C.
A palm fruit dropped to the ground from the top of
a tree 45m tall. How long does it take to reach the
ground? A. 9s B. 4.5s C. 6 D. 7.5s E. 35
(g = 10ms2).​

Answer:

the correct answer is B

Explanation:

The quantum numbers are the constants obtained when solving the Schrodinger equation, the first quantum number or principal quantum number (n), can take values ​​from zero to infinity.

This quantum number is placed as a coefficient in the quantum distribution.

In this case for phosphorus, the number is n = 3

the correct answer is B

Answer:

68.6 J

Explanation:

Applying,

P.E = mgh............... Equation 1

Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket

From the question,

Given: m = 3.5 kg, h = 2.00 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 3.5×2×9.8

P.E = 68.6 J

Hence the potential energy of the basket is 68.6 J

Answer:

6.4 m/s

Explanation:

Given that :

The average width of the Colorado river = 100 m

Average depth of the river is = 8 m

Therefore, area = [tex]$A_1= 100 \ m \times 8 \ m$[/tex]

Speed of the river, [tex]$v_1 = 3 \ m/s$[/tex]

After the lava falls on the river,

Width of the river becomes = 25 m

Depth of the river became = 15 m

Therefore, area = [tex]$A_2= 25 \ m \times 15 \ m$[/tex]

Now, since the volume flow rate of the Colorado river is same, then from the Continuity equation,

[tex]$Q_1=Q_2$[/tex]

[tex]$A_1v_1=A_2v_2$[/tex]

∴ [tex]$100 \times 8 \times3 = 25 \times 15 \times v_2$[/tex]

[tex]$v_2=\frac{100 \times 8 \times 3}{25 \times 15}$[/tex]

    = 6.4 m/s

Therefore, the speed of the river in this location is 6.4 m/s

Answer:

0.5 A

Explanation:

Applying,

V = IR.................. Equation 1

Where V = Voltage, I = current, R = Resistance.

make I the subject of the equation

I = V/R............... Equation 2

From the question,

Given: V = 75 V, R = 150 Ω

Substitute these values into equation 2

I = 75/150

I = 0.5 A.

Hence the cuurent through the resistor is 0.5 A

Answer:

a)   p = 4.167 cm, b)   m = + 6

Explanation:

a) For this exercise we must use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm

Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.

       [tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]

      [tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]

      [tex]\frac{1}{ p}[/tex] = 024

      p = 4.167 cm

b) angular magnification

     m = h ’/ h = - q / p

     m = - (-25) /4.167

     m = + 6

the positive sign indicates that the image is straight and enlarged

Answer:

E = 16581.6 J

Explanation:

Given that,

Current, I = 4.9 A

Time for which the current is set up, I = 4.7 min = 282 s

The voltage of the battery, V = 12 V

We need to find how much chemical energy of the battery reduced. Let It is E. We know that,

E = P t

Where

P is power of battery, P = VI

So,

[tex]E=VIt[/tex]

Put all the values,

[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]

So, 16581.6 J of chemical energy of the battery is reduced.

Answer:

Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.  

Answer:

The rotation rate is 15.3 rad/s.

Explanation:

maximum voltage, V = 120 V

Magnetic field, B = 1.5 T

length, L = 10.2 cm

width, W = 12.8 cm

Number of loops, N = 400

Let the rate of rotation is w.

The maximum voltage is given by

V = N B A w

120 = 400 x 1.5 x 0.102 x 0.128 x w

w = 15.3 rad/s

Answer:

sieving is when you separate particles of different sizes.

Explanation:

separating sand mixtures

separating chaffs from local garri

area= length × length

area = 4m × 9m

ans 36

SOLVED DOWN BELOW

Explanation:

In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law

I = 25/200

I= .125 amps or 125 ma

__________

R= r1 * r2 / r1 +r2

R= 100 * 100 / 100 + 100

R= 10000 / 200

R= 50 ohms

Answer:

F = 312 N

Explanation:

Given that,

The mass of a crate, m = 40 kg

Acceleration of the crate, a = 2 m/s²

As the carte is falling downward, the net force exerted by the rope on the carte is given by :

F = m(g-a)

Put all the values,

F = 40(9.8-2)

F = 312 N

Hence, the required force exerted by the rope on the crate is equal to 312 N.

Answer:

6.24 x 1018 electrons.

Explanation:

So I think C

Answer:

36.11 degrees

Explanation:

index of refraction n = sin i/sinr

i is the angle of incidence

r is the angle of refraction

Substitute into the expression

1.39 = sin55/sin(r)

1.39 = 0.8191/sin(r)

sin(r) = 0.8191/1.39

sin(r) = 0.5893

r = arcsin(0.5893)

r = 36.11

hence the angle of refraction of light is 36.11 degrees

Answer:

a) Ink X is likely to be pure because it only contain 1 spot.

b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.

c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.

The different spots from Y are found at various heights because they represent different compounds.

The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.

We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.

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Answer:

[tex]E=6.86\times 10^6\ N/C[/tex]

Explanation:

Given that,

Mass of the sphere, m = 2.1 g = 0.0021 kg

Charge, q = 3 nC

We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,

qE = mg

[tex]E=\dfrac{mg}{q}[/tex]

Put all the values,

[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]

So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].

The resulting wave has the largest possible amplitude when Wave-1 and Wave-2 are exactly in step ... their peaks both happen at the same time and their troughs both happen at the same time.

This means that Wave-1 and Wave-2 have the same frequency, and the phase shift from one wave to the other is zero.

