The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 90I I2 + I + 9 where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?

Answer: Ions are formed when atoms from group 1/3/2/17 gains or loses a small number of electrons.

Explanation: Let's take Na(Sodium) and Cl(chlorine), their atomic number are consecutively 11 and 17. Sodium's closest noble element is Neon(10) and Cl's is Argon(18). Now all atoms want to gain stability like the noble gases, they want their outer shell to be filled so that they don't have to wander around with other atoms. So to have stability Na will lose one of its electrons and Cl will take in one. Thus they consecutively will become plus charged and minus charged as the balance between their proton and electron numbers are gone

An ion is formed when a neutral atom gains or loses an electron.

A neutral atom has the same number of protons (+) and electrons (-).

If it gains another electron, it has too many negative charges, so it's a negative ion.

If it loses an electron, it has too many positive charges, so it's a positive ion.

=================================

What if it gains or loses a proton ?

That's a great question.  I'm glad you asked.

Under normal circumstances, this hardly ever happens.  But if it did ...

- If it somehow lost a proton from its nucleus, first of all, it would change the number of protons in the nucleus, so the atom would immediately become an atom of a different element.

- If it didn't lose an electron at the same time, then it would have too many electrons for the number of protons in the nucleus, so it would be a negative ion of the new element.

-  If it somehow added a proton to its nucleus, it would change the number of protons in the nucleus, so the atom would  become an atom of a different element.  If it didn't add another  electron at the same time, then it would have too few electrons for the number of protons in the nucleus, so it would be a  positive ion of the new element.

Answer:

s = 23.72 m

v = 21.56 m/s²  

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

   u = 0 m/s

   g = 9.8 m/s²

   s = 0 + 0.5 × 9.8 × 2.2²

  s = 23.72 m

b) impact velocity

      v = √(2gh)

      v = √(2× 9.8 × 23.72)    

      v =  √464.912

      v = 21.56 m/s²  

Answer:

A. d ≈ 23.72 m.

B. v ≈ 21.56 m/s.

Explanation:

d = height

v = final velocity

v0 = Initial velocity

g = gravity

t = time

A. 

Given that the time is 2.2 seconds and the initial velocity is 0, we can replace the values in to the formula.

d = v0 * t + 0.5 * g * t^2

d = 0 * 2.2 + 0.5 * 9.8 * 2.2^2

d = 0 + 23.716

d ≈ 23.72 m.

B.

​

​Sincec the initial speed is 0 m/s,

v^2 = 2 * g * d

v = sqrt(2 * g * d)

v = sqrt(2 * 9.8 * 23.72)

v = sqrt(464.912)

v ≈ 21.56 m/s

Answer:

D) 0.33 m/s² to the right

Explanation:

Apply Newton's second law.  Take right to be positive and left to be negative.

∑F = ma

20 N − 5 N = (45 kg) a

a = 0.33 m/s²

The couch accelerates at 0.33 m/s² to the right.

Answer: acceleration = 0.33m/s² to the right.

Explanation: when two forces act in opposite direction along same axis thier resultant direction is always with that force of higher magnitude in this case 20N

Resultant force F = -5N + 20N = 15N to the right.

But

F = Mass * acceleration

acceleration = F/ mass

= 15N/45kg

= 0.33N/kg

Which is also 0.33m/s² to the right.

Answer:

Determining planetary laws of motion

Explanation

Kepler worked for Tycho Brahe who was recording planetary motion for 20 years. After Brahe's demise Kepler inherited those records of planetary motion. After analyzing those records he put forth the three laws of planetary motion.

Law of Orbits: Every planet in our solar system move in elliptical orbit with sun at one focus It is also called the law of Ellipses

Law of Areas: The imaginary line drawn between the planet and sun will sweep out equal areas in equal period

Law of periods: The square of the time period of the planet is directly proportional to the cube of the semi major axis of the orbit. It is also called the law of Harmonies

Johannes Kepler is known for determining the planetary laws of motion in astronomy. Therefore option 2 is correct.

"determining planetary laws of motion," is the correct answer. Johannes Kepler was a German astronomer and mathematician who made significant contributions to our understanding of the motion of planets.

His three laws of planetary motion, known as Kepler's laws, revolutionized the field of astronomy and laid the foundation for Isaac Newton's later work on universal gravitation.

Kepler's first law, known as the law of ellipses, states that planets orbit the Sun in elliptical paths, with the Sun at one of the foci of the ellipse. This challenged the prevailing notion that planetary orbits were perfect circles.

Know more about Kepler's first law:

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Answer:

-99.1 V

Explanation:

For a free electron accelerated from rest, the final kinetic energy of the electron is equal to the change in electric potential energy of the electron:

[tex]K=\Delta U\\\frac{1}{2}mv^2 = q\Delta V[/tex]

where

[tex]m=9.11\cdot 10^{-31}kg[/tex] is the mass of the electron

[tex]v=5.9\cdot 10^6 m/s[/tex] is the final speed

[tex]q=-1.6\cdot 10^{-19} C[/tex] is the charge of the electron

[tex]\Delta V[/tex] is the potential difference

Solving for [tex]\Delta V[/tex], we find

[tex]\Delta V=\frac{mv^2}{2q}=\frac{(9.11\cdot 10^{-31})(5.9\cdot 10^6)^2}{2(-1.6\cdot 10^{-19})}=-99.1 V[/tex]

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus.

