Two subway stops are separated by 1.1 km. If a subway train accelerates at +1.2 m/s2 from rest
through the first half of the distance and decelerates at −1.2 m/s2 through the second half, what are
(a) its travel time and (b) its maximum speed? (c) Graph x, v and a versus t for the trip.

Answer:

0.14 J

Explanation:

Use the force to calculate the spring constant.

F = k Δx

27 N = k (0.32 m − 0.28 m)

k = 675 N/m

Work is the change in energy:

W = PE

W = ½ k (Δx)²

W = ½ (675 N/m) (0.30 m − 0.28 m)²

W = 0.135 Nm

W = 0.135 J

Rounding to two decimal places, W = 0.14 J.

Your answer was correct, but it was in units of Ncm, and you needed to answer in units of J.

The work done to stretch the spring from 28 cm to 30 cm is 39.00 Joules, computed using the principles of Hooke's Law and the concept of work done.

In this problem, we are dealing with the concept of work done on a spring, which falls under Physics principles. Hooke's Law states that the force to compress or extend a spring by a distance x from its natural length is proportional to x. It can be written as F = kx, where F is the force, k is the spring constant, and x is the distance.

In this case, the force (F) is 27 N, and the length of stretch (x) is 32 cm - 28 cm = 4 cm. We can find the spring constant (k) using the formula k = F / x = 27 N / 4 cm = 6.75 N/cm.

The work done (W) to stretch the spring from 28 cm to 30 cm is the area under the force/displacement graph from '28 cm' to '30 cm'. Since the force is linear with displacement for a spring, this area can be found using the formula for the area of a trapezoid: W = ½ (F1 + F2) x d. F1 is the initial force (k*28 cm), F2 is the final force (k*30 cm), and d is the displacement (30 cm - 28 cm). Substituting the values, W = ½ [(6.75 N/cm*28 cm)+(6.75 N/cm*30 cm)]*(2 cm) = 39.00 J.

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Answer:

The average normal stress is 5 ksi.

Explanation:

Given that,

Diameter = 2 inch

Load = 15.71 kip

We need to calculate the average normal stress

Using formula of stress

Average normal stress [tex]\sigma =\dfrac{F}{A}[/tex]

Where, F = load

A = area

Put the value into the formula

[tex]\sigma=\dfrac{15.71}{\pi\times(\dfrac{2}{2})^2}[/tex]

[tex]\sigma = 5\ kip/inc^2[/tex]

[tex]\sigma=5\ ksi[/tex]

Hence, The average normal stress is 5 ksi.

Answer:

E. 5

Explanation:

N₀ = initial total number of radioactive elements number

N = Number of atoms of radioactive element after "n" half lives = N₀ /32

n = number of half lives

Number of atoms of radioactive element after "n" half lives is given as

[tex]N = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]

inserting the values

[tex]\frac{N_{0}}{32} = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]

[tex]\frac{1}{32} = \left ( \frac{1}{2} \right )^{n}[/tex]

n = 5

Answer:

Magnetic field = 1.41 T

Explanation:

Magnetic field of solenoid, B = μnI

μ = 4π x 10⁻⁷N/A²

Number of turns per meter, [tex]n=\frac{N}{L}=\frac{5601}{9.1\times 10^{-2}}=6.15\times 10^4turns/m[/tex]

Current, i = 18.2 A

B = μnI = 4π x 10⁻⁷ x 6.15 x 10⁴ x 18.2 = 1.41 T

Magnetic field = 1.41 T

Answer:

Explanation:

The efficiency of a refrigerator is defined in the terms of coefficient of performance (COP).

The ratio of amount of heat in cold reservoir to the work done is termed as the COP.

COP = QL / W

COP = T2 / (T1 - T2)

Where, T1 be the temperature of hot reservoir, T2 be the temperature of cold reservoir.

The distance at which the puck of mass 0.13 kg, traveling across the ice at speed 17.4 m/sec will stop is 131.2 meters.

Given to us

Mass of the puck, m = 0.13kg

The velocity of the ice, u = 17.4 m/sec

Friction force, f = 0.15 N

We know we want to stop the puck, therefore, the final velocity of the puck will be 0.

v = 0

We know that according to the first law of motion,

Force = mass x acceleration

F = m x a

Substitute the value,

[tex]0.15 = 0.13 \times a[/tex]

[tex]a = 1.1538\rm\ m/s^2[/tex]

As we know that the final velocity of the puck will be 0, therefore, there will be a deceleration in the puck.

a = -1.1538 m/s².

Thus, the acceleration of the ice puck is -1.1538 m/s².

We know that according to the third equation of the motion,

[tex]v^2-u^2 = 2as[/tex]

substitute the values,

[tex]0^2-(17.4)^2 = 2(-1.1538)s[/tex]

s = 131.2012 = 131.2 meters

Hence, the distance at which the puck of mass 0.13 kg, traveling across the ice at speed 17.4 m/sec will stop is 131.2 meters.

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Final answer:

The stopping distance of a hockey puck  is 131.1 m after rounding to the nearest tenth of a meter.

Explanation:

To calculate the stopping distance of a hockey puck, we first need to determine the deceleration caused by the frictional force.

The formula for deceleration due to friction is a = f/m, where a is the acceleration (deceleration, in this case, as it's negative), f is the frictional force, and m is the mass of the puck. Given that the frictional force f is 0.15 N and the mass m of the puck is 0.13 kg, we can calculate the deceleration as follows:

a = f/m = 0.15 N / 0.13 kg ≈ 1.15 m/s2

Now, to find the stopping distance we can use the formula d = v2 / (2×a), where d is the stopping distance, v is the initial speed, and a is the deceleration. Plugging in the initial speed v = 17.4 m/s and the deceleration a = 1.15 m/s2, we get:

d = (17.4 m/s)2 / (2 × 1.15 m/s2) ≈ 131.06 m

To round to the nearest 0.1 m, the stopping distance of the puck is 131.1 m.

Answer:

i think 7

Explanation:

Answer:

0.7

Explanation:

f = 50 hz, R = 40 ohm, L = 0.3 h, C = 60 uC = 60 x 10^-6 c

XL = 2 x 3.14 x f x L = 2 x 3.14 x 50 x 0.3 = 94.2 ohm

Xc = 1 / ( 2 x 3.14 x 50 x 60 x 10^-6) = 53.078 ohm

Impedance is Z.

Z^2 = R^2 + ( XL - Xc)^2

Z^2 = 40^2 + (94.2 - 53.078)^2

Z^2 = 1600 + 1691.019

Z = 57.37 ohm

The power factor is given by

CosФ = r / Z = 40 / 57.37 = 0.697 = 0.7

The power factor of a circuit is the ratio of real power to apparent power and is calculated by the cosine of the phase angle between voltage and current. With the information given, we cannot calculate the power factor of the RLC series circuit because the rms voltage of the source is required but not provided.

The power factor of a circuit represents the cosine of the phase angle between the voltage and the current in an AC circuit. In an RLC circuit, like the one described, the power factor can be calculated by dividing the real power (measured in watts) by the apparent power (volt-amps). To find the power factor, we need to first calculate the impedance (Z) of the circuit using the formula Z = √(R² + (XL - XC)²), where XL is the inductive reactance and XC is the capacitive reactance.

Inductive reactance (XL) is given by XL = 2πfL, and capacitive reactance (XC) by XC = 1/(2πfC). Since we know that f = 50 Hz, L = 0.30 H,  and C = 60 μF, we can calculate XL and XC. Substituting R for the resistance in the circuit, we can find the impedance. After finding the impedance, we can calculate the real power (P) using P = I²R, where I is the rms current.

From the real power and the apparent power, which is IZ, we can find the power factor by calculating P/(IZ). However, we need the actual values of all elements, including the voltage, to complete these calculations. With the information provided in the question, we can't calculate the power factor because the rms voltage of the source isn't given.

Explanation:

It is given that,

Mass of the bullet, m₁ = 0.093 kg

Initial speed of bullet, u₁ = 100 m/s

Mass of block, m₂ = 2.5 kg

Initial speed of block, u₂ = 0

We need to find the speed of the block after the bullet embeds itself in the block. Let it is given by V. On applying the conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]

[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]

[tex]V=\dfrac{0.093\ kg\times 100\ m/s+0}{(0.093\ kg+2.5\ kg)}[/tex]

V = 3.58 m/s

So, the speed of the bullet is 3.58 m/s. Hence, this is the required solution.

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed [tex]\omega[/tex], since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

[tex]v=\omega r[/tex]

where

[tex]\omega[/tex] is the angular speed

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

Answer:

The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

Explanation:

Given that,

Velocity of car relative to ground = 15.3 m/s

Velocity of truck relative to car = 22.5 m/s

We need to calculate the magnitude of the truck's velocity relative to the ground

We need to calculate the x component of the velocity

[tex]v_{x}=22.5\cos\theta[/tex]

[tex]v_{x}=22.5\cos52^{\circ}[/tex]

[tex]v_{x}=13.852\ m/s[/tex]

We need to calculate the y component of the velocity

[tex]v_{y}=15.3+22.5\sin\theta[/tex]

[tex]v_{y}=15.3+22.5\sin52^{\circ}[/tex]

[tex]v_{y}=33.030\ m/s[/tex]

Using Pythagorean theorem

[tex]|v|=\sqrt{v_{x}^2+v_{y}^2}[/tex]

[tex]|v|=\sqrt{(13.852)^2+(33.030)^2}[/tex]

[tex]|v|=35.82\ m/s[/tex]

Hence, The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

Answer:

Explanation:

The change in temperature may be calculated from the formula:

Where:

Then, you can make these assumptions or inferences from the conditions stated in the problem:

Then, you can state that, for those samples, ΔT = k / C, i.e. the increase in temperature is inversely related to the specific heat.

That means that the higher the specific heat the lower ΔT, and the lower the specific heat the higher ΔT.

The ranking in decrasing order of specific heat is:

Ranking in increasing order of ΔT:

And since all of them started at the same temperature, that is the ranking in resulting temperature from lowest to highest:

That means that the sample of water, the matter with the highest specific heat capacity (4.19 J/g°C), will reach the lowest temperature, and the sample of iron, the matter with the lowest heat capacity (0.46 J/g°C) will reach the highest temperature.

Answer:

ur mom

hsheu7shrbrjxbfbbrndnifidfjf

Answer:

Part a)

charge through the battery is

[tex]Q = 9.72 \times 10^5 C[/tex]

Part b)

Maximum current is

i = 386 A

Explanation:

Part a)

As we know that the rating of battery is given as

[tex]R = 270 A h[/tex]

here we also know that the charge given by the battery is same as the capacity of the battery

so we will have

[tex]Q = i t[/tex]

[tex]Q = (270 A)(3600 s)[/tex]

[tex]Q = 9.72 \times 10^5 C[/tex]

Part b)

Now we know that current in the wire is given by

[tex]i = \frac{Q}{t}[/tex]

now plug in all values in it

[tex]i = \frac{9.72 \times 10^5}{42\times 60}[/tex]

[tex]i = 386 A[/tex]

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]

[tex]\dfrac{d\phi}{dt}[/tex] is the rate of change if magnetic flux.

[tex]\phi=BA\ cos\theta[/tex]

[tex]\theta[/tex] is the angle between the magnetic field and the normal to area vector.

[tex]\theta=90-60=30[/tex]

[tex]\epsilon=NA\dfrac{dB}{dt}\times cos30[/tex]

[tex]\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)[/tex]

[tex]\epsilon=0.0598\ T[/tex]

[tex]\epsilon=59.8\ mT[/tex]

or

EMF = 60 mT

So, the magnitude of  emf induced in the coil is 60 mT. Hence, this is the required solution.

Answer:

[tex]1.0672\times 10^{-9}N[/tex]

Explanation:

[tex]G[/tex] = Gravitational constant = 6.67 x 10⁻¹¹

[tex]F[/tex]  = Gravitational force between these spheres

[tex]m_{1}[/tex] = mass of first sphere = 1.60 kg

[tex]m_{2}[/tex]  = mass of second sphere = 16 g = 0.016 kg

[tex]r[/tex]  = distance between the centers of the sphere = 4 cm = 0.04 m

Gravitational force between these spheres is given as

[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]

[tex]F = \frac{(6.67\times 10^{-11})(1.60)(0.016)}{0.04^{2}}[/tex]

[tex]F = 1.0672\times 10^{-9}N[/tex]

Answer:

[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Explanation:

As we know that heat flux is given by the formula

[tex]q^n = K\frac{dT}{dx}[/tex]

here we know that

K = thermal conductivity

[tex]\frac{dT}{dx}[/tex] = temperature gradient

now we know that

[tex]q^n = 11 W/m^2[/tex]

also we know that

K = 1.7 W/mK

now we have

[tex]11 = 1.7 \frac{dT}{dx}[/tex]

so temperature gradient is given as

[tex]\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m [/tex]

also in other unit it will be same

[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

To find the total mass of chloroform in the second sample, the mass of carbon is scaled proportionally to find the corresponding masses of hydrogen and chlorine. By summing these, the total mass of chloroform is calculated to be 298.525 grams.

To determine the total mass of chloroform in the second sample, we must first understand that chloroform has a known molecular formula of CHCl3. Given that the first sample contains 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen, we can deduce the mass ratios of the elements within chloroform. Using the molecular mass of chloroform, which is 119.37 g/mol, we can calculate the masses of hydrogen and chlorine in the second sample based on the given mass of carbon.

For the second sample: If 12.0 g of carbon is accompanied by 1.01 g of hydrogen and 106.4 g of chlorine, then 30.0 g of carbon should be accompanied by:

Thus, the total mass of chloroform in the second sample would be the sum of the masses of carbon, hydrogen, and chlorine: 30.0 g C + 2.525 g H + 266.0 g Cl = 298.525 g of chloroform.

Answer:

N = 768.4 N

Explanation:

As per given FBD we can see that the person inside the elevator have two forces on it

1) Normal force upwards

2) weight downwards

Now from Newton's law of motion we can say

[tex]F_{net} = ma[/tex]

[tex]N - mg = ma[/tex]

[tex]N = mg + ma[/tex]

now plug in all values in it

[tex]N = 68(1.5) + 68(9.8)[/tex]

[tex]N = 768.4 N[/tex]

Answer:

option (c) - 10 j

Explanation:

F = (6 i + 4 j - 2 k) N

r1 = (1.5, 3, -4.5) m = (1.5 i + 3j - 4.5 k) m

r2 = (4, -2.5, - 3) m = (4 i - 2.5 j - 3 k) m

displacement, r = r2 - r1 = ( 2.5 i - 5.5 j + 1.5 k) m

Work done is defined as the dot product of force vector and teh displacement vector.

[tex]W = \overrightarrow{F}.\overrightarrow{r}[/tex]

W = (6 i + 4 j - 2 k) . ( 2.5 i - 5.5 j + 1.5 k)

W = 15 - 22 - 3 = - 10 J

Answer:

[tex]strain = 1.4 \times 10^{-3} [/tex]

Explanation:

As we know by the formula of elasticity that

[tex]E = \frac{stress}{strain}[/tex]

now we have

[tex]E = 110 GPA[/tex]

[tex]E = 110 \times 10^9 Pa[/tex]

Area = 15.2 mm x 19.1 mm

[tex]A = 290.3 \times 10^{-6}[/tex]

now we also know that force is given as

[tex]F = 44500 N[/tex]

here we have

stress = Force / Area

[tex]stress = \frac{44500}{290.3 \times 10^{-6}}[/tex]

[tex]stress = 1.53 \times 10^8 N/m^2[/tex]

now from above formula we have

[tex]strain = \frac{stress}{E}[/tex]

[tex]strain = \frac{1.53 \times 10^8}{110 \times 10^9}[/tex]

[tex]strain = 1.4 \times 10^{-3} [/tex]

Answer:

a)

1.73 m/s

b)

6.43 x 10⁻⁴ m

Explanation:

m = mass of the flea = 1.8 x 10⁻⁴ kg

v₀ = initial speed of the flea = 0 m/s

v = final speed of the flea

W = work done by the force on the flea = 2.7 x 10⁻⁴ J

Using work-change in kinetic energy, Work done is given as

W = (0.5) m (v² - v₀²)

Inserting the values

2.7 x 10⁻⁴ = (0.5) (1.8 x 10⁻⁴) (v² - 0²)

v = 1.73 m/s

b)

d = distance moved by the flea while pushing off

F = Upward force applied on the flea by ground = 0.42 N

Work done is also given as

W = F d

2.7 x 10⁻⁴ = (0.42) d

d = 6.43 x 10⁻⁴ m

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

T2 = 32 C

T3 = 32 K

Convert all the temperatures in degree C

The relation between F and C is given by

(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

C = 0

So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

Answer:

a) 0.124 m

b) 0.93 ms⁻¹

c) 0.5 k A² cos ² ( ωt )  

Explanation:

1) Potential energy = U = 0.5 k A² , where A is the amplitude and k = 850 N/m is the spring constant.

