Which of the following statements are true?
1. Liquid water expands with increasing temperature above 4°C.
2. Liquid water expands with increasing temperature between 0°C and 4°C.
3. Water contracts as it freezes at 0°C.
4. Solid ice is less dense than liquid water.

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s

Answer:

10 m/s²

Explanation:

Acceleration: This the rate of change of velocity. The unit of acceleration is m/s²

From the question,

a = (v-u)/t.................... Equation 1

Where a = acceleration of the cheetah, v = final velocity of the cheetah, u = initial velocity of the cheetah, t = time.

Given: u = 0 m/s, v = 25 m/s, t = 2.5 s.

Substitute these values into equation 1

a = (25-0)/2.5

a = 25/2.5

a = 10 m/s²

Hence the acceleration of the cheetah = 10 m/s²

Answer:

-15 m/s

Explanation:

The computation of the velocity of the 4.0 kg fragment is shown below:

For this question, we use the correlation of the momentum along with horizontal x axis

Given that

Weight of stationary shell = 6 kg

Other two fragments each = 1.0 kg

Angle = 60

Speed = 60 m/s

Based on the above information, the velocity = v is

[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]

[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]

[tex]v = \frac{60}{4}[/tex]

= -15 m/s

Answer:

Explanation:

(a)

The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric

pressure in the tube then the resulting vapour pressure is

Patm - pgh = Prapor

The final reading ion the barometer is

pgh = Palm - Proper

Hence, the true atmospheric pressure is greater.

you can find the answer in this book

physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p

Answer:

  AB = CD = √17

Explanation:

The distance formula is used to find the length of a line segment between two points. Here, we want to show the distance AB is the same as the distance CD.

  d = √((x1 -x1)² +(y2 -y1)²)

__

AB: d = √((1 -2)² +(3 -(-1))²) = √((-1)² +4²) = √17

CD: d = √((7-6)² +(1-5)²) = √(1² +(-4)²) = √17 . . . . same as AB

Segment AB is congruent to segment CD.

Answer:

Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.

Answer:

0.303s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

Answer:

Ec = 6220.56 kcal

Explanation:

In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.

You use the following formula:

[tex]W_c=Mgh[/tex]        (1)

Wc: work done by the climber

g: gravitational constant = 9.8 m/s^2

M: mass of the climber = 78.4 kg

h: height reached by the climber = 5.42km = 5420 m

You replace in the equation (1):

[tex]W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J[/tex]     (2)

Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:

[tex]0.16(E_c)=4,164,294.4J[/tex]

Ec: Calories

You solve for Ec and convert the result to Cal:

[tex]E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal[/tex]

The amount of Calories needed by the climber was 6220.56 kcal

Answer:

a_total = 14.022 m/s²

Explanation:

The total acceleration of a uniform circular motion is given by the following formula:

[tex]a=\sqrt{a_c^2+a_T^2}[/tex]         (1)

ac: centripetal acceleration

aT: tangential acceleration

Then, you first calculate the centripetal acceleration by using the following formula:    

[tex]a_c=r\omega^2[/tex]

r: radius of the circular trajectory = 2.0m

w: final angular velocity  of the ball = 7.0 rad/s

[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]        

Next, you calculate the tangential acceleration. aT is calculate by using:

[tex]a_T=r\alpha[/tex]    (2)

α: angular acceleration

The angular acceleration is:

[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]

wo: initial angular velocity = 13 rad/s

t: time = 15 s

Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:

[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]

Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:

[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]

The total acceleration of the ball is 14.022 m/s²

Answer:

  60 m

Explanation:

After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.

__

The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.

Answer:

% increase in resistance of tungsten = 9.27%

Explanation:

We are given:

Co-efficient of resistivity for the metal gold; α_g = 0.0034 /°C

Co-efficient of resistivity for tungsten;α_t = 0.0045 /°C

% Resistance change of gold wire due to temperature change = 7%

Now, let R1 and R2 be the resistance before and after the temperature change respectively.

Thus;

(R2 - R1)/R1) x 100 = 7

So,

(R2 - R1) = 0.07R1

R2 = R1 + 0.07R1

R2 = 1.07R1

The equation to get the change in temperature is given as;

R2 = R1(1 + αΔt)

So, for gold,

1.07R1 = R1(1 + 0.0034*Δt)

R1 will cancel out to give;

1.07 = 1 + 0.0034Δt

(1.07 - 1)/0.0034 = Δt

Δt = 20.59°C

For this same temperature for tungsten, let Rt1 and Rt2 be the resistance before and after the temperature change respectively and we have;

Rt2 = Rt1(1 + α_t*Δt)

So, Rt2/Rt1 = 1 + 0.0045*20.59

Rt2/Rt1 = 1.0927

From earlier, we saw that;

(R2 - R1)/R1) x 100 = change in resistance

Similarly,

(Rt2 - Rt1)/Rt1) x 100 = change in resistance

Simplifying it, we have;

[(Rt2/Rt1) - 1] × 100 = %change in resistance

Plugging in the value of 1.0927 for Rt2/Rt1, we have;

(1.0927 - 1) × 100 = %change in resistance

%change in resistance = 9.27%

Answer:

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Explanation:

Firstly, there is supposed to be a diagram attached in order  to complete this question;

I have attached  the diagram below in order to solve this question.

From the data given;

The radius of the car R = 600 ft

Velocity of the car  B, [tex]V_B = 45 mi / hr[/tex]

We are to determine  the magnitude of the acceleration of A and the rate at which the speed of B is changing.

To start with the magnitude of acceleration A;

We all know that

1 mile = 5280 ft and an hour =  3600 seconds

Thus for ; 1 mile/hr ; we have :

5280 ft/ 3600 seconds

= 22/15 ft/sec

However;

for the velocity of the car B = 45 mi/hr; to ft/sec, we have:

= (45 × 22/15) ft/sec

= 66 ft/sec

A free body diagram is attached in the second diagram showing how we resolve the vector form

Now; to determine the magnitude of the acceleration of A; we have:

[tex]^ \to {a_A} = a_A sin 45^0 ^{\to} + a_A cos 45^0 \ j ^{\to} \\ \\ ^\to {a_B} = -(a_t)_B \ i ^ \to + (a_c )_B cos 45 ^0 \ j ^{\to}[/tex]

Where;

[tex](a_c)_B[/tex] = radial acceleration of B

[tex](a_t)_B[/tex] = tangential acceleration of B

From observation in the diagram; The acceleration of B is 0 from A

So;

[tex]a_B ^\to - a_A ^\to = a_{B/A} ^ \to[/tex]

[tex](-(a_t)_B - a_A sin 45^0 ) ^\to i+ ((a_t)_B-a_A \ cos \ 45^0) ^ \to j = 0[/tex]

[tex](a_c)_B = \dfrac{V_B^2}{R}[/tex]

[tex](a_c)_B = \dfrac{(66)^2}{600}[/tex]

[tex](a_c)_B = 7.26 ft/s^2[/tex]

Equating the coefficient of i and j now; we have :

[tex](a_t)_B = -a_A \ sin 45^0 --- (1)\\ \\ (a_c)_B = a_A cos \ 45^0 --- (2)\\ \\[/tex]

From equation (2)

replace [tex](a_c)_B[/tex] with 7.26 ft/s^2; we have

[tex]7.26 \ ft/s^2 = a_A cos \ 45^0 \\ \\ a_A = \dfrac{7.26 \ ft/s^2}{co s \ 45^0}[/tex]

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

Similarly;

From equation (1)

[tex](a_t)_B = -a_A \ sin 45^0[/tex]

replace [tex]a_A[/tex] with 10.267 ft/s^2

[tex](a_t)_B = -10.267 \ ft/s^2 * \ sin 45^0[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Answer:

Friction

Explanation:

There are tiny bumps and grooves on every object, which make them rough and more difficult to rub against each other. Even though the crate remains at rest at first, the frictional force causes it to stay in place and accelerate with the truck. Hope this helps!

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

[tex]\Sigma \tau=0[/tex]

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

[tex]-\tau_l+\tau_p+\tau_{cm}=0[/tex]          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

[tex]\tau=Fd[/tex]           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

[tex]-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N[/tex]

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

[tex]F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N[/tex]

The force applied by the left support is 2123.33 N

Explanation:

(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:

[tex]T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s[/tex]

(b) The frequency of a wave is inversely proportional to its period:

[tex]f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz[/tex]

(c) The wavelength is the distance between two successive crests, so:

[tex]\lambda=30.2m[/tex]

(d) The speed of a wave is defined as:

[tex]v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}[/tex]

Answer:

A.) 78.4 J for both

B.) 78.4 J for both

C.) 8.85 m/s for both

D.) 17.7 kgm/s

Explanation:

Given information:

Mass m = 2 kg

Distance d = 20 m

High h = 4 m

A.) Gravitational potential energy can be calculated by using the formula

P.E = mgh

P.E = 2 × 9.8 × 4

P.E = 78.4 J

Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J

B.) According to conservative energy,

Maximum P.E = Maximum K.E.

Therefore, the kinetic energy of the two blocks will be 78.4 J

C.) Since K.E = 1/2mv^2 = mgh

V = √(2gh)

Solve for velocity V by substituting g and h into the formula

V = √(2 × 9.8 × 4)

V = √78.4

V = 8.85 m/s

The velocities of both block will be 8.85 m/s

D.) Momentum is the product of mass and velocity. That is,

Momentum = MV

Substitute for m and V into the formula

Momentum = 2 × 8.85 = 17.7 kgm/s

Both block will have the same value since the ramp Is frictionless.

Hope this helps....

Good luck on your assignment.....

Answer:

0.05 N

Explanation:

Data provided in the question

The Wire carries a current of 4A to the left direction

The constant magnetic field of magnitude = 0.05 T

Pointing upward i.e Z direction

The wire is in the magnetic field = 25 cm

Based on the above information, the force on the current is

[tex]= Current \times constant\ magnetic\ field\ of\ magnitude \times magnetic\ field[/tex]

[tex]= 4 \times 0.05 \times 0.25[/tex]

= 0.05 N

The direction will be the negative Y direction

Answer:

0 - Then, the ray is totally reflected

Explanation:

The ray reaches the boundary between the two mediums at 49.2°.

If the ray is totally reflected it is necessary that the crictical angle is lower that the incidet angle.

You use the following to calculate the critical angle:

[tex]\theta_c=sin^{-1}(\frac{n_2}{n_1})[/tex]       (1)

n2: index of refraction of the second medium (air) = 1.00

n1: index of refraction of the first medium (glass) = 1.51

You replace the values of the parameters in the equation (1):

[tex]\theta_c=sin^{-1}(\frac{1.00}{1.51})=41.47\°[/tex]

The critical angle is 41.47°, which is lower than the incident angle 49.2°.

Then, the ray is totally reflected.

0

4.Moving with constant speed

Complete question is;

After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.

Answer:

Tangential speed of foot just before it lands is; v = 5.37m/s

Explanation:

Let U (potential energy) be zero on the ground.

So, initially, U = mgh

where, h = 0.98/2 = 0.49m (midpoint of the leg)

Now just before the leg hits the floor it would have kinetic energy as;

K = ½Iω²

where ω = v/r and I = ⅓mr²

So, K = ½(⅓mr²)(v/r)²

K = (1/6) × (mr²)/(v²/r²)

K = (1/6) × mv²

From principle of conservation of energy, we have;

Potential energy = Kinetic energy

Thus;

mgh = (1/6) × mv²

m will cancel out to give;

gh = (1/6)v²

Making v the subject, we have;

v = √6gh

v = √(6 × 9.81 × 0.49)

v = √28.8414

v = 5.37m/s

Answer:

The maximum amount of meat that the butcher can sell is  [tex]3\frac{1}{12}\:lb[/tex]

Explanation:

The maximum amount can be found by taking the difference of mixed numbers.

[tex]5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\[/tex]

Best Regards!

Answer:

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.

Explanation:

The maximum height h is;

h = height covered during acceleration plus height covered when the motor stops.

h = h1 + h2 .......1

height covered during acceleration h1 can be derived using the equation of motion;

h1 = ut + 0.5at^2

Initial speed u = 0

h1 = 0.5at^2

acceleration a = 20 m/s^2

Time t = 6.0 s

h1 = 0.5×(20 × 6^2)

h1 = 0.5(20×36)

h1 = 360 m

height covered when the motor stops h2 can be derived using equation of motion;

h2 = ut + 0.5at^2 .......2

Where;

a = g = acceleration due to gravity = -9.8 m/s^2

The speed when the motor stops u;

u = at = 20 m/s^2 × 6.0 s = 120 m/s

Time t2 can be derived from;

v = u - gt

v = 0 (at maximum height velocity is zero)

u = gt

t = u/g

t = 120m/s / 9.8m/s^2

t = 12.24 seconds.

Substituting the values into equation 2;

h2 = 120(12.24) - 0.5(9.8×12.24^2)

h2 = 734.69376 m

h2 = 734.69 m

From equation 1;

h = h1 + h2 . substituting the values;

h = 360m + 734.69m

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

[tex]A_1V_1=A_2V_2[/tex]

[tex]A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}[/tex]

the diameter decreases 86% so

[tex]d_2 = 0.86d_1[/tex]

[tex]v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}[/tex]

Thus, speed in smaller section is 48.6 m/s

Answer: not sure

Explanation:

Answer:

(a) T = 1.35 s

(b) vmax = 0.17 m/s

(c) v = 0.056 m/s

Explanation:

(a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:

[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]          (1)

m: mass of the cart = 350 g = 0.350kg

k: spring constant = 7.5 N/m

[tex]T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s[/tex]

The period of oscillation of the car is 1.35s

(b) The maximum speed of the car is given by the following formula:

[tex]v_{max}=\omega A[/tex]       (2)

w: angular frequency

A: amplitude of the motion = 3.8 cm = 0.038m

You calculate the angular frequency:

[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}[/tex]              

Then, you use the result of w in the equation (2):

[tex]v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}[/tex]

The maximum speed if 0.17m/s

(c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:

[tex]x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s[/tex]

The speed is the value of the following function for t = 0.069s

[tex]|v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}[/tex]

The speed of the car is 0.056m/s

Answer:

The inner diameter is 57.3 cm

Explanation:

The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):

[tex] W_{s} = W_{w} [/tex]

[tex] m_{s}*g = m_{w}*g [/tex]            

[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]    

Where D is the density and V is the volume

[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]    

Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter    

[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]                    

[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]          

[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]  

[tex] d_{i} = 57.3 cm [/tex]                  

Therefore, the inner diameter is 57.3 cm.    

I hope it helps you!    

Answer;

The pond's water level will fall.

Explanation;

Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.

When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.

Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.

When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.

Hope this Helps!!!

Buoyancy is an upward force exerted by a fluid on a body partially or completely immersed in it

The pond water level will lower slightly

According to Archimedes principle, the up thrust on the boat by the water is given by the volume of the water displaced

Learn more here;

https://brainly.com/question/24529607

Answer:

KE = 135,000 j or 135 KJ

Explanation:

KE=0.5mv^2

KE=0.5*1200*15^2

KE = 135,000 joules or 135 Kilo Joules