The Pearson correlation coefficient, r, measures the strength and direction of the linear relationship between two variables. In this case, a correlation coefficient of r = 0.43 was obtained for a sample of 20 participants. The correlation is not statistically significant at both alpha levels of 0.01 and 0.05.
Explanation:The Pearson correlation coefficient, r, measures the strength and direction of the linear relationship between two variables. In this case, the researcher obtained a correlation coefficient of r = 0.43 for a sample of n = 20 participants. To determine the significance of the correlation, a hypothesis test can be conducted at different alpha levels. At an alpha level of 0.01, the critical value for a two-tailed test with 18 degrees of freedom is 2.898.
The calculated t-value using the formula r * sqrt((n-2)/(1-r^2)) is compared to the critical value. If the t-value is less than the critical value, the correlation is not statistically significant. In this example, the calculated t-value is less than 2.898, indicating that the correlation is not statistically significant at the 0.01 level.
At an alpha level of 0.05, the critical value for a two-tailed test with 18 degrees of freedom is 2.101. Since the calculated t-value is less than 2.101, the correlation is not statistically significant at the 0.05 level as well.
To determine if the Pearson correlation of r = 0.43 is significant for n = 20, one must calculate the t-statistic and compare it to the critical t values from a t distribution table for df = 18 for each alpha level (α). These critical values determine whether the correlation is significant at the .05 or .01 level.
Explanation:To determine the significance of a Pearson correlation coefficient r for a given alpha level (α), one can use the t distribution as a reference. In the given scenario, the researcher has obtained a Pearson correlation of r = 0.43 with a sample size of n = 20 participants. Before evaluating significance, one must calculate the t-statistic using the formula given:
t = r√n-2/√(1-r²)
The degrees of freedom (df) for this calculation could be determined by df = n - 2, which here would be 20 - 2 = 18. Upon comparing the calculated t value to the critical t value from a t distribution table for df = 18, one can establish the significance of the correlation coefficient.
Looking at a standard t distribution table or using statistical software, one would find the critical t values for a two-tailed test. For α = 0.05, the critical t value for df = 18 would typically be around 2.101. For α = 0.01, this value would be higher, representing a stricter criterion for significance. If the calculated t value is greater than the critical value, the null hypothesis, which proposes no significant correlation, is rejected.
Given the details of the Pearson correlation in the question, without the exact t value calculated, it is not possible to definitively say whether the r = 0.43 is significant at the α = 0.05 or 0.01 level. However, provided that the calculated t value is larger than the critical t value at either α level, the null hypothesis would be rejected, indicating a significant correlation.
Why is the halo of a galaxy the most important part of a galaxy?
A car of mass 960 kg is free-wheeling down an incline at a constant speed of 9.0 m s–1.
The slope makes an angle of 15° with the horizontal.
Deduce that the average resistive force acting on the car is 2.4×103N
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1. This question is about the breaking distance of a car and specific heat capacity. (a) A car of mass 960 kg is free-wheeling down an incline at a constant speed of 9.0 m s–1. The slope makes an angle of 15° with the horizontal. (i) Deduce that the average resistive force acting on the car is 2.4×103N. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) (ii) Calculate the kinetic energy of the car. ........................................................................................................................... ........................................................................................................................... (1) (b) The driver now applies the brakes and the car comes to rest in 15 m. Use your answer to (a)(ii) to calculate the average braking force exerted on the car in coming to rest. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (c) The same braking force is applied to each rear wheel of the car. The effective mass of each brake is 5.2 kg with a specific heat capacity of 900 J kg–1 K–1....
Attachments:
Homework-10-d....rtf
i) The average resistive force acting on the car is approximately 2.4 × 10³ N. ii) The kinetic energy of the car is 3,888 J. b) The average braking force exerted on the car is coming to rest is approximately 259.2 N. ii) The average braking force exerted on the car is coming to rest is approximately 259.2 N. 2) c) The change in temperature of each rear brake is approximately 0.907 Kelvin.
i) Given information:
Mass of the car (m) = 960 kg
Speed of the car (v) = 9.0 m/s
The angle of the slope (θ) = 15°
Resistive force (Fr) = m × g × sin(θ)
Fr = 960 × 9.8 × sin(15°)
Fr = 2.4 × 10³ N
Therefore, the average resistive force acting on the car is approximately 2.4 × 10³ N.
(ii) To calculate the kinetic energy of the car, we can use the formula:
Kinetic energy (KE) = (1/2) × m × v²
KE = (1/2) × 960 × (9.0)²
KE = 3,888 J
Therefore, the kinetic energy of the car is 3,888 J.
(b) The average braking force exerted on the car is:
Average braking force = Change in kinetic energy/braking distance
Average braking force = KE / braking distance
Average braking force = 3,888 / 15
Average braking force = 259.2 N
Therefore, the average braking force exerted on the car in coming to rest is approximately 259.2 N.
(ii) The work done by a force can be calculated using the formula:
Work = Force × Distance × cos(θ)
3,888 = Force × 15 × cos(0°)
Since cos(0°) = 1, the equation simplifies to:
3,888 = Force × 15
Force = 3,888 ÷ 15
Force = 259.2 N
Therefore, the average braking force exerted on the car in coming to rest is approximately 259.2 N.
2) (c) To calculate the change in temperature of each rear brake, we can use the formula:
Change in heat energy = mass × specific heat capacity × change in temperature
Change in heat energy = Work
The work done by the braking force is equal to the product of the force and the braking distance (15 m),
Therefore, the change in heat energy for each brake is:
Change in heat energy = Work = Force × braking distance
Change in heat energy = 259.2 × 15 = 3,888 J
Now, we can use the formula to find the change in temperature:
Change in heat energy = mass × specific heat capacity × change in temperature
3,888 = 5.2 × 900 × change in temperature
change in temperature ≈ 0.907 K
Therefore, the change in temperature of each rear brake is approximately 0.907 Kelvin.
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Four 100 Watt bulbs connected in series will approximately consume in ten hours an energy of
A 10cm x 10cm x 10cm wood block with a density of 700 kg/m^3 floats in water.
What is the distance from the top of the block to the water in cm if the water is fresh, what if it is sea water?
The distance from the top of the wood block to the water can be calculated using the buoyancy force. For fresh water, the distance is 0.1 cm and for sea water, it is 0.09709 cm.
Explanation:To determine the distance from the top of the wood block to the water, we need to consider the buoyancy force acting on the block. When an object is submerged in a fluid, it experiences an upward force called buoyant force equal to the weight of the fluid displaced by the object.
The density of the wood block is given as 700 kg/m^3. Since the block floats, the density of the wood is less than the density of water (1000 kg/m^3). Therefore, the volume of the wood block is the same as the volume of water it displaces.
To find the distance from the top of the block to the water, we first need to calculate the volume of the block. The volume formula for a cube is V = s^3, where s is the length of a side. In this case, the side length of the block is 10 cm, so the volume is 10 cm x 10 cm x 10 cm = 1000 cm^3.
Next, we can calculate the weight of the water displaced by the block. The weight of an object is given by the formula W = m x g, where m is the mass and g is the acceleration due to gravity. The mass of the displaced water can be found using the density formula m = d x V, where d is the density and V is the volume. In this case, the density of water is 1000 kg/m^3 and the volume of the block is 1000 cm^3. Converting the volume to m^3, we get 0.001 m^3.
Finally, we can calculate the distance from the top of the block to the water using the formula h = (W/Water density) x (100 cm/1 m). Plugging in the values, we get h = (0.001 m^3 x 1000 kg/m^3) / (1000 kg/m^3) x (100 cm/1 m) = 0.001 m x 100 cm = 0.1 cm.
Therefore, the distance from the top of the block to the water is 0.1 cm when the water is fresh. For sea water, the density is higher at around 1030 kg/m^3. Using the same calculations, we get h = (0.001 m^3 x 1000 kg/m^3) / (1030 kg/m^3) x (100 cm/1 m) = 0.001 m x 97.09 cm = 0.09709 cm.
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Two equal-mass stars maintain a constant distance apart of 8.0 x 1010 m and rotate about a point midway between them at a rate of one revolution every 12.6 yr.
(a) Why don’t the two stars crash into one another due to the gravitational force between them?
(b) What must be the mass of each star?
Stars do not crash into each other because the gravitational force acts as a centripetal force, maintaining circular orbits and keeping them apart. The masses of the stars can be calculated using Kepler's third law for circular orbits.
The scenario provided describes a system in which two equal-mass stars maintain a constant distance apart and revolve around their common center of mass due to the gravitational force between them. The system is in balance when the gravitational force provides the necessary centripetal force to keep the stars in their circular orbits without them crashing into each other.
Part (a)
The two stars don't crash into one another because the gravitational attraction is providing the centripetal force that keeps them in stable, circular orbits around their common center of mass. This balance between gravitational forces and the stars' inertia maintains the separation between them.
Part (b)
To determine the mass of each star, we can use Kepler's third law and Newton's version of it for two bodies orbiting each other, which states:
[tex]T^2 = (4*pi^2*r^3)/(G*(m1+m2))[/tex]
Since m1 = m2 (equal masses), we can solve for a single mass. By substituting the given values (radius and period), we can calculate the mass of each star.
(a) The two stars do not crash into each other because they are in a stable orbit, rotating around their common center of mass (barycenter). (b) The mass of each star must be approximately [tex]9.61 \times 10^{11} \, \text{kg}[/tex].
