Which equation describes the fastest runner?
A. distance=0.5⋅time
B. distance=0.33⋅time
C. distance=time
D. distance=2⋅time

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.

Learn more about the mass of an object here:

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The gravitational potential energy of a two-particle system can be calculated with a specific formula that uses the gravitational constant and both masses. The work done by the gravitational force and an external force when separation is tripled can be found by calculating the change in gravitational potential energy.

The gravitational potential energy between two particles can be calculated with the formula: U = -G (M₁M₂/R) , where U is the gravitational potential energy, G is the gravitational constant (6.67 × 10^-11 Nm²/kg²), M₁ and M₂ are the masses of the two bodies, and R is the distance of separation between them.

To find the work done by gravitational force and by you when the separation is tripled, we will need to calculate the change in gravitational potential energy which is given by ΔU = U_final - U_initial. The negative of this value gives the work done by the gravitational force, while the magnitude of ΔU gives the work done by an external force (you in this case).

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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.

I Hope It's Helpful :)

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.

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:

[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:

The answer to your question is: Ф = 81.1° = 278.9°

Explanation:

Data

angle, x, = ?

height = 170 ft

distance 1084 ft

Process

tan Ф = opposite side / adjacent side

tan Ф = 1084 / 170 = 6.37

Ф = tan ⁻¹ (6.37)

Ф = 81.1° or 360 - 81.1 = 278.9°

Answer:

a) Minimum acceleration is [tex]a=2.31\frac{m}{s^{2} }[/tex].

b) It will take [tex]t_{f}=14.41s[/tex].

Explanation:

Let's order the information.

Initial velocity: [tex]v_{i}=0m/s[/tex]

Final velocity: [tex]v_{f}=33.3m/s[/tex]

Initial position: [tex]x_{i}=0m[/tex]

Final position: [tex]x_{f}=240m[/tex]

a) We can use velocity's equation:

[tex]v_{f}^{2} = v_{i}^{2} +2a(x_{f}-x_{i})[/tex]

⇒ [tex]a=\frac{v_{f}^{2}-v_{i}^{2}}{2(x_{f}-x_{i})}[/tex]

⇒ [tex]a=2.31\frac{m}{s^{2} }[/tex].

b) For this, equation for average acceleration will be helpful. Taking [tex]t_{i}=0[/tex] and having [tex]t_{f}[/tex] as the unknown time it becomes airborne:

[tex]a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}} =\frac{v_{f} }{t_{f}}[/tex]

⇒ [tex]t_{f}=\frac{v_{f}}{a}=\frac{33.3\frac{m}{s}}{2.31\frac{m}{s^{2}}}[/tex]

⇒ [tex]t_{f}=14.41s[/tex].

To take off, the Cessna aircraft needs a minimum acceleration of 2.31 m/s² and it will take approximately 14.4 seconds to become airborne after a run of 240 meters.

To find the minimum constant acceleration required by a Cessna aircraft to become airborne after a takeoff run of 240 meters, and the time it takes to become airborne, follow these steps:

(a) Minimum Constant Acceleration

We use the kinematic equation: v² = u² + 2as, where:

Substitute the known values:

[tex](33.3 m/s)^2 = 0 + 2a(240 m)\\1108.89 = 480a\\a = 2.31 m/s^2[/tex]

(b) Time to Become Airborne

We use the kinematic equation: v = u + at, where:

Substitute the known values:

[tex]33.3 m/s = 0 + (2.31 m/s^2)t\\t = 33.3 / 2.31\\t = 14.4 seconds[/tex]

Answer:

a = 4.054 m/ s2

Explanation:

[tex]safety\ Time\ for\ cockroach = \frac{Distance\ travelled\ by\ cockroach}{ speed of cockroach}[/tex]

[tex]t = \frac{1.3}{1.5}[/tex]

t = 0.86 seconds

total distance travelled by you = 0.8 + 1.3 = 2.1 m

Initial velocity u = 0.7 m/s

let a  is acceleration,

from equation of motion we have

s = ut + 0.5 at^2

[tex]2.1 = 0.7 \frac{1.3}{1.5} + 0.5 a [\frac{1.3}{1.5}]^2[/tex]

2.1 = 0.60 + 0.37 a

1. 5 = 0.37a

a = 4.054 m/ s2

Answer:

Kinetic energy

Solids 1, liquids 2, gases 3

Temperature

Solids 1, liquids 2, gases 3

Density

Solids 3, liquids 2, gases 1

Explanation:

Most of elements can be in the three different states at different temperatures, solid in lower temperatures then liquid and gas in higher temperatures.

When it's gaseous it is more expanded, it's volume is bigger then it's density is lower. As the molecules are more separated they can move more easily.

On the opposite hand in solid state, it's more compressed, the volume is smaller and the density is higher. The molecules are more compressed an they can't move easily.

Answer:

28.8 meters

Explanation:

We must first determine at which velocity the ball hits the water. To do so we will:

1) Assume no air resistance.

2) Use the Law of conservation of mechanical energy: E=K+P

Where

E is the mechanical energy (which is constant)

K is the kinetic energy.

P is the potential energy.

With this we have [tex]\frac{m}{2} *v^{2}  = m*g*h[/tex]

Where:

m is the balls's mass <- we will see that it cancels out and as such we don't need to know it.

v is the speed when it hits the water.

g is the gravitational constant (we will assume g=9.8[tex]\frac{m}{s^{2} }[/tex].

h is the height from which the ball fell.

Because when we initially drop the ball, all its energy is potential (and [tex]P = - m*g*h[/tex]) and when it hits the water, all its energy is kinetic ([tex]K=\frac{m}{2} *v^{2}[/tex]. And all that potential was converted to kinetic energy.

Now, from [tex]\frac{m}{2} *v^{2}  = m*g*h[/tex] we can deduce that [tex]v=\sqrt{2*g*h}[/tex]

Therefore v=9.6[tex]\frac{m}{s}[/tex]

Now, to answer how deep is the lake we just need to multiply that speed by the time it took the ball to reach the bottom.

So D=9.6[tex]\frac{m}{s}[/tex]*3[tex]s[/tex]=28.8[tex]m[/tex]

Which is our answer.

Final answer:

The depth of the lake can be calculated by determining the time it takes for the ball to reach the bottom. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity, and t is the time taken, we can solve for t. The time taken to fall from the diving board is subtracted from the total time taken to sink to the bottom, and then the distance fallen from the diving board to the bottom of the lake is calculated using the equation d = 1/2gt^2. The depth of the lake is found to be 19.42 meters.

Explanation:

To determine the depth of the lake, we first need to calculate how long it takes for the ball to reach the bottom. Since the ball sinks with a constant velocity, the time it takes to reach the bottom is the same as the time it takes to fall from the diving board. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken, we can solve for t: 5 = 1/2 * 9.8 * t^2. Rearranging the equation, we have t^2 = 5 / (1/2 * 9.8), which simplifies to t^2 = 1.02. Taking the square root of both sides, we find t = 1.01 s. Since the ball reaches the bottom after 3.0 s, we subtract the time taken to fall from the diving board to get the time taken to sink to the bottom: 3.0 s - 1.01 s = 1.99 s.

Next, we calculate the distance fallen from the diving board to the bottom of the lake using the equation d = 1/2gt^2. The time taken is 1.99 s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values, we get d = 1/2 * 9.8 * (1.99)^2. Simplifying the equation, we find d = 19.42 m. Therefore, the depth of the lake is 19.42 meters.

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

Answer:

1) Vx=0.8 m/s

2) V=0.94339 m/s

Explanation:

We know that the speed to cross the river is 0.5 m/s. This is our y axis. Vy

And we don't know the speed of the river. This is in our x axis. Vx

Since we know that the shore is 25m away, and we have a 0.5m/s of speed.

We can find with y=v.t

Since we know y=25m and v=0.5m/s

We can find that 50 seconds is the time we take to cross the river.

Now we need to calculate the velocity of the river, for that we use the same equation, x=v.t where x is 40m at downstream, and t now we know is 50 seconds.

We can find that v of the river, or Vx is 0.8 m/s.

With the two components of the velocity we use this equation to calculate the module or the velocity of the swimmer respect a fix point on the ground.

[tex]V=\sqrt{(Vx)^{2}+(Vy)^{2}  }[/tex]

We replace the values of Vx and Vy and we find. V=0.94339 m/s.

The river current is flowing at a speed of 0.8 m/s concerning the ground, and the swimmer's speed concerning a friend on the ground is approximately 0.94 m/s.

An athlete swims across a 25-m-wide river with a speed of 0.5 m/s perpendicular to the water current. The swimmer is carried 40 m downstream, indicating the presence of a river current. To find the speed of the river current, we use the Pythagorean theorem since the swimmer's motion relative to the river and the river's motion relative to the ground are perpendicular to each other.

The time taken to cross the river is the width of the river divided by the swimmer's speed concerning the water:
Time = 25 m / 0.5 m/s = 50 s. Now, the speed of the river current can be calculated using the downstream distance covered (40 m) and time (50 s): Speed of river = Distance downstream / Time = 40 m / 50 s = 0.8 m/s.

To calculate the swimmer's speed relative to the ground, we consider both the swimmer's speed across the river and the river's current speed. Using the Pythagorean theorem:
Swimmer's speed relative to the ground = √((0.5 m/s)^2 + (0.8 m/s)^2) = √(0.25 + 0.64) m/s = √(0.89) m/s ≈ 0.94 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]

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:

Half life of the sample, [tex]t_{\frac{1}{2}} = 12.15 min[/tex]

Given:

Initial amount, N = 1679

Final count of amount, N' = 1336

Time elapsed, t = 4 min = 240 s

Solution:

Now, To calculate the half life, using the relation:

[tex]N' = N(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]

Now, substituting the given values in the above mentioned formula:

[tex]\frac{1336}{1679} = (\frac{1}{2})^{\frac{4}{t_{\frac{1}{2}}}}[/tex]

[tex]0.796 = 0.5^{\frac{4}{t_{\frac{1}{2}}}}[/tex]

Taking log on both the sides:

ln(0.796) = \frac{4}{t_{\frac{1}{2}}}ln(0.5)

[tex]0.329 = 4t_{\frac{1}{2}}[/tex]

[tex]t_{\frac{1}{2}} = 12.15 min[/tex]

The half-life of a substance, in nuclear physics, is the time it takes for half of the original substance to decay. The half-life for the unknown sample in the question can be calculated using the radioactive decay formula. The given values of the initial and final counts, along with the time taken, are plugged into this formula.

The subject of this question is related to the concept of half-life in Physics, particularly Nuclear Physics. In nuclear physics, the half-life of a substance refers to the time taken for half the original elements to decay. To calculate the half-life of the unknown sample, we make use of the radioactive decay formula which says A = A0e^(-0.693/T1/2)T, where A is the final amount, A0 is the initial amount, T is the time elapsed, and T1/2 is the half-life we're trying to find.

In the given case, A0 = 1679 counts (the initial reading), A = 1336 counts (the reading after 4 minutes) and T = 4 minutes. Putting these into the equation, we would solve for T1/2, the half-life.

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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.

Answer:

Option 2. and 3.

Explanation:

Anode heel effect or simply heel effect is observed in X-ray tubes.

It is the change in the intensity of the emitted X-rays from the anode and this intensity is direction dependent and depends on the direction of the emitted X-rays along the axis of anode-cathode.

This effect results from the absorption of X-rays prior to leaving the anode, their production source.

The absorption of photons is more in the anode heel than the toe of the anode which results in the anode heel effect.

Explanation:

From Newton's second law:

F = ma

Given that m = 4 kg and a = 8 m/s²:

F = (4 kg) (8 m/s²)

F = 32 N

If m is reduced to 1 kg and F stays at 32 N:

32 N = (1 kg) a

a = 32 m/s²

So the acceleration increases by a factor of 4.

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.

Answer:

The false statement is a. ΔErxn is a measure of heat.

Explanation:

ΔErxn represents the energy change in a chemical reaction. However, in real processes the energy change is not only a measure of heat but also other forms like work and potential and kinetic energy.

Enthalpy is actually defined like the sum of a system's internal energy and the product of pressure and volume, so b is correct.

ΔHrxn is also the heat of reaction, which is also true

The last ones are also true

The false statement is a. ΔErxn is a measure of heat. ΔErxn refers to the change in energy in a reaction, not specifically heat.

The false statement is a. ΔErxn is a measure of heat. This statement is not correct because ΔErxn refers to the change in energy in a reaction, which can be measured in different ways, such as heat, light, or electric potential. However, it is not solely a measure of heat.

Just to clarify the other options, b. Enthalpy is indeed the sum of a system's internal energy and the product of its pressure and volume, c. ΔHrxn is indeed the heat of reaction, d. An exothermic reaction does give off heat to the surroundings, and e. Endothermic reactions do have a positive ΔH, which represents the amount of energy needed for the reaction to occur.

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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.

The magnitude of the average acceleration during the collision is 368.41 m/s^2. When this value is expressed in terms of 'g's, the acceleration becomes 37.6g.

Part A: The magnitude of the average acceleration of the driver during the collision can be found using one of the equations of motion: a = v2 / 2d where v is the final velocity, and d is the distance traveled. Given that the car was traveling at 93 km/hr (or 25.83 m/s after conversion) and the driver came to rest after 0.90 m, the equation becomes a = (25.832)/ (2*0.90) which equals 368.41 m/s2.

Part B: To express the answer in terms of "g's", we divide the acceleration by the acceleration due to gravity (9.80 m/s2). Hence, the acceleration in g's = 368.41 / 9.80 = 37.6g

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Part A: Average acceleration is 370.6 m/s².

Part B: Acceleration is 38 g's.

We start by converting the car's speed from kilometers per hour to meters per second. The conversion factor is 1 km/h equals 0.27778 m/s. Thus, 93 km/h equals 93 multiplied by 0.27778, which equals approximately 25.83334 m/s.

Part A : Next, we apply the kinematic equation to find the acceleration:

Final velocity (v) equals initial velocity (u) plus acceleration (a) multiplied by time (t), where the driver comes to rest, so v equals 0. The displacement (s) is given as 0.90 m. The formula rearranges to:

a = (v - u) / t.

We can also use another kinematic equation:

v² = u² + 2as.

Substituting the known values, we get:

0 = (25.83334)² + 2a(0.90).

Solving for a gives us:

a = - (25.83334)² / (2 * 0.90).

Calculating this yields:

a = - (667.92856) / 1.80,

a = -370.6 m/s².

The magnitude is 370.6 m/s².

For Part B, to express the acceleration in terms of g's:

a in g's = 370.6 m/s² / 9.80 m/s²

= 37.8 g.

Answer:

Wavelnght

Explanation:

He says undesired wavelengths can be eliminated with the new sound mixer. Musical sounds are spread over a medium through waves of different characteristics, one of them is wavelenght, that is the distance between begining and the end of an oscilation, normally measured on small distance units such as milimeters or even less. Sound may be higher or deeper depending on its wavelenght, so this sound mixer is able to filter all the sound  the user does not want to include in his work, for example, noises from enviroment. Now the producer can make music focused on high notes, like the ones from violins, or focused on low ones like the ones from a bass.

Answer:

[tex]a = 0.1137 m/s^2[/tex]

Explanation:

Let Vc be the velocity of the car and Vm the velocity of the motorcycle. If we convert their given values, we get:

Vc = 87 km/h * 1000m / 1km * 1h / 3600s = 24.17m/s

Vm = 95 km/h * 1000m / 1km * 1h / 3600s = 26.38m/s

Since their positions are equal after 17s we can stablish that:

[tex]Xc = d + Vc*t  = Xm = Vm*t + \frac{a*t^2}{2}[/tex]

Where d is the initial separation distance of 54m. Solving for a, we get:

[tex]a = \frac{d+Vc*t-Vm*t}{t^2}*2[/tex]   Repacing the values:

[tex]a = 0.1137 m/s^2[/tex]

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:

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:

volume of water in the kettle, V = [tex]774 cm^{3}[/tex]

Given:

Power output of burner, P = 2.0 kW = 2000 W

Mass of kettle, m = 810 g = 0.81 kg

Temperature of water in the kettle, T = [tex]20^{\circ}C[/tex]

Time taken by water to boil, t = 2.4 min = 144 s

Temperaturre at boiling, T' = [tex]100^{\circ}C[/tex]

Solution:

Now, we know that:

Iron's specific heat capacity, [tex]c = 0.45 J/g ^{\circ}C[/tex]

Water's specific heat capacity, [tex]c' = 4.18 J/g ^{\circ}C[/tex]

Now,

Total heat, q = Pt

q = [tex]2000\times 144 = 288 kJ[/tex]

Now,

q = (mc +m'c')(T' - T)

[tex]288\times 10^{3} = (0.81\times 0.45 + m'\times 4.18)(100^{\circ} - 20^{\circ})[/tex]

Solving the above eqn m', we get:

m' = 774 g

Now, the volume of water in the kettle, V:

[tex]V = \frac{m'}{\rho}[/tex]

where

[tex]\rho = density of water = 1 g/cm^{3}[/tex]

Now,

[tex]V = \frac{774}{1}[/tex]

Volume, V = [tex]774 cm^{3}[/tex]

At their highest point, grasshoppers A and B have the same mechanical and gravitational potential energy. Grasshopper B has a non-zero total velocity due to its horizontal motion at its highest point, while grasshopper A does not. They both reach the same height, so they have the same gravitational potential energy.

To answer your question about the two identical grasshoppers that jump into the air with the same initial speed, experiencing no air resistance: At their highest point, grasshopper A going straight up and grasshopper B going up at a 66° angle, we need to understand some basic concepts of physics first.

At the highest point of their jump, the vertical component of their velocity becomes zero. There is no kinetic energy associated with vertical motion for both of them.

Therefore, the answer to your multiple-choice question is:

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Answer:

Part (a) a + b = (8.0i - 2.0j) and [tex]\theta = 14.03^o[/tex] from the x-axis

Part (b) b - a = (2.0i - 6.0j) and [tex]\theta = -71.06^o[/tex] from the x- axis

Explanation:

Given,

From the addition of the two vectors,

[tex]\vec{a}\ +\ \vec{b}\ =\ (3.0i\ +\ 4.0j)\ +\ (5.0i\ -\ 2.0j)\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ (3.0\ +\ 5.0)i\ +\(4.0\ -\ 2.0)j\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ 8.0i\ +\ 2.0j[/tex]

Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).

[tex]\therefore Tan\theta\ =\ \dfrac{2.0}{8.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{2.0}{8.0}\ \right )\\\Rightarrow \theta\ =\ 14.03^o[/tex]

Part (b)

From the subtraction of the two vectors,

[tex]\vec{b}\ -\ \vec{a}\ =\ (5.0i\ -\ 2.0j)\ -\ (3.0i\ +\ 4.0j)\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ (5.0\ +\ 3.0)i\ +\(-2.0\ -\ 4.0)j\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ 2.0i\ -\ 6.0j[/tex]

Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).

[tex]\therefore Tan\theta\ =\ \dfrac{-6.0}{2.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{-6.0}{2.0}\ \right )\\\Rightarrow \theta\ =\ -71.06^o[/tex]