A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the maximum height reached by the ball? ​

Answer:

If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. ... Some energy is dissipated in the compression and decompression phases.

Explanation:

Answer:

Vy = 80.5 [m/s]

Explanation:

In order to solve this problem we must use the Pythagorean theorem.

V = 90 [m/s]

The components are Vx and Vy:

Therefore:

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

where:

Vy = 2*Vx ; because one is twice of the other.

[tex]90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s][/tex]

and the bigger vector is:

Vy = 40.25*2

Vy = 80.5 [m/s]

Answer:

 E ’= E / 8

therefore the correct answer is A

Explanation:

Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface

       Фi = ∫ E. dA = [tex]q_{int}[/tex] /ε₀

       E (4πr²) = q_{int} / ε₀

density is

      ρ = q_{int} / V

      q_{int} = ρ V = ρ 4/3 π r³

we substitute        

        E (4π r²) = ρ 4/3 π r³ /ε₀

        E = 1 /3ε₀   ρ r

let's change the density by

       ρ = Q / V = ​​Q / (4/3 π R³)

         E = 1 / 4πε₀ Q r / R³

if we now distribute the same charge on a sphere of radius R' = 2R

        E ’= 1 / 4pieo Q r / (2R)³

        E ’= 1 / 4ft Qr / R³ ⅛

        E ’= E / 8

therefore the correct answer is A

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Answer:

The answer is Top-Down processing

Explanation:

I had this question on a apex quiz and i got it correct.

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   [tex]COP  =  \frac{T_i}{T_o - T_i}[/tex]

Here [tex]T_i[/tex] is the inside temperature

while  [tex]T_o[/tex] is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   [tex] \frac{T_i}{T_o - T_i}[/tex] heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         [tex]Q \ \alpha \ (T_o - T_i)[/tex]

=>        [tex]Q= k (T_o - T_i)[/tex]

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   [tex] \frac{T_i}{T_o - T_i}[/tex]  amount of heat

   E  unit of electricity will remove  [tex]Q= k (T_o - T_i)[/tex]

So

      [tex]E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }[/tex]

=>   [tex]E = \frac{k}{T_i} (T_o - T_i)^2[/tex]

given that  [tex]\frac{k}{T_i}[/tex] is constant

    =>  [tex]E \  \alpha  \  (T_o - T_i)^2[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  [tex]T_i   =  30 ^oC[/tex]

 and      [tex]T_o  =  40 ^oC[/tex]

Hence  

     [tex]E = K (T_o - T_i)^2[/tex]

Here K stand for a constant

So  

        [tex]E = K (40 -  30)^2[/tex]

=>      [tex]E = 100K [/tex]

Now if  the  [tex]T_i   =  20 ^oC[/tex]

Then

       [tex]E = K (40 -  20)^2[/tex]

=>      [tex]E = 400 \ K[/tex]

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       [tex]Q = k (T_o - T_i )^{\frac{1}{2} }[/tex]

So

       [tex]E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }[/tex]

=>   [tex]E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }[/tex]

Assuming [tex]\frac{k}{T_i}[/tex] is a constant

Then  

     [tex]E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

Answer:

The Basic Concept

When we look at an object at rest, we say that it is in equilibrium since all the forces being applied on it, cancel out

now, if one of the force was slightly more than the other one in the same case. The object will start to move and NOT be in equilibrium

BUT

if an object is in a vacuum and moving on a frictionless surface, the object will attain equilibrium after some seconds since it will be moving with constant speed and all the forces acting on it will be equal

Hence, if the object is accelerating. we can say with surety that the object is not is equilibrium since from the second law of motion,

F = ma ; when a is a non-zero value, there is definitely some net force being applied on the object

Looking at the given case

in the question, we are given that the object is 'accelerating'  upwards

we proved above that if an object is accelerating, there is some net force on that object and hence the object is NOT in equilibrium

Since the object is accelerating, from the second law of motion:

F = ma; m cannot be zero and if a is a non-zero value as given in the question, there is definitely some net force on the object

Since there is some force being applied on the object, the object is NOT in equilibrium

Conclusion

Since we found that the object is NOT in equilibrium and that there is some net force on the object,

The first option is correct

Answer:Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.