Answer is A.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus. Option a is correct.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus. Young's modulus is a measure of the material's stiffness or rigidity. The larger the Young's modulus, the more stress is required to stretch the material to the same extent. For example, steel has a high Young's modulus, while rubber has a low Young's modulus.

Young’s Modulus and Material Rigidity

Young’s modulus, also known as the elastic modulus, is a measure of the stiffness of a material. It quantifies how much a material will deform (i.e., stretch or compress) under a given amount of force. A very rigid material is one that exhibits minimal deformation when subjected to large forces, and this characteristic is directly related to the value of Young’s modulus.

Young’s Modulus and Material Behavior

When a material is subjected to an external force, it deforms in response to that force. This deformation can take the form of stretching (tensile deformation) or compression (compressive deformation). Young’s modulus specifically measures the material’s response to tensile or compressive forces. A high value of Young’s modulus indicates that the material experiences minimal elongation or compression when subjected to these forces, signifying its rigidity.

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The IKAROS spacecraft, propelled by the force of light from the sun, would have its speed increased by 30 m/s after 3 months of flight under the assumption that this is the only force acting on the 290 kg spacecraft.

The increase in speed of the IKAROS spacecraft can be calculated using the formula force = mass x acceleration. The force exerted by the sunlight is given as 1.12 mN or 0.00112 N. The mass of the spacecraft is given as 290 kg. So, the acceleration can be calculated as force/mass which is (0.00112 N)/(290 kg) = 3.86 x 10⁻⁶ m/s². This acceleration is continuous, so over time it imparts a speed increase to the spacecraft.

The time over which the acceleration is acting is 3 months or 3 x 30 x 24 x 60 x 60 seconds = 7.776 x 10⁶ s. Using the formula speed = acceleration x time, the increase in speed can be calculated as (3.86 x 10⁻⁶ m/s²) x (7.776 x 10⁶ s) = 30 m/s. So, the speed of the spacecraft would increase by 30 m/s after 3 months of flight assuming this force is the only one acting on it.

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The speed of the spacecraft would increase by 100800 m/s after 3.0 months of flight.

To calculate the increase in speed of the spacecraft after 3.0 months of flight, we can use the equation:

F = m * a

Where F is the force, m is the mass, and a is the acceleration. Rearranging the equation, we get:

a = F / m

Plugging in the force value of 1.12 mN and the mass of the spacecraft which is 290 kg, we can calculate the acceleration:

a = 1.12 mN / (290 kg)

Next, we need to convert the time of 3.0 months into seconds. Since there are 30 days in each month, and each day has 24 hours and 60 minutes, we can calculate:

3.0 months = 3.0 * 30 days * 24 hours * 60 minutes * 60 seconds

After converting the time into seconds, we can use the equation:

v = u + at

Where v is the final velocity, u is the initial velocity (assumed to be 0 since the spacecraft is starting from rest), a is the acceleration, and t is the time. Rearranging the equation, we get:

v = at

Plugging in the calculated acceleration and the converted time, we can calculate the final velocity:

v = (1.12 mN / (290 kg)) * (3.0 * 30 days * 24 hours * 60 minutes * 60 seconds)

Simplifying the calculation, we get:

v = 100800 m/s

Therefore, the speed of the spacecraft would increase by 100800 m/s after 3.0 months of flight, assuming the force from the solar sails is the only force acting on the spacecraft.

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Answer:

91.5 m/s

Explanation:

m = mass of clay = 12 g = 0.012 kg

M = mass of wooden block = 100 g = 0.1 kg

d = distance traveled by the combination before coming to rest = 7.5 m

μ = Coefficient of friction = 0.65

V = speed of the combination of clay and lock just after collision

V' = final speed of the combination after coming to rest = 0 m/s

acceleration caused due to friction is given as

a = - μ g

a = - (0.65) (9.8)

a = - 6.37 m/s²

Using the kinematics equation

V'² = V² + 2 a d

0² = V² + 2(- 6.37) (7.5)

V = 9.8 m/s²

v = speed of clay just before collision

Using conservation of momentum

m v = (m + M) V

(0.012) v = (0.012 + 0.100) (9.8)

v = 91.5 m/s

To find the speed of the clay immediately before the impact, we can use the principle of conservation of momentum and the equation for friction. By solving the two equations simultaneously, we can determine the speed.

To find the speed of the clay immediately before the impact, we can use the principle of conservation of momentum. Since there is no external force acting on the clay-block system in the horizontal direction, the total momentum before the impact is equal to the total momentum after the impact.

Before the impact, the clay has a mass of 12 g and a velocity of v. The wooden block has a mass of 100 g and is initially at rest. After the impact, the clay and block stick together and move with a common velocity.

Using the conservation of momentum:

(mass of clay) × (velocity of clay before impact) + (mass of block) × (velocity of block before impact) = (mass of clay and block) × (velocity of clay and block after impact)

(0.012 kg) × v + (0.100 kg) × 0 = (0.112 kg) × v'

Then substituting the given distance the block slides before coming to rest and the coefficient of friction into the equation:

μ × (force of friction) × (distance) = (0.112 kg) × (final velocity after impact)² / 2

By substituting the given values and solving the two equations simultaneously, we can find the speed of the clay immediately before the impact.

Therefore, the speed of the clay immediately before the impact is approximately 2.89 m/s.

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Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

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

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

[tex]u^2-0^2 = 2\times9.8\tiimes1.08[/tex]

or

[tex]u = \sqrt{21.168}[/tex]

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed ([tex]v_f[/tex]) = v = 4.6 m/s

Then conservation of energy says  

[tex]\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2[/tex]  

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s  

u₂ = velocity of the basketball before collision= 4.6 m/s  

v₁ =  velocity of the tennis ball after collision  

v₂ = velocity of the basketball  after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

[tex]v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2[/tex]

substituting the values in the above equation we get

[tex]v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6[/tex]

or

[tex]v = -8.565-3.965[/tex]

or

[tex]v = -12.53m/s[/tex]

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

[tex]K.E = \frac{1}{2}mv^2[/tex]

or

[tex]K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2[/tex]

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

The magnitude of the downward velocity with which the basketball reaches the ground is 4.51 m/s. The height to which the tennis ball rebounds after an elastic collision with the basketball can be found by solving for the velocity of the tennis ball after the collision and using the equation h = (v2')² / (2g).

To find the magnitude of the downward velocity with which the basketball reaches the ground, we can use the equation v = gt, where g is the acceleration due to gravity. Since both the basketball and tennis ball are released from rest and fall through the same distance, they will reach the ground at the same time. Therefore, the time it takes for the basketball to reach the ground is the same as the time it takes for the tennis ball to fall and bounce back up:

t = sqrt(2h/g)

where h is the distance the balls fall. We can substitute the given values into the equation to find the time:

t = sqrt(2 * 1.08m / 9.8m/s²) = 0.46 s

Therefore, the magnitude of the downward velocity with which the basketball reaches the ground is given by v = gt:

v = 9.8m/s² * 0.46s = 4.51 m/s

For part (b), to find the height to which the tennis ball rebounds after an elastic collision with the basketball, we can use the conservation of momentum and the conservation of kinetic energy. Since the collision is elastic, the total momentum and the total kinetic energy of the two balls before the collision are equal to the total momentum and total kinetic energy after the collision:

m1v1 + m2v2 = m1v1' + m2v2'

1/2m1v1² + 1/2m2v2² = 1/2m1v1'² + 1/2m2v2'²

We can substitute the given values into the equations and solve for v2', the velocity of the tennis ball after the collision:

0.044kg * 0m/s + 0.594kg * 4.51m/s = 0.044kg * v1' + 0.594kg * v2'

1/2 * 0.044kg * 0m/s² + 1/2 * 0.594kg * (4.51m/s)² = 1/2 * 0.044kg * v1'² + 1/2 * 0.594kg * v2'²

From these equations, we can solve for v2', the velocity of the tennis ball after the collision. The height to which the tennis ball rebounds can be found using the equation:

h = (v2')² / (2g)

We can substitute the calculated value of v2' into the equation to find the height:

h = (v2')² / (2 * 9.8m/s²)

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1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

[tex]W=K_f - K_i[/tex]

Since the cart was initially at rest, [tex]K_i = 0[/tex], so

[tex]W=K_f = \frac{1}{2}mv^2[/tex] (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, [tex]F_g = 99.5 N[/tex]:

[tex]m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg[/tex]

So solving eq.(1) for v, we find the final speed of the cart:

[tex]v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s[/tex]

2) [tex]2.51\cdot 10^7 J[/tex]

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

[tex]F=4.28 \cdot 10^5 N[/tex]

[tex]d=586 m[/tex]

So the work done is

[tex]W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J[/tex]

3)  [tex]2.51\cdot 10^7 J[/tex]

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

[tex]W=\Delta K = K_f - K_i[/tex]

where

W is the work done

[tex]\Delta K[/tex] is the change in kinetic energy

Therefore, the change in kinetic energy is

[tex]\Delta K = W = 2.51\cdot 10^7 J[/tex]

4) 37.2 m/s

According to the work-energy theorem,

[tex]W=\Delta K = K_f - K_i[/tex]

where

[tex]K_f[/tex] is the final kinetic energy of the train

[tex]K_i = 0[/tex] is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

[tex]W=K_f = \frac{1}{2}mv^2[/tex]

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

[tex]v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s[/tex]

Answer:

15.8 cm

Explanation:

In 20.8 seconds, the number of vibrations = 42

In 1 second, the number of vibrations = 42 / 20.8 = 2.019

Frequency is defined as the number of vibrations in one second.

f = 2.019 Hz

Crest travel 371 cm in 11.6 s

v = 3.71 / 11.6 = 0.319 m / s

Use

v = f x λ

λ = v / f = 0.319 / 2.019 = 0.158 m = 15.8 cm

Answer:

[tex]p_2 = 1.76 atm[/tex]

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

[tex]\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}[/tex]

Putting the values

[tex]\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}[/tex]

[tex]9.733*10^{-3} = \frac{p_2 *1.68}{304}[/tex]

[tex]9.733*10^{-3}*304 = p_2*1.68[/tex]

[tex]\frac{9.733*10^{-3}*304}{1.68} =p_2[/tex]

[tex]p_2= 1.76 atm[/tex]

Answer:

The new pressure inside the container is 394.48 Pa.

Explanation:

Given that,

Pressure = 550.0 kPa

Temperature = 25.0°C

Initial volume = 5.20 L

Final volume = 7.25 L

We need to calculate the new pressure inside the container

Using Boyle's law

PV = constant

[tex]P_{1}V_{1}=P_{2}V_{2}[/tex]

Where, P₁ = Pressure

V₁ = Initial volume

V₂ = Initial volume

Put the volume into the formula

[tex]550\times5.20=P_{2}\times7.25[/tex]

[tex]P_{2}=\dfrac{550\times5.20}{7.25}[/tex]

[tex]P_{2}=394.48\ Pa[/tex]

Hence, The new pressure inside the container is 394.48 Pa.

Final answer:

The new pressure inside the container after expansion to 7.25 L at a constant temperature of 25.0 °C is approximately 544.28 kPa.

Explanation:

The subject of the question at hand involves the behavior of gases and is related to the laws governing gas behavior, specifically Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when temperature and the amount of gas are held constant.

To generate an accurate answer, we must assume that the temperature remains constant at 25.0 °C, as per the question's conditions and use the values given to apply Boyle's Law.

Initial conditions are given as:
P1 = 550.0 kPa
V1 = 5.20 L
The final volume V2 is 7.25 L.
By Boyle's Law (P1V1 = P2V2) we can calculate the new pressure (P2) upon expansion:

P2 = (P1 × V1) / V2
P2 = (550.0 kPa × 5.20 L) / 7.25 L
P2 = 3948 kPa ÷ 7.25 L
P2 ≈ 544.28 kPa

Thus, the new pressure inside the container after expansion to 7.25 L at a constant temperature of 25.0 °C is approximately 544.28 kPa.

Answer: Because its period of rotation is 243.01 days (retrograde)

Explanation:

In comparison, Earth rotates once in about 24 hours with respect to the Sun, but once every 23 hours, 56 minutes, and 4 seconds with respect to other, distant, stars. Earth's rotation is slowing slightly with time, so in the past a day was shorter.

Venus's weak magnetic field can be attributed to its slow retrograde rotation period, lack of a convective liquid metal core, and lack of significant tectonic activity. These conditions are all contrary to those that generally contribute to generating a strong magnetic field.

The weak magnetic field of Venus could be linked to its retrograde rotation period and lack of a convective liquid metal core. Unlike planets with strong magnetic fields, Venus has a retrograde rotation period of 243 days, compared to Earth's 24-hour rotation period. The rotation period plays a key role in generating a planet's magnetic field. A faster rotation aids the creation of a dynamo effect, generating a stronger magnetic field.

Furthermore, the composition of Venus' core could also contribute to its weak magnetic field. A planet's magnetic field is typically caused by the movement of liquid metal within its core, which serves as a good conductor of electricity. It seems that Venus lacks this convective liquid metal core, thereby lacking the necessary conditions for a strong magnetic field.

Additionally, Venus may lack tectonic activity that is known to influence the generation of a magnetic field. While the surface of Venus has been modified by tectonics driven by mantle convection, it does not exhibit the same kind of plate tectonics as Earth.

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Answer:

A-2

B-1

Explanation:

Cylindrical projection is a type of map projection that projects the earth's surface on a cylindrical tangent. It is useful in projecting the equatorial region and the near-polar region but it gets distorted at the poles. This distortion occurs in terms of both the scale and the shape.

Planar projection is a type of map projection in which the three-dimensional earth's surface is projected on a flat surface. It is more efficient in mapping the polar regions. The North pole or the South Pole in this type of map are located at the center of the map.

Thus, the correct answers are being matched.

Answer:

The ball is in the air for 3.628 sec

Explanation:

We can find speed in the "y" direction when the ball lands with this equation:

[tex]V_{fy} ^{2} = V_{oy} ^{2} + 2*g*(y_{f} - y_{o})[/tex]

Where:   [tex]V_{oy} = 0 m/sec[/tex]   because the ball is thrown horizontally

              [tex]g = - 9.8 m/sec^{2}[/tex]   because gravity is a vector that points down in the "y" direction

              [tex]y_{o} = 64.5 m[/tex]

              [tex]y_{f} = 0 m[/tex]   because the ball lands on the ground

Then:      [tex]V_{fy} ^{2} = (0 m/sec)^{2} + 2(-9.8 m/sec^{2})(0 m - 64.5 m)[/tex]

              [tex]V_{fy} ^{2} = 2(- 9.8 m/sec^{2})(- 64.5 m)[/tex]

              [tex]V_{fy} ^{2} = 1264.2 m^{2}/sec^{2}[/tex]

              [tex]V_{fy} = \sqrt{1264.2 m^{2}/sec^{2}}[/tex]

              [tex]V_{fy} = - 35.5556 m/sec[/tex]   because the ball is falling and that means speed in the "y" direction is a vector that points down

Now we can calculate the time with next equation:

[tex]V_{fy} = V_{oy} + g*t[/tex]

[tex]V_{fy} - V_{oy} = g*t[/tex]

[tex]t = \frac{V_{fy} - V_{oy}}{g}[/tex]

Then:       [tex]t = \frac{-35.5556 m/sec - 0 m/sec}{- 9.8 m/sec^{2}}[/tex]

Finally      t = 3.628 sec

The ball is in the air for approximately 3.63 seconds.

Separate Motions: We can analyze the horizontal and vertical motions of the ball independently since air resistance is neglected.

Horizontal Motion: This is a constant velocity motion because there's no horizontal acceleration (ignoring air resistance).

Vertical Motion: This is a free-fall motion with constant acceleration due to gravity (g ≈ 9.81 m/s²).

Horizontal Distance (x):

Given: x = 103.8 m (horizontal distance from the building base)

Time (t): We need to find the total time (t) the ball spends in the air.

Vertical Motion:

Unknown: Time (t) for the vertical motion (which is the same as the total time in the air)

Known: Vertical distance (y) = -64.5 m (negative since downward) and acceleration due to gravity (g)

Vertical Kinematics Equation: We can use the kinematic equation for constant acceleration to solve for the time (t) for the vertical motion:

y = 1/2 * g * t² (We've plugged in the negative value for y since it's moving downward)

Solve for Time (t) in Vertical Motion:

Rearrange the equation to isolate t: t² = 2 * y / g

Substitute the known values: t² = 2 * (-64.5 m) / (9.81 m/s²) ≈ 13.19 s²

Take the square root of both sides to find t (remember there can be positive and negative time solutions, but we're interested in the positive time for the ball to be in the air).

t ≈ ± 3.63 s (We discard the negative solution)

However, this time represents only the vertical motion.

Relating Horizontal and Vertical Motion: Since the ball is thrown horizontally, the horizontal time (t_horizontal) is the same as the vertical time (t) we just calculated (t_horizontal = t_vertical).

Therefore, the ball is in the air for approximately 3.63 seconds.

Answer:T=[tex]\frac{-26}{r}[/tex]+36

Explanation:

Given Temperature at r=1m is [tex]10^{\circ}C [/tex]

Temperature at r=2m is [tex]28^{\circ}C[/tex]

[tex]\frac{\mathrm{d^2} T}{\mathrm{d} r^2}+\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=0[/tex]

Let [tex]\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=S[/tex]

[tex]\frac{\mathrm{d^2} T}{\mathrm{d} r^2}=\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}[/tex]

therefore [tex]\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}+\frac{2}{r}[/tex]S=0

[tex]\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}=-\frac{-2S}{r}[/tex]

solving

[tex]r^2[/tex] S=constant

substitute S value

[tex]\frac{\mathrm{d}T}{\mathrm{d}r}[/tex]=[tex]\frac{c}{r^2}[/tex]

Solving it we get

T=[tex]\frac{-c}{r}+c_2[/tex]

Now using given condition

10=[tex]\frac{-c}{1}+c_2[/tex]

28=[tex]\frac{-c}{2}+c_2[/tex]

[tex]c_2[/tex]=36,c=26

putting c values

T=[tex]\frac{-26}{r}[/tex]+36

The expression for the temperature T(r) for the two spheres one inside other is given as,

[tex]T=-\dfrac{26}{r}+36[/tex]

Differential equation is the equation in which there is one or more number of unknown variable exist to find the rate of change of one variable with respect to other.

The radius of the sphere is 1 m and the temperature of the sphere is 10 degree Celsius. The sphere lies inside a concentric sphere with radius 2 m and temperature 28°C.

The differential equation which spheres satisfies is given as,

[tex]\dfrac{d^2T}{dr^2}+\dfrac{2}{r}\dfrac{dT}{dr}=0[/tex]

Let the above equation is equation one,

Suppose S = dT/dr, then

[tex]\dfrac{dS}{dt}=\dfrac{2}{r}\dfrac{d^2T}{dr^2}[/tex]

Put the values in the equation one as,

[tex]\dfrac{2}{r}\dfrac{dS}{dt}+\dfrac{-2}{r}S=0\\\dfrac{2}{r}\dfrac{dS}{dt}=-\dfrac{-2}{r}S\\r^2S=C[/tex]

Here, (C) is the constant value. Now put the value of S as,

[tex]r^2\dfrac{dT}{dr}=C\\\dfrac{dT}{dr}=\dfrac{C}{r^2}\\T=-\dfrac{C}{r}+C_2[/tex]

Put the value of radius as 1 m,

[tex]10=-\dfrac{C}{1}+C_2\\10=-C+C_2[/tex]

Similarly, for the radius 2 meters and temperature 28 degree Celsius,

[tex]28=-\dfrac{C}{2}+C_2[/tex]

On solving above equation, we get,

[tex]C=26\\C_2=36[/tex]

Put the values of constant for the required expression as,

[tex]T=-\dfrac{26}{r}+36[/tex]

Thus, the expression for the temperature T(r) for the two spheres one inside other is given as,

[tex]T=-\dfrac{26}{r}+36[/tex]

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Final answer:

The magnitude of the magnetic field at the origin when t = 0.15 μs can be found using the formula for the magnetic field produced by a moving charge.

Explanation:

The magnitude of the magnetic field at the origin when t = 0.15 μs can be found using the formula for the magnetic field produced by a moving charge:

B = (μ₀Iv)/(2πr)

Where B is the magnetic field, μ₀ is the permeability of free space, I is the current, v is the velocity of the charge, and r is the distance from the charge to the point where we want to find the magnetic field. In this case, the charge is moving parallel to the x-axis, so the current can be calculated using the formula:

I = (Qv)/L

Where Q is the charge and L is the length of the moving charge. Plugging in the values and solving the equations will give the magnitude of the magnetic field at x = 4 m when t = 0.15 μs.

Answer:

560 m

Explanation:

"A projectile is fired at time t = 0.0 s from point 0 at the edge of a cliff, with initial velocity components of v₀ₓ = 30 m/s and v₀ᵧ = 100 m/s.  The projectile rises, and then falls into the sea at point P. The time of flight of the projectile is 25 s.  Assume air resistance is negligible.  What is the height of the cliff?"

Use constant acceleration equation in the y direction.

y = y₀ + v₀ t + ½ gt²

0 = h + (100 m/s) (25 s) + ½ (-9.8 m/s²) (25 s)²

h = 560 m

The height of the cliff is 560 m.

To find the energy in joules equivalent to one jelly doughnut, multiply the Calorie value by 4.184. The number of stairs required to perform the same work as the energy in one jelly doughnut is calculated by dividing the work done to move up the stairs by the work done to move one stair height. To account for the body's efficiency in converting energy, divide the number of stairs calculated in part (b) by the efficiency.

(a) To find the number of joules of energy equivalent to one jelly doughnut, we need to convert the Calorie value to joules. One Calorie (capital C) is equal to 1000 calories (lowercase c), which is equal to 4.184 joules. Therefore, one jelly doughnut contains 625 calories × 4.184 joules/calorie = 2615 joules.

(b) The work done by climbing stairs is equal to the product of the force applied and the distance moved. The force can be calculated using the weight of the woman, which is equal to her mass multiplied by the acceleration due to gravity (66 kg × 9.8 m/s2). Dividing the work done to move up the stairs by the work done to move one stair height (15 cm) gives the number of stairs that the woman must climb. This is equal to 1764 joules ÷ (66 kg × 9.8 m/s2 × 0.15 m) = 181 stairs.

(c) Since the human body is only 26% efficient in converting chemical energy to mechanical energy, we need to divide the work calculated in part (b) by the efficiency to account for this loss. Therefore, the number of stairs the woman must climb to work off her breakfast is 181 stairs ÷ 0.26 = 696 stairs.

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Answer:

Explanation:

A) mass of displaced water = loss of mass in water = 63 - .0975 = 62.9025 kg

B) her volume = volume of displaced water

= mass of displaced water / density of water = 62.9025 / 1000 m³.

= 62.9025 x 10⁻³ m³

C) her average density = her mass / her volume = 63 / 62.9025 x 10⁻³

= 1.00155 x 10³ kg / m³

D) lung capacity = 1.75 L = 1.75 x 10⁻³ m³

buoyant force created by air in the lungs = 1.75 x 10⁻³ ( 1000- 1.29 ) =  1.74 kg

As it is more than 0.0975 kg , the apparent weight  in water so woman will float in water .

A. Mass of Water Displaced: ≈ 62.99 kg

B. Volume of Woman: ≈ 0.06299 m³

C. Average Density: ≈ 1000 kg/m³

D. Yes, she can float with lungs filled with air.

We can solve this problem using the concepts of buoyancy and density. Here's how to find the answers:

A. Mass of Water Displaced:

1. Buoyant Force: The difference between her mass in air and apparent mass underwater represents the buoyant force exerted by the water.

2. Buoyant Force = Weight in Air - Apparent Weight = (63 kg * 9.81 m/s²) - (0.0975 kg * 9.81 m/s²) (We can multiply mass by gravitational acceleration to convert it to weight, but for this problem, it cancels out)

3. Buoyant Force ≈ 618.27 N (Newtons)

Since buoyant force equals the weight of the water displaced, the mass of water displaced is:

Mass of Displaced Water = Buoyant Force / Gravitational Acceleration

Mass of Displaced Water ≈ 618.27 N / 9.81 m/s² ≈ 62.99 kg (rounded to two decimal places)

B. Volume of Woman:

1. Density of Water: We can assume the density of water to be 1000 kg/m³ (a standard value).

2. Volume of Displaced Water = Mass of Displaced Water / Density of Water

3. Volume ≈ 62.99 kg / 1000 kg/m³ ≈ 0.06299 m³ (rounded to five decimal places)

C. Average Density of Woman:

1. Woman's Volume (from part B): 0.06299 m³

2. Woman's Mass: 63 kg

3. Density = Mass / Volume

4. Average Density ≈ 63 kg / 0.06299 m³ ≈ 1000 kg/m³ (rounded to three decimal places)

D. Ability to Float:

1. Density of Air: 1.29 kg/m³

2. Volume of Air in Lungs: 1.75 L = 0.00175 m³ (convert liters to cubic meters)

3. When lungs are filled with air, her effective volume increases by the volume of air in her lungs.

4. Effective Volume with Air: 0.06299 m³ (woman's volume) + 0.00175 m³ (air in lungs) ≈ 0.06474 m³

5. When considering air in her lungs, her effective density becomes:

Effective Density = Mass / Effective Volume

Effective Density ≈ 63 kg / 0.06474 m³ ≈ 973.3 kg/m³ (rounded to one decimal place)

Since her effective density (with air in lungs) is lower than the density of water (1000 kg/m³), she would be able to float with her lungs filled with air.

Answers:

A. Mass of Water Displaced: ≈ 62.99 kg

B. Volume of Woman: ≈ 0.06299 m³

C. Average Density: ≈ 1000 kg/m³

D. Yes, she can float with lungs filled with air.

Answer:

100 rad/s

Explanation:

Angular speed = linear speed / radius

ω = v / r

ω = (5.0 m/s) / (0.050 m)

ω = 100 rad/s

The angular speed of the pulley when the string has a linear speed of 5.0 m/s is found using the formula v = rω. So the angular speed is 100 rad/s.

To find the angular speed of the pulley when the string has a linear speed of 5.0 m/s, we can use the relationship between linear speed (v), angular speed (ω), and radius (r) of the pulley. This relationship is given by the formula v = rω, where v is the linear speed, r is the radius of the pulley, and ω is the angular speed we want to find. Given that the radius r is 5.0 cm (which we need to convert to meters by dividing by 100, so r = 0.05 m), and v is 5.0 m/s, we rearrange the formula to solve for ω:

ω = v / r

ω = (5.0 m/s) / 0.05 m

ω = 100 rad/s

Therefore, the angular speed of the pulley is 100 radians per second (rad/s).

The current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

What is the current when t = 0.7 s?

Recall the relationship between current and charge:

Current (I) is the rate of change of charge (Q) over time. Mathematically, it's represented as:

I(t) = dQ(t)/dt

Differentiate the charge function:

Given the function for Q(t) = t³ - 2t² + 4t + 1, we can find the current I(t) by differentiating it with respect to time t:

I(t) = 3t² - 4t + 4

Calculate the current at t = 0.7 s:

Now, plug t = 0.7 into the expression for I(t):

I(0.7) = 3(0.7)² - 4(0.7) + 4

I(0.7) ≈ 2.67 C/s

Therefore, the current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

Answer:

a) 3.94 rad/s

b) 11.84 rad/s

Explanation:

Given:

Distance, s = 78 cm = 0.78 m

Now the time taken, t

we know

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

where,

s  = distance

u = initial speed

g = acceleration due to gravity

since it is a free fall. thus, u = 0

thus, we get

[tex]0.78 = 0\times t +\frac{1}{2}9.8\times t^2[/tex]

or

[tex]t=\sqrt{\frac{2\times 0.78}{9.8}}=0.398 s[/tex]

a) now, the smallest angle will be 1/4 of the revolution so as to fall on the butter side i.e 90° or ([tex]\frac{\pi}{2}[/tex])

also,

[tex]angular\ speed\ (\omega) = \frac{Change\ in\ angle}{time}[/tex]

thus, we have

[tex]angular\ speed\ (\omega) = \frac{\frac{\pi}{2}}{0.398} = 3.94rad/s[/tex]

b)now, the largest angle will be 3/4 of the revolution so as to fall on the butter side i.e 270° or ([tex]\frac{3\pi}{2}[/tex]) contributing to the largest angular speed

[tex]angular\ speed\ (\omega) = \frac{Change\ in\ angle}{time}[/tex]

thus, we have

[tex]angular\ speed\ (\omega) = \frac{\frac{3\pi}{2}}{0.398} =11.84rad/s[/tex]

All points in a rigid body move with the same velocity and acceleration if the rigid body is subjected to pure translational motion.

When a rigid body is subjected to pure translational motion, all points in the body move with the same velocity and acceleration. This means each point in the body experiences the same linear velocity and acceleration at any given instant.

Answer:

The mass of the puck is 0.166 kg

Explanation:

It is given that,

Speed of the ice hockey puck, u = 12 m/s

Impulse delivered by the hockey stick, J = 4 kg-/s

After delivering an impulse, the puck move off in the opposite direction with the same speed, v = -12 m/s

We need to find the mass of the puck. The impulse is equal to the change in momentum i.e.

[tex]J=m(v-u)[/tex]

m is the mass of the puck

[tex]4=m(-12-12)[/tex]

[tex]4=-24\ m[/tex]

[tex]m=\dfrac{1}{6}=0.166\ kg[/tex]

So, the mass of the puck is 0.166 kg. Hence, this is the required solution.

When the mass of the puck that moves with the same speed in opposite direction is equal to 0.1667 kg.

Impulse is a large force that is applied to an object for a very short period of time. It is given by the formula,

[tex]J =m(v-u) = F(\delta t)[/tex]

where v and u are the final and initial velocity, F is the force that is been applied, m is the mass of the object, t is the time for which the force is been applied.

We know that the impulse can be written as,

[tex]J = m(v-u)[/tex]

Given to us

J  =  4 kg⋅m/s,

u = 12 m/s

v = -12 m/s(opposite in direction with the same speed)

Substitute the values,

[tex]4 = m(-12-12)\\\\4 = m(-24) \\\\m = -0.1666\\\\m = 0.1667 \rm\ kg[/tex]

Since the mass of the punk can not be negative, therefore, the mass of the puck is 0.1667 kg.

