Answer: C. Wind
Explanation:
An erosion is a phenomena in which a physical agent takes away along with it soil and deposit it to some other place.
Sand dune is a kind of wind erosion of the loose soil called as sand. It is typically seen in desert regions. The high speed wind currents blow away huge amount of soil.
If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, 18 m away? for this estimate, ignore air resistance and spin on the ball.
A kite 100 ft above the ground moves horizontally at a speed of 12 ft/s. at what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
Final answer:
The angle between the string and the horizontal is decreasing at a rate of -0.4 radians/second.
Explanation:
In order to solve this problem, we can use trigonometry and the chain rule from calculus. Let's denote the angle between the string and the horizontal as θ. We need to find the rate at which this angle is decreasing (dθ/dt) when 200 ft of string have been let out.
First, we can find the length of the string using the Pythagorean theorem: 100^2 + r^2 = (100+r)^2, where r is the length of the horizontal portion of the string. Solving this equation gives us r = 50 ft.
Next, we can differentiate the equation with respect to time using the chain rule: d/dt (100^2 + r^2) = d/dt ((100+r)^2). Simplifying the equation and solving for dθ/dt gives us dθ/dt = -100/(r+r^2/100).
Plugging in r = 50 ft, we can find dθ/dt = -0.4 radians/second.
A flowerpot falls off a balcony 85m above the street how long does it take to hit the ground
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
Time (t) =?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:
H = ½gt²85 = 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]
t = 4.12 sTherefore, it will take 4.12 s for the flowerpot to fall to the ground.
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a toy projectile is fired from the ground vertically upward with an initial velocity of 26.5 m/s. The project arrives at its maximum altitude in 2.7s.
Determine the greatest height the projectile reaches. How do you know?
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]
A horizontal pipe of diameter 0.985 m has a smooth constriction to a section of diameter 0.591 m . the density of oil flowing in the pipe is 821 kg/m3 . if the pressure in the pipe is 8100 n/m2 and in the constricted section is 6075 n/m2 , what is the rate at which oil is flowing
To find the rate at which oil is flowing through a constricted pipe with different diameters and pressures, apply the principle of continuity.
A horizontal pipe of diameter 0.985 m has a smooth constriction to a section of diameter 0.591 m. The pressure in the pipe is 8100 N/m2, and in the constricted section, it is 6075 N/m2. The density of oil flowing in the pipe is 821 kg/m3.
To find the rate at which oil is flowing, we can apply the principle of continuity, which states that the product of the cross-sectional area and the fluid velocity is constant in a pipe with steady flow.
By applying the principle of continuity, you can calculate the rate at which oil is flowing through the pipe.
How are stars important to the Milky Way
Which of the following is an example of velocity?
A. 25 east
B. 50 mph east
C. 100 mph
D. 100 miles east
50 mph east,because Velocity is a physical vector quantity both magnitude and direction are needed to define it.
What is velocity?
When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
50 mph east,because Velocity is a physical vector quantity both magnitude and direction are needed to define it.
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Part a consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 w (this rate represents the power consumption of the bird). assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. how far will the bird fly before feeding again?
A common small-molecular weight (and therefore fast diffusing for an organic molecule) ingredient in perfumes is vanillin, the primary component of vanilla bean extract (molecular weight = 152). the d for vanillin in air is 0.114 cm2/s. if i open a bottle of vanilla on the other side of the room 3 meters away, and the air is still so there is no convection, about how long would i have to wait before i could expect to smell the vanilla?
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.
Does a person standing motionless in the aisle of a moving bus have kinetic energy
Twenty students were surveyed to find out how many hours of tv they watch during a school week
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.
A mirror faces a cliff located some distance away. mounted on the cliff is a second mirror, directly opposite the first mirror and facing toward it. a gun is fired very close to the first mirror. the speed of sound is 336 m/s. how many times does the flash of the gunshot travel the round-trip distance between the mirrors before the echo of the gunshot is heard?
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.
Which characteristic does an object with a constant acceleration always have?
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.
A toy projectile is fired from the ground vertically upward with an initial velocity of 26.5 m/s. The project arrives at its maximum altitude in 2.7s.
What is the velocity of the projectile when it hits the ground? How?
a hockey player has an acceleration of -3.1m/s2 how long does it take him come to a complete stop from a speed of 13m/s
Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. these are called geosynchronous orbits. the radius of the earth is 6.37 * 106 m, and the altitude of a geosynchronous orbit is 3.58 * 107 m 1â22,000 miles2. what are (a) the speed and (b) the magnitude of the acceleration of a satellite in a geosynchronous orbit?
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.
Explanation: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.
What is the wavelength of a photon whose energy is twice that of a photon with a 600 nm wavelength?
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.
A head-on, elastic collision between two particles with equal initial speed v leaves the more massive particle (mass m1) at rest. find the ratio of the particle masses
The ratio of the particle masses is \boxed{\frac{1}{3}} or \boxed3 .
Further explain:
We have to calculate the ratio of the particle masses.
As we know, in the elastic collision between two masses the momentum and the energy both are conserved.
Here, the collision between the masses the head-on it means head to head.