To address the questions about the two stars, we need to use concepts from orbital mechanics and Newtonian gravity. Here's the step-by-step approach:
(a) Why don’t the two stars crash into one another due to the gravitational force between them?
The two stars do not crash into each other because they are in a stable orbit, rotating around their common center of mass (barycenter). The gravitational force between the stars provides the necessary centripetal force to keep them in this orbit.
(b) What must be the mass of each star?
To find the mass of each star, we can use the following relationship between gravitational force and centripetal force. First, let's summarize the given data:
- Distance between the stars, [tex]d = 8.0 \times 10^{10} \, \text{m}[/tex]
- Period of revolution, T = 12.6 yr
Convert the period T into seconds:
[tex]T = 12.6 \, \text{yr} \times \left(365.25 \, \text{days/yr}\right) \times \left(24 \, \text{hours/day}\right) \times \left(3600 \, \text{seconds/hour}\right) \\\\T \approx 12.6 \times 365.25 \times 24 \times 3600 \, \text{s} \approx 3.97 \times 10^8 \, \text{s}[/tex]
Each star rotates around the common center of mass with a radius r that is half the distance between the stars, so:
[tex]r = \frac{d}{2} = \frac{8.0 \times 10^{10} \, \text{m}}{2} = 4.0 \times 10^{10} \, \text{m}[/tex]
The gravitational force F providing the centripetal force [tex]F_c[/tex] can be equated as follows:
F = [tex]F_c[/tex]
The gravitational force between the two stars is given by Newton's law of gravitation:
[tex]F = \frac{G m_1 m_2}{d^2}[/tex]
Since the stars have equal mass ( m₁ = m₂ = m ):
[tex]F = \frac{G m^2}{d^2}[/tex]
The centripetal force required to keep each star in orbit is:
[tex]F_c = m \frac{v^2}{r}[/tex]
Where v is the orbital velocity of each star. The orbital velocity v can be related to the period T and the radius r by:
[tex]v = \frac{2 \pi r}{T}[/tex]
Now substituting v into the centripetal force equation:
[tex]F_c = m \left( \frac{2 \pi r}{T} \right)^2 \frac{1}{r} \\\\F_c = m \frac{4 \pi^2 r^2}{T^2 r} \\\\F_c = m \frac{4 \pi^2 r}{T^2}[/tex]
Since F = [tex]F_c[/tex] :
[tex]\frac{G m^2}{d^2} = m \frac{4 \pi^2 r}{T^2}[/tex]
We can cancel one m from each side:
[tex]\frac{G m}{d^2} = \frac{4 \pi^2 r}{T^2}[/tex]
Solving for m :
[tex]m = \frac{4 \pi^2 r d^2}{G T^2}[/tex]
We know:
[tex]r = 4.0 \times 10^{10} \, \text{m} \\\\d = 8.0 \times 10^{10} \, \text{m} \\\\G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2} \\\\T = 3.97 \times 10^8 \, \text{s}[/tex]
Substitute these values in:
[tex]m = \frac{4 \pi^2 (4.0 \times 10^{10} \, \text{m}) (8.0 \times 10^{10} \, \text{m})^2}{(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2}) (3.97 \times 10^8 \, \text{s})^2}[/tex]
First calculate [tex]r \cdot d^2[/tex]:
[tex]r \cdot d^2 = 4.0 \times 10^{10} \, \text{m} \times (8.0 \times 10^{10} \, \text{m})^2 = 4.0 \times 10^{10} \, \text{m} \times 64.0 \times 10^{20} \, \text{m}^2 = 256.0 \times 10^{30} \, \text{m}^3[/tex]
Then calculate T² :
[tex]T^2 = (3.97 \times 10^8 \, \text{s})^2 = 15.76 \times 10^{16} \, \text{s}^2[/tex]
Now, plug in the values:
[tex]m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{6.674 \times 10^{-11} \times 15.76 \times 10^{16}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{1.05136 \times 10^6 \times 10^{16}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{1.05136 \times 10^{22}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{8}}{1.05136} \\\\m = \frac{4 \times 9.8696 \times 256.0 \times 10^{8}}{1.05136} \\\\m = \frac{10102.4 \times 10^{8}}{1.05136} \\\\m \approx 9.61 \times 10^{11} \, \text{kg}[/tex]
Thus, the mass of each star must be approximately [tex]9.61 \times 10^{11} \, \text{kg}[/tex].
When blue light of wavelength 470 falls on a single slit, the first dark bands on either side of center are separated by 45.0. determine the width of the slit
The width of the slit can be determined using the formula θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. By rearranging the formula and substituting the given values, we find that the width of the slit is approximately 244 nm.
Explanation:The width of the slit can be