0.5 ( 850) (A² ) = 6.5

⇒ A = 0.124 m = Amplitude.

b) From energy conservation,  0.5 m v² =  6.5

⇒ speed = v = 0.93 ms⁻¹

c) If x = A cos ωt ,

Potential energy = 0.5 k A² = 0.5 k A² cos ² ( ωt )  

Answer:

1.4719 m per sec

Explanation:

Hello

Kinetic energy is the energy associated with the movement of objects. Although there are many forms of kinetic energy  

the formula to use is

[tex]E=\frac{mv^{2} }{2}[/tex]

where m is the mass of the object and v the velocity

lets see the kinetic energy of the sprinter running

[tex]E=\frac{65 Kg*10(\frac{m}{s} ^)){2} }{2} \\\\E=\frac{65 *100 }{2} \\E=3250 Joules\\\\[/tex]

Now, the elephant must have the same kinetic energy

[tex]E=\frac{m*v_{2} ^{2} }{2} \\\\E*2=m*v_{2} ^{2}\\ \frac{2E}{m} =v_{2} ^{2} \\\sqrt{\frac{2E}{m} } =v_{2}  \\\\\\v_{2} =\sqrt{\frac{2*3250}{3000} }\\ \\v_{2} =1.4719 \frac{m}{s} \\\\[/tex]

it works only the positive root, so the elephant must to  walk  to 1.4719 m/s to have the same kinetic energy.

Have a great day

A 3000-kg elephant needs to move at a speed of approximately 1.183 m/s to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s.

The subject of the question falls into the category of physics, specifically dealing with the concept of kinetic energy and motion. Kinetic energy is given by the equation KE = 1/2 m v², where m is the mass and v is the velocity of the object. In this case, we want the elephant and the sprinter to have the same kinetic energy.

To find the velocity at which the elephant must move, we set the kinetic energy of the elephant (1/2 * 3000 kg * v²) equal to the kinetic energy of the sprinter (1/2 * 65.0 kg * (10.0 m/s)²) and solve for v (velocity of the elephant).

This gives us v = sqrt((1/2 * 65 kg * (10 m/s)²) / (1/2 * 3000 kg)) = 1.183 m/s. Therefore, a 3000-kg elephant must move at a speed of approximately 1.183 m/s to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s.

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Answer:

Option C is the correct answer.

Explanation:

Heat required to melt solid in to liquid is calculated using the formula

            H = mL, where m is the mass and L is the latent heat of fusion.

Latent heat of fusion for water = 333.55 J/g

Mass of ice = 0.3 kg = 300 g

Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

          H = mL = 300 x 333.55 = 100,375 J

Option C is the correct answer.

Answer:

Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.

Explanation:

Let line of action be horizontal axis , mass of ball be m and unknown velocity be v.

Here momentum is conserved.

Initial momentum =Final momentum

Initial momentum = m x 6i + m x 0i = 6m i

Final momentum = m x (5.21cos 29.7 i + 5.21sin 29.7 j) + m x v = 4.26 m i + 2.58 m j + m v

4.26 m i + 2.58 m j + m v = 6m i

v = 1.74 i - 2.58 j

Magnitude of velocity [tex]=\sqrt{1.74^2+(-2.58)^2}=3.11m/s[/tex]

Direction,

         [tex]\theta =tan^{-1}\left ( \frac{-2.58}{1.74}\right )=--56^0[/tex]

Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.

Answer:

2440.24 J

Explanation:

Moment of inertia, I1 = 5 kg m^2

frequency, f1 = 3 rps

ω1 = 2 x π x f1 = 2 x π x 3 = 6 π rad/s

Moment of inertia, I2 = 2 kg m^2

Let the new frequency is f2.

ω2 = 2 x π x f2

here no external torque is applied, so the angular momentum remains constant.

I1 x ω1 = I2 x ω2

5 x 6 π = 2 x 2 x π x f2

f2 = 7.5 rps

ω2 = 2 x π x 7.5 = 15 π

Initial kinetic energy, K1 = 1/2 x I1 x ω1^2 = 0.5 x 5 x (6 π)² = 887.36 J

Final kinetic energy, K2 = 1/2 x I2 x ω2^2 = 0.5 x 3 x (15 π)² = 3327.6 J

Work done, W = Change in kinetic energy = 3327.6 - 887.36 = 2440.24 J

Answer:

The speed of the water shoot out of the hole is 20 m/s.

(d) is correct option.

Explanation:

Given that,

Height = 20 m

We need to calculate the velocity

Using formula Bernoulli equation

[tex]\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]

Where,

v₁= initial velocity

v₂=final velocity

h₁=total height

h₂=height of the hole from the base

Put the value into the formula

[tex]v_{1}^2=2g(h_{2}-h_{1})[/tex]

[tex]v_{1}=\sqrt{2g(h_{2}-h_{1})}[/tex]

[tex]v_{1}=\sqrt{2\times9.8\times(20-0.005)}[/tex]

[tex]v_{1}=19.7\ m/s= approximate\ 20\ m/s[/tex]

Hence, The speed of the water shoot out of the hole is 20 m/s.