77. Two forces act on a parachutist falling in air: the force of gravity and air resistance. If the fall is steady, with no gain or loss of speed, then the parachutist is in dynamic equilibrium. How do the magnitudes of the gravitational force and air resistance compare?

Answer:

Explanation:

We use the harmonic motion position equation:

[tex]x(t) = A\cos(\omega t+\phi)[/tex]

where A = 0.350 and for t = 0

[tex]x(0) A = A\ cos(\phi)[/tex]

so: [tex]\phi = 0[/tex]

and also:

[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{4.10} = 1.532 rad/s[/tex]

so we have:

x(t)=0.350cos(1.532 t)

For t = 3.403 s

x(3.403)=0.350cos(1.532 (3.403)) = 0.348 m

The block attached to a spring is undergoing simple harmonic motion. By fitting the provided variables into the displacement equation for such a motion, it is found that the block's position 3.403 seconds after release is -0.279 meters.

The problem describes a scenario involving simple harmonic motion (SHM). For such a motion, the displacement, x(t) from equilibrium position could be expressed as x(t) = A * cos(ωt + φ), where 'A' is the amplitude, 'ω' = 2π/T is the angular frequency, 'T' is the period of the motion, t is the given time, and 'φ' is the phase constant. Since the motion begins at the amplitude, φ = 0.

Amplitude given is 0.35 m and period is 4.10 s. Thus, 'ω' = 2π / 4.10 = 1.5 rad/s. Putting these values into the displacement equation, x(t) = 0.350 m * cos(1.5 rad/s * 3.403 s) = -0.279 m. Therefore, the position of the mass 3.403 s after it's released is -0.279 m.

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Answer:

[tex]d = 0.018 m[/tex]

Explanation:

Charge on the plates of the capacitor due to transfer of electrons is given as

[tex]Q = Ne[/tex]

here we know that

[tex]N = 2.1 \times 10^9[/tex]

so we have

[tex]Q = (2.1 \times 10^9)(1.6 \times 10^{-19})[/tex]

[tex]Q = 3.36 \times 10^{-10} C[/tex]

now we have electric field between the plates is given as

[tex]E = \frac{Q}{A\epsilon_0}[/tex]

here we have

[tex]1.5 \times 10^5 = \frac{3.36 \times 10^{-10}}{A(8.85 \times 10^{-12})}[/tex]

[tex]A = 2.53 \times 10^{-4} m^2[/tex]

now we have

[tex]A = \frac{\pi d^2}{4}[/tex]

[tex]2.53 \times 10^{-4} = \frac{\pi d^2}{4}[/tex]

[tex]d = 0.018 m[/tex]

The  the diameters of the disks of the parallel-plate capacitor are  0.018 m.

Given parameters:

Number of electron transferring from one disk to another: n = 2.1×10⁹.

Electric field strength: E =  1.5×10⁵ N/C .

We have to find: the diameters of the disks: d = ?

Charge of an electron is: e = 1.6 × 10⁻¹⁹ C.

So, charge transferring from one disk to another:

Q = ne = 2.1×10⁹×1.6 × 10⁻¹⁹ C = 3.36×10⁻¹⁰ C

For parallel-plate capacitor: electric field strength is given by:

E = Q/ε₀A

Where:

A = area of each disks = πd²/4

ε₀ = permittivity of free space = 8.85 × 10⁻¹² Si units.

Hence,

 A = Q/ε₀E

⇒ πd²/4 =  Q/ε₀E

⇒ d = √( 4Q/πε₀E)

Putting numerical values we get:

d = √( 4 × 3.36×10⁻¹⁰/π×8.85 × 10⁻¹² × 1.5×10⁵ ) m.

= 0.018 m.

Hence, the diameters of the disks are 0.018 m.

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Answer:

y = k/x

Explanation:

y = k/x is a graph of a hyperbola that has been rotated about the origin.

The equation y = k/x  represents a generic equation suggested by a graph showing a hyperbola

A right circular cone and a plane are intersected at an angle in analytical geometry to form a hyperbola, which is a conic section that intersects both cone halves. It results in two distinct unbounded curves that are mirror images of one another.

While a set of points in a plane that are equally spaced apart from a directrix or focus is known as parabolas. The difference of distances between a group of points that are present in a plane to two fixed points and is a positive constant is how the hyperbola is defined.

