Which is the best example of translational motion?

A. a soccer ball passed between two players.

B.a powered up rocket on the launch pad.

C.an ice skater spinning in place.

D.water molecules in a glass of water.

It will take 4.12 s for the flowerpot to fall to the ground.

From the question given above, the following data were obtained:

Height (h) = 85 m

NOTE: Acceleration due to gravity (g) = 10 m/s²

The time taken for the flowerpot to fall to the ground can be obtained as follow:

85 = ½ × 10 × t²

85 = 5 × t²

Divide both side by 5

[tex]t^{2} = \frac{85}{5}\\\\t^{2} = 17[/tex]

Take the square root of both side

[tex]t = \sqrt{17}[/tex]

Therefore, it will take 4.12 s for the flowerpot to fall to the ground.

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The maximum possible speed of the ball at the top of the loop is 4.50 m/s

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)

v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Centripetal Acceleration of circular motion could be calculated using following formula:

[tex]\large {\boxed {a_s = v^2 / R} }[/tex]

a = centripetal acceleration ( m/s² )

v = velocity ( m/s )

R = radius of circle ( m )

Let us now tackle the problem!

Given:

mass = m = 0.34 kg

length of string = R = 0.52 m

maximum tension = Tmax = 9.9 N

Unknown:

v = ?

Solution:

[tex]mg + T = ma[/tex]

[tex]mg + T = m\frac{v^2}{R}[/tex]

[tex]0.34 \times 9.8 + 9.9 = 0.34 \times \frac{v^2}{0.52}[/tex]

[tex]13.232 = \frac{0.34}{0.52} \times v^2[/tex]

[tex]v^2 = 20.2372[/tex]

[tex]\large {\boxed {v \approx 4.50 ~ m/s} }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

The maximum possible speed of the baseball at the top of the loop is approximately 3.17 m/s. This is calculated by using the maximum tension the string can support, and the gravitational force acting on the baseball.

To find the maximum possible speed of the baseball at the top of the loop without breaking the string, we need to consider the forces acting on the baseball. Two key forces are at play here: the gravitational force pulling the ball downward and the tension in the string that counteracts this pull. At the top of the loop, for minimum speed, the tension in the string can be zero because the gravitational force provides the necessary centripetal force. However, the question states that the string can only support a maximum tension (Tmax) before breaking which means we must find the speed where the tension does not exceed Tmax.

The maximum tension is the sum of the centripetal force needed to keep the ball moving in a circular path and the force due to gravity. Mathematically, this is expressed as Tmax = m * v^2 / l + m * g, where v is the velocity, m is the mass of the baseball, l is the length of the string, and g is the acceleration due to gravity (9.8 m/s^2).

Rearranging the formula to solve for v gives us v = sqrt((Tmax - m * g) * l / m). Plugging in the values Tmax = 9.9 N, m = 0.34 kg, l = 0.52 m, we get:

v = sqrt((9.9 N - (0.34 kg * 9.8 m/s^2) * 0.52 m) / 0.34 kg)

Calculating the above expression, we find the maximum velocity:

v = sqrt((9.9 - 3.332) * 0.52 / 0.34)

v = sqrt(6.568 * 0.52 / 0.34)

v = sqrt(3.4152 / 0.34)

v = sqrt(10.0447)

v ≈ 3.17 m/s

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Answer:

v =  3321.85 mph

Explanation:

Equivalences :

1 mile = 1609.34 m

1 hour = 3600 seconds

Data

v= 1485 m/s : speed of sound in water

Problem Development

To calculate the speed of sound in mph (mile / hour), we multiply by the conversion factors using the equivalences:

[tex]v= (1485 \frac{m}{s} )*(\frac{1mile}{1609.34 m} )*(\frac{3600s}{hour})[/tex]

We cancel the units in seconds (s) and meters (m) to get the answer in miles per hour (miles / hour or mph)

[tex]v=\frac{1485*3600}{1609.34} \frac{mile}{hour}[/tex]

v= 3321.85 mile/hour

v =  3321.85 mph

The speed of a satellite in a geosynchronous orbit is approximately 2.98 km/s, and the magnitude of the acceleration is approximately 1.92 x 10^-3 m/s^2.

