What is the dependent variable in this
experiment?
DONE
Biologists designed an experiment to test
the effect of compost on the development
of root crops. They tested several different
crops, including carrots, potatoes, beets,
and onions. They grew most of the plants
in the greenhouse, but due to space issues,
they had to grow some outdoors. They gave
all the plants the same amount of compost.
They obtained the compost from a local
farmer and from the local hardware store.
They ran out of the farmer's compost, so
some of the plants received that compost
when the seeds were planted and other
plants got hardware store compost after
the plants had already started growing.
What is the independent variable in this
experiment?
DONE

Answer:

The magnetic field is doubled.

Explanation:

What happens to the strength of the magnetic field if one more 1.5 v dry cell is added?

The magnetic field due to a current carrying conductor is directly proportional to the current flowing in the wire.

If we connect one more battery of 1.5 V so the voltage is doubled and according to the Ohm's law, as the resistance of wire is constant, so the current in the wire is also doubled.

When the current doubles, the magnetic field produced by the wire is also  doubled.

Speed = distance /  time

Speed = 12km /  5 minutes

Speed = 2.4 km per minute

Speed is measured per hour, 1 hour has 60 minutes.

2.4 km per minute x 60 minutes = 144 km per hour

Answer:

-8.036 kV/m

Explanation:

The electric field E = -ΔV/Δx where ΔV = change in electric potential = V - V' where V = electric potential at x = 5.6 cm =  450 V and V' = electric potential at x = 0 cm, = 0 V . So, ΔV = V - V' = 450 V - 0 V = 450 V.

Δx = distance between the 0 V plate and the 450 V point = 5.6 cm = 0.056 m

So, E = -ΔV/Δx

Substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -450 V/0.056 m

E = -8035.7 V/m

E = -8.0357 kV/m

E ≅ -8.036 kV/m

Since the electric field between two parallel conducting plates is constant, the electric field between the plates is E = -8.036 kV/m

Answer:

D. bacteria feeding on the large volume of carp feces depleted the oxygen

Explanation:

In the context, it is given that in a small farm pond, the owner added 20 grass carp and feed them with the abundant plants and algae that is found on her pond. The carp grew large after many years but one summer the owner found the grass carps along with other fishes were dead.

The most possible explanation for the dead of the fishes in the pond because the bacteria feeds on the carp feces which depletes the dissolved oxygen present in the water. Thus the fishes could not breathe and were finally dead.

Therefore, the correct option is (D).

Answer:

U = - ∫ F dx

U = - 3F₀

Explanation:

Force and potential energy are related

         F = [tex]- \frac{dU}{dx}[/tex]

         dU = - F dx

        ∫ dU = - ∫ F dx

        U- U₀ = - ∫ F dx

to solve the problem we must know the form of the force, if we assume that F = F₀ and U₀ = 0 for x = 0

         U = - F₀ 3

The answer would be B..

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

Answer:

Make a nice powerpoint and think about the question

Explanation:

Its easier than you think :)

Answer:

1kg = 1000g

67kg therefore would be equal to

67 x 1000 = 67,000g = 6.7 x 10⁴g.

Answer:

(a) 1.5 nF, 1.2 nF, 1 nF

(b) 0.4 nF

Explanation:

V = 150 V

V' = 40 V, V'' = 50 V, V''' = 60 V, q = 6 x 10^-8 C

(a) C' = q/V' = 6 x 10^-8 / 40  = 1.5 x 10^-9  F

C'' = q/V'' = 6 x 10^-8 / 50 = 1.2 x 10^-9 F

C''' = q/V''' = 6 x 10^-8 / 60 = 1 x 10^-9 F

(b) The effective capacitance is

[tex]\frac{1}{C}=\frac{1}{C'}+\frac{1}{C''}+\frac{1}{C'''}\\\\\frac{1}{C}=\frac{10^9}{1.5}+\frac{10^9}{1.2}+\frac{10^9}{1}\\\\C = 0.4\times 10^{-9} F[/tex]  

Answer:

Explanation:

The PE equation for a mass/spring system is

[tex]PE=\frac{1}{2}k[/tex]Δx² and filling in:

[tex]130=\frac{1}{2}k(.10)^2[/tex] and

[tex]k=\frac{2(130)}{(.10)^2}[/tex] so

k = 26000 N/m

If the displacement from equilibrium changes more, the PE needed to compress it will also change.

[tex]PE=\frac{1}{2}(26000)(.20)^2[/tex] gives us that

PE = 520J

the order of vector addition doesn't affect the resultant vector and grouping or order of pair doesn't effect the sum.

Answer:

(n) alluvial fan sandbar

Explanation:

Answer:

Part A

a)  F = -16x + 4,  b)  x = 0.25 m, c) STABLE

Explanation:

Part A

a) Potential energy and force are related

          F = [tex]- \frac{dU}{dx}[/tex]- dU / dx

          F = - (8 2x -4)

          F = -16x + 4

b) The object is in equilibrium when the forces are zero

          0 = -16x + 4

          x = 4/16

          x = 0.25 m

c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.

In this case there is only one equilibrium point

by changing the position a bit

           x ’= x + Δx

we substitute

          F ’= - 16 x’ + 4

          F ’= - 16 (x + Δx) + 4

          F ’= (-16x +4) - 16 Δx

at equilibrium position F = 0

          F ’= 0 - 16 Δx

we can see that the body returns to the equilibrium position, therefore it is STABLE

PART B

This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved

initial instant. Before the shock

        p₀ = m v

final instant. After the crash

        p_f = (m + M) v_f

We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic

          p₀ = pf

          mv = mv ’+ M v_f

in the case of the elastic collision, the kinetic energy is conserved

        K₀ = K_f

        ½ m v² = ½ m v’² + ½ M v_f²

we write the system of equations

        mv = mv ’+ M v_f (1)

         m (v² -v'²) = M v_f ²

         m (v - v ’) = M v_f

         m (v-v ’) (v + v’) = M v_f

        v + v ’= v_f

we substitute in equation 1 and solve

         v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]

         v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]

the mechanical energy of the neutron is

  initial

          Em₀ = K = ½ m v²

final moment

          Em_f = K + U = ½ m v_f ² + U

U is the energy lost in the collision

total energy is conserved

          Em₀ = Em_f

          ½ m v² = ½ m v_f ² + U

         U = ½ m (v² -v_f ²)

         U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex]  v)² ]

       U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]

       U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]

       U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]

Let's do the same calculations for the nucleus

initial     Em₀ = 0

final        Em_f = K + U = ½ M v_f ² + U

            Em₀ = Em_f

            0 = K + U

            U = -K

            U = - ½ M v_f ²

            U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²

            U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]  

We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.

b) the fraction of energy lost

          f = U / Em₀

          f = 4 m M / m + M        

c) let's calculate the fraction of energy lost in a collision

          m = 1.67 10⁻²⁷ kg

          M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg

         f = 4 1.6 20 / (1.6+ 20)    10⁻²⁷

         f = 5.92 10⁻²⁷ J

the energy of a fast neutron is greater than 1 eV

         Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J

Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo

         #_collisions = 0.95 Eo / f

         #_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷

         #_collisions = 2.7 10⁷ collisions

Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

Answer:

A. to reduce the loss of energy in the form of heat across the transformer

Explanation:

The core of the transformer is laminated to minimise the energy as they interfere with the efficient transfer of energy from the primary coil to the secondary one. The eddy currents cause energy to be lost from the transformer as they heat up the core - meaning that electrical energy is being wasted as heat.

Scalar quantity are physical quantities that have just magnitude, not direction.

The English Bill of Rights helped to shape the colonists' views of government by promoting the ideas of individual liberty, representative democracy, and limited government.

The English Bill of Rights is a document that was passed by the Parliament of England in 1689. It was a key moment in the development of modern democratic government, as it established certain rights and protections for English citizens, such as the right to bear arms, the right to a fair trial, and the prohibition of cruel and unusual punishment.

The English Bill of Rights also limited the power of the monarchy, requiring the king or queen to obtain the consent of Parliament before levying taxes or making other important decisions. Its influence can still be seen today in many democracies around the world.

Learn more about English bill of rights, here:

https://brainly.com/question/20986147

#SPJ7

Answer:

[tex]{ \bf{power \: { \tt{(watts)}} = \frac{workdone \: { \tt{(joules)}}}{time \: { \tt{(seconds)}}} \: }}[/tex]

Answer:

1. 45 m

2. 50 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 20 m/s

Time (t) = 3 s

1. Determination of the position (i.e height)

Time (t) = 3 s

Acceleration due to gravity (g) = 10 m/s²

Height (h) =?

h = ½gt²

h = ½ × 10 × 3²

h = 5 × 9

h = 45 m

2. Determination of the velocity.

Initial velocity (u) = 20 m/s

Time (t) = 3 s

Acceleration due to gravity (g) = 10 m/s²

Velocity (v) =?

v = u + gt

v = 20 + (10 × 3)

v = 20 + 30

v = 50 m/s

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂  = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

Answer:

8m in 1 second

480m = 1 hour

11,520m in one day

Explanation:

Hope this is helpful

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

[tex]F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}[/tex] I'm going to do the math on the top and then on the bottom and divide at the end.

[tex]F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}[/tex] and now when I divide I will express my answer to the correct number of sig dig's:

[tex]Fg=[/tex] 6.45 × 10¹⁶ N

Explanation:

Hey there!

Given;

Mass of one object (m1) = 250kg

Mass of another object (m2) = 400 kg

Distance (d) = 100 m

Gravitational constant (g) = 6.67*10^-11

Now;

[tex]f = \frac{g.m1.m2}{ {d}^{2} } [/tex]

Keep all values;

[tex]f = \frac{6.67 \times {10}^{ - 11} \times 250 \times 400}{ {(100)}^{2} } [/tex]

Simplify

[tex]f = \frac{6.67 \times {10}^{ - 11} {10}^{5} }{10000} [/tex]

[tex]f = \frac{6.67 \times {10}^{ - 6} }{10000} [/tex]

Therefore, gravitational force is 6.67*10^-10.

Hope it helps!

Answer:

[tex]v=-4.93\times 10^{5}~m/s[/tex] [negative sign denotes that it is moving away]

Explanation:

Actual wavelength of hydrogen spectral lines, [tex]\lambda=121.6~nm[/tex]

Apparent wavelength of hydrogen spectral lines from a star, [tex]\lambda'=121.8~nm[/tex]

So, shift in wavelength:

[tex]\Delta \lambda=\lambda'-\lambda[/tex]

[tex]\Delta \lambda=121.8-121.6[/tex]

[tex]\Delta \lambda=0.2~nm[/tex]

Using Doppler's shift:

[tex]\frac{\Delta \lambda}{\lambda} =-\frac{v}{c}[/tex]

[tex]\frac{0.2}{121.6} =-\frac{v}{3\times 10^8}[/tex]

[tex]v=-4.93\times 10^{5}~m/s[/tex] [negative sign denotes that it is moving away]

Explanation:

distance travelled = average speed x time

                               =30m/s*100s

                                =3000m

Answer:

3000m

Explanation:

30m/s*100s

3000m

Answer:

588n is the answer may be

Answer:

Explanation:

First job is to convert 72 km/hr to m/s:

[tex]72\frac{km}{hr}[/tex] × [tex]\frac{1000m}{1km}[/tex] × [tex]\frac{1hr}{3600s}[/tex] = 2.0 × 10¹ m/s

Now to find the acceleration which is

[tex]a=\frac{v_f-v_0}{t}[/tex] and filling in:

[tex]a=\frac{2.0*10^1-0}{11.5}=1.7\frac{m}{s^2}[/tex] That's part a. Part b want to know how far the car can get in 11.5 seconds (because that's the time it takes for the car to get to 72 km/hr). Since we know that the car can get 2.0 × 10¹ meters in 1 second, that means that in 11.5 seconds, the car can get 11.5(2.0 × 10¹) which is 230 meters.

hope it is helpful to you