The general equation representing a hyperbola

y= k/x

Thus,a graph with a hyperbola suggests the equation y = k/x as a general equation.

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Answer:

a) v = 2.733 m/s

b) t = 6.58s

c) a = -0.415 m/s^2

d) v = 0.789 m/s

Explanation:

We can see it in the pics

A motorcycle has a constant acceleration of 2.5 m/s², the motorcycle takes 4 seconds to change its speed by 10 m/s.

The kinematic equation can be used to determine how long it will take the motorcycle to alter its speed. Initial velocity ([tex]v_0[/tex]), final velocity (v), acceleration (a), and time (t) are related by the following equation:

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

Given that the acceleration (a) is 2.5 m/s² and both velocity and acceleration are in the same direction, we can use this equation to solve for time (t).

(a) Changing speed from 21 m/s to 31 m/s:

Initial velocity ([tex]\(v_0\)[/tex]) = 21 m/s

Final velocity (v) = 31 m/s

Acceleration (a) = 2.5 m/s²

31 = 21 + 2.5t

2.5t = 31 - 21

2.5t = 10

[tex]\[t = \dfrac{10}{2.5} = 4 \, \text{seconds}\][/tex]

(b) Changing speed from 51 m/s to 61 m/s:

Initial velocity ([tex]\(v_0\)[/tex]) = 51 m/s

Final velocity (v) = 61 m/s

Acceleration (a) = 2.5 m/s²

61 = 51 + 2.5

2.5t = 61 - 51

2.5t = 10

[tex]\[t = \dfrac{10}{2.5} = 4 \, \text{seconds}\][/tex]

Thus, in both cases, the motorcycle takes 4 seconds to change its speed by 10 m/s.

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The time required for the motorcycle to change its speed by 10 m/s, under a constant acceleration of 2.5 m/s², is 4 seconds in both scenarios: from 21 m/s to 31 m/s, and from 51 m/s to 61 m/s.

In order to answer your question about the time required for the motorcycle to change its speed, we use the formula for acceleration, which is change in velocity divided by time (a = Δv/Δt). Thus, we can solve for time with Δt = Δv / a.

(a) For a speed change from 21 m/s to 31 m/s, Δv = 31 m/s - 21 m/s = 10 m/s. Substituting into the formula, we find Δt = 10 m/s / 2.5 m/s² = 4 seconds.

(b) For a speed change from 51 m/s to 61 m/s, Δv = 61 m/s - 51 m/s = 10 m/s. Again substituting into the formula, we find Δt = 10 m/s / 2.5 m/s² = 4 seconds.

Therefore, the time required for the motorcycle to change its speed by 10 m/s, under a constant acceleration of 2.5 m/s², is 4 seconds in both cases.

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The equation which has the fastest runner is Distance = 2 * Time.

The correct answer is Option D.

Given data:

The equation that describes the fastest runner among the given options is D. distance = 2⋅time.

This equation suggests that the distance traveled by the runner is directly proportional to twice the time taken. In other words, if two runners start at the same time and one of them has a faster speed, they will cover a greater distance in the same amount of time. This equation aligns with the principles of speed and distance in physics.

Option D reflects that the distance covered by a runner is determined by the product of time and speed. When the speed is greater (as indicated by the coefficient of 2), the distance covered in a given time will be larger, making this equation representative of the fastest runner. The other options, A, B, and C, do not consider the concept of speed and do not accurately describe the behavior of a fastest runner in relation to distance and time.

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Final answer:

The fastest runner is described by Equation D, which represents a speed of 2 meters per second (2 m/s), as it has the highest coefficient of time in the distance = speed × time equation.

Explanation:

The equation that describes the fastest runner is the one where the coefficient of time (t) in the distance = speed × time equation is the greatest, as this coefficient represents the speed of the runner. The faster the runner, the greater the distance they would cover in a given time. Comparing the given equations:

A. distance = 0.5×time (speed of 0.5 m/s)

B. distance = 0.33×time (speed of 0.33 m/s)

C. distance = time (speed of 1 m/s)

D. distance = 2×time (speed of 2 m/s)

The fastest runner is described by Equation D, where the distance is twice the time, indicating a speed of 2 meters per second (2 m/s).

[tex]\boxed{n=2.92\times 10^{10} \ stars}[/tex]

Order-of-magnitude estimates are useful when we want to get a very crude estimate of a quantity. Sometimes, to get the exact calculation of a problem is very difficult, or impossible, so we use order of magnitud in order to get a rough approximation. In this exercise, we need to find the order of magnitude of the number of stars in the Milky Way. We know:

The distance from the Sun to the nearest star is:

[tex]4 \times 10^{16}m[/tex]

The Milky Way galaxy is roughly a disk of diameter:

[tex]10^{21}[/tex]

The Milky Way galaxy is roughly a disk of thickness:

[tex]10^{19}[/tex]

So we can approximate volume of the Milky Way:

[tex]V=\pi r^2 h \\ \\ r=\frac{10^{21}}{2}m=5 \times 10^{20}m \\ \\ h=10^{19}m \\ \\ \\ V=\pi(5 \times 10^{20})^2(10^{19}) \\ \\ V=7.85\times 10^{60}m^3[/tex]

Now, let's estimate a rough density for the Milky Way. It is well known that in a sphere with a radius of  [tex]4\times 10^{16}m[/tex] there is a star, which is the sun. So the density of the Milky way is:

[tex]\rho =\frac{n}{V} \\ \\ \rho:Density \ of \ Milky \ way \\ \\ n: Number \ of \ stars \\ \\ V:Volume[/tex]

For one star [tex]n=1[/tex] so we know the data at the neighborhood around the Sun, so the volume is a sphere:

[tex]V=V_{sphere}=\frac{4}{3}\pi r^3 \\ \\ V_{sphere}=\frac{4}{3}\pi (4\times 10^{16})^3 \\ \\ V_{sphere}=2.68\times 10^{50}m^3 \\ \\ So: \\ \\ \rho=\frac{1}{2.68\times 10^{50}}=3.73\times 10^{-51}stars/m^3[/tex]

Finally, the numbers of stars can be found as:

[tex]n=\rho V \\ \\ V:Volume \ of \ the \ Milky \ Way \\ \\ p:Density \ of \ the \ Milky \ Way \\ \\ n:Number \ of \ stars \\ \\ n=(3.73\times 10^{-51})(7.85\times 10^{60}) \\ \\ \boxed{n=2.92\times 10^{10} \ stars}[/tex]

Answer:

[tex]E_{k}=15204Joules[/tex]

[tex]v=22.2m/s[/tex]

Explanation:

We can apply energy conservation law:

Initial Energy = Final Energy

In the beginning there is only potential energy, and in the end there is only kinetic energy.

[tex]E_{k}=E_{p}=mgh=61.5*9.81*25.2=15204Joules[/tex]

And:

[tex]E_{k} = \frac{1}{2}mv^2[/tex]

[tex]v=\sqrt{2*E_{k}/m}=\sqrt{2*15204/61.5}=22.2m/s[/tex]

Answer:

b .at the outer edge of the Galactic Disk

Explanation:

The sun is located  on the outeredge of the galaxy. So it has an orbit within our Milky Way galaxy that is also located arround the outer edge of the galactic disk, and goes around in about 30 million years.

In reallity, allong its trayectory, the sun  goes above and below the gallactic disk, but it stays in the outer edge of the galaxy, far away from the galactic bulge and the galactic center, but closer to it than the galactic halo.

Answer:

a) t=8.19s; x=44.2m

b) v=1.401 m/s

c) see attachment

d) The second solution is a later time at which the bus catches the student. v=9.40 m/s

e) No, she won't.

f) v=3.63m/s;t=21.2; x=77m

Explanation:

a) The motion of both the bus and the student can be explained by the equation [tex]x= v_{0} +\frac{1}{2}at^{2}[/tex]. Since the student is not accelerating, but rather maintaining a constant speed; the particular equation that describes the motion of the student is: [tex]x_{student} = 5.4 \frac{m}{s} * t[/tex]. Meanwhile, since the bus starts its motion at an initial velocity of zero, the equation that describes its motion is: [tex]x_{bus} =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2}[/tex]. The motion of the student relative to that of the bus can be described by the equation: [tex]x_{s} =x_{bus} +38.5m[/tex]. By replacing terms in the last equation we end up with the following quadratic equation: [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(5.4\frac{m}{s} *t)+38.5m =0[/tex]. Solving the quadratic equation will yield two solutions; t1=8.19s and t2=55.0s. By plugging in t1 onto the equation that describes the motion of the student we will find the distance runned by her, x=44.2m.

b) The velocity of the bus can be modeled by the equation [tex]v^{2} = v_{0} ^{2} +2ax[/tex]. Since the initial velocity of the bus is zero, the first term of the equation cancels. Next, we solve for v and plug in the acceleration of 0.171 m/s2 and the distance 5.74m (traveled by the bus, note: this is equal to the 44.2m travelled by the student minus the 38.5m that separated the student from the bus at the beginning of the problem).

c) The equations that make up the x-t graph are: [tex] x = 5.4 \frac{m}{s} * t[/tex] and  [tex]x =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2} + 38.5[/tex]; as described in part a.

d) The first solution states that the student would have to run 44.2 m in 8.19 s in order to catch the bus. But, at that point, the student has a greater speed than that of the bus. So if both were to keep he same specified motion, the student would run past the bus until it reaches a velocity greater than that of the student. At which point, the bus will start to narrow the distance with the student until it finally catches up with the student 55 seconds after both started their respective motion.

e) No, because the quadratic equation [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(3.5\frac{m}{s} *t)+38.5m =0[/tex] has no solution. This means that the two curves that describe the distance vs time graph for both student and bus do not intersect.

f) This answer is reached by finding b of the quadratic equation. The minimum that b can be in order to find a real answer to the quadratic equation is found by solving [tex]b^{2} -4ac=0[/tex]. If we take a = 0.0855 and c = 38.5, then we find that the minimum speed that the student has to run at is 3.63 m/s. If we then solve the quadratic equation [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(3.63\frac{m}{s} *t)+38.5m =0[/tex],we will find that the time the student will run is 21.2 seconds. By pluging in that time in the equation that describes her motion: [tex]x_{student} = 3.63 \frac{m}{s} * t[/tex] we find that she has to run 77 meters in order to catch the bus.

To determine if and when the student catches the bus, one needs to use kinematic equations to calculate time and distance, as well as understand that the second solution of these equations is hypothetical and signifies the student overtaking the bus. The answer involves solving systems of equations and applying principles of kinematics.

We need to solve the kinematics equations for the student and the bus to find out when and if the student will catch the bus. From kinematics, the position of an object moving with constant acceleration is given by x = x0 + v0t + ½at2, where x0 is the initial position, v0 is the initial velocity, a is acceleration, and t is time.

Given the student's speed 5.4 m/s and the bus's acceleration of 0.171 m/s2, we set up two equations:

We find the time at which the student catches the bus by equating xs and xb and solving for t.

After finding the time, we can calculate the distance the student runs by plugging it back into the student's position equation. Then we find the speed of the bus at that point by using the equation of motion v = v0 + at.

The second solution represents another point in time where the student and bus would meet if they continued at the same speeds and acceleration, which physically represents the student overtaking the bus. Finally, if the student's top speed is 3.5 m/s, we again set up the equations similarly and solve to determine if she can catch the bus. The minimum speed required is found by setting the time derivative of the position equations equal and solving for the student's speed.

Look through your class notes and your textbook and check out all the formulas and equations for the distance, speed and acceleration of falling objects. You won't find the weight of the object in any of those formulas because it doesn't matter. Without air resistance, all objects fall with the same speed and acceleration. If two objects are dropped from the same height at the same time, they both hit the ground at the same time. It doesn't matter what either one weighs. One can be a feather and the other one can be a school bus. Without air resistance it doesn't matter.

The answer is choice-(e).

Two small heavy balls have the same diameter but one weighs twice as much as the other. The time to reach the ground below will be nearly the same for both balls.

The mass of an object refers to the weight the object occupies in space.

When these two objects with different masses are dropped from a second-story building, the gravitational force acts on the objects in space.

In space, the objects fall with nearly the same acceleration regardless of the weight of their mass, the diameter, or the size of the object.

Therefore, we can conclude that for the two small heavy balls, the time to reach the ground below will be nearly the same for both balls.

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Answer:

C) The amount of water released through the dam and the dam's height account for the potential energy

Explanation:

The gravitational potential energy of an object is given by:

[tex]U=mgh[/tex]

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object relative to the ground

As we see from the equation, the potential energy depends on two factors related to the object:

- its mass

- its height above the ground

So for the water in the dam, its potential energy depends on

- the mass of the water (= the amount of water released)

- the height of the water (= the dam's height)

So the correct answer is

C) The amount of water released through the dam and the dam's height account for the potential energy

Final answer:

The slope of the catenary curve where it meets the right pole is -1. The angle between the line and the pole is -45 degrees.

Explanation:

The shape of the telephone line between two poles is known as a catenary. To find the slope where it meets the right pole, we can use the formula:

slope = -c/a

where 'c' is the distance from the vertex to the right pole and 'a' is half the distance between the poles. In this case, c = 7m and a = 7m. Substituting these values into the formula, we get:

slope = -7/7 = -1

Therefore, the slope of the curve where the line meets the right pole is -1.

To find the angle between the line and the pole, we can use the formula:

angle = arctan(slope)

Substituting the slope (-1) into the formula, we get:

angle = arctan(-1)

Using a calculator, we find that the angle is approximately -45 degrees.

Answer:

(D) 0.250 mole

Explanation:

From molecular weight we know that 1 mole of P4O10 wights 284 grams. So

[tex]14.2 g P4O10 \times \frac{1 mole P4O10}{284 g P4O10} = 0.05 moles P4O10 [/tex]

In order to form 1 mole of P4O10 it is necessary 5 moles of O2,  so that the number of oxygen atoms are the same.

Knowing that 0.05 moles of P4O10 is formed we need 5 x 0.05 = 0.25 moles of O2

The question is weird. It doesn't matter whether it was a man or a woman, or whether he went on a diet, or whether he gained or lost mass or stayed the same, or what his mass was later. In fact, it doesn't even matter WHAT the number 100.6 represents. The answer would be the same in any case.

100.6 = 1.006 x 10^2

If a man finds that he has a mass of 100.6 kg. He goes on a diet, and several months later he finds that he has a mass of 96.4 kg, then his mass in the scientific notation would be 1.006 × 10² kg.

In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.

Only the digits permitted by the measurement resolution are trustworthy, therefore if a number expressing the result of a measurement (such as length, pressure, volume, or mass) has more digits than the number of digits permitted by the measurement resolution, only these can be significant figures.

Thus, If a man discovers that he weighs 100.6 kg. He starts a diet, and after a few months discovers he weighs 96.4 kg, his mass would be written as 1.006 10² kg in the scientific notation.

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Answer:

[tex]\frac{dQ}{dt}= 4312 W[/tex]

Explanation:

As we know that base of the slab is given as

[tex]A = 11 \times 8[/tex]

[tex]A = 88 m^2[/tex]

now we know that rate of heat transfer is given as

[tex]\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)[/tex]

here we know that

[tex]k = 1.4 W/m k[/tex]

Also we have

[tex]x =0.20[/tex]

[tex]\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)[/tex]

[tex]\frac{dQ}{dt}= 4312 W[/tex]

The rate of heat loss through the concrete slab of the basement is 4312 watts, calculated using Fourier's Law of thermal conduction with the given dimensions and temperatures.

To calculate the rate of heat loss through the concrete slab of a basement, we can use the formula for thermal conduction, which is given by Fourier's Law of heat conduction:
Q = \frac{k \cdot A \cdot (T_{hot} - T_{cold})}{d}
Where:
Q is the rate of heat transfer (W),
k is the thermal conductivity of the material (W/m·K),
A is the cross-sectional area through which heat is being transferred (m²),
T_{hot} is the temperature on the hot side of the material (°C),
T_{cold} is the temperature on the cold side (°C),
and d is the thickness of the material (m).
Using the given values:
k = 1.4 W/m·K,
A = 11 m * 8 m = 88 m²,
T_{hot} = 17°C,
T_{cold} = 10°C,
d = 0.20 m,
the rate of heat loss Q can be calculated as follows:
Q = \frac{1.4 \cdot 88 \cdot (17 - 10)}{0.20} = \frac{1.4 \cdot 88 \cdot 7}{0.20}
Q = 4312 W
The basement's heat loss through the concrete slab is 4312 watts. If the gas furnace operates at an efficiency of 0.9 (90%), the actual energy output required from the furnace would be higher.

Using the equation of motion, it is calculated that the electron will take approximately 9.18 x 10^-12 seconds to reach the specified speed.

The question relates to the fundamental physics concept of motion, where the final speed of an object can be calculated from its initial speed, the acceleration it undergoes, and the time it accelerates. Specifically, we can use the formula v = u + at, where v is the final speed, u is the initial speed, a is the acceleration, and t is the time.

Here, the final speed v of the electron is 7.07 × 105 m/s, its initial speed u is 4.04 × 105 m/s, and its acceleration a is 3.3 × 1014 m/s2. We can rearrange the formula to solve for time t, which gives t = (v - u) / a.

Substituting the given values into this formula, we get t = (7.07 × 105 - 4.04 × 105) / 3.3 × 1014 = 9.18 x 10-12 seconds. Therefore, it will take the electron approximately 9.18 x 10-12 seconds to reach a speed of 7.07 × 105 m/s.

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Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)  

vo² = 630               simplifying

vo = 25 m/s            result

Answer:

1.5mi/hr, 4.5mi/hr

Explanation:

Given:

45 min = 0.75h, 15 min = 0.25h

(1) 0.75v₁ = 0.25v₂

(2) v₁ = v₂ - 3

Solve for v₁ and v₂:

0.75(v₂ - 3) = 0.25v₂ = 0.75v₂ - 2.25 = 0.25v₂

0.5v₂ = 2.25

v₂ = 4.5

v₁ = 1.5

Answer:

uphill speed = 1.5 mph, downhill speed = 4.5 mph

Explanation:

We are asked to find Julian’s uphill speed and downhill speed. Let’s let r represent Julian’s uphill speed in miles per hour. In 45min=34hr, he rides uphill a distance rt=34r miles. Since his downhill speed is 3 miles per hour faster, we represent that as r+3. In 15min=14hr, he rides downhill a distance (r+3)t=r+34 miles. We can set these distances equal to find

34r=r+34

Multiplying both sides by 4 and then subtracting r from both sides yields

3r2r=r+3=3

Solving for r=32, Julian’s uphill speed is 1.5 miles per hour and his downhill speed is r+3=4.5 miles per hour.

Answer:

Part a)

[tex]\delta y = 0.85 m[/tex]

Part b)

[tex]\theta = 0.65 degree[/tex]

Explanation:

Part a)

As we know that the target is at distance 75 m from the hunter position

so here we will have

[tex]x = v_x t[/tex]

here we know that

[tex]v_x = 180 m/s[/tex]

so we have

[tex]75.0 = 180 (t)[/tex]

[tex]t = 0.42 s[/tex]

now in the same time bullet will go vertically downwards by distance

[tex]\delta y = \frac{1}{2}gt^2[/tex]

[tex]\delta y = \frac{1}{2}(9.81)(0.42^2)[/tex]

[tex]\delta y = 0.85 m[/tex]

Part b)

In order to hit the target at same level we need to shot at such angle that the range will be 75 m

so here we have

[tex]R = \frac{v^2 sin(2\theta)}{g}[/tex]

[tex]75 = \frac{180^2 sin(2\theta)}{9.81}[/tex]

[tex]\theta = 0.65 degree[/tex]

Answer:

(a)  1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

[tex]\overrightarrow{d_{1}}=580\widehat{j}[/tex]

[tex]\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )[/tex]

[tex]\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}[/tex]

[tex]\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )[/tex]

[tex]\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}[/tex]

The resultant displacement is given by

[tex]\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]

[tex]\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}[/tex]

[tex]\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}[/tex]

magnitude of the displacement

[tex]d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m[/tex]

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

[tex]tan\theta =\frac{1294.18}{35.36}=36.6[/tex]

θ = 88.44°

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

[tex]x=V_{o}cos\theta t[/tex]   (1)

Where:

[tex]V_{o}=48.1 m/s[/tex] is the cannonball's initial velocity

[tex]\theta=0[/tex] because we are told the cannonball is shot horizontally

[tex]t[/tex] is the time since the cannonball is shot until it hits the ground

y-component:

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (2)

Where:

[tex]y_{o}=1.5m[/tex]  is the initial height of the cannonball

[tex]y=0[/tex]  is the final height of the cannonball (when it finally hits the ground)

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it [tex]X_{max}[/tex] and this occurs when [tex]y=0[/tex].

So, firstly we will find the time with (2):

[tex]0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2[/tex]   (3)

Rearranging the equation:

[tex]0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m[/tex]   (4)

[tex]-(4.9 m/s^{2})t^2+(48. 1 m/s)  t+1.5 m=0[/tex]   (5)

This is a quadratic equation (also called equation of the second degree) of the form [tex]at^{2}+bt+c=0[/tex], which can be solved with the following formula:

[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (6)

Where:

[tex]a=-4.9 m/s^{2}[/tex]

[tex]b=48.1 m/s[/tex]

[tex]c=1.5 m[/tex]

Substituting the known values:

[tex]t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}[/tex] (7)

Solving (7) we find the positive result is:

[tex]t=9.847 s[/tex] (8)

Substituting this value in (1):

[tex]x=(48.1 m/s)cos(0\°) (9.847 s)[/tex]   (9)

[tex]x=473.640 m[/tex]  This is the horizontal distance the cannonball traveled before it landed on the ground.

Answer:

Explanation:

Applied force, F = 18 N

Coefficient of static friction, μs = 0.4

Coefficient of kinetic friction, μs = 0.3

θ = 27°

Let N be the normal reaction of the wall acting on the block and m be the mass of block.

Resolve the components of force F.

As the block is in the horizontal equilibrium, so

F Cos 27° = N

N = 18 Cos 27° = 16.04 N

As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .

The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N   .... (1)

The vertically downward force acting on the block is mg - F Sin 27°

                                                      = mg - 18 Sin 27° = mg - 8.172    ... (2)

Now by equating the forces from equation (1) and (2), we get

mg - 8.172 = 6.42

mg = 14.592

m x 9.8 = 14.592

m = 1.49 kg

Thus, the mass of block is 1.5 kg.  

Answer:

In 1838, Matthias Schleiden, a German botanist, concluded that all plant tissues are composed of cells and that an embryonic plant arose from a single cell. He declared that the cell is the basic building block of all plant matter.

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Explanation:

Final answer:

Matthias Schleiden, along with Theodor Schwann, contributed greatly to cell theory by proposing that all plants are composed of cells and are the basic unit of life, which was fundamental in stating that all living things are made up of cells.

Explanation:

Matthias Schleiden, a notable German botanist, had a pivotal role in the development of the cell theory. In 1838, Schleiden made significant microscopic observations of plant tissues and proposed that plants are composed of cells, which he believed to be the fundamental unit of life. Along with Theodor Schwann, a zoologist who studied animals, Schleiden advanced the concept that all living things are made of cells. Although Schleiden initially thought cells formed through crystallization, later discoveries, including those by Rudolf Virchow, confirmed that new cells arise from the division of existing cells. This collaborative work laid the foundations for our understanding that cells are the basic building blocks of life, constituting all organisms, and perpetuating life through cell division.

The cornerstone principles that Schleiden helped to formulate include: all living things are composed of one or more cells, the cell is the basic unit of life, and all new cells are produced from preexisting cells. These principles remain central to modern biology and medicine and provide a critical framework for understanding the structure and function of all living organisms.

A maximum of four bonds can be used between two atoms.

Explanation:

Ionic bond is one of the strongest bond used in chemical reaction where the valence electrons are bonded and shared between the atoms. Sometimes four covalent bonds between two atoms might make them unstable as the bonds are formed in their outermost shell to make them to form or complete a total of 8 in their shell. For example carbon has 3 as maximum.

Answer:

[tex]0.283\ m^3/s[/tex]

Explanation:

Given that,

1.00 in = 2.54 cm

We know that,

1 feet = 12 inches

[tex]1\ feet =12\times0.0254[/tex]

[tex]1\ feet =0.3048\ m[/tex]

We calculate the number

We need to converted [tex]ft^3/min[/tex] to [tex]m^3/sec[/tex]

The number required is

[tex]600ft^3/min = \dfrac{600\times(0.3048)^3}{60}[/tex]

[tex]600ft^3/min=0.283\ m^3/s[/tex]

Hence, This is the required solution.

The average speed of air in the duct is 11.23 m/s.

To find the average speed of air in the duct, we need to calculate the volume flow rate. The volume of the house's interior is given as 13.0 m × 20.0 m × 2.75 m = 715.0 m³. The air is carried through the duct every 15 minutes, so we need to convert the time to seconds: 15 minutes × 60 seconds/minute = 900 seconds. Therefore, the volume flow rate is 715.0 m³ / 900 s = 0.794 m³/s.

