10. Complete each of the following radioactive decay equations.
235U → Th+a
92
90
87
37Rb Sr +
6C
→
F
e
1¹B ++ie

The correct answer is Self-critical.

Why is self-improvement?

Self-critical:

Self-critical causes them to always strive for self-improvement and never to see accomplishing a goal as good enough.

The characteristic of a turbulent person causes them to always strive for self-improvement and to never see accomplishing a goal as good enough is Self-critical.

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A sine bar is used to get the angular measurement of a part feature. The angle of the part feature is 18.24°

From the question

Given that,

Length of the sine bar, L = 8.000 in

Diameter of the rolls = 1.000 in

Height under the roll, H = 2.0000 + 0.5000 + 0.0050

                                     = 2.505 in

From sine bar formula,

we know that,

H = sin A x L

sin A = H ÷ L  

where,

A ⇒ angle of part feature

H ⇒ height under the roll

L ⇒ length of the sine bar

Substituting values in the above equation,

sin A = H / L

A = sin⁻¹ ( 2.505 ÷ 8 )

A = sin⁻¹ (0.3131)

A = 18.24⁰

Hence the angle of the part feature = 18.24°

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a.

b.

c.

Pressure is the force per unit area on a surface.

Speed is the distance moved per unit time.

Pressure

Since pressure, P = hρg where

Since ρ and g are constant

P ∝ h

So, we see that pressure is directly proportional to depth.

Since U is lower than Q, Pressure at U is greater than pressure at Q.

So,PU is greater than PQ.

Using the continuity equation

VUAU = VTAT where

So, VUAU = VTAT

VUπ(dU)² = VTπ(dT)²

VU = VT(dT)²/(dU)²

VU = VT(2.0)²/(1.0)²

VU = VT(4)

VU = 4VT

Since VU = 4VT,VU is Greater than 2VT

Since R is at the same depth as U, Pressure at R is equal to pressure at U.

So,PR is Equal to PU.

Using the continuity equation

VRAR = VSAS where

So, VRAR = VSAS

VRπ(dR)² = VSπ(dS)²

VR = VS(dS)²/(dS)²

VR = VS(2.0)²/(2.0)²

VR = VS(1)

VR = VS

Since VR = VS,VR is Equal to VS

Using the continuity equation

VQAQ = VUAU where

So, VQAQ = VUAU

VQπ(dQ)² = VUπ(dU)²

VQ = VU(dU)²/(dQ)²

VQ = VU(1.0)²/(1.0)²

VQ = VU(1)

VQ = VU

Since VQ = VU, VQ is Equal to VU

Since R is lower than S, Pressure at R is greater than pressure at S.

So,PR is Greater than PS.

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yes electrons will be emitted with the kinetic energy of

[tex]30.87 \times 10 { }^{ - 19} j[/tex]

what is photoelectric effect?

In photoelectric effect, the electrically charged particles are released from or within a material when it absorbs electromagnetic radiation.

what is kinetic energy?

It is a form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.

Given:

wavelength,λ

[tex] = 5.32 \times 10 {}^{ - 8} m[/tex]

Frequency

[tex] = 9.7 \times {10}^{14} [/tex]

Einstein's photoelectric equation

€= w + KE

€ = hc/ λ

[tex] = \frac{6.63 \times {10}^{ - 34} \times 3 \times {10}^{8} } {5.32 \times 10 {}^{ - 8} } [/tex]

[tex]€= 3.73 \times {10}^{ - 18} [/tex]

w=h.f

[tex] = 6.63 \times {10}^{ - 34} \times 9.7 \times {10}^{14} [/tex]

[tex] = 6.43 {10}^{ - 19} j[/tex]

KE = € - W

[tex]37.3 \times {10}^{ - 19} - 6.43 \times {10}^{ - 19} [/tex]

[tex]kE = 30.87 \times {10}^{ - 19} [/tex]

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We push a car to the distance of 33.939m if we do 28000J work , exerting a force of 825N.

F = applied force

S = distance moved

=28000J/825N

=33.939meter

Thus, we can conclude that the distance moved by the car is 33.939m.

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(a) The amplitude of the magnetic field produced is 2.667 mV/m.

(b) The intensity of the laser is 2.832 W/m².

(c) The  power delivered by the laser is 2.22 x 10⁻⁶ W.

In electromagnetic waves, the amplitude of magnetic field is the maximum field strength of the magnetic fields.

B₀ = E₀/c

where;

B₀ = (0.8 MV/m) / (3 x 10⁸)

B₀ = (0.8  x 10⁶ V/m) / (3 x 10⁸)

B₀ = 2.667 x 10⁻³ V/m

B₀ = 2.667 mV/m

The intensity of the laser is calculated as follows;

I = ¹/₂ε₀E₀²

I = (0.5)(8.85 x 10⁻¹²)(0.8 x 10⁶)²

I = 2.832 W/m²

P = IA

where;

A = πd²/4

where;

A = π(1 x 10⁻³)²/4

A = 7.854 x 10⁻⁷ m²

Power = (2.832 W/m²) x (7.854 x 10⁻⁷ m²)

Power = 2.22 x 10⁻⁶ W

Thus, the amplitude of the magnetic field produced is 2.667 mV/m.

The intensity of the laser is 2.832 W/m².

The  power delivered by the laser is 2.22 x 10⁻⁶ W.

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[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

We would calculate the magnitude by applying pythagorean theorem:

[tex]\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}[/tex]

[tex]\longrightarrow \sf{Magnitude = 33.12}[/tex]

[tex]\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}[/tex]

The angle between two vectors is given by the formula:

[tex]\sf{\longrightarrow \small \cos \emptyset = \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }[/tex]

In two dimensional, the x axis of vector form is:

[tex]\small\sf{\longrightarrow (b1, b2) = (1, 0) }[/tex]

[tex]\sf{\longrightarrow \small \cos \: \emptyset = \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }[/tex]

[tex]\small\longrightarrow \sf{ \dfrac{14}{33.12} }[/tex]

[tex]\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}[/tex]

[tex]\small\longrightarrow \sf{\emptyset= 115^\circ}[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

[tex] \small\bm{The \: angle \: between \: the \: vector \: }[/tex]

[tex]\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}[/tex]

The information is missing here but data that may support the existence of life in an exoplanet may include the discovery of atmospheric biological gases.

An exoplanet is any planet that surrounds another star beyond the sun, and therefore they are discovered by analyzing other galaxies in the Universe.

The existence of life in these exoplanets can be indirectly supported by the presence of gases that are only produced by organisms, which are detected by suitable telescopes that analyze emission spectra.

In conclusion, the information is missing here but data that may support the existence of life in an exoplanet include biological gases.

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Sol^n :

aA=6ai-8aj

bB=-8bi-3bj

C=26i-19j

Given aA+bB+C=0i+0j

(6a-8b+26)i-(8a+3b+19)j=0i+0j

comparing coefficient of i and j

6a-8b=-26

2a-4b=-13.....(1)

8a+3b=-19......(2)

making the coefficient of a similar

8a-16b=-52....(3)

8a+3b=-19....(4)

equation 4-3 implies

19b=33

b=33/19

from (1)

2a=-13+4(33/19)

2a=-115/19

a=-115/38

Answer:

work done = ( force × displacement)

(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.

work done by gravity = (250 × 10) = 2500 joule

(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.

Four are known: s, p, d, and f.

Thank you,

Eddie

Answer:

A i think...

Explanation:

Sorry if its wrong

The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex] , radial component is 0.016 m/[tex]s^{2}[/tex], tangential component is 0.9347*[tex]10^{-5} m/s^{2}[/tex].

Angular acceleration,

ω=1/60=0.016 rev/s

[tex]\alpha[/tex]=0.016/(6*60)=4.44*[tex]10^{-5} rev/s^{2}[/tex]

The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex]

Radial component of the linear acceleration=[tex]\alpha_{r}[/tex]

v=ωr=0.016*4.75=0.076 m/s

[tex]\alpha_{r}[/tex]=0.076 /4.75=0.016 m/[tex]s^{2}[/tex]

The tangential component = [tex]\alpha_{t}[/tex]=4.44*[tex]10^{-5} rev/s^{2}[/tex]/4.75=0.9347*[tex]10^{-5} m/s^{2}[/tex]

The temporal rate at which angular velocity changes is known as angular acceleration. Due to the fact that there are two different types of angular velocity—spin angular velocity and orbital angular velocity—there are also two different types of angular acceleration, referred to as spin angular acceleration and orbital angular acceleration, respectively. The terms "orbital angular acceleration" and "spin angular acceleration" describe the angular acceleration of a point particle about a fixed origin and, respectively, the angular acceleration of a rigid body about its center of rotation.

The unit of measurement for angular acceleration is the angle per unit time squared, or, in SI units, radians per second squared. It is typically denoted by the symbol alpha ().

In traveling to the Moon, astronauts aboard the Apollo spacecraft put spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis.

a)Determine the angular acceleration of the ship.

Express your answer using two significant figures.

b)Determine the radial component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.

Express your answer to two significant figures and include the appropriate units.

c)Determine the tangential component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.

Express your answer to two significant figures and include the appropriate units.

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D. The minimum braking torque that must be applied is -11.2 Nm.

The moment of inertia of the uniform cylinder is calculated as follows;

I = ¹/₂MR²

where;

I = (0.5)(8)(0.1²)

I = 0.04 kgm²

τ = -Iα

where;

α = ω/t

α = (80,000 x 2π/rev x 1 min/60s) / (30 s)

α =  (80,000 x 2π)/(60 x 30)

α = 279.25 rad/s²

τ = - ( 0.04 kgm²) x (279.25)

τ =  -11.2 Nm

Thus, the minimum braking torque that must be applied is -11.2 Nm.

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By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,

To find the answer, we need to know about the Bernoulli's equation as well as the Equation of continuity.

1) We have to find the relation between pressure at U and Q.

                  [tex]P+\frac{1}{2}dV^2+dgh=constant.[/tex]

where, P is the pressure, V is the velocity, d is density, g is acceleration due to gravity and h is the height of the flow.

                       [tex]P_u+\frac{1}{2}dV_u^2+dg*0= P_q+\frac{1}{2}dV_q^2+dgh\\where,\\V_u=V_q, \\since, D_p=D_q\\Thus,\\P_u=P_q+dgh[/tex]

                       [tex]P_u > P_q[/tex]

2) We have to find the relation between velocity at U and T.

                           [tex]AV=constant\\A_1V_1=A_2V_2[/tex]

                          [tex]A_u=\pi r^2=\pi *(0.5)^2=0.25\pi *mm^2\\A_t=\pi *1=\pi mm^2\\V_u=V\\V_t=?[/tex]

                      [tex]V_t=\frac{A_uV_u}{A_t}=\frac{V}{4}[/tex]

                      [tex]2V_t=2*\frac{V}{4} =\frac{V}{2}\\V_u=V\\[/tex]

                       [tex]V_u > 2V_t[/tex]

3) We have to find the relation between pressure at R and U

                 [tex]P_u+\frac{1}{2}dV_u^2+dg*0= P_r+\frac{1}{2}dV_r^2+dg*0\\\\V_u=V , then\\V_r=V_t=\frac{V}{4}=\frac{V_u}{4} \\\\P_u=P_r+\frac{1}{16}[/tex]

                   [tex]P_u > P_r[/tex]

4)  We have to find the relation between velocity at R and S

                                     [tex]V_r=V_s[/tex]

5)  We have to find the relation between velocity at Q and U

                                    [tex]V_q=V_u[/tex]

6)  We have to find the relation between pressure at R and S

                          [tex]P_r=P_s+dgh[/tex]

                           [tex]P_r > P_s[/tex]

Thus, we can conclude that, By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,

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The magnitude of the electric force that one particle exerts on the other is 8.59 x [tex]10^{-8}[/tex] N and the force is attractive.

Coulomb law states that the attractive or repulsive force between two point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. That is,

F = KQq / r²

Given that two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.02 nC.

Where

(A) The magnitude of the electric force that one particle exerts on the other will be

F = (9 x [tex]10^{9}[/tex] x 7.03 x [tex]10^{-9}[/tex] x 4.02 x [tex]10^{-9}[/tex]) / 1.72²

F = 2.54 x [tex]10^{-7}[/tex] / 2.96

F = 8.59 x [tex]10^{-8}[/tex] N

(B) Since our answer is positive, the force is therefore attractive.

Therefore, the magnitude of the electric force that one particle exerts on the other is 8.59 x [tex]10^{-8}[/tex] N and the force is attractive.

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A. The magnitude (in N) of the electric force that one particle exerts on the other is 8.60×10⁻⁸ N

B. The force is repulsive

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

The magnitude of the electric force can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = (9×10⁹ × 7.03×10¯⁹ × 4.02×10¯⁹) / (1.72)²

F = 8.60×10⁻⁸ N

From the question given, we were told that:

Since both charge are positive, then the force attraction between them is repulsive as like charges repels and unlike charges attracts

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The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

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A ship sounds its foghorn at a frequency of 178Hz when it is 2.40km away from a sizable rock face. (Audio travels at 343 m/s.) Reconstruction succeeded in resolving the federalism controversy that had dogged the country since the 1790s.

By most other standards, reconstruction was a failure: radical Republican legislation eventually failed to shield former slaves from white persecution and to bring about significant alterations to the South's socioeconomic structure. The federalism question, which had been a topic since the 1790s practically immediately, was at hand when President Rutherford B. Former Confederate officials and slaves returned to the South after Hayes withdrew federal soldiers from the region in 1877. These newly powerful white southern legislators established anti-progressive laws like voter ID requirements and black codes with the help of a conservative Supreme Court in an effort to roll back the rights that blacks had won during Radical Reconstruction. With its rulings in the Slaughterhouse Cases, the Civil Rights Cases, and United States v., the U.S. Supreme Court strengthened this anti-progressive movement federalism.

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The energy becomes 0.50 times in 6.72 s.

Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.

Et=0.50E

1

k(4₂)² = (0.5) - kA²

(4₂)² = (0.5) A²

At = 0.71A

So, the amplitude of the oscillator becomes 0.71 times its initial ar

0.71A = = A(0.96)¹2

log(0.71)

log(0.96)

8.4

n=

So, the time taken for n oscillation is obtained as,

t = n (0.800 s)

= (8.4) (0.800)

= 6.72 s

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Assuming point K is held in equilibrium, by Newton's second law we have

• net horizontal force

[tex]F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0[/tex]

• net vertical force

[tex]F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0[/tex]

where the angle [tex]\theta[/tex] that rope C makes with the horizontal axis satisfies

[tex]\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57[/tex]

Solve the first equation for [tex]F_C[/tex].

[tex]F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\sec(x)=\frac1{\cos(x)}[/tex].)

Substitute this into the second equation and solve for [tex]F_B[/tex].

[tex]F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}[/tex].)

[tex]F_B = \dfrac57 F_A[/tex]

[tex]\boxed{F_B \approx 36.1\,\rm N}[/tex]

Answer:

Patterns in science are a little different. Data doesn't have to follow a trend, always going up or down over time. A pattern is a when data repeats in a predictable way. A good example of a pattern in science comes from the father of genetics, Gregor Mendel.After data is collected, it can be analyzed by looking for trends, patterns, and relationships. Trends are general directions of data, such as an overall increase in global temperature. Patterns don't necessarily involve data going one way or the other, but rather describe a repeating observation.In order to interpret and understand scientific data, one must be able to identify the trends, patterns, and relationships in it. Examine the importance of scientific data and recognize how understanding its trends, patterns, and relationships can lead a researcher to support or refute a hypothesis. Updated: 01/06/2022 What Is Scientific Data?

Explanation:

The hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

The independent variable which is can be changed by increasing the rate of salt added to the equation.

The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.

Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.

Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.

If  the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.

Hence, the  hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

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The reason behind the heavier hydrometer bulb is the sinking of hydrometer is inversely proportional to the density of hydrometer, hence hydrometers is made heavier.

We all know that hydrometers float in liquid hence to maintain the centre of gravity while floating the hydrometer is made heavier using lead shots.

(1) The effective spring constant of the system is 7.14 N/m.

(2) The maximum x-acceleration of the glider is 0.9 m/s².

(3) The x-coordinate of the glider at time t= 0.650T is 0.28 m.

(4) The kinetic energy of the glider at x=0.00 m is zero.

The effective spring constant of the system is calculated as follows;

F = kx

where;

k = F/x

k = 0.5/0.07

k = 7.14 N/m

a = ω²x

where;

ω = √k/m

ω = √(7.14/0.55)

ω = 3.6 rad/s

a =  (3.6)² x 0.07

a = 0.9 m/s²

T = 2πx/v

T = 2πx/(ωx)

T = 2π/ω

T = 2π/(3.6)

T = 1.75 seconds

t = 0.65T

t = 0.65 x 1.75

t = 1.14 seconds

x = vt

x = (ωx)t

x = (3.6 x 0.07) x 1.14

x = 0.28 m

At position x = 0, the glider is at rest, the velocity is zero and the kinetic energy will be zero.

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Since the displacement is completely perpendicular to the direction of the applied force, the work done on the bag is zero.

The work is done on an object when the force applied is multiply by the distance moved by the object in the direction of the force applied.

Given that a man carries a hand bag by hanging on his hand moves horizontally where the bag does not up or down.

What is work if the displacement is not in the direction of force ?

The work done can only be zero if the displacement is perpendicular to the direction of force. otherwise, it will not be equal to zero.

Also, the work done will be zero, if the displacement is zero.

In the question above, the displacement is completely perpendicular to the direction of the applied force.

Therefore, the work done on the bag is zero.

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Answer:

M V R = constant      angular momentum is constant because  no forces act in the direction of V

Since M (mass) = constant

V R = constant

The force is directed along the gravitational force vector (towards the center of rotation)

The magnitude of the Coulomb force (in N) on one of the spheres, given the data is 4.59×10⁻⁴ N

Sphere 1 losses 2.40×10¹² electrons

But

1 electron = 1.6x10¯¹⁹ C

Thus,

Charge on sphere 1 = +1.6x10¯¹⁹ × 2.40×10¹² = +3.84×10¯⁷ C

Sphere 2 gains 2.40×10¹² electrons

But

1 electron = 1.6x10¯¹⁹ C

Thus,

Charge on sphere 2 = -1.6x10¯¹⁹ × 2.40×10¹² = -3.84×10¯⁷ C

Using the Coulomb's law equation, the force can be obtained as illustrated below:

F = Kq₁q₂ / r²

F = (9×10⁹ × 3.84×10¯⁷ × 3.84×10¯⁷) / (1.7)²

F = 4.59×10⁻⁴ N

Thus, the magnitude of the Coulomb's force is 4.59×10⁻⁴ N

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The values of the required solutions are

What is Force?

In the field of physics, an influence that can alter the motion of an object is referred to as a force. An object having mass can experience a change in its velocity, often known as an acceleration, when subjected to a force. Intuitively, force can also be conceptualized as either a push or a pull. Because it may be measured in both magnitude and direction, a force is considered a vector quantity.

What is atmospheric pressure?

The pressure that is exerted within the atmosphere of the Earth is referred to as barometric pressure as well as atmospheric pressure. As a unit of pressure, the standard atmosphere is defined as having a value of 101,325 Pa. This value is equivalent to 1013.25 millibars, 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.

In most situations, the equation for force can be expressed numerically as

Force = pressure * A

Therefore

[tex]F= (970 - 15 B)(\pi * (0.430 )^2)[/tex]

F= 53731 N

In conclusion, If each horse can pull with a force of 1450N

The number of horses required is

[tex]N=\frac{60754 }{1450}[/tex]

N=37.05

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The work done by this gravitational force during a fast time interval in which the Earth drives through a displacement in its orbital path is (a) zero.

Gravitational Force

The kilogram force (kgf) is a gravitational unit of force. The force exercised by the earth on a body of mass 1 kg is regarded as 1 kgf. Force = Mass × Acceleration. 1 kgf = 9.8 N.

The gravitation force and perpendicular to the path of movement

hence angle between gravitational force [ F ]and displacement (d)=90°

work done (W)= F d cos 90°

           = 0

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Question: 5

The length of the pendulum is 7.6 m.

So, L = T²g/4π²

= 7.6 m

Thus, we can conclude that the length of the pendulum is 7.6 m i.e option C is correct.

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Question: 6

The object takes 2.55 seconds to reach the ground.

t= √(2h/g)

t= √(2×31/9.8)

= 2.55s

Thus, we can conclude that the option A is correct.

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