Explanation:

Answer: its C

Explanation:

Answer:

 N = 136.77 N

Explanation:

This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.

Let's use trigonometry to decompose the force F1

           cos θ = F₁ₓ / F₁

           sin θ = F_{1y} / F₁

           F₁ₓ = F₁ cos θ

           F_{1y} = F₁ sin θ

now let's apply Newton's second law to each axis

X axis

        F₁ₓ - F4 = m a

Y axis

        N + F3 + F_{1y} -F₂ -W = 0

the acceleration can be calculated with kinematics

        v = v₀ + a t

since the object starts from rest, the initial velocity is zero v₀ = 0

        a = v / t

        a = 20/3

        a = 6.667 m / s²

we substitute in the equation

        F₁ₓ = F₄ + m a

        F₁ₓ = 42 + 15  6,667

        F₁ₓ = 142 N

        F₁ cos θ = 142

        cos θ = 142/206 = 0.6893

        θ = cos⁻¹ 0.6893

        θ = 46.42º

now let's work the y axis

        N = W + F₂ - F₃ - F_{1y}

        N  = 15 9.8 + 144 -5 - 206 sin 46.42

        N = 286 - 149.23

        N = 136.77 N

The velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south would be  10.35 meters / seconds.

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

Note that these equations are only valid for a uniform acceleration.

As given in the problem we have to find the velocity of the car we have to find the velocity of the car  after 6.9 s if its acceleration is 1.5 m/s² due south,

The acceleration of the car = 1.5 m/s²

The time taken by the car = 6.9 seconds

By using the first equation of the motion,

v = u + at

v = 0 + 1.5*6.9

v = 10.35 meters / seconds

Thus, the velocity of the car after 6.9 s, if its acceleration is 1.5 m/s² due south, would be  10.35 meters / seconds.

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Answer:

As the earth rotates, it tries to drag/bring the tidal bulges with it. When a large amount of friction is applied, the earth spin will gradually and slow down but not all the way down.

Answer:

If the Earth spun slower the rate of tides will be higher because the moon will start revolving faster tan the Earth, creating more tides as the moon will revolve more around the Earth in a month.

If my answer helped, kindly mark me as the brainliest !!

Thank You!!

Answer: sorry here’s the answers, I didn’t feel like typing it all

Explanation:

The correct answer to the 5 questions are;

1) Option A; The current stays the same throughout the circuit.

2) Option A;  The current will stay the same at 1 amp after passing through both resistors.

3) Option C; R1: 12 V, R2: 12 V

4) Option A; I1 is equal to I2

5) Option C; The sum of the current flowing in is equal to the sum of current flowing out.

1) In an electrical circuit, usually as current moves through each component, it stays the same. Thus, option A is correct.

2) Formula for current is;

I = V/R

We are told that there are two resistors in series each having a resistance of 5Ω. Thus; Total resistance = 5 + 5 = 10 Ω.

Thus; Current = 10/10 = 1 A.

The current will stay same at 1 A after passing through both resistors.

3) From the circuit we are given, we see that the Voltage is 12 V. Now, the same voltage would be transmitted through both resistor R1 and R2.

Option C is correct

4) The current splits upon passing resistor 1 and as such it means the current I2 going through the second resistor would be the same. Thus; I1 = I2.

5) Kirchoff's junction rule states that all the incoming currents to a particular junction must be equal to sum of all currents going out of that same junction. Thus, option C is correct.

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Answer:

The thermal conductivity  [tex]k  = 1.4094   W/ m\cdot K[/tex]

Explanation:

From the question we are told that

  The  depth of the thermocouple from the surface is  x =  10 mm  = 0.01 m

   The  temperature is  [tex]T_f  =  100 ^o C[/tex]

   The  initial temperature is   [tex]T_i  =  30 ^o C[/tex]

    The  temperature of the thermocouple after t =  2 minutes( 2 * 60 =  120 \ seconds)  is   [tex]T_t  =  65 ^o C[/tex]

    The  density of the material  is  [tex]\rho =  2200 kg/m^3[/tex]

     The specific heat of the solid [tex]c_s  =  700 J/kg \cdot K[/tex]

Generally  the  equation for  semi -infinite medium  is mathematically as  

    [tex]\frac{T_s - T }{T_i - T} =  erf [\frac{x}{2 \sqrt{\alpha  * t} } ][/tex]