To determine the speed of a satellite in a geosynchronous orbit, we can use the formula:

speed = 2 x π x radius / period

Given that the radius of the Earth is 6.37 * 10^6 m and the altitude of a geosynchronous orbit is 3.58 * 10^7 m, we can use the formula to calculate the speed:

speed = 2 x 3.14 x (6.37 * 10^6 + 3.58 * 10^7) / (24 x 60 x 60)

The magnitude of the acceleration of a satellite in a circular orbit can be calculated using the formula:

acceleration = (velocity)^2 / radius

Using the calculated speed and the radius of the orbit, we can find the magnitude of the acceleration:

acceleration = (2.98 x 10^3)^2 / (6.37 * 10^6 + 3.58 * 10^7)

Therefore, the speed of a satellite in a geosynchronous orbit is approximately 2.98 km/s and the magnitude of the acceleration is approximately 1.92 x 10^-3 m/s^2.

Using Fick's Law of Diffusion, it would take about 394,737 seconds for the scent of vanilla (vanillin) to travel 3 meters in still air, considering its diffusivity in the given conditions.

The time it takes for a scent to travel through air can be estimated using Fick's Law of Diffusion, which relates diffusion time to the diffusivity of the substance, the distance it needs to travel, and the area through which it diffuses.

Diffusion time = (Distance^2) / (2 * Diffusivity)

Given that the distance is 3 meters and the diffusivity (D) of vanillin in air is 0.114 cm^2/s, we need to convert the distance to centimeters before applying the formula:

Diffusion time = (300 cm)^2 / (2 * 0.114 cm^2/s)

Diffusion time ≈ 90,000 cm^2 / 0.228 cm^2/s

Diffusion time ≈ 394,737 seconds

So, it would take approximately 394,737 seconds for the smell of vanilla to reach you from a distance of 3 meters in still air.

Based on the survey of twenty students, the average number of hours watched during a school week is four.

Step 1: Gather Data

- Let's assume we have the following data from the survey:

| Student | Hours of TV watched (per week) |

|---------|--------------------------------|

| 1       | 3                              |

| 2       | 2                              |

| 3       | 4                              |

| ...     | ...                            |

| 20      | 5                              |

Step 2: Calculate the Total Hours of TV Watched

- Add up all the hours reported by each student.

Total Hours = 3 + 2 + 4 + ... + 5

Step 3: Calculate the Average Hours

- Divide the total hours by the number of students surveyed.

Average Hours = Total Hours / Number of Students

Now, let's perform the calculations.

Given:

Number of students surveyed (N) = 20

Hours of TV watched by each student:

Student 1: 3 hours

Student 2: 2 hours

Student 20: 5 hours

Step 2: Calculate the Total Hours

Total Hours = 3 + 2 + 4 + ... + 5

Total Hours = (3 + 2 + 4 + ... + 5) (20 times)

We can simplify this by realizing that we're adding the same number (the hours of TV watched by each student) 20 times:

Total Hours = (3 + 2 + 4 + ... + 5) (20 times)

           = (3 + 2 + 4 + ... + 5) * 20

Step 3: Calculate the Average Hours

Average Hours = Total Hours / Number of Students

             = (Total Hours) / 20

Now, let's find the sum of the hours:

Sum of hours = 3 + 2 + 4 + ... + 5

To find the sum, we can use the formula for the sum of an arithmetic series:

[tex]\[S = \frac{n}{2}(a_1 + a_n)\][/tex]

where:

- (S) is the sum of the series,

- (n) is the number of terms in the series,

- (a_1) is the first term in the series, and

- (a_n) is the last term in the series.

In our case:

(n = 20 (number of students surveyed),

a_1 = 3 (hours of TV watched by the first student), and

a_n = 5 (hours of TV watched by the last student).

[tex]\[S = \frac{20}{2}(3 + 5)\][/tex]

S = 10(8)

S = 80

Now, let's plug this sum into the formula for the average:

Average Hours = Total Hours / Number of Students

             = 80 / 20

             = 4

So, on average, the students surveyed watch 4 hours of TV during a school week.

complete question :

Twenty students were surveyed to determine the number of hours they watch TV during a school week. The data collected from the survey are as follows (in hours):

3, 5, 8, 2, 4, 6, 7, 5, 3, 9, 10, 1, 4, 7, 8, 6, 5, 2, 3, 7.

Explanation:

By definition, the word acceleration is equal to the rate of change of velocity. Mathematically, it is given by :

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]dv=a.dt[/tex]

[tex]v=\int\limits^t_0 {a.dt}[/tex]

Since, it is given that acceleration is constant

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

v₀ is the constant of integration and it corresponds to initial velocity

From above equation, it is clear that when acceleration is constant the speed varies linearly. Hence, when an object move with constant acceleration, it always changes its velocity.

Answer:

35.8 m

Explanation:

Given:

Initial Velocity u = 26.5 m/s

Time period t = 2.7 s

To find:

Maximum height H = ?

Solution:

The toy is projected vertically upward. So the motion is happening in y axis

When a projectile reaches its maximum height, at that point its velocity vill be zero

Using equations of motion we can find the height

[tex]v^{2} =u^{2} -2gH\\\\0^{2} =26.5^{2} -2\times 9.8 \times H\\\\19.6H = 702.25\\\\H = 35.8 m[/tex]

Verification

[tex]H = ut - \frac{1}{2} gt^{2}\\\\H = 26.5 \times 2.7- 0.5 \times 9.8 \times 2.7^{2}\\\\H = 35.8 m[/tex]

This physics problem involves the principle of kinetic energy and work-energy. Given the situation presented, the increase in the car's kinetic energy due to a doubling of initial speed means that the braking stopping distance quadruples from 3 meters to 12 meters.

This Physics problem concerns the relationship between velocity, mass, and stopping distance under braking conditions. It's dealing with the principle of kinetic energy (1/2*m*v²) and the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

If the initial speed is doubled, as it is in this case from 70 m/s to 140 m/s, the kinetic energy (and thus the work needing to be done to stop the vehicle) quadruples, assuming the mass stays constant. This means, due to the direct relationship between work done and distance when force is held constant, the stopping distance will also quadruple from the original 3 meters.

Therefore, the 1000-kilogram car, when moving at 140 m/s, will take 12 meters to stop under full braking in similar conditions.

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The biker's change in velocity from point B to point C is 0 m/s, indicating that there is no change in velocity. Consequently, the biker's acceleration between these points is also 0 m/s
to the power of 2; there is no acceleration.

The change in velocity of the biker as they travel from point B to point C is determined by subtracting the initial velocity at point B from the final velocity at point C. As given, the biker's velocity at point B is 8 m/s, and at point C, it remains 8 m/s. Therefore, the change in velocity (Δv) is:

Δv = final velocity - initial velocity

Δv = 8 m/s - 8 m/s

Δv = 0 m/s

Since the velocity does not change, the acceleration
a) from point B to point C is:

a = Δv/Δt

a = 0 m/s ÷ 1 s

a = 0 m/s²

Thus, there is no change in velocity and no acceleration as the biker moves from point B to point C.

The normal force is calculated by adding the weight of the suitcase and the vertical component of pulling force, while total resistance force is equal to the horizontal component of pulling force. Both forces play a significant role in Renee's effort to move the suitcase at a constant speed.

Here's how to find the normal force and the total resistance force for Renee's suitcase:

Remember, while calculating remember to convert the angle to radians if your calculator is set to radian mode.

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To find the normal force on Renee's suitcase, we resolve the pulling force into its components and subtract the vertical component from the suitcase's weight. As the suitcase is moving at a constant speed, the horizontal component equals the resistance force, which includes both friction and any air resistance.

Renee is pulling her 21-kg suitcase at a constant speed of 0.47 m/s through the airport. To determine the normal force acting on the suitcase, we need to consider the components of the pulling force. The force has a magnitude of 120 N and is exerted at an angle of 38° above the horizontal. We must resolve this force into vertical and horizontal components. The vertical component (Fy) helps support the weight of the suitcase and is calculated as Fy = 120 N × sin(38°). The weight of the suitcase is W = m × g, where m is the mass of the suitcase and g is the acceleration due to gravity (9.8 m/s²).

The normal force is given by N = W - Fy since the vertical component of the pulling force acts upwards, reducing the normal force exerted by the ground. As the suitcase is moving at a constant speed, the net horizontal force must be zero. Therefore, the horizontal component of the pulling force, which is Fx = 120 N × cos(38°), must be equal to the total resistance force (friction + air resistance).

The equations to find the normal force and resistance force are:

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Knowing how to graph motion helps in understanding kinematics properties by deriving motion characteristics from the graph, visualizing equations in a comprehendible form, and revealing underlying relationships between physical quantities.

Knowing how to graph movement can be practically beneficial for several reasons, these include:

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These are your answers:

A is Insulin (7)

B is Obesity (5)

C is Diabetes (1)

D is Type 1 (6)

E is Hyperglycemia (3)

F is Regular aerobic exercise (10)

G is Pancreas (9)

H is HDL (4)

I is Type 2 (8)

J is atherosclerosis(2)

Here is why:

A. Hormone that helps the body control the level of glucose in the blood.

Insulin is a hormone. It helps regulate the levels of glucose in the blood by turning glucose into energy. This is why it plays an important role in metabolism. This hormone is produced by the pancreas.

B. Main cause of Type 2 Diabetes

Obesity is the main cause of Type 2 diabetes. Unhealthy eating and lack of exercise are often listed as causes of Diabetes 2, and this kind of lifestyle collectively leads to obesity.

C. Condition that makes it hard for the body to control the level of glucose in the blood.

Diabetes is a condition where the levels of glucose in the blood is high. This happens because the body cannot produce enough insulin, which is the hormone that controls glucose levels.

D. Damage to the pancreas caused by ones own antibodies.

In Diabetes Type 1, the immune system attacks the panceatic beta cells, which produce insulin. Unlike Type 2, Type 1 Diabetes is unavoidable and hereditary. So if you have it, you have it.

E. The elevation of glucose levels in the blood.

Hyperglycemia - Hyper means high or elevated. Gly means glucose or sugar. -cemia means blood. Put together, elevated glucose in the blood.  

F. Found to help with treatment of clinical depression

Studies have shown that aerobic exercise can help with clinical depression. It helps elevate moods and lessen tension. This helps relieve stress.

G. Organ where insulin is produced

Like mentioned above, insulin is produced by the pancreas.

H. "Good" Cholesterol

HDL is High-density Lipoprotein. HDL is considered as good cholesterol because it actually assists in removing other forms of cholesterol from the blood.

I. 90% to 95% of the case of diabetes in America

Studies have shown that in America Diabetes 2 is the most common case. Like mentioned above, cause of Diabetes type 2 is eating habits and lack of exercise and many foods today are full of processed sugars and are consumed in great amounts because of convenience.

J. Hardening of the arteries caused by a build up of fatty materials.

Fatty materials create plaque and they accumulate in the blood vessels. This leads to constriction and hardening in arteries specifically. This constriction makes the vessel more narrow and it can limit the flow of oxygen to the other organs of the body.

The energy required to remove second electron from lithium is more as compared to removing fourth electron from carbon

EXPLANATION:

The amount of energy required to remove an electron from an isolated atom is called as ionization energy of the electron.

The second ionization energy of lithium atom is more as compared to fourth ionization energy of carbon atom. It is so because the second electron which is to be emitted from the K-shell of lithium atom, is tightly bound by the nucleus as the orbit is very closer to the nucleus.

In case of carbon, the fourth electron is present in the valence shell.The radius of valence shell is not so close as compared to lithium.The screening effect is also more for carbon as compared to lithium.

Hence, the energy required to remove a second electron from lithium is more that the energy required to remove fourth electron from carbon.

It is easier to take out 4th electron from carbon than 2nd electron from lithium because 2nd electron of lithium is closer to the nucleus.

The electronic configuration for,

[tex]\rm \bold{ Li_3 - 1s^2 2s^2}\\\\\rm \bold{ C_6- 1s^2 2s^2 2p^2}[/tex]

Therefore, we can conclude that the it is easier to take out 4th electron from carbon than 2nd electron from lithium.

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The flash travels the round-trip distance approximately 1,000,000 times.

The speed of sound is 336 m/s, and the speed of light (which represents the speed at which the flash travels) is approximately [tex]3\times 10^8 m/s[/tex].

Let's denote the distance between the mirrors as d. The time it takes for the sound to travel the round trip (to the cliff and back) is [tex]2d/336[/tex]seconds. During this time, the flash of light travels at [tex]3\times10^8m/s.[/tex]

To find out how many times the flash of light can travel the round-trip distance before the sound is heard, we calculate:

[tex]\text{Number of round trips}=(3\times10^8\times2d/336)/2d=(3\times10^8)/336\approx1000,000[/tex]

Thus, the flash of the gunshot travels the round-trip distance approximately 1,000,000 times before the echo of the gunshot is heard.

The wavelength of the photon having twice the energy as that of the photon of wavelength [tex]600\,{\text{nm}}[/tex] is [tex]\boxed{300\,{\text{nm}}}[/tex] .

Further Explanation:

The photons are the small packets of energy that move at the speed of light. The photons are considered to remain always in motion. The energy associated with a moving photon is given by:

[tex]E = \dfrac{{hc}}{\lambda }[/tex]

Here,  [tex]E[/tex]  is the energy associated with the photon, [tex]h[/tex] is the Planck’s constant, [tex]c[/tex] is the speed of light and [tex]\lambda[/tex] is the wavelength of the moving photon.

The value of the Planck’s constant is [tex]6.6 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}[/tex] .

The wavelength of the photon is [tex]600\,{\text{nm}}[/tex] .

The energy associated with the photon of wavelength [tex]600\,{\text{nm}}[/tex] is:

[tex]\begin{aligned}{E_1}&=\frac{{\left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{600 \times {{10}^{ - 9}}}}\\&=\frac{{1.98 \times {{10}^{ - 25}}}}{{6 \times {{10}^{ - 7}}}}\\&= 3.3 \times {10^{ - 19}}\,{\text{J}}\\\end{aligned}[/tex]

The wavelength of photon having energy double of this:

[tex]\begin{aligned}E' &= 2{E_1}\\&= 2 \times\left( {3.3 \times {{10}^{ - 19}}} \right)\,{\text{J}}\\&{\text{ = 6}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 19}}\,{\text{J}}\\\end{aligned}[/tex]

The new wavelength of the photon will be:

 [tex]\lambda ' = \dfrac{{hc}}{{E'}}[/tex]

Substitute [tex]6.6 \times {10^{ - 19}}\,{\text{J}}[/tex] for [tex]E'[/tex] in above expression.

[tex]\begin{aligned}\lambda ' &= \frac{{\left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{6.6 \times {{10}^{ - 19}}}}\\&=\frac{{1.98 \times {{10}^{ - 25}}}}{{6.6 \times {{10}^{ - 19}}}}\,{\text{m}}\\&= 3.0 \times {10^{ - 7}}\,{\text{m}}\\&= 300\,{\text{nm}}\\\end{aligned}[/tex]

The wavelength of the photon having twice the energy as that of the photon of wavelength [tex]600\,{\text{nm}}[/tex] is [tex]\boxed{300\,{\text{nm}}}[/tex].

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Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Photon and Energy

Keywords:  Wavelength, photon, energy, E=hc/lamda, 600nm, twice the energy, Planck’s constant, small packets of energy, 300nm, speed of light.