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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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Is the ratio between the sine of an angle of incidence to the sine of an angle of refraction is called the refractive index?
motikmotik

Answer:

Yes

Explanation:

Yes it is called the refractive index denoted by n

n=sin<i/sin<r

6 0
3 years ago
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A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

                                 = 21.99kN

At Equilibrium the summation of the of force on the horizontal axis is zero

                  i.e   \sum F_x =0

=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

               Since The speed at both A and B nozzle are the same then v_2 remains the same

 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

6 0
2 years ago
Read 2 more answers
How are musical instruments grouped?
rosijanka [135]

A and B are equivalent.  That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")

C is another way instruments are often grouped; (the "woods", the "brass")

D is another way instruments are often grouped; (the "strings", the "percussions")

6 0
3 years ago
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Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

y=\frac{\lambda (m+\frac{1}{2})D}{d}

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

\lambda is the wavelength of the light used

In this problem we have:

\lambda=683 nm = 683\cdot 10^{-9} m is the wavelength of the light

d=1.1 mm = 0.0011 m is the width of the slit

m = 13 is the order of the minimum

y=8.57 cm = 0.0857 m is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m

6 0
3 years ago
An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average speed for the tri
Anettt [7]

Answer:

270 mi/h

Explanation:

Given that,

To the south,

v₁ = 300 mi/h, t₁ = 2 h

We can find distance, d₁

d_1=v_1\times t_1\\\\d_1=300\times 2\\\\d_1=600\ \text{miles}

To the north,

v₂ = 250 mi/h, d₂ = 750 miles

We can find time, t₂

t_2=\dfrac{d_2}{v_2}\\\\t_2=\dfrac{750\ \text{miles}}{250\ \text{mi/h}}\\\\t_2=3\ h

Now,

Average speed = total distance/total time

V=\dfrac{d_1+d_2}{t_1+t_2}\\\\V=\dfrac{600+750}{2+3}\\\\V=270\ \text{mi/h}

Hence, the average speed for the trip is 270 mi/h.

3 0
2 years ago
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