EXAMPLE
BIKE/STROLLER - WHEEL AND AXLE
SCISSORS/HEDGE CUTTER - LEVER
STAIRS - INCLINED PLANED
SCREW DRIVER/SCREW/NAILS- SCREW
AXE - WEDGE
WELL - PULLEY
Its not much but hope this help you
What does FAI stand for?
Answer:
Femoroacetabular impingement
It occurs when an area of your hip joint becomes pinched when you move in certain directions. :)
A ball is being rolled by a normal push of 180N. It is opposed by friction which has a force of 61N and air resistance which has a force of 23N. Calculate the resultant force.
Resultant force is basically the force left after everything is added.
if a ball is being pushed one one side with 180N, and being pushed on teh opposite side with 84N (I added friction and air resistance since they're acting on the same side), then the resultant force would be:
180N - 84N = 96N (you can determine whether it's positive or negative based on the direction of the vector)
a 0.350 kg block at -27.5 °C is added to 0.217 kg of water at 25.0 °C. they come to equilibrium at 16.4 °C. what is the specific heat of the block?
Answer:
[tex]C_{pb}=0.501\ kJ/kg.K[/tex]
Explanation:
Given that
[tex]m_1=0.35 kg[/tex]
[tex]T_1=-27.5^oC[/tex]
[tex]m_2=0.214 kg[/tex]
[tex]T_2=25^oC[/tex]
[tex]T=16.4^oC[/tex]
We know that
[tex]C_{pw}=4.187 kJ/kg.K[/tex]
By using energy conservation
Heat lost by water = Heat gain by block
[tex]m_2\times C_{pw}\times (T_2-T)=m_1\times C_{pb}\times (T-T_1)[/tex]
[tex]0.214\times 4.187\times (25-16.4)=0.35\times C_{pb}\times (16.4+27.5)[/tex]
[tex]C_{pb}=0.501\ kJ/kg.K[/tex]
Therefore the specific heat of the block will be 0.501 kJ/kg.K
Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine the net force charge acting at q1 (+ 2.0 × 10^-5C), caused by q2 (-4.0 × 10-5 C) and q3 (-4.0 × 10^-5C). They create a right angles triangle, where q1 is at the 90° corner
Determine the net electric field acting at q1
Answer:
E≅1.2×10^7 N/C
Explanation:
First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!
Edit: I ended up adding it anyways ;P
Since we are observing the net electric field acting at q1, we need to use the formula: [tex]E=k\frac{q}{r^{2} }[/tex]
And since we are observing the effects of multiple charges at once...
E=ΣE, which just means wee need to add all the observed electric fields together:
ΣE= [tex]k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }[/tex]
Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:
q2=q3they are the same distance from q1These are actually really nice to have, because now we can simplify our expression to:
[tex]E=k\frac{2q}{r^{2} }[/tex]
Now let's plug in our values and get an answer out.
E= 2(8.99×10^9)(4×10^-5)/(0.24)
Plugging all that in, I get:
E≅1.2×10^7 N/C
If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.
Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!
Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.
Answer the following questions regarding the equation:
N₂ + 3H₂ → 2NH₃
1) indicates what type of reaction is
2) what represents the coefficients 3 and 2 in the previous reaction is done for
3) What would be missing in the previous equation to make it more accurate is
Explanation:
1) This is a synthesis reaction (two or more reactants combine to form a single product).
2) The coefficients are added to balance the reaction.
3) Adding the states of matter (solid, liquid, gas) will make the reaction more precise.
Which statement accurately describes this atomic spectrum? There is a horizontal rectangle with the colors of the rainbow from violet to red. There are lines in purple, blue, green and orange. The black lines represent the energy emitted by the electrons. The black lines represent the energy absorbed by the electrons. The colored lines represent the energy emitted by the electrons. The colored lines represent the energy absorbed by the electrons.
Answer:
The black lines represent the energy absorbed by the electrons.
Explanation:
Atoms emit lights when they are excited. These lights are of particular wavelengths that match with different colors. A series of colored lines appear along with spaces in the middle of the two colors. The middle of the colors is filled with dark spaces. Each spectral line of an element represents a specific characteristic of the element. These colored lines appearing in the form of series are termed to be the atomic spectrum of the element. Identification of the elements is done through the line of the spectrum they possess.
Answer:
(B) The black lines represent the energy absorbed by the electrons.
Explanation:
What makes a clinical thermometer suitable for measuring small changes in body temperature? *
Answer: Because of the fine bore of the tube.
Explanation:
Temperature is the degree of hotness and coldness. And thermometer is the instrument use to measure temperature.
The two most common types of themometric fluids for thermometer are alcohol and mercury.
What makes a clinical thermometer suitable for measuring small changes in body temperature is because of the fine bore of the tube which makes it possible for small temperature changes to cause large changes in the length of mercury columns, making the thermometer very sensitive to temperature changes.
The most prominent feature of the thermometer is the kink or constriction of bore near the bulb.
Answer:
xxx
Explanation:
when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how much time was it in contact with the floor?
Answer:
Time, t = 3.2 ms
Explanation:
It is given that,
Mass of basketball, m = 0.622 kg
Initial velocity, u = 4.23 m/s
Final velocity, v = 3.85 m/s
Average force acting on the ball, F = 72.9 N
We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,
[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms[/tex]
So, the ball is in contact with the floor for 3.2 ms.
Choose all correct statements: Refractive index of medium A with respect to medium B is 4/3 and refractive index of medium A with respect to medium C is 3/2. Medium B is denser than medium C When light travels from B to C light bends away from the normal Refractive index of C with respect to B is 8/9 Ratio of speed of light in medium B to the speed of light in medium A = 4/3
Answer:
Medium B is denser than medium C
When light travels from B to C light bends away from the normal .
Refractive index of C with respect to B is 8/9
Ratio of speed of light in medium B to the speed of light in medium A = 4/3
Explanation:
Refractive index of medium A / Refractive index of medium B = 4/3
Refractive index of medium A / Refractive index of medium C = 3/2
Dividing ,
Refractive index of medium B / Refractive index of medium C
= 3/2 x 3/4
= 9 / 8
Refractive index of medium B > Refractive index of medium C
Refractive index of C with respect to B is 8/9
speed of light in medium B / the speed of light in medium A
= refractive index of A / refractive index of B = 4 / 3
A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.15 s
Answer:
The magnitude of the average force on the wall during the collision is 6 N.
Explanation:
Given;
mass of snowball, m = 120 g = 0.12 kg
velocity of the snowball, v = 7.5 m/s
duration of the collision between the snowball and the wall, t = 0.15 s
Magnitude of the average force can be calculated by applying Newton's second law of motion;
F = ma
where;
a is acceleration = v / t
a = 7.5 / 0.15
a = 50 m/s²
F = ma
F = 0.12 x 50
F = 6 N
Therefore, the magnitude of the average force on the wall during the collision is 6 N.
Which measurement is a potential difference?
O A. 115 N
O B. 115 C
O C. 115 J
O D. 115 V
Answer:
option d is answer because pd is measured in volt.
What type of image does a concave lens form? A. real image B. magnified image C. virtual image D. reverse image
Assessment started: undefined. Item 1 Which statement accurately describes what happens when ice melts in terms of energy? A:The ice releases energy which causes chemical bonds to break, changing the ice to water.;B:The ice releases energy which causes the water molecules to have less kinetic and potential energy, changing their configuration from solid to liquid. C:The ice absorbs energy which causes the water molecules to have more kinetic and potential energy, changing their configuration from a solid to a liquid. D:The ice absorbs energy which causes chemical bonds to break, changing ice to water.
Answer:
A.
Explanation:
it's needed top release energy due to the breaking of chemical bonds.
Answer:
A: the ice releases energy which causes chemical bonds to break, changing the water to ice.
Explanation:
An object of mass 800g hangs on a spring. Calculate the force exerted by the object if acceleration due to gravity is 10/s2
Answer:
8 N
Explanation:
Using the equation F=ma (F: force/ m: mass in kg/ a: acceleration),
F = (800/1000)(10)
F = 8 N
What us meant by Velocity ratio of simple machine is 4?
a) Under what circumstances would a constant force result in increasing acceleration on a body? b) Under what circumstances would a constant force result in zero acceleration on a body?
Answer:
Remember the equation:
F = m*a
where F is force, m is mass and a is acceleration.
If we have F constant. and we want that increases, then we can have the case where m decreases.
The mass can decrease in cases like a rocket, where as the fuel of the rocket consumes, the mass of the rocket decreases and the acceleration increases.
b) The cases where a constant force results in a constant acceleration of zero, is when the force is canceled, an example of this is the constant force of the gravity in all the objects. The objects that are in the ground are being affected by this force, but the gravitational force is canceled with the normal force of the ground. Then we have a constant force that does not cause any acceleration.
10 advantages of friction
Answer:
1. it helps to change the direction.
2. it helps us to walk on ground.
3. it helps the vechils to break while moving.
4. helps in changing one form of enegry to another form. eg when we rub our hands we feel heat energy.
5. it opposites the force.
6. it helps us to change shape of objects.eg we roll the dough to make it roti.
7. it changes the state of body from rest motion.eg when we push any obj from inclined plane it moves.
i all know is just 7..
Q.) Miscellaneous conversations. a) mass=120*10^8 g (Convert this value in mg and kg Write in standard form after converting) b) length=200000*10^3 Convert this value in micrometres cm and km Write in standard form after converting
Answer:
a. Convert 120 × 10⁸ g to i mg = 1.2 × 10¹³ mg ii. to g = 1.2 × 10⁷ kg
b. Convert 200000 × 10³ m to i. micrometers = 0.2 × 10³ μm ii. to cm = 2 × 10⁶ cm iii. to km = 2 × 10⁵ km
Explanation:
a. i. To convert the mass = 120 × 10⁸ g to mg, We know that 1000 mg = 10³ mg = 1 g, Since we are converting to mg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10³ mg = 120 × 10¹¹ mg = 1.2 × 10² × 10¹¹ mg = 1.2 × 10¹³ mg
ii. To convert the mass = 120 × 10⁸ g to kg, We know that 1000 g = 10³ g = 1 kg, 1 g = 10⁻³ kg. Since we are converting to kg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10⁻³ kg = 120 × 10⁵ kg = 1.2 × 10² × 10⁵ kg = 1.2 × 10⁷ kg
b. i.To convert the length = 200000 × 10³ m to micrometers, We know that 1/1000000 μm = 10⁻⁶ mg = 1 m, Since we are converting to micrometers, μm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/1000000 μm = 200000/1000000 × 10³ μm = 0.2 × 10³ μm
ii. To convert the length = 200000 × 10³ m to cm, We know that 100 cm = 10² cm = 1 m, 1 m = 10⁻² cm = 1/100 cm. Since we are converting to cm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/100 cm = 200000/100 × 10³ cm = 2000 × 10³ cm = 2 × 10³ × 10³ cm = 2 × 10⁶ cm
iii. To convert the length = 200000 × 10³ m to km, We know that 1000 m = 10³ m = 1 km, 1 m = 10⁻³ km = 1/1000 km Since we are converting to km, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/1000 km = 200000/1000 × 10³ km = 200 × 10³ km = 2 × 10² × 10³ km = 2 × 10⁵ km
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).
Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
[tex]C=2\pi r[/tex] (1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
[tex]r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m[/tex]
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
[tex]L=Lo[1+\alpha \Delta T][/tex] (2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
[tex]L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m[/tex]
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
[tex]r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m[/tex]
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m
Explanation:
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).
During or after exercise,it is normal for a student to comment that his/her heart feels like it is "beating out of my chest"?
Answer:
Yes it is normal
Explanation:
When you exercise, your heart beat goes up, resulting in people saying that their heart feels like it is "beating out of their chests".
What is the last step in creating an argumentative essay?
outlining
prewriting
researching
revising
Answer:
The answer is Revising
Answer:
D
Explanation:
Revising
Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de sección por el que sale el agua con un chorro continuo. A) ¿Qué cantidad de líquido saldrá del depósito expresada en m3/s?
Answer:
a) El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].
Explanation:
a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:
[tex]\Delta z = \frac{v_{out}^{2}}{2\cdot g}[/tex]
Donde:
[tex]\Delta z[/tex] - Diferencia de altura, medida en metros.
[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.
[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.
Se despeja la rapidez de salida del chorro:
[tex]v_{out} = \sqrt{2\cdot g \cdot \Delta z}[/tex]
Si [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]\Delta z = 0.3\,m[/tex], entonces la rapidez de salida del chorro es:
[tex]v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}[/tex]
[tex]v_{out} \approx 2.426\,\frac{m}{s}[/tex]
Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:
[tex]\dot V_{out} = v_{out}\cdot A_{t}[/tex]
Donde:
[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.
[tex]A_{t}[/tex] - Área transversal del orificio, medido en metros cuadrados.
[tex]\dot V_{out}[/tex] - Caudal de salida del chorro, medido en metros cúbicos por segundo.
Dado que [tex]v_{out} = 2.426\,\frac{m}{s}[/tex] y [tex]A_{t} = 5\,cm^{2}[/tex], el caudal de salida del chorro es:
[tex]\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)[/tex]
[tex]\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex]
El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].
What kind of substance can you pour from one container into another without a change in volume
Answer:
Liquids
Explanation:
Liquids take up the shape of the container it is poured into but will never change its volume.
Choose the friction which opposes the relative motion between surfaces in motion a.Static friction b.Kinetic friction c.Sliding friction d.Both kinetic and sliding friction
Answer:
d. Both kinetic and sliding friction
Explanation:
Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.
What is the meaning of wwe
Answer:
World Wrestling Entertainment, Inc., d/b/a WWE, is an American integrated media and entertainment company that is primarily known for professional wrestling. WWE has also branched out into other fields, including movies, football, and various other business ventures.
Answer:
This means WORLD WRESTLING ENTERTAINMENT.
Explanation:
This is an american integrated media and entertainment company that is primarily known for PROFESSIONAL WRESTLING.
A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499cm3499cm 3 of air at atmospheric pressure (1.01×105Pa)(1.01×10 5 Pa) and a temperature of 27.0∘C27.0 ∘C. At the end of the stroke, the air has been compressed to a volume of 46.2cm 346.2cm 3 and the gauge pressure has increased to 2.72×106Pa2.72×10 6Pa. Compute the final temperature.
Answer:
503°C
Explanation:
According to the given situation, the computation of the final temperature is shown below:
In this question we use the law of ideal gas i.e
pV = nRT
i.e
[tex]\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}[/tex]
Therefore
[tex]T_2 = T_1 (\frac{p_2}{p_1}) (\frac{V_2}{V_1})[/tex]
[tex]= 300\ k (\frac{2.72 \times 10^{6} Pa + 1.01 \times 10^{5} Pa}{1.01 \times 10^{5} Pa})(\frac{46.2 cm^3}{499 cm^3})[/tex]
= 776 k
= (776 - 273)° C
= 503°C
Therefore the final temperature is 503°C
We simply applied the above formulas so that the final temperature could arrive
The skier starts from rest. The total distance travelled by the skier during the descent is 2800 m. The average resistive force on the skier is 220 N. Calculate the work done against the resistive force
Answer:
Explanation:
Force equal to resistive force will be applied for movement . So force applied
F = 220 N .
displacement = 2800 m
work done against resistive force
= force x displacement
= 220 x 2800 J
= 6.16 x 10⁵ J .
I need a story about kepler's laws for a child, something a child would understand. And i'd like it to be okay for someone if I copy anhd paste it or else my parents are gonna kill me and my teacher will be very disappointe, this is for summer school.
Answer:
(1) The orbits are ellipses, with focal points ƒ1 and ƒ2 for the first planet and ƒ1 and ƒ3 for the second planet. The Sun is placed in focal point ƒ1.
(2) The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2.
(3) The total orbit times for planet 1 and planet 2 have a ratio a13/2 : a23/2
Answer:
g the day off and I will be there at work and I will be there at work and I will
Explanation:
you are not going to be able to make it to the meeting tonight but I can tomorrow
identify units for volts
Answer:
Volts itself is SI Unit of Electric Potential.
Explanation:
Two projectiles are thrown from the same point with the velocity of49ms-1. First is
projected making an angle with the horizontal and the second at an angle of (90- ). The
second is found to rise 22.5m higher than the first. Find the heights to which each will rise?
Answer:
Height of the first projectile = 49.98 m
Height of the second projectile = 72.52 m
Explanation:
From the given information;
Two projectiles are thrown from the same point with the velocity of49m/s
First is projected making an angle θ with the horizontal
and the second at an angle of 90 - θ.
Thus; for the first height to the horizontal; we have;
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex] ----- (1)
the second height in the vertical direction is :
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex] -----(2)
However; the second is found to rise 22.5 m higher than the first; so , we have :
[tex]\dfrac{v^2 cos^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
Let's recall that :
Cos²θ = 1 - Sin²θ
Replacing it into above equation; we have:
[tex]\dfrac{v^2}{2g} - \dfrac{v^2 sin^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]\dfrac{v^2}{2g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } \dfrac{v^2}{g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } - \dfrac {9.8 \times 22.5}{(49)^2} = sin^2 \theta[/tex]
[tex]\dfrac{1}{2 } - \dfrac {220.5}{2401} = sin^2 \theta[/tex]
[tex]sin^2 \theta= 0.408[/tex]
From (1);
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]H_1 = \dfrac{49^2 \times 0.408}{2*9.8}[/tex]
[tex]H_1 = \dfrac{979.608}{19.6}[/tex]
[tex]\mathbf{H_1 =49.98 \ m }[/tex]
Height of the first projectile = 49.98 m
Similarly;
From(2)
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex]
[tex]H_2 = \dfrac{v^2 (1-sin^2 \theta)}{2g}[/tex]
[tex]H_2 = \dfrac{49^2 (1-0.408 )}{2 \times 9.8}[/tex]
[tex]H_2 = \dfrac{2401 (0.592 )}{19.6}[/tex]
[tex]H_2 = \dfrac{1421.392}{19.6}[/tex]
[tex]\mathbf{H_2 = 72.52 \ m}[/tex]
Height of the second projectile = 72.52 m