     [tex]\frac{65 - 100 }{30 - 100} =  erf [\frac{x}{2 \sqrt{\alpha  * t} } ][/tex]

        [tex]0.5 =  erf [\frac{0.01}{2 \sqrt{\alpha  * 120} } ][/tex]

Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]

  [tex]\frac{0.01}{ 2 (\sqrt{\alpha  *  120 } )}[/tex]   this is from the Gaussian function table  

    [tex]0.0 1 =  0.954 * (\sqrt{\alpha * 120  } )[/tex]

=>   [tex]\sqrt{\alpha  * 120  } =  \frac{0.01 }{0.954 }[/tex]

=>   [tex]\alpha =  9.1525 *10^{-7} \  m^2 /s[/tex]

Generally the thermal  conductivity is mathematically represented as

     [tex]k  =  \alpha  *  \rho * c_s[/tex]

      [tex]k  = 9.1525 *10^{-7}   *  2200 * 700[/tex]

     [tex]k  = 1.4094   W/ m\cdot K[/tex]

Answer:

a) Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b) Amount of heat transfer is 5320 kJ  

c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Explanation:

Given that;

Diameter D = 30 cm

Height H = 4m

heat transfer coeff h = 14 W/m².°C

thermal conductivity k = 0.79 W/m.°C

thermal diffusivity α  = 5.94 × 10⁻⁷ m²/s

Density p = 1600 kh/m³

specific heat Cp = 0.84 Kj/kg.°C

a)

the Biot number is

Bi = hr₀ / k

we substitute

Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C

Bi = 2.658

From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,  

λ₁ = 1.7240

A₁ = 1.3915

Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁

the Fourier number is determined to be  

[ T(r₀, t) -T∞ ] / [ Ti - T∞]  = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)

(27 - 28) / (14 - 28)   = (1.3915)e^-(17240)²t (0.3841)  

t' = 0.6771

Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes  

t =  t'r₀² / ₐ

= (0.6771 × 0.15 m)² /  (5.94 x 10⁻⁷ m²/s)

= 23,650 s

= 7.1 hours

Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b)

The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.  

Maximum heat transfer between the ambient air and the column is

m = pV

= pπr₀²L

= (1600 kg/m³ × π × (0.15 m)² × (4 m)

= 452.389 kg

Qin = mCp [T∞ - Ti ]

= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C

= 5320 kJ  

Amount of heat transfer is 5320 kJ  

(c)

the amount of heat transfer until the surface temperature reaches to 27°C is

(T(0,t) - T∞) / Ti - T∞  = A₁e^(-λ₁²t')

= (1.3915)e^-(1.7240)² (0.6771)

= 0.1860

Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes  

(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)

= 1 - 2 × 0.1860 × (0.5787  / 1.7240)  

= 0.875

Q = 0.875Qmax

Q = 0.875(5320 kJ)  

Q = 4660 kJ

Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Answer:

At least [tex]2\; \rm m[/tex].

Explanation:

The torque [tex]\tau[/tex] that a force exerts on a lever is equal the product of the following:

[tex]\tau = F\cdot r \cdot \sin\theta[/tex].

The force in this question is (at most) [tex]50\; \rm N[/tex]. That is: [tex]F = 50\; \rm N[/tex].

[tex]\sin \theta[/tex] is maximized when [tex]\theta = 90^\circ[/tex]. In other words, the force on the door gives the largest-possible torque when that force is applied perpendicular to the door. When [tex]\theta = 90^\circ\![/tex], [tex]\sin \theta =1[/tex].

If the force here is applied at a distance of [tex]r[/tex] meters away from the hinge (the fulcrum of this door,) the torque generated would be:

[tex]\begin{aligned}\tau &= F \cdot r \cdot \sin \theta \\ &= (50\, r)\; \rm N \cdot m\end{aligned}[/tex].

That torque is supposed to be at least [tex]100\; \rm N\cdot m[/tex]. That is:

[tex]50\, r \ge 100[/tex].

[tex]r \ge 2[/tex].

In other words, the force needs to be applied at a point a minimum distance of [tex]2\; \rm m[/tex] away from the hinge of this door.

Given parameters:

Mass of puck = 162g  = 0.162kg (1000g = 1kg)

Force exerted on puck = 171N

Final velocity  = 42.3m/s

Unknown

A.  time of the acceleration

B. radius of the curve?

Solution:

A. time of the acceleration

the initial velocity of the puck = 0m/s

    We know that;

              Force  = mass x acceleration

      Acceleration  = [tex]\frac{Final velocity - Initial velocity}{time taken}[/tex]

       Acceleration  = [tex]\frac{42.3 - 0}{t}[/tex]

  So force  = mass x [tex]\frac{42.3 }{t}[/tex]

 Input the parameters and solve for time;

             171 = 0.162 x [tex]\frac{42.3 }{t}[/tex]

             171 = [tex]\frac{6.85}{t}[/tex]  

                 t = [tex]\frac{6.85}{171}[/tex]   = 0.04s

The time of acceleration is 0.04s

B. radius of the curve;

      to solve this, we apply the centripetal force formula;

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

  where;

      F is the centripetal force

     m is the mass

      v is the velocity

      r is the radius

               Since the force exerted on the puck is 151;

      input the parameters and solve for r²;

             151 = [tex]\frac{0.162 x 42.3^{2} }{r}[/tex]

               151r  = 0.162 x 42.3²

                r  = 1.92m

The radius of the circular curve is 1.92m

Answer:

Making an observation

Explanation:

The use of senses as a tool to gather information is described as making an observation.

While making an observation, the senses must be at alert.

Therefore, the act of using senses or tools to gather information is called making an observation.

Answer:

A soil

Explanation:

Sand is a granular material composed of finely divided rock and mineral particles. It is defined by size, being finer than gravel and coarser than silt. Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.

Answer:

mixture

sand is a mixture

Answer:

Explanation:

To find the momentum of an object given it's mass and velocity we use the formula

From the question

mass = 20 kg

velocity = 5 m/s

We have

momentum = 20 × 5

We have the final answer as

Hope this helps you

Answer:

transition metal, and inner transition metals groups are numbered 1-18 from left to right

Answer:

The tension in the web is 0.017738 N

Explanation:

Net Force

The net force exerted on an object is the sum of the vectors of each individual force applied to an object.

If the net force equals 0, then the object is at rest or moving at a constant speed.

The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.

A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:

T=W=m.g

The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:

[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]

[tex]T=0.017738\ N[/tex]

The tension in the web is 0.017738 N

The heat transfer takes place from the Ice to lemonade (ice → lemonade).

The term “heat transfer” refers to the movement of heat. The flow of heat across a system's boundary is due to a temperature differential between the system and its surroundings.

When a temperature difference exists between states of matter, heat transfer happens solely in the direction of decreasing temperature, that is, from a hot object to a cold item.

The temperature of ice is increasing while the lemonade is decreasing. Heat transfer happens solely in the direction of decreasing temperature,

Hence, the heat transfer takes place from the Ice to lemonade (ice → lemonade.

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Answer: C

Explanation:

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:

[tex]E = U_{g} + K[/tex] (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the projectile, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.

[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.

If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:

[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 498556.296\,J[/tex]

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)

Where:

[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.

[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.

[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]

[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)

Where:

[tex]m[/tex] - Mass of the projectile, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.

[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.

If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:

[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]

[tex]W_{loss} = 125960.4\,J[/tex]

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)

[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]

Where:

[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.

[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.

[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.

[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)

[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]

[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]

[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]

If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:

[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]

[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Answer:

volume is 0.1 L

Explanation:

you can use the equation density=mass/volume

100 = 1000 / v

divide by 1000 on both sides

0.1 = v

Answer:

it is whatever the temperature is at 5(I cant seem to see it clearly)

Explanation:

Answer:

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Answer:

40 m/s^2

Explanation:

Mass= 0.40 kg

Force= 16 N

Therefore the acceleration can be calculated as follows

F = ma

16= 0.40 × a

16= 0.40 a

a= 16/0.40

a= 40 m/s^2

Hence the acceleration is 40 m/s^2

Answer:

is a guideline to help an individual write and achieve well-specified goals.

Explanation:

An action plan  is a guideline to help an individual write and achieve well-specified goals.

Answer:

yes, the guideline is to help an individual write and achieve well-specified goals.

Explanation: