A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how far would the cannonball travel before it lands on the ground? (Show all work)

Answer:

10,200 Cal. per day

Explanation:

The mouse consumes 3.0 Cal each day, and has a mass of 20 grams. We can use this data to obtain a ratio of energy consumption per mass

[tex]\frac{3.0 \ Cal}{20 g} = 0.15 \frac{Cal}{g}[/tex].

For the human, we need to convert the 68 kilograms to grams. We can do this with a conversion factor. We know that:

[tex]1 \ kg = 1000 \ g[/tex],

Now, we can divide by 1 kg on each side

[tex]\frac{1 \ kg}{1 \ kg} = \frac{1000 \ g}{1 \ kg}[/tex],

[tex] 1 = \frac{1000 \ g}{1 \ kg}[/tex].

Using this conversion factor, we can obtain the mass of the human in grams, instead of kilograms. First, lets take:

[tex]mass_{human} = 68 \ kg[/tex]

We can multiply this mass for the conversion factor, we are allowed to do this, cause the conversion factor equals 1, and its adimensional

[tex]mass_{human} = 68 \ kg * \frac{1000 \ g}{1 \ kg} [/tex]

[tex]mass_{human} = 68,000 g [/tex]

Now that we know the mass of the human on grams, we can multiply for our ratio of energy consumption

[tex]68,000 \ g * 0.15 \frac{Cal}{g} = 10,200 \ Cal[/tex]

So, we would need 10,200  Cal per day.

If a 68 kg human used the same energy per kg of body mass as a mouse, they would need approximately 10,200 Calories each day. This high energy requirement is due to the greater relative metabolic rates of smaller mammals, as they experience higher heat loss and require more energy to maintain body temperature.

The question asks for the approximate energy a 68 kg human would need each day if they used the same energy per kg of body mass as a mouse. To find this, we first calculate the energy per gram of the mouse, which is 3.0 Calories/20g = 0.15 Cal/g.

Converting this to per kg, we get 0.15 Cal/g * 1000g/kg = 150 Cal/kg. Then, multiplying this by the human's mass gives us the daily energy requirement for the human: 150 Cal/kg * 68 kg = 10,200 Calories a day, a significantly higher amount than the Basal Metabolic Rate (BMR) of a typical human.

This high energy requirement can be attributed to the higher metabolic rate of smaller mammals. Smaller animals have a greater surface area to mass ratio, which contributes to higher heat loss and thus higher energy demands to maintain body temperature. This results in a higher BMR per body weight in smaller mammals, as demonstrated by the mouse in the question.

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Answer:

a) It takes 6,37 s b) The Velocity is -59,43 m/s

Explanation:

The initial variables of the balloon are:

Xo = 0 m

Vo = 3 m/s

After one minute the situation is the following:

t= 60 s

X1 = Xo + Vo*t

X1= 0 m + 3 m/s * 60s

X1= 180m

So when the bag falls, its initial variables are the following:

Xo = 180m

X1 = 0m

Vo = 3 m/s

V1= ?

a= -9,8 m/s2

The ecuation of movement for this situation is:

X = Vo*t + 1/2 a*[tex]t^{2}[/tex]

So:

-180m = 3m/s*t+ 1/2*-9,8 m/s2 * [tex]t^{2}[/tex]

To solve this we have

a=-9,8/2

b=3

c=180

The formula is:[tex](-b +/- \sqrt{b^{2} -4ac}) /2a[/tex]

Replacing, we get to 2 solutions, where only the positive one is valid because we are talking about time.

So the answer a) is t= 6,37 s

With that answer we can find the question b), with the following movement formula.

Vf = Vo + at

Vf = +3 m/s + (-) 9,8 m/s2 *6,37s

b) Vf = -59,43 m/s

(a) The time taken for the sandbag to reach the ground is 6.37 s.

(b) The velocity of the sandbag when it hits the ground is -59.42 m/s.

(a) The initial velocity of the balloon is, [tex]u = 3\,m/s[/tex].

Given that the motion of the balloon is constant, i.e.; [tex]a = 0\,m/s[/tex]

So, the height of the balloon after [tex]t = 1\, min = 60\,s[/tex] can be calculated using the second kinematics equation given by;

[tex]s = ut +\frac{1}{2} at^2[/tex]

Substituting the known values, we get;

[tex]s = 3\,m/s \times 60\,s = 180\,m[/tex]

Now, a sandbag is dropped.

The initial velocity of the sandbag will be 3m/s in the upward direction.

i.e.; [tex]u_s = 3\,m/s[/tex]

The acceleration of the bag is given by the gravitational force of the earth.

[tex]a_s = -g = - 9.8\,m/s^2[/tex]

The second kinematics equation is given by;

[tex]s = ut +\frac{1}{2} at^2[/tex]

Substituting the known values, we get;

[tex]-180= 3t -(\frac{1}{2} \times 9.8\times t^2)\\\\\implies 4.9t^2 -3t -180=0\\\\\implies t = \frac{3\pm\sqrt{9 - (4\times 4.9\times -180)}}{9.8} = 6.37\,s[/tex]

(b) The final velocity of the sandbag can be found using the equation;

[tex]v=u+at[/tex]

Substituting the known values, we get;

[tex]v_s = 3m/s -(9.8m/s^2 \times 6.37\,s)=-59.42\,m/s[/tex]

Learn more about kinematics equations here:

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Answer:

I'll answer below

Explanation:

22.- HCl + Na₂S

23.- H₂S + NaCl

24.- 2HCl + Na₂S ⇒ H₂S + 2NaCl

26.- alfa They don't penetrate very deeply into the skin, in fact, clothing can stop alpha particles.

       beta  penetrate clothing and skin.

       gamma can travel through most forms of matter because they have no mass. It takes several inches of lead or several feet of concrete.

27.- Fission and fusion are two processes that release too much amounts of energy.

28.- Well, I could be because in a formula we write first the metal and later the non metal.

29.- The elements that are most likely to have similar properties are Y and Z, that's because they have the same valence electron. All the elements with they same valence electron have similar properties.

Answer:

22 - Reactant 1: HCl

       Reactant 2: Na₂S

23 - Product 1: H₂S

       Product 2: NaCl

24 - 2HCl + Na₂S ⇒ H₂S + 2NaCl

26 - alpha (Lowest)

       beta

       gamma (Highest)

27 - Statement "c" is False

The true statement is:

Fission and fusion are two processes that release high amounts of energy.

28 - From its formula we know that each Mg atom is bounded with 2 Br atoms. From periodic table we can see that Br has 7 electrons in its outer shell and require 1 electron to complete its outer shell. MgBr₂ each Br has taken one electron from Mg to complete its outer shell which means that Mg has given away 2 electrons forming ion with positive 2 (+2) charge. Metals are the ones that form positive ions by giving away electrons therefor we know that Mg is a metal.

29 - Y and Z are most likely to have similar properties

Explanation:

22 - Hydrogen has 1 electrons in its outer shell and Chlorine requires 1 electrons to complete its outer shell because it lies in 7th group of the periodic table. Hydrogen gives away 1 electron while Chlorine excepts 1 electron. Therefore the formula is HCl (one hydrogen atom bonded with one Cl atom)

Similarly Na lies in the first group of the periodic table so it has 1 electron in its outer shell. It gives away that electron but Sulfide requires 2 electrons so 2 Na atoms form bond with Sulfide.  

23 - same logic as mentioned above in the explanation of 22 can be applied.

24 - Balancing of the equation means equal number of atoms for each element must be present on reactant and product side of equation.

26 - Three types of radiations are alpha (Lowest energy) beta and gamma (Highest energy)

27 - Statement "c" is False

The true statement is:

Fission and fusion are two processes that release high amounts of energy.

28 - From its formula we know that each Mg atom is bounded with 2 Br atoms. From periodic table we can see that Br has 7 electrons in its outer shell and require 1 electron to complete its outer shell. MgBr₂ each Br has taken one electron from Mg to complete its outer shell which means that Mg has given away 2 electrons forming ion with positive 2 (+2) charge. Metals are the ones that form positive ions by giving away electrons therefor we know that Mg is a metal.

29 - Y and Z are most likely to have similar properties because same number of electrons in their outer shell.

The text looks incomplete. Here the complete question found on google:

"The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?"

A) 6.88 m/s

The velocity of the bus can be found by integrating the acceleration. Therefore:

[tex]v(t) = \int a(t) dt = \int (\alpha t)dt=\frac{1}{2}\alpha t^2+C[/tex] (1)

where

[tex]\alpha = 1.28 m/s^3[/tex]

C is a constant term

We know that at [tex]t_1 = 1.13 s[/tex], the velocity is [tex]v=5.09 m/s[/tex]. Substituting these values into (1), we can find the exact value of C:

[tex]C=v(t) - \frac{1}{2}\alpha t^2 = 5.09 - \frac{1}{2}(1.28)(1.13)^2=4.27 m/s[/tex]

So now we can find the velocity at time [tex]t_2 = 2.02 s[/tex]:

[tex]v(2.02)=\frac{1}{2}(1.28)(2.02)^2+4.27=6.88 m/s[/tex]

B) 11.2 m

To find the position, we need to integrate the velocity:

[tex]x(t) = \int v(t) dt = \int (\frac{1}{2}\alpha t^2 + C) dt = \frac{1}{3}\alpha t^3 + Ct +D[/tex] (2)

where D is another constant term.

We know that at [tex]t_1 = 1.13 s[/tex], the position is [tex]v=5.92 m[/tex]. Substituting these values into (2), we can find the exact value of D:

[tex]D = x(t) - \frac{1}{6}\alpha t^3 -Ct = 5.92-\frac{1}{6}(1.28)(1.13)^3 - (4.27)(1.13)=0.79 m[/tex]

And so now we can find the position at time [tex]t_2 = 2.02 s[/tex] using eq.(2):

[tex]x(2.02)=\frac{1}{6}(1.28)(2.02)^3 + (4.27)(2.02) +0.79=11.2 m[/tex]

The question deals with calculating the time-dependent acceleration of a bus using the equation ax(t)=αt, applicable to Physics (specifically mechanics), and is suitable for a high school level.

The question pertains to the acceleration of an object, specifically a bus, which makes this a Physics problem. Acceleration is defined as the change in velocity over time and is a vector quantity having both magnitude and direction. Given the equation ax(t)=αt, where α is a constant equal to 1.28 m/s3, we can recognize that the acceleration is linearly increasing over time because for each second, the acceleration increases by 1.28 m/s2. This kind of problem typically appears in high school Physics.

For instance, to calculate the acceleration of a bus at a specific time point, we simply multiply the constant α by the time t at which we want to know the acceleration. So if we wanted to know the acceleration at t=15 s, we would calculate it as ax(15s) = 1.28 m/s3 × 15 s = 19.2 m/s2. This demonstrates that as time progresses, the acceleration value increases.

The pilot first flies 1.5 hours at 675 mi/h resulting in a distance of 1012.5 miles. Then 'turns' right and flies 2 hours at the same speed for a distance of 1350 miles. Using the law of cosines with these as vectors, the pilot is approximately 1340 miles from the start.

This problem involves trigonometry and the concept of vectors to solve. During the initial 1.5 hours of flight, the pilot would cover a distance of 1.5 hours x 675 mi/h = 1012.5 miles. When she makes a course correction and heads 10° to her right, she then flies another distance of 2 hours x 675 mi/h = 1350 miles.

We can use the law of cosines, which in this case is c² = a² + b² - 2abcos(θ), where θ is the angle between vectors a and b. With a=1012.5 miles, b=1350 miles, and θ=10°, the calculation results in c = 1339.73 miles. Therefore, the distance from her initial position is approximately 1340 miles.

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Answer:

x = +7, x = -7

Explanation:

The equation to solve is:

[tex]\left|\frac{x}{7}\right|=1[/tex]

Which means that the absolute value of the fraction [tex]\frac{x}{7}[/tex] must be equal to 1. Since we have an absolute value, we have basically two equations to solve:

1) [tex]\frac{x}{7}=+1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = 1\cdot 7 \rightarrow x = +7[/tex]

2) [tex]\frac{x}{7}=-1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = -1\cdot 7 \rightarrow x = -7[/tex]

So the two solutions are +7 and -7.

Explanation:

1. Rotational speed : It is defined as the number of revolutions per unit time i.e. revolution per minute.

2. Tangential speed : Instantaneous linear speed of an object in circular motion.

3. Centripetal force : This force acts on object when it moves in a circle. It is a force that produces uniform circular motion.

4. Centripetal acceleration : center-seeking change in direction of velocity

5. Newton's Second Law : This law defines the magnitude of force acting on the object. It is given by, F = m × a

6. Newton's First Law : An object in motion stays in motion unless an external force acts on it.  

Answer:

1. Rotational speed :.............................  revolutions per minute

2. Tangential speed : Instantaneous linear speed of an object in circular motion.

3. Centripetal force :  ......................force that produces uniform circular motion

4. Centripetal acceleration : center-seeking change in direction of velocity

5. Newton's Second Law :  F = ma

6. Newton's First Law : An object in motion stays in motion unless an external force acts on it.  

Explanation:

1. Rotational speed :.............................................  revolutions per minute

2. Tangential speed : Instantaneous linear speed of an object in circular motion.

3. Centripetal force :  ......................force that produces uniform circular motion

4. Centripetal acceleration : center-seeking change in direction of velocity

5. Newton's Second Law : This law defines the magnitude of force acting on the object. It is given by, F = m × a

6. Newton's First Law : An object in motion stays in motion unless an external force acts on it.  

Answer:

Arrival times of P and S waves

Explanation:

Seismological recording station has a seismometer  that senses that motion in the ground, a clock that records time and a data recorded.

The distance between beginning of the first P wave and the first S wave will give you the time the waves are apart.This time value will be used to find the distance between the seismograph and the epicenter of earthquake and you mark it.This is corresponding distance in km to the time in seconds obtained  before.You then find the amplitude of the strongest wave and mark it on the right side of chart.Amplitude is the height on paper of the strongest wave.Using a ruler join the amplitude point and the point where you marked the distance to epicenter.This line will cross the magnitude chart at a point which represents the magnitude of the Earthquake.

Answer:

350x

Explanation:

In a microscope the objective has higher magnification than the eyepiece so, this is a microscope

The magnification of a microscope is given by the product of the magnifications of the eyepiece and and the objective.

Objective lens magnification = 35x =[tex]m_o[/tex]

Eyepiece magnification = 10x =[tex]m_e[/tex]

Total magnification

[tex]M=m_o\times m_e\\\Rightarrow M=35\times 10\\\Rightarrow M=350[/tex]

Total magnification is 350x

The total magnification of the microscope with a 35X objective lens and a 10X ocular lens is 350X.

The subject of the question is the calculation of total magnification in a compound light microscope. The total magnification of a microscope can be determined by multiplying the magnification of the objective lens by the magnification of the ocular (or eyepiece) lens. For the specified microscope with a 35X objective lens and a 10X ocular lens, the total magnification is simply the product of these two numbers.

To calculate the total magnification, use the following formula:

Total magnification = (Objective lens magnification) x (Ocular lens magnification)

Total magnification = (35X) x (10X)

Total magnification = 350X

Therefore, when using a 35X objective lens and a 10X ocular lens, the total magnification of the system is 350X.

Answer:

a) 15.68 m/s

b) 130.34 m

Explanation:

Let's consider the origin of the coordinates at the ground, therefore, the equation of motion of the rock and its velocity are

x(t)= h + vt - (1/2)gt^2

v(t) = v - gt

Where h is the height of the building and v is the initial velocity.

The maximum height is reached when v=0, that is v(1.6s) = 0, and we know that x(7s) = 0

Therefore

0 = v(1.6s) = v - (9.8 m/s^2)(1.6s)

v = 15.68 m/s

and

0 = x(7s) = h + (15.68 m/s )(7 s) - (1/2)(9.8 m/s^2)(49s^2)

h = (1/2)(9.8 m/s^2)(49s^2) -  (15.68 m/s )(7 s)

h = 130.34 m

Final answer:

Using kinematic equations, the initial upward velocity is found to be 15.7 m/s, the maximum height attained above the building is 12.6 m, and the height of the building is calculated to be 172.9 m.

Explanation:

To solve the problem of a rock thrown vertically upward, we can use the physics of kinematic equations for uniformly accelerated motion. We'll assume the acceleration due to gravity (g) is -9.81 m/s², acting downward.

Part (a): Upward Velocity

At the maximum height, the rock's velocity is 0 m/s. The time it takes to reach this point is half of the total time in the air before coming back to the starting height. Therefore, using the equation v = u + g*t (where v is the final velocity, u is the initial velocity, g is acceleration due to gravity, and t is time), we get u = -g*t. Plugging in the values, we find the initial upward velocity (u) to be 15.7 m/s.

Part (b): Maximum Height Above Building

To find the maximum height, use the equation h = ut + (1/2)*g*t². Substituting the values, we get a maximum height of 12.6 m above the building.

Part (c): Height of the Building

The total time the rock is in the air is 7.00 s. We can calculate the total height from which the rock was thrown (the height of the building plus the maximum height reached) using the equation s = ut + (1/2)*g*t² where s stands for displacement. Finally, we find the height of the building to be 172.9 m.

Answer: If T is the time you wait to trim the lawn since you last did it and B is the lenght at which you trim, then the speed will be:

[tex]v=\frac{B}{T}[/tex]

Explanation:

Let's say T is the time you wait to trim the lawn since you last did it (1 month for example) and B is the lenght at which you trim (10cm for example).

If v is the speed at which the lawn grows, then the lenght of the lawn as a function of time will be:

[tex]b(t)=b_{0}+ vt[/tex]

where [tex]b_{0}[/tex] is an arbitrary initial lenght. Let's say [tex]b_{0}=0[/tex].

[tex]b(t)=vt[/tex]

Then we have:

[tex]v=\frac{b(t)}{t}[/tex]

So, at time T the lawn has grown a lenght B. And the speed is:

[tex]v=\frac{B}{T}[/tex]

Answer:

A.[tex]a=203.14\ \frac{m}{s^2}[/tex]

B.s=397.6 m

Explanation:

Given that

speed  u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

0 = 284.4 -a x 1.4     (here we take negative sign because this is the case of de acceleration)

[tex]a=203.14\ \frac{m}{s^2}[/tex]

So the acceleration  while stopping will be [tex]a=203.14\ \frac{m}{s^2}[/tex].

Lets take distance travel before come top rest is s

We know that

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

[tex]0=284.4^2-2\times 203.14\times s[/tex]

s=397.6 m

So the distance travel while stopping is 397.6 m.

Answer:

[tex]a) acceleration = -203.14 m/s^2\\\\b) distance = 199.1 m[/tex]

Explanation:

Given

initial speed u = 284.4 m/s

final speed v = 0 m/s

duration t = 1.4 s

Solution

a)

Acceleration

[tex]a = \frac{v - u}{t} \\\\a = \frac{0-284.4}{1.4} \\\\a = -203.14 m/s^2[/tex]

b)

Distance

[tex]v^2  - u^2 = 2as\\\\0^2 - 284.4^2 = 2 \times  (-203.14) \times S\\\\S = 199.1 m[/tex]

Answer:

[tex]-0.44 m/s^2[/tex]

Explanation:

First of all, we need to calculate the distance covered by the locomotive during the reaction time, which is

t = 0.46 s

During this time, the locomotive travels at

v = 13 m/s

And the motion is uniform, so the distance covered is

[tex]d_1 = vt = (13)(0.46)=6.0 m[/tex]

The locomotive was initially 200 m from the crossing, so the distance left to stop is now

[tex]d=200 - 6.0 = 194.0 m[/tex]

And now the locomotive has to slow down to a final velocity of [tex]v=0[/tex] in this distance. We can find the minimum deceleration needed by using the suvat equation:

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

where

v = 0 is the final velocity

u = 13 m/s is the initial velocity

a is the deceleration

d = 194.0 m is the distance to stop

Solving for a,

[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-13^2}{2(194)}=-0.44 m/s^2[/tex]

Answer:

The answer to your question is:

a) t = 3.64 s

b) vox = 31.59 m/s

c) vy = 35.71 m/s

d) v = 47.67 m/s

Explanation:

a) To calculate the time, we know that voy = 0 m/s so we this this formula

   h = voy + 1/2(gt²)

   65  = 0 + 1/2(9.81)(t²)

   65 = 4.905t²

   t² = 65/4.905 = 13.25

   t = 3.64 s

b) To calculate vox  we use this formula vox = d/t ; vox is constant

   vox = 115/3.64 = 31.59 m/s

c) To calculate voy we use the formula    vy = voy + gt

but voy = 0

               vy = gt = 9.81 x 3.64 = 35.71 m/s

d) To calculate v  we use the pythagorean theorem

            c2 = a2 + b2

            c2 = 31.59² + 35.71² = 997.92 + 1275.20

            c2 = 2273.12

              c = 47.67 m/s

Answer:

The plane needs 1,56 seconds to clear the intersection.

Explanation:

This is a case of uniformly accelerated rectilinear motion.

[tex]V_0^{2} = V_f^{2} - 2ad[/tex]

[tex]V_0=\sqrt{V_0^{2} } = ?[/tex]

Vf=50 m/s

[tex]V_f^{2}  = (50 m/s)^{2} = 2500  m^{2}/s^{2}[/tex]

a = -5.4 [tex]m/s^{2}[/tex] (Negative because is decelerating)

d = displacement needed to clear the intersection. It should be the width of the intersection plus the lenght of the plane.

d= 59,7m + 25 m = 84.7 m

Calculating [tex]V_0[/tex]:

[tex]V_0^{2} = V_f^{2} - 2ad[/tex]

[tex]V_0^{2}= 2500 \frac{m^{2} }{s^{2} } - 2(-5.4\frac{m}{s^{2} })(84.7 m)[/tex]

[tex]V_0^{2}= 3,414.76 \frac{m^{2} }{s^{2} }[/tex]

[tex]V_0= \sqrt{3,414.76} = 58.44 \frac{m}{s}[/tex]

Otherwise:

[tex]t = \frac{V_f-V_0}{a} =\frac{50\frac{m}{s} - 58.44\frac{m}{s}  }{-5.4 \frac{m}{s^{2} } } = 1.56 s[/tex]

The time needed for the plane to clear the intersection is approximately 1.88 s.

Given data:

- Length of the plane (L) = 59.7 m

- Width of the intersection (d) = 25.0 m

- Deceleration (a) = 5.4 m/s²

- Final speed (v) = 50 m/s

The plane decelerates through the intersection. The final speed v and the deceleration a are given. The total distance the plane covers while decelerating through the intersection is the sum of the length of the plane and the width of the intersection:

[tex]\[ \text{Total distance} = L + d = 59.7 \, \text{m} + 25.0 \, \text{m} = 84.7 \, \text{m} \][/tex]

Determine the initial speed [tex](v_0)[/tex]

Using the kinematic equation:

[tex]\[ v^2 = v_0^2 + 2a \Delta x \][/tex]

We rearrange to solve for [tex]\( v_0 \)[/tex]:

[tex]\[ v_0^2 = v^2 - 2a \Delta x \][/tex]

[tex]\[ v_0^2 = (50 \, \text{m/s})^2 - 2 \times 5.4 \, \text{m/s}^2 \times 84.7 \, \text{m} \]\[ v_0^2 = 2500 - 2 \times 5.4 \times 84.7 \]\[ v_0^2 = 2500 - 913.56 \]\[ v_0^2 = 1586.44 \]\[ v_0 = \sqrt{1586.44} \]\[ v_0 \approx 39.83 \, \text{m/s} \][/tex]

Calculate the time (t)

Using the kinematic equation:

[tex]\[ v = v_0 + at \][/tex]

Rearrange to solve for t:

[tex]\[ t = \frac{v - v_0}{a} \]\[ t = \frac{50 \, \text{m/s} - 39.83 \, \text{m/s}}{5.4 \, \text{m/s}^2} \]\[ t = \frac{10.17 \, \text{m/s}}{5.4 \, \text{m/s}^2} \]\[ t \approx 1.88 \, \text{s} \][/tex]

Answer:

The magnitude of force on P is 2.75 N

The magnitude of force on Q is 7.15 N

Solution:

As per the question:

Mass of the object, M = 11.0 kg

Acceleration of the object when the forces are directed leftwards, a = [tex]0.900 m/s^{2}[/tex]

Acceleration when the forces are in opposite direction, a' = [tex]0.400 m/s^{2}[/tex]

Now,

The net force on the object in first case is given by:

[tex]F_{net} = |\vec{F_{P}}| + |\vec{F_{Q}}| = Ma[/tex]       (1)

The net force on the object in second case is given by:

[tex]F_{net} = |\vec{F_{P}}| - |\vec{F_{Q}}| = Ma'[/tex]       (2)

Adding both eqn (1) and (2):

[tex]2|\vec{F_{Q}}| = M(a + a')[/tex]

[tex]|\vec{F_{Q}}| = \frac{11.0(0.900 + 0.400)}{2} = 7.15 N[/tex]

Putting the above value in eqn (1):

[tex]|\vec{F_{P}}| = 11\times 0.900 - |\vec{F_{Q}}|[/tex]

[tex]|\vec{F_{P}}| = 9.900 - 7.15 = 2.75[/tex]

Explanation:

Given

launch angle[tex]=32^{\circ}[/tex]

ball lands exactly 224 m

Range of projectile =224 m

[tex]Range =\frac{u^2sin2\theta }{g}[/tex]

[tex]224=\frac{u^2sin64}{9.8}[/tex]

[tex]u^2sin64=224\times 9.8=2195.2[/tex]

[tex]u^2=2442.383[/tex]

u=49.42 m/s

(b)Maximum height reached

[tex]H_{max}=\frac{u^2sin^2\theta }{2g}[/tex]

[tex]H_{max}=\frac{49.42^2\times sin^{2}32}{2\cdot 9.8}[/tex]

[tex]H_{max}=34.99 m\approx 35 m[/tex]

it takes approximately [tex]\(1.3863 \times 10^{-4}\)[/tex] seconds for the copper rod to cool to an average temperature of 25°C when exposed to an air stream at 20°C with a heat transfer coefficient of [tex]\(20,000 \, \text{W/m}^2 \cdot \text{K}\)[/tex].

The rate at which the copper rod loses heat can be described by Newton's Law of Cooling, which is given by the formula:

[tex]\[ Q(t) = Q_0 e^{-ht} \][/tex]

where:

[tex]\( Q(t) \)[/tex] is the heat at time[tex]\( t \)\\[/tex],

[tex]\( Q_0 \)[/tex]is the initial heat content,

[tex]\( h \)\\[/tex] is the heat transfer coefficient,

[tex]\( t \)[/tex] is the time.

The heat transfer coefficient [tex]\( h \)[/tex] can be calculated using the formula:

[tex]\[ h = \frac{k}{D} \][/tex]

where:

[tex]\( k \)[/tex] is the thermal conductivity of copper,

[tex]\( D \)[/tex] is the diameter of the rod.

The time ( t ) it takes for the rod to cool to a certain temperature can be determined by solving for ( t) when [tex]\( Q(t) \)[/tex]reaches a specific value. In this case, we want to find ( t ) when the average temperature of the rod is 25°C.

Let's proceed with the calculations:

1. **Calculate [tex]\( h \)[/tex]:**

  The thermal conductivity of copper[tex](\( k \)) is approximately \( 400 \, \text{W/m} \cdot \text{K} \)[/tex].

[tex]\[ h = \frac{k}{D} \][/tex]

 [tex]\[ h = \frac{400 \, \text{W/m} \cdot \text{K}}{0.02 \, \text{m}} \] \[ h = 20,000 \, \text{W/m}^2 \cdot \text{K} \][/tex]

2. **Plug in values and solve for ( t ):**

  The initial temperature of the rod[tex](\( T_0 \))[/tex] is 100°C, the ambient temperature [tex](\( T_{\text{air}} \))[/tex] is 20°C, and the average temperature[tex](\( T \))[/tex] is 25°C. The temperature difference[tex]\( \Delta T \) is \( T_0 - T_{\text{air}} \)[/tex].

[tex]\[ \Delta T = T_0 - T_{\text{air}} = 100°C - 20°C = 80°C \][/tex]

 Now, [tex]\( h \) and \( \Delta T \) are known, and we can rearrange the formula \( Q(t) = Q_0 e^{-ht} \) to solve for \( t \)[/tex]:

[tex]\[ t = -\frac{1}{h} \ln\left(\frac{T - T_{\text{air}}}{T_0 - T_{\text{air}}}\right) \][/tex]

Substituting the values:

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{25°C - 20°C}{100°C - 20°C}\right) \][/tex]

Calculate ( t ) using the natural logarithm function, and you'll get the time it takes for the copper rod to cool to an average temperature of 25°C.

let's calculate the time it takes for the copper rod to cool to an average temperature of 25°C.

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{25°C - 20°C}{100°C - 20°C}\right) \][/tex]

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{5}{80}\right) \][/tex]

Now, calculate the natural logarithm:

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(0.0625\right) \][/tex]

[tex]\[ t \approx -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \times (-2.7726) \][/tex]

[tex]\[ t \approx \frac{2.7726}{20,000 \, \text{W/m}^2 \cdot \text{K}} \][/tex]

[tex]\[ t \approx 1.3863 \times 10^{-4} \, \text{s} \][/tex]

So, it takes approximately [tex]\(1.3863 \times 10^{-4}\)[/tex] seconds for the copper rod to cool to an average temperature of 25°C when exposed to an air stream at 20°C with a heat transfer coefficient of [tex]\(20,000 \, \text{W/m}^2 \cdot \text{K}\)[/tex].

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The time it would take for the copper rod to cool to an average temperature of 25°C is approximately 22.17 minutes.

To solve this problem, we will use the concept of Newton's law of cooling, which states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The formula for the heat transfer rate is given by:

[tex]\[ \frac{dQ}{dt} = hA(T_s - T_{\infty}) \][/tex]

where:

[tex]\( \frac{dQ}{dt} \)[/tex] is the rate of heat transfer (W),

h is the heat transfer coefficient [tex](W/m^2\·K)[/tex],

A is the surface area of the object [tex](m^2)[/tex],

[tex]\( T_s \)[/tex] is the surface temperature of the object (°C), and

[tex]\( T_{\infty} \)[/tex] is the temperature of the surrounding air (°C).

The copper rod is a cylinder, so its surface area A can be calculated using the formula for the surface area of a cylinder:

[tex]\[ A = 2\pi rL \][/tex]

where:

r is the radius of the rod, and

L is the length of the rod.

Given that the diameter of the rod is 2.0 cm, the radius r is half of that, which is 1.0 cm or 0.01 m. The length L of the rod is not given, so we will assume it to be L meters.

The initial temperature of the rod is [tex]\( T_{initial} = 100 \°C \)[/tex], and the final temperature we want to reach is [tex]\( T_{final} = 25 \°C \)[/tex]. The temperature difference between the rod and the surrounding air is [tex]\( T_s - T_{\infty} \)[/tex].

The heat lost Q by the rod as it cools from [tex]\( T_{initial} \)[/tex] to [tex]\( T_{final} \)[/tex] can be found using the specific heat capacity of copper (c) and the mass (m) of the rod:

[tex]\[ Q = mc(T_{initial} - T_{final}) \][/tex]

The mass (m) of the rod can be calculated from its density \rho and volume V:

[tex]\[ m = \rho V \][/tex]

[tex]\[ V = \pi r^2 L \][/tex]

The density of copper [tex]\( \rho \)[/tex] is approximately [tex]\( 8930 \) kg/m^3[/tex].

Now, we can equate the heat lost to the heat transfer over time:

[tex]\[ mc(T_{initial} - T_{final}) = hA(T_s - T_{\infty}) \cdot \Delta t \][/tex]

Solving for [tex]\( \Delta t \)[/tex], we get:

[tex]\[ \Delta t = \frac{mc(T_{initial} - T_{final})}{hA(T_s - T_{\infty})} \][/tex]

The specific heat capacity of copper (c) is approximately 385 J/kg·K.

Substituting the expressions for (m) and (A) into the equation for [tex]\( \Delta t \)[/tex], we have:

[tex]\[ \Delta t = \frac{\rho \pi r^2 L c(T_{initial} - T_{final})}{h(2\pi rL)(T_s - T_{\infty})} \][/tex]

Simplifying, we get:

[tex]\[ \Delta t = \frac{\rho r c(T_{initial} - T_{final})}{2h(T_s - T_{\infty})} \][/tex]

Now we can plug in the values:

[tex]\( \rho = 8930 \) kg/m^3[/tex],

r = 0.01 m,

c = 385 J/kg·K,

[tex]T_{initial}[/tex] = 100°C ,

[tex]T_{final}[/tex] = 25°C,

[tex]h = 200 W/m^2\·K[/tex],

[tex]T_{\infty}[/tex] = 20°C.

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times (100 - 25)}{2 \times 200 \times (25 - 20)} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 75}{2 \times 200 \times 5} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 15}{2 \times 200} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 15}{400} \][/tex]

[tex]\[ \Delta t = \frac{532095}{400} \][/tex]

[tex]\[ \Delta t = 1330.2375 \text{ seconds} \][/tex]

To express this in minutes, we divide by 60:

[tex]\[ \Delta t = \frac{1330.2375}{60} \text{ minutes} \][/tex]

[tex]\[ \Delta t \approx 22.170625 \text{ minutes} \][/tex]

Answer:

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chromatography chamber.

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures.

A strip of moist filter paper is put into the chromatography chamber so that its bottom touches the solvent and the paper lies on the chamber wall and reaches almost to the top of the container.

The container is closed and left for a few minutes to let the solvent vapors ascend the moist filter paper and saturate the air in the chamber.

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chamber.

Answer:

0.006 N

Explanation:

First find the mass of the horse which is the only unknown variable.

Then use it with the new data. Working out is in the attachment.

Answer:

[tex]v_{i} =28.86\frac{ft}{s}[/tex]

Explanation:

Conceptual analysis

We apply the kinematic formula for an object that moves vertically upwards:

[tex](v_{f} )^{2} =(v_{i} )^{2} -2*g*y[/tex]

Where:

[tex]v_{f}[/tex] : final speed in ft/s

[tex]v_{i}[/tex] : initial speed in ft/s

g: acceleration due to gravity in ft/s²

y: vertical position at any time in ft

Known data

For [tex]v_{f} = 25\frac{ft}{s}[/tex] ,[tex]y=\frac{1}{4} h[/tex]; where h is the maximum height

for y=h,  [tex]v_{f} =0[/tex]

Problem development

We replace  [tex]v_{f} = 25\frac{ft}{s}[/tex]  , [tex]y=\frac{1}{4} h (ft)[/tex] in the formula (1),

[[tex]25^{2} =(v_{i} )^{2} -2*g*\frac{h}{4}[/tex]   Equation (1)

in maximum height(h): [tex]v_{f} =0[/tex], Then we replace in formula (1):

[tex]0=(v_{i} )^{2} - 2*g*h[/tex]

[tex]2*g*h=(v_{i} )^{2}[/tex]

[tex]h=\frac{(v_{i})^{2}  }{2g}[/tex]   Equation(2)

We replace (h) of Equation(2) in the  Equation (1) :

[tex]25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2}  }{2g} }{4}[/tex]

[tex]25^{2} =(v_{i} )^{2} -\frac{(v_{i})^{2}  }{4}[/tex]

[tex]25^{2} =\frac{3}{4} (v_{i} )^{2}[/tex]

[tex]v_{i} =\sqrt{\frac{25^{2}*4 }{3} }[/tex]

[tex]v_{i} =28.86\frac{ft}{s}[/tex]

Answer:

Normal force: 167.48 N

Explanation:

        ∑[tex]F_{x}[/tex]:  [tex]F{x}-20 = 0[/tex]  

        ∑[tex]F_{y}[/tex]: [tex]N -W+F_{y}=0[/tex]

       [tex]N = W - Fy[/tex]

        [tex]N = mg -Fsin(\theta)[/tex]

        [tex]Fcos(\theta) -20 = 0[/tex]   ⇒   [tex]Fcos(\theta) = 20[/tex][tex]\theta                  

        [tex]\theta=cos^{-1}(20/F)[/tex]

       [tex]N = 20 x 9.81 - 35sin(55.15)[/tex]  ⇒ [tex]N = 167.48 N[/tex]

Answer:

[tex]\Delta S = 2.11 J/K[/tex]

Explanation:

As we know that entropy change for phase conversion is given as

[tex]\Delta S = \frac{\Delta Q}{T}[/tex]

Here we know that heat required to change the phase of the water is given as

[tex]\Delta Q = mL[/tex]

here we have

[tex]m = 0.349 g = 3.49\times 10^{-4} kg[/tex]

L = 2250000 J/kg

now we have

[tex]\Delta Q = (3.49 \times 10^{-4})\times 2250000[/tex]

[tex]\Delta Q = 788.9 J[/tex]

also we know that temperature is approximately same as boiling temperature

so we have

[tex]T = 373 k[/tex]

so here we have

[tex]\Delta S = \frac{788.9}{373}[/tex]

[tex]\Delta S = 2.11 J/K[/tex]

Final answer:

To calculate the angle of the second displacement, typically one would use vector addition and trigonometry, but the question, as presented, does not provide sufficient information to determine this angle.

Explanation:

To find the angle of the second displacement for the particle, we need to consider the information given: the first displacement is 11 m at an angle of 82° from the positive x-axis, and the resulting displacement after the second move is 8.7 m at 135° to the positive x-axis, using the counterclockwise direction as positive.

Using vector addition, the resulting displacement's direction can help us determine the second displacement angle. However, without more information about the specifics of the second displacement (like its magnitude), we cannot calculate the exact angle or vector using typical methods such as vector components or the law of cosines.

Given the problem as stated, assuming standard vector addition, a common approach would involve drawing a vector diagram and applying trigonometry; but in this case, we seem to be missing details to perform accurate calculations.

Answer:

[tex]1.40\times 10^{4}Joules [/tex]

Explanation:

The mechanical energy  is conservated in the absence of non conservative forces. In this case there is an air drag, wich will acount for the lost of energy. In other words:

[tex]\frac{1}{2}mv_{i} ^{2} +mgh_{i}-E_{lost}=\frac{1}{2}mv_{f} ^{2} +mgh_{f}[/tex]

Solving for the Energy lost:

[tex]E_{lost}=\frac{1}{2}mv_{i} ^{2} +mgh_{i}-\frac{1}{2}mv_{f} ^{2} -mgh_{f}[/tex]

And so

[tex]E_{lost}=\frac{1}{2}m(v_{i} ^{2}-v_{f} ^{2}) +mg(h_{i}-h_{f}) [/tex]

substituting with data

[tex]E_{lost}=\frac{1}{2}(64)((27) ^{2}-(25) ^{2}) +(64)(9.8)(17)=1.40\times 10^{4}Joules [/tex]

Hope my answer helps.

Have a nice day!

Answer:

Fx = 98 lb

Fy = 69 lb

Explanation:

We can see it in the pic

Answer:

98 lb, 69 lb

Explanation:

Tension in wire, T = 120 lb

Angle made with the horizontal, θ = 35°

The horizontal component is given by

[tex]T_{x}=T Cos35[/tex]

[tex]T_{x}=120 Cos35=98.3 lb[/tex]

[tex]T_{x}=98 lb[/tex]

The vertical component is given by

[tex]T_{y}=T Sin35[/tex]

[tex]T_{y}=120 Sin35=68.8 lb[/tex]

[tex]T_{y}=69 lb[/tex]

Thus, the horizontal component of tension is given by 98 lb and the vertical component of tension is given by 69 lb.

Final answer:

A person can produce approximately 85 W of useful electric power using a bicycle-powered generator when factoring in the typical 25% body efficiency and an 85% efficient generator.

Explanation:

A typical person can maintain a steady energy expenditure of 400 W on a bicycle. Assuming a typical efficiency for the body and a generator that is 85% efficient, the useful electric power produced can be calculated as follows:

First, consider the efficiency of the human body. If we assume that the body is 25% efficient (which means that only 25% of the energy consumed by the body is converted to mechanical work, while the rest is lost as heat), then the actual mechanical work a person can produce is 25% of their energy expenditure.

Therefore, the mechanical work output would be:

Next, we account for the efficiency of the generator:

To conclude, a person can produce approximately 85 W of useful electric power with a bicycle-powered generator when considering typical human body efficiency and an 85% efficient generator.

The work function of the metal is approximately 3.67 eV.

The work function, or binding energy, of a metal is the minimum energy required to eject an electron from the metal surface. In the photoelectric effect, the maximum kinetic energy of ejected electrons (photoelectrons) is given by the equation KE = hf - BE, where hf is the photon energy and BE is the work function. To find the work function W0 of the metal, we can use Equation 6.16, which states that the threshold wavelength for observing the photoelectric effect is given by λ = (hc)/(W0), where h is Planck's constant, c is the speed of light, and W0 is the work function.

Given that the maximum wavelength that can still eject electrons is 542 nm, we can rearrange the equation to solve for W0:

W0 = (hc)/λ

Plugging in the values, we have:

W0 = (6.626 × 10^-34 J·s)(3 × 10^8 m/s)/(542 × 10^-9 m)

Simplifying the calculation, we find that the work function W0 of the metal is approximately 3.67 eV.

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Final answer:

The maximum wavelength for ejecting electrons from the metal surface is 542 nm. Using the energy equation for photons, we calculate the work function (W0) to be approximately 2.29 electron volts.

Explanation:

The question relates to the photoelectric effect and the calculation of the work function of a metal when electromagnetic radiation of a certain wavelength is incident upon it. The work function (W0) is the minimum energy needed to eject an electron from the metal surface. According to the question, the maximum wavelength that can eject an electron is 542 nm.

To find the work function in electron volts, we can use the equation:

E(photon) = hc / λ

Where:

E(photon) is the energy of the photon

h is Planck's constant (4.135667696 × [tex]10^-^1^2[/tex] eV·s)

c is the speed of light (3 × 108 m/s)

λ is the wavelength of the photon in meters (542 nm = 542 × [tex]10^-^9[/tex] m)

First, convert the energy into electron volts:

E(photon) = (4.135667696 × [tex]10^-^1^2[/tex] eV·s × 3 × 108 m/s) / (542 × [tex]10^-^9[/tex] m)

After calculation, we find that E(photon) is approximately 2.29 eV. For a photon to eject an electron, its energy must be equal to or greater than the work function of the metal. Therefore, the work function of the metal is also 2.29 eV.

Hey!

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Answer:

B. level of statistical support

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Explanation:

When you create an incorrect conclusion within a study typically means that the study had something to do with a hypothesis. Well, if you have a statistical support and you have error in it, then the the lab is incorrect. One little thing can get someone confused.

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Hope This Helped! Good Luck!

Answer:

building footprint is 21000 sf

Explanation:

given data

floors = 8

total square footage = 168000 sf

to find out

building footprint

solution

we know building footprint is area where building built up same as perimeter of any building plan

so here

we have calculate here total perimeter

building footprint = [tex]\frac{total square footage}{total floors}[/tex]   ........1

put here value

here all floor are equal

building footprint =  [tex]\frac{168000}{8}[/tex]

building footprint = 21000 sf

so building footprint is 21000 sf