}{V_{B} - V_{A}} = -\frac{6340 eV - 9970 eV}{19.0 V - 55.0 V} = -100.83 e \cdot \frac{1.6 \cdot 10^{-19} C}{1 e} = -1.61 \cdot 10^{-17} C                                              

Therefore, the charge of the particle is -1.61x10⁻¹⁷ C.

I hope it helps you!

7 0
3 years ago
Question 8 Unsaved
julia-pushkina [17]
The first one:
Like poles repel each other and opposite poles attract each other
3 0
3 years ago
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