When all of that happens, the amplitude of the resulting wave is the sum of the amplitudes of Wave-1 and Wave-2.  If Wave-1 and Wave-2 have the same amplitude, then the resulting wave will have double that amplitude.

Answer:

b. is the correct answer ....

Answer:

 F_net = 9.87 10⁻⁴ N

Explanation:

Let's use that force is a vector magnitude

         ∑ F = F₁₃ + F₂₃

De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract

the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m

       ∑ F = F₁₃ - F₂₃

       F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]

in this case q₁ = q₂ = q

       F_net = k q q₃  (  )

      let's look for the distance

      r₂₃ = y₂ - y₃

      r₂₃ = -7 -16

      r₂₃ = - 23 m

      r₁₃ = 38 - 16

      r₁₃ = 22 m

let's calculate

      F_net = 9 10⁹ 24 26 10⁻¹² ( )

      F_net = 5.616 ( 1.758 10⁻⁴ )

      F_net = 9.87 10⁻⁴ N

Answer:

Newtons third law of motion: Balanced forces

Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs.   Once more, tug-of-war is a prime illustration.

The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.

The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.

Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.

Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.

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Answer:

(a) I = 1650000 A

(b) 4.125 T

Explanation:

Magnetic field, B = 5.5 T

distance, r = 0.06 m

(a) Let the current is I.

The magnetic field due to a long wire is given by

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]

(b) Let the magnetic field is B' at distance r = 0.08 m.

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]

(a) 4.20 J

(b) 16.74 J

For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;

C = A∈₀ / d              --------------------(i)

Where;

∈₀ = constant called permittivity of vacuum.

The energy U stored in such capacitor is given by;

U = [tex]\frac{1}{2}[/tex]CV²             ----------------------(ii)

or

U =  [tex]\frac{1}{2}[/tex](Q²/C)        -------------------(**)

Where;

V = potential difference or voltage across the plates.

Q = charge on the plates.

(a) If the charge is held constant

Combine equations (i) and (**) to give;

U =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)     -----------------------(iii)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iii)

8.40 =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)

8.40 =  [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)

Multiply through by 2

2 x 8.40 = Q² x (0.023 / A∈₀)

16.80 = Q² x (0.023 / A∈₀)

Divide through by 0.023

16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023

730.4 = Q² / (A∈₀)

Make Q² subject of the formula

Q² = 730.4(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

Q = constant [this means that Q² still remains 730.4(A∈₀) ]

The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;

U =   [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)

U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)

U = 4.20J

Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J

(b) If the voltage is held constant

Combine equations (i) and (ii) to give;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / d)V²     -----------------------(iv)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iv)

8.40 =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²

Multiply through by 2 x 0.023

2 x 0.023 x 8.40 = (A∈₀)V²

2 x 0.023 x 8.40 = (A∈₀)V²

0.385 = (A∈₀)V²

Make V² subject of the formula

V² = 0.385/(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

V = constant [this means that V² still remains 0.385/(A∈₀) ]

The energy stored is found by substituting these values of d and V² into equation (iv) as follows;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]  

U = [tex]\frac{1}{2}[/tex](0.385/0.0115)

U = 16.74

Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J

Answer:

This is because the age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

Explanation:

The age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

This is due to the fact that the expansion rate in the coasting model is constant and never changes. Because the cosmos is growing faster now than during the old days, recollapsing and critical models give shorter ages. According to the accelerating model, the universe is growing at a slower rate currently than in the past, implying an older age.

Answer:

Less

Explanation:

Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.

Answer:

Z-number

Explanation:

The Z number is the number of protons in an atom, and this does not change when an isotope is created. I got it right on the test.

Answer:

[tex]Weight\ loss=1.6321N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=85.9kg[/tex]

Altitude [tex]h= 6.33 km[/tex]

Let

Radius of Earth [tex]r=6380km[/tex]

Gravity [tex]g=9.8m/s^2[/tex]

Generally the equation for Gravity at altitude is mathematically given by

 [tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]

 [tex]g_s=9.781m/s^2[/tex]

Therefore

Weight at sea level

 [tex]W_s=9.8*85.9[/tex]

 [tex]W_s=841.82N[/tex]

Weight at 6.33 altitude

 [tex]W_a=9.781*85.9[/tex]

 [tex]W_a=840.2N[/tex]

Therefore

 [tex]Weight loss=W_s-W_b[/tex]

 [tex]Weight loss=841.82-840.2[/tex]

 [tex]Weight loss=1.6321N[/tex]

Answer:

Explanation:

In a Solid sphere; the moment of inertia around its geometrical axis can be expressed by using the formula:

[tex]\mathtt{I_s = \dfrac{2}{5} M_s R^2_s}[/tex]

For the solid disk; the moment of inertia around the central axis is:

[tex]\mathtt{I_D= \dfrac{1}{2}M_DR_D^2}[/tex]

Suppose [tex]M_D = M_S[/tex]; then we can say both to be equal to M

As well as [tex]R_D = R_S[/tex]; then that too can be equal to R

Now;

[tex]\mathtt{I_s = \dfrac{2}{5} M R^2} --- (1)[/tex]

[tex]\mathtt{I_D= \dfrac{1}{2}MR^2}---(2)[/tex]

Multiplying equation (1) by 2, followed by dividing it by 2; we have:

[tex]\mathtt{I_s= \dfrac{2}{5}MR^2} \times \dfrac{2}{2}[/tex]

[tex]I_s = \dfrac{4}{5} \times \dfrac{1}{2}MR^2 \\ \\ I_s = \dfrac{4}{5}\times I_D \\ \\ I_s > I_D[/tex]

Answer:yeah it A

Explanation: