The answer is 1350 N
How much of earths water is found in our oceans??
Compared to gamma rays, X–rays have relatively
less energy and short wavelengths.
more energy and short wavelengths.
less energy and long wavelengths.
more energy and long wavelengths.
A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 m from the bulb, the light intensity is I0, the average energy density of the waves is u0, and the rms electric and magnetic field values are E0 and B0, respectively.
1. At 2.0 m from the bulb, what is the light intensity?
2. At 2.0 m from the bulb, what is the rms magnetic field value?
3. At 2.0 m from the bulb, what is the average energy density of the waves?
The light intensity at 2.0 m from the bulb would be I0/4. The rms magnetic field value at 2.0 m from the bulb would be B0/2. The average energy density of the waves at 2.0 m from the bulb would be u0/4.
Explanation:1. The light intensity follows the inverse square law, which means that the intensity decreases as the distance squared increases. So at 2.0 m from the bulb, the light intensity would be I0/4.
2. The rms magnetic field value is related to the light intensity through the equation B0 = sqrt((2u0cI0)/(ε0c^2)), where c is the speed of light and ε0 is the vacuum permittivity. Therefore, at 2.0 m from the bulb, the rms magnetic field value would be B0/sqrt(4) = B0/2.
3. The average energy density of the waves is equal to the energy per unit volume. It can be calculated using the formula u0 = ε0E0^2/2, where E0 is the rms electric field value. At 2.0 m from the bulb, the average energy density of the waves would be u0/4.
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At 2.0 m from the 60-W light bulb, the light intensity is one-fourth of the intensity at 1.0 m, the RMS magnetic field value is half of the initial value, and the average energy density also becomes one-fourth of the initial value.
To solve the problem involving a 60-W light bulb radiating electromagnetic waves:
Light intensity at a distance of 2.0 m: Considering that intensity (I) varies inversely with the square of the distance (r) from the source, we apply the formula: [tex]\( I_2 = \frac{I_0}{4} \)[/tex]. Thus, at 2.0 m, the intensity [tex]\( I_2 = \frac{I_0}{4} \)[/tex]RMS Magnetic Field Value at 2.0 m: The RMS magnetic field value B is inversely proportional to the distance r. Therefore,[tex]\( B_2 = \frac{B_0}{2} \)[/tex] at 2.0 m.Average Energy Density at 2.0 m: The energy density u is proportional to the intensity. Hence, at 2.0 m, u2 = u0 / 4.In deep space, there is very little friction. Once they launch a probe into deep space, where there are no external forces acting on it, scientists shut the probe’s engines off because the scientists want the probe to stop immediately. speed up. slow down. move at constant velocity.
Answer:
move at constant velocity.
Explanation:
Newton's first law (also known as law of inertia) states that:
"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".
In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.
through which medium would sound travel the fastest, water, a steel bar, or nitrogen gas explain
a car has a speed of 2m/s and a mass of 1500 kg. what is the car's kinetic energy
two cars are each traveling at 72 km/h one car is traveling northeast, and the other is traveling south the two cars have different ____
Answer:
Velocities
Explanation:
Given that, two cars are each traveling at 72 km/h one car is traveling northeast, and the other is traveling south. Since, both objects are moving with same speeds but the direction of both cars is opposite. In this case, both cars will have different velocities. Velocity of an object is vector quantity i.e. it will have same magnitude but different velocity.
Hence, two cars have different velocities.
In deep space there is very little friction once they are launched into a probe into deep space where there are no external forces acting on it scientists shut the probe's engines off because the scientists want the probe to
Answer:
move at a constant velocity
Explanation:
When the friction is present the engine helps move the probe by constantly doing work against the friction. But when the friction is absent then there is no need for the engine to work all the time. According to newton's first law, no object can change its state of rest or uniform motion with constant velocity without an external force. When the engine is shut off, the probe will continue to move at a constant speed due to inertia.
Coughing forces the trachea to contract, which affects the velocity v of the air passing through the trachea. Suppose the velocity of the air during coughing is v = k(R-r)r2 where k and R are constants, R is the normal radius of the trachea, and r is the radius during coughing. What radius will produce the maximum air velocity?
To find the radius that maximizes air velocity during coughing, we differentiate the given velocity equation, set it to zero, and solve for the radius. The maximum air velocity occurs when the radius r is two-thirds of the normal radius R. Therefore, the radius that maximizes air velocity is 2R / 3.
To find the radius[tex]\( r \)[/tex]that produces the maximum air velocity v during coughing, we need to maximize the function v = [tex]k(R - r)r^2 \),[/tex] where k and R are constants.First, let's rewrite the function for clarity:
[tex]\[ v(r) = k(R - r)r^2 \][/tex]
To find the maximum value, we need to take the derivative of [tex]\( v(r) \)[/tex] with respect to r , set it equal to zero, and solve for r .
Take the derivative:[tex]\[ \frac{dv}{dr} = k \frac{d}{dr}[(R - r)r^2] \][/tex]
Using the product rule:
[tex]\[ \frac{dv}{dr} = k \left[ (R - r) \cdot \frac{d}{dr}(r^2) + r^2 \cdot \frac{d}{dr}(R - r) \right] \][/tex]
[tex]\[ \frac{dv}{dr} = k \left[ (R - r) \cdot 2r + r^2 \cdot (-1) \right] \][/tex]
[tex]\[ \frac{dv}{dr} = k \left[ 2r(R - r) - r^2 \right] \][/tex]
[tex]\[ \frac{dv}{dr} = k \left[ 2rR - 2r^2 - r^2 \right] \][/tex]
[tex]\[ \frac{dv}{dr} = k \left[ 2rR - 3r^2 \right] \][/tex]
Set the derivative equal to zero:[tex]\[ 0 = k \left[ 2rR - 3r^2 \right] \][/tex]
Since k is a constant and not equal to zero, we can divide both sides by k :
[tex]\[ 0 = 2rR - 3r^2 \][/tex]
Factor out of the r :
[tex]\[ r(2R - 3r) = 0 \][/tex]
So, the solutions are:
[tex]\[ r = 0 \][/tex]
[tex]\[ 2R - 3r = 0 \][/tex]
Solve for r :
[tex]\[ 2R = 3r \][/tex]
[tex]\[ r = \frac{2R}{3} \][/tex]
The solution [tex]\( r = 0 \)[/tex] is not physically meaningful in this context since it would imply the trachea is completely closed. Thus, the radius that produces the maximum air velocity is:
[tex]\[ r = \boxed{\frac{2R}{3}} \][/tex]
Bobby tries to push his new big screen TV into the living room. However, Bobby does not push hard enough and cannot move the TV. List and describe the forces that would be included on the free body diagram of Bobby's TV. Be sure to include the name, direction and brief description for each force. ...?
Calculate the change in potential energy of 8,000,000 kg of water dropping 50.0 m over Niagara Falls
If a boulder has a mass of 50 kg and a potential energy of 490 j what is the height of the boulder
Kathy is changing the tire of her car on a steep hill 20m high. She trips and drops the 10kg spare tire which rolls down the hill. What is the speed of the tire at the top of the next hill if the height of the hill is 5m high?
Final answer:
The speed of the tire at the top of the 5m hill, calculated using conservation of energy principles and ignoring any work done by friction, is approximately 17.15 m/s.
Explanation:
To solve this problem, we can use the conservation of energy principle, which states that if no external work is done on the system (like work by friction), the total mechanical energy remains constant. This means that the potential energy lost by the tire as it rolls down from the higher hill will be converted into kinetic energy.
The potential energy at the top of the 20m hill is given by PE = mgh, where m is mass, g is acceleration due to gravity (9.8 m/s2), and h is the height of the hill. At the 20m hill, PE = 10kg × 9.8 m/s2 × 20m. When the tire reaches the top of the next hill, its potential energy will be PE = 10kg × 9.8 m/s2 × 5m.
We can then equate the initial potential energy minus the final potential energy to the kinetic energy at the top of the 5m hill: KE = ½ mv2, and solve for the speed v.
Conservation of energy: mgh1 - mgh2 = ½ mv2
Calculation:
PE at 20m: (10 × 9.8 × 20) J = 1960 J
PE at 5m: (10 × 9.8 × 5) J = 490 J
Kinetic energy at 5m hill: 1960 J - 490 J = 1470 J
1470 J = ½ × 10kg × v2
v2 = (1470 J × 2) / 10kg
v2 = 294 m2/s2
v = √294 m2/s2
v ≈ 17.15 m/s
Therefore, the speed of the tire at the top of the 5m hill is approximately 17.15 m/s.
What is the acceleration of an object if it goes from a velocity of 25 m/s to rest in 5.0 s?
a.–5 m/s2b. 5 m/s2c.–25 m/s2d. 25 m/s2
The force component along the displacement varies with the magnitude of the displacement, as shown in the graph. (a) 0 to 1.0 m,
(b) 1.0 to 2.0 m, and
(c) 2.0 to 4.0 m. The force component along the displacement varies with the magnitude of the displacement, as shown in the graph. (a) 0 to 1.0 m,
(b) 1.0 to 2.0 m, and
(c) 2.0 to 4.0 m.
The question pertains to the concept of work done by a force over a displacement, emphasized by the calculation methods for constant and variable forces and illustrated through the area under a force vs. displacement graph.
Explanation:The force component along the displacement varying with the magnitude of the displacement refers to the physical concept of work done by a force along a certain displacement. The work done is calculated by integrating the force component in the direction of displacement over the path taken. When the force component (F cos θ) is constant, the work done is simply the product of this force component and the displacement (d), represented as W = Fd cos θ. However, when the force varies along the displacement, the calculation involves dividing the area under the force vs. displacement graph into strips, calculating the work done for each strip as (F cos θ)i(ave) di, and summing these values to find the total work done. This method highlights that the total work done is equivalent to the area under the curve in a force versus displacement graph, which is a core principle in physics for understanding work and energy.
(a), The force component remains relatively constant
(b), There's a discernible increase in the force component
(c), The force component exhibits a steeper incline
The graph depicts how the force component changes concerning displacement magnitude across three intervals: (a) 0 to 1.0 m, (b) 1.0 to 2.0 m, and (c) 2.0 to 4.0 m.
In segment (a), the force component remains relatively constant, suggesting a consistent force acting within this range.
Transitioning to segment (b), there's a discernible increase in the force component, indicating a proportional rise in force with displacement.
However, in segment (c), the force component exhibits a steeper incline, suggesting a nonlinear relationship where the force increases more rapidly concerning displacement magnitude.
Such variation implies complex interactions between the force and displacement, possibly influenced by factors like material properties, external forces, or system dynamics.
Analyzing these intervals aids in understanding the system's behavior and optimizing its performance within different displacement ranges.
Is it true or false that at 40 mph, your response time for steering is ½ of a second and you will travel 29 feet during that time
Answer:
"At 40 mph, your response time for steering is ½ of a second and you will travel 29 feet during that time." The statement is true.
Explanation:
Speed, s = 40 mph
Converting mph to m/s :
1 mph = 0.44704 m/s
40 mph = 17.8816 m/s
Time taken, t = 1/2 seconds
Distance covered, d = speed × time
d = 17.8816 m/s × (1/2 s)
d = 8.9408 meters
Now converting meters to feet :
1 meter = 3.28084 foot
So, 8.9408 meters = 29.4 feets
or d = 29 feets
Hence, the given statement is true.
High energy waves have
Choose one answer.
a. long wavelengths and low frequencies.
b. long wavelengths and high frequencies.
c. short wavelengths and low frequencies.
d. short wavelengths and high frequencies
How would the acceleration of a chain of three shopping carts compare with the acceleration of a single cart if the same force acted on both?
A. 1/3 as much the single
B. 1/2 as much the single
C. 3 times as much the single
D. 2 times as much the single
a = F/m
a' = F/3m
a'/a = 1/3
Based on this, the correct answer would be option A.
1/3 as much the single Hope this helps.a 20kg rock is on the edge of a 100m cliff. what gravitational energy does the rock process relative to the base of the cliff
The rock has 19600 J of gravitational potential energy relative to the base of the cliff.
The gravitational potential energy [tex]\( E_p \)[/tex] of an object relative to a reference point (in this case, the base of the cliff) can be calculated using the formula:
[tex]\[ E_p = mgh \][/tex]
Where:
m is the mass of the object (20 kg in this case)
g is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \) on Earth)[/tex]
h is the height of the object relative to the reference point (100 m in this case)
Substituting the values:
[tex]\[ E_p = (20 \, \text{kg}) \times (9.8 \, \text{m/s}^2) \times (100 \, \text{m}) \]\[ E_p = 20 \times 9.8 \times 100 \, \text{J} \]\[ E_p = 19600 \, \text{J} \][/tex]
So, the gravitational potential energy that the rock possesses relative to the base of the cliff is [tex]\( 19600 \, \text{J} \).[/tex]
Which best compares AC and DC?
AC flows in one direction, and DC repeatedly switches direction.
DC flows in one direction, and AC repeatedly switches direction
AC is used only in generators, and DC is used only in motors
DC is used only in generators, and AC is used only in motors
Answer: DC flows in one direction, and AC repeatedly switches direction
Explanation:
DC stands for direct current.
AC stands for alternating current.
When current flows only in single direction, it is known as direct current. When current changes direction i.e. it alternates direction, it is known as alternating current.
There are both AC generators and DC generators.
AC generators supply power to home appliances and small motors. DC generators are used to power large electric motors.
AC flows in one direction, and DC repeatedly switches direction.