Hence, when the mass of the puck that moves with the same speed in opposite direction is equal to 0.1667 kg.

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Answer:

CONCAVE

Explanation:

Convex lenses bend light rays so they come together at focus. Concave lenses spread light rays apart so they do not come together at a focus.

hope this answer help u!!

Answer:

Concave lens

Explanation:

Concave lenses are thicker at ends and thinner at middle. When parallel rays of light pass through the concave lens, they are refracted and spread out so that they appear to come from a common point known as principle focus. Due to this reason, Concave lens also called as diverging lens.

Final answer:

Using the principle of buoyancy and Archimedes' principle, we can determine the volume of the rock from the change in tension when submerged in water. Then, we use the volume of the rock to calculate the density of the unknown liquid where the tension is 11.4 N.

Explanation:

The problem described involves the principle of buoyancy and Archimedes' principle in particular, which states that the upward buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

When the rock is submerged in water and the tension in the string is 51.9 N, this is the weight of the rock in the air. When the rock is immersed in water and the tension becomes 31.6 N, the apparent weight is reduced due to the buoyant force of water. The difference in tension (51.9 N - 31.6 N = 20.3 N) is the weight of the water displaced, which can be converted into mass (20.3 N / 9.8 m/s2 = 2.07 kg). Knowing that the density of water is 1,000 kg/m3, we can calculate the volume of water displaced and thus the volume of the rock. When the rock is submerged in the unknown liquid and the tension is 11.4 N, a similar calculation for the buoyant force allows us to determine the density of the unknown liquid based on the known volume of the rock.

To solve for the density of the unknown liquid, we first need to determine the volume of the rock using the reduced tension in water (Archimedes' principle). Then we use the reduced tension in the unknown liquid to calculate the density of the unknown liquid. As we're given tensions rather than masses, we translate tension into weight (force) and use the equation for buoyant force (buoyant force = weight in air - apparent weight in fluid) to find the volume displaced. Since the rock displaces an equal volume of fluid, we can find the density of the unknown liquid using the buoyant force equation with the known volume of the rock.

A. The bullet's speed is [ ( M + m ) / m ] √ ( 2 μ g d )

B. The initial speed of the 9.0 g bullet is about 610 m/s

[tex]\texttt{ }[/tex]

Let's recall Impulse formula as follows:

[tex]\boxed {I = \Sigma F \times t}[/tex]

where:

I = impulse on the object ( kg m/s )

∑F = net force acting on object ( kg m /s² = Newton )

t = elapsed time ( s )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

mass of bullet = m = 9.0 g = 9.0 × 10⁻³ kg

mass of block = M = 12 kg

sliding distance = d = 5.4 cm = 5.4 × 10⁻² m

coefficient of kinetic friction = k = 0.20

Asked:

initial bullet's speed = u₁ = ?

Solution:

Firstly, we will use Conservation of Energy formula to find the speed of the block:

[tex]W = \Delta Ek[/tex]

[tex]fd = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu N d = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu (M + m)g d = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu g d = \frac{1}{2} v^2[/tex]

[tex]\boxed {v = \sqrt{2 \mu g d}}[/tex] → Equation A

[tex]\texttt{ }[/tex]

Next, we will use Conservation of Momentum formula to find the initial speed of the bullet:

[tex]\texttt{Total Momentum Before Collision = Total Momentum After Collision}[/tex]

[tex]m u_1 + M u_2 = ( m + M ) v[/tex]

[tex]m u_1 + M (0) = ( m + M ) v[/tex]

[tex]m u_1 = ( m + M ) v[/tex]

[tex]m u_1 = ( m + M ) \sqrt { 2\mu g d}[/tex] ← Equation A

[tex]\boxed {u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}}[/tex]

[tex]\texttt{ }[/tex]

[tex]u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}[/tex]

[tex]u_1 = \frac { 9.0 \times 10^{-3} + 12 }{ 9.0 \times 10^{-3} } \sqrt { 2 \times 0.20 \times 9.80 \times 5.4 \times 10^{-2}}[/tex]

[tex]\boxed{u_1 \approx 610 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

(a) The expression for the speed of the bullet is [tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]

(b) The speed of the bullet at the given parameters is 613.78 m/s.

The given parameters;

Apply the principle of conservation of linear momentum to determine the initial speed of the bullet;

[tex]m_1 u_1 + m_2 u_2 = V(m_1 + m_2)\\\\m_1 u_1 + 0 = V(m_1 + m_2)\\\\u_1 = \frac{V(m_1 + m_2)}{m_1}[/tex]

Apply the principle of work-energy theorem to determine the speed of the bullet-block system;

[tex]K.E - P.E = W_f\\\\\frac{1}{2} MV^2 - 0 = \mu_k (Mg)d\\\\V^2 = 2\mu_k gd\\\\V = \sqrt{2\mu_k gd[/tex]

The expression for the speed of the bullet is written as;

[tex]u_1 = \frac{V(m_1 + m_2)}{m_1} \\\\u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]

The speed of the bullet at the given parameters is calculated as follows;

[tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1} \\\\u_1 = \frac{(0.009 + 12) \sqrt{2\times 0.2 \times 9.8 \times 0.054} }{0.009} \\\\u_1 = 613.78 \ m/s[/tex]

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