The main uptake air duct has a diameter of 0.300 m, which we can use to find the cross-sectional area of the duct: A = (π/4) × (0.300 m)² = 0.0707 m².

Now, we can calculate the average speed of air in the duct: average speed = volume flow rate / cross-sectional area = 0.794 m³/s / 0.0707 m² = 11.23 m/s.

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Answer:

weight of sealed tube that is filled with iodine gas is 27 gm

Explanation:

As tube is closed therefore mole of gas is enclosed in a tube, thus it will only converted into gas ( Iodine Gas). In the gas form it will going to exert pressure on the wall of a tube. From conservation of mass principle weight of tube remain same, there will be no change in the weight of gas. therefore weight of sealed tube that is filled with iodine gas is 27 gm

The weight of the sealed tube filled with iodine gas will remain 27.0 grams, as the principle of conservation of mass dictates that the total mass in a closed system does not change.

The principle of conservation of mass states that matter cannot be created or destroyed. Therefore, the total mass before and after the iodine sublimates remains the same in a closed system. The total mass of the tube and iodine before heating is 27.0 grams, including the 1.0-gram iodine sample and the tube. After the iodine sublimates, it remains within the sealed tube; thus, there is no change in mass. So, the weight of the sealed tube filled with iodine gas will remain 27.0 grams.

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Answer:

a) [tex]t=6.37s[/tex]

b) [tex]t=3.3333s[/tex]

Explanation:

The knowable variables are the initial hight and initial velocity

[tex]s_{o}=80ft[/tex]

[tex]v_{os}=64ft/s[/tex]

The equation that describes the motion of the ball is:

[tex]s=80+64t-16t^{2}[/tex]

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

[tex]0=80+64t-12t^{2}[/tex]

a) Solving for t, we are going to have two answers

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

a=-16

b=64

c=80

[tex]t=-1.045 s[/tex] or [tex]t=6.378s[/tex]

Since time can not be negative the answer is t=6.378s

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

[tex]v_{y}=v_{o}+gt[/tex]

at maximum point the velocity is 0

[tex]0=64-32.2t[/tex]

Solving for t

[tex]t=1.9875 s[/tex]

Now, we must know how much distance does it take to reach maximum point

[tex]s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft[/tex]

So, the ball pass the top of the building on its way down at 160 ft

[tex]160=80+64t-16t^{2}[/tex]

Solving for t

[tex]t=2s[/tex] or [tex]t=3.333s[/tex]

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

Answer:

(a) 5 s

(b) 4s

Explanation:

height of building, s = 80 feet

The equation of motion is given by

[tex]s=80+64t-16t^{2}[/tex]

(a) Let it takes t time to reach the ground. A it reach the ground, s = 0

So, put s = 0 in the given equation, we get

[tex]0=80+64t-16t^{2}[/tex]

[tex]t^{2}-4t-5=0[/tex]

(t + 1)(t - 5) = 0

t = -1 s, t = 5 s

As time cannot be negative, so t = 5 s

Thus, the time taken by the ball to reach the ground is 5 s.

(b) Let it crosses the building in t second, s put s = 80 feet in the above equation, we get

[tex]80+64t-16t^{2}=80[/tex]

t(4 - t)= 0

t = 0s or t  4 s

So, it takes 4 second to cross the building.

Answer:

34.5 m at 55.5 degrees above the positive x-axis

Explanation:

Resolving a vector means finding its components along two perpendicular axis: most commonly, they are chosen as the x and y axis.

In this problem, we have a vector whose components are:

x -component: 19.5 m

y- component: 28.4 m

The two components are perpendicular to each other: this means that we can find the magnitude of the vector by using the Pythagorean theorem

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{19.5^2+28.4^2}=34.5 m[/tex]

The direction, instead, can be found by using the following formula:

[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1} (\frac{28.4}{19.5})=55.5^{\circ}[/tex]

Answer:

Negative acceleration

Explanation:

We know that

v= u+at

Where v is the final velocity

u is the initial velocity velocity

a is the acceleration

t is time

When final velocity of train will be zero .

v=0

0= u+at

a=- u/t

So from above we can say that acceleration will be negative .When acceleration is positive then train will never come in rest position.