Explanation:AC flows in one direction, and DC repeatedly switches direction. This is incorrect. AC, or alternating current, periodically changes direction, while DC, or direct current, flows in one direction only. Examples of AC include household electrical outlets and power generated by generators, while DC is commonly used in batteries and electronic devices.
DC flows in one direction, and AC repeatedly switches direction. This is the correct answer. As mentioned earlier, DC flows in one direction, while AC repeatedly switches direction.
Therefore, the best comparison between AC and DC is that DC flows in one direction, and AC repeatedly switches direction.
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Substances X and Y are both nonpolar. If the volatility of X is higher than that of Y, what is the best explanation?
X’s molecules experience stronger dipole-dipole forces than Y’s molecules.
Y’s molecules experience stronger dipole-dipole forces than X’s molecules.
X’s molecules experience stronger London dispersion forces than Y’s molecules.
Y’s molecules experience stronger London dispersion forces than X’s molecules. r
The answer is,
D. Y’s molecules experience stronger London dispersion forces than X’s molecules.
(:
Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 50 km/h. After six hours, the velocity of the car is 54 km/h.
Part A: Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used.
Part B: How can you graph the equation obtained in Part A for the first six hours?
MgBr2 2 is a subscript what does the subscript indiacate
what is the average salt content of seawater?
A 4000kg truck is parked on a 15.0∘ slope.
How big is the friction force on the truck?
A 4000kg truck is parked on a 15.0∘ slope, the friction force on the truck is approximately 10,293 N.
We must take into account the truck's weight and the slope's angle in order to calculate the friction force on a truck that is parked on a hill.
The following formula can be used to determine the truck's weight:
Weight = mass × gravitational acceleration
Weight = 4000 kg × 9.8 m/s²
= 39,200 N
Parallel Component = Weight × sin(angle)
The angle of the slope is given as 15.0 degrees. Converting this to radians, we get:
Angle in radians = 15.0 degrees × (π/180)
≈ 0.2618 radians
Now we can calculate the parallel component:
Parallel Component = 39,200 N × sin(0.2618)
≈ 10,293 N
Therefore, the friction force on the truck is approximately 10,293 N.
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Which energy-level change shown in the diagram below emits electromagnetic radiation with the longest wavelength?
a) an electron moving from 4 to 5
b) an electron moving from 5 to 2
c) an electron moving from 6 to 1
d) an electron moving from 2 to 1
Which of the following is not a property of cells:
ability to reproduce
using energy for growth
all cells are the same
adapting to their environment
its timed
A volume of 229 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 Torr and the temperature is 25°C. Calculate the atomic mass of the metal.
To calculate the atomic mass of the metal, we can use the ideal gas law. Given the pressure, volume, and temperature, we can determine the number of moles of hydrogen gas. By dividing the mass of hydrogen by the number of moles, we can calculate the atomic mass of the metal.
Explanation:To calculate the atomic mass of the metal, we need to use the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the given pressure from Torr to atm by dividing it by 760. So, the pressure becomes 0.995789 atm.
Next, we convert the volume from mL to L by dividing it by 1000. So, the volume becomes 0.229 L.
Now, we can use the ideal gas law to calculate the number of moles of hydrogen. Rearranging the equation, we get n = (PV) / (RT).
Plugging in the values, we have n = (0.995789 atm * 0.229 L) / (0.08205 L atm /(K mol) * (25 + 273.15)K).
Simplifying the equation gives us n = 0.01012 mol.
Since hydrogen gas has a molar mass of 2.02 g/mol, the atomic mass of the metal can be calculated by dividing the mass of hydrogen by the number of moles. So, the atomic mass of the metal is (2.02 g/mol) / (0.01012 mol) = 199.60 g/mol.
Suppose that a sled is accelerating at a rate of 2 m/s^2. if the net force is tripled and the mass is doubled, then what is the the new acceleration of the sled?
By using Newton's second law of motion, we can determine that the new acceleration of the sled when the net force is tripled and the mass is doubled is 3 m/s².
Explanation:To calculate the new acceleration, we will use Newton's second law of motion which is F = m * a, where F is the net force, m is the mass of the object, and a is the acceleration. Initially, consider the force as F = m * a. After the changes, the new force and mass become F' = 3F = 2m * a', where F' is the new force, m' is the new mass, and a' is the new acceleration.
So, now you have the equation 3F = 2m * a'. Substitute the initial force F (m * a) into the equation and you get 3 * m * a = 2m * a'. Now you can solve for the new acceleration a', a' = (3/2) * a. Therefore, the new acceleration of the sled is 1.5 times the original, in this case 1.5 * 2 m/s² = 3 m/s². So the new acceleration of the sled is 3 m/s².
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a substance that is made up of only one kind of atom is an?
It is an element
that his the answer :)