For head on head collision the masses will travel parallel but opposite in the direction.
We have two masses one is heavier and another is lighter.
The mass of massive or heavier particle is [tex]{m_1}[/tex].
The mass of the lighter particle is [tex]{m_2}[/tex].
From the conservation of linear momentum total initial momentum is equal to the total final momentum.
Therefore,
[tex]\boxed{\left( {{m_1}v - {m_2}v} \right) = \left( {{m_1}{v_1} + {m_2}{v_2}} \right)}[/tex]
Here, after the collision the massive particle comes into rest.
So, final expression will be,
[tex]\left( {{m_1}-{m_2}}\right)v={m_2}{v_2}[/tex] …… (1)
From the conservation of the energy,
Total kinetic energy before collision is equal to the total kinetic energy after collision.
Therefore,
[tex]\begin{aligned}\frac{1}{2}{m_1}{v^2}+\frac{1}{2}{m_2}{v^2}&=\frac{1}{2}{m_2}{\left( {{v_2}} \right)^2}\\{m_1}{v^2}+{m_2}{v^2}&={m_2}{\left( {{v_2}}\right)^2}\\\left( {{m_1}+{m_2}}\right){v^2}&={m_2}{\left( {{v_2}}\right)^2}\\\end{aligned}[/tex]
Simplify the above equation,
[tex]\begin{aligned}{m_2}{\left( {{v_2}} \right)^2}&=\frac{{\left( {{m_1}+{m_2}} \right){v^2}}}{{{m_2}}}\\{v_2}&=\left( {\sqrt {\frac{{\left( {{m_1}+{m_2}} \right)}}{{{m_2}}}} }\right)v\\\end{aligned}[/tex]
Substitute the value of [tex]{v_2}[/tex] in equation (1).
[tex]\begin{aligned}\left( {{m_1} - {m_2}} \right)v&={m_2}\left( {\sqrt {\frac{{\left( {{m_1} + {m_2}}\right)}}{{{m_2}}}} } \right)v \\\left( {{m_1} - {m_2}} \right)&=\sqrt {{m_2}\left( {{m_1} + {m_2}}\right)}\\{m_2}\left( {\frac{{{m_1}}}{{{m_2}}} - 1}\right)&={m_2}\sqrt {\left( {\frac{{{m_1}}}{{{m_2}}} + 1} \right)}\\\left( {\frac{{{m_1}}}{{{m_2}}}-1}\right)&=\sqrt {\left( {\frac{{{m_1}}}{{{m_2}}}+ 1}\right)}\\\end{aligned}[/tex]
Substitute [tex]x[/tex] for[tex]\dfrac{{{m_1}}}{{{m_2}}}[/tex] in above equation.
[tex]\left( {x - 1} \right)=\sqrt {\left( {x + 1} \right)}[/tex]
Squaring both the sides in above equation,
[tex]\begin{aligned}{\left( {x - 1} \right)^2}&=\left( {x + 1}\right)\\{x^2} - 2x + 1&=x + 1\\{x^2}-3x&=0\\\end{aligned}[/tex]
Taking [tex]x[/tex] as a common in the above equation.
[tex]x\left( {x - 3} \right)=0[/tex]
On solving above equation
We get,
[tex]x = 3[/tex]
Replace the value of [tex]x[/tex]
[tex]\boxed{\frac{{{m_1}}}{{{m_2}}} = 3}[/tex]
Or,
[tex]\boxed{\frac{{{m_2}}}{{{m_1}}} = \frac{1}{3}}[/tex]
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Answer details:
Grade: Senior School
Subject: Physics
Chapter: Impulse and Momentum
Keywords:
Head on collision, two particles, equal speed, ratio of particle masses, momentum, conservation of momentum, energy, conservation of energy, masses, ratio.
Is the 3 in the molecule a coefficient, subscript, or element? 3H₂O₂
Renee is on Spring Break and pulling her 21-kg suitcase through the airport at a constant speed of 0.47 m/s. She pulls on the strap with 120 N of force at an angle of 38° above the horizontal. Determine the normal force and the total resistance force (friction and air resistance) experienced by the suitcase.
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.
Explanation:Here's how to find the normal force and the total resistance force for Renee's suitcase:
Normal Force: The normal force is the force exerted by a surface that supports the weight of an object resting on it. It acts perpendicular to the surface. In this case, there are two components to Renee's force - the horizontal and the vertical. The vertical component of the force can be calculated using the formula F = F_pull*sin(angle), where F_pull is the force with which Renee pulls the suitcase (120 N) and the angle is 38°. After calculating the vertical component of the force, you subtract it from the weight of the suitcase (mass*gravity) to get the normal force. Total Resistance Force: The total resistance force is the force that opposes the motion of the suitcase. In this case, at constant speed, the total resistance force is equal to the horizontal component of the pulling force. This can be calculated using the formula F = F_pull*cos(angle).
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.
Explanation: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:
N = (21 kg × 9.8 m/s²) - (120 N × sin(38°))resistance force = 120 N × cos(38°)Learn more about Resolving Forces here:https://brainly.com/question/32037721
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Microwave ovens emit microwave energy with a wavelength of 12.4 cm. what is the energy of exactly one photon of this microwave radiation?
The energy of the photon emitted is about 1.60 × 10⁻²⁴ Joule
Further explanationThe term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :
[tex]\large {\boxed {E = h \times f}}[/tex]
E = Energi of A Photon ( Joule )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
f = Frequency of Eletromagnetic Wave ( Hz )
The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
E = Energi of A Photon ( Joule )
m = Mass of an Electron ( kg )
v = Electron Release Speed ( m/s )
Ф = Work Function of Metal ( Joule )
q = Charge of an Electron ( Coulomb )
V = Stopping Potential ( Volt )
Let us now tackle the problem !
Given:
λ = 12.4 cm = 12.4 × 10⁻² m
h = 6.63 × 10⁻³⁴ Js
c = 3 × 10⁸ m/s
Unknown:
E = ?
Solution:
[tex]E = h \times \frac{c}{\lambda}[/tex]
[tex]E = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{12.4 \times 10^{-2}}[/tex]
[tex]E \approx 1.60 \times 10^{-24} ~ Joule[/tex]
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Subject: Physics
Chapter: Quantum Physics
Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light
The speed of sound in water is measured to be 1485 m/s. what is this in mph?
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
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker is slipping with a velocity of 8.6 m/s north and the halfback is sliding with a velocity of 7.4 m/s east.
The question is about the conservation of momentum. By calculating each player's momentum before the collision and combining them, we find the resultant momentum. The velocity after the collision can then be found.
Explanation:This is an example of conservation of momentum, a fundamental concept in physics. Whenever objects interact and there is no external force, the total momentum of the system of objects is conserved.
In the scenario given, the linebacker and the halfback can be regarded as a closed system because the only significant forces are their mutual ones. Before the collision, we can calculate the momentum for each player: the linebacker's momentum is mass x velocity = 120 kg x 8.6 m/s = 1032 kg.m/s (north), and the halfback's momentum is 75 kg x 7.4 m/s = 555 kg.m/s (east).
Using the law of vector addition, we can combine them to find the resultant momentum. We then divide the resultant momentum by the total mass (120 kg + 75 kg) to find the velocity of the resulting 'player blob' immediately after the collision.
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A balloon is inflated from 0.0100 l to 0.500 l against an external pressure of 10.00 atm. how much work is done in joules? 101.3 j = 1 l atm
Answer:
-496.37 J
Explanation:
P(V2-V1) = 10(.5-.01)
10(.49) =4.9
L x ATM = 4.9 x 101.3= 496.37 J
External pressure means negative therefore its -496.37J
The work done in this scenario can be calculated by multiplying the change in volume, external pressure, and a conversion factor. In this case, the work done is 494.9 J.
Explanation:The work done in this scenario can be calculated using the formula:
Work = change in volume * external pressure * conversion factor
Given:
Initial volume (V1) = 0.0100 l
Final volume (V2) = 0.500 l
External pressure = 10.00 atm
First, we need to find the change in volume:
Change in volume = V2 - V1 = 0.500 l - 0.0100 l = 0.490 l
Next, we can calculate the work done:
Work = change in volume * external pressure * conversion factor
= 0.490 l * 10.00 atm * 101.3 J/l atm
= 494.9 J
Therefore, the work done in joules is 494.9 J.
What is the frequency of radiation whose wavelength is 2.40 x 10-5 cm?
In general, which of the following is NOT a critical skill for participating in a sport? A. agility and balance B. endurance and muscle control C. hand-eye coordination and acuity D. none of the above
The answer is :D. None of the above
Agility and balance is very important in sport activities such as soccer, especially seen when players dribble the ball at high speed. Endurance and muscle control is very important in sport such as gymnastic or boxing. Hand-eye coordination and acuity is important in sports such as golf.
Answer:
the answer is d none of the above
Explanation:
What is tarzan's speed vf just before he reaches jane? express your answer in meters per second to two significant figures?
A 500 lb steel beam is lifted up by a crane to a height of 100 ft and is held there.
A) How much work is being done to hold the beam in place?
B) How much work was done to lift the beam?
C) How much work would it take if the steel beam were raised from 100 ft to 200ft?
A. How much work is being done to hold the beam in place?
Work is the product of Force and Displacement. Since there is no Displacement involved in just holding the beam in place, hence the work is zero.
B. How much work was done to lift the beam?
In this case, force is simply equal to weight or mass times gravity. Hence the work is:
Work = weight * displacement
Work = 500 lbf * 100 ft
Work = 50,000 lbf * ft
C. How much work would it take if the steel beam were raised from 100 ft to 200ft?
The displacement is still 100 ft since 200 – 100 = 100 ft, hence the work done is still similar in B which is:
Work = 50,000 lbf * ft
What is the change in the ball's position from t=0s to t=1.0s? express your answer using two significant figures?
People watching their team lose important games often make faces, yell at the screen, and demonstrate general agitation. Although they aren't playing the game themselves, they may be able to experience the frustration felt by the players because of ________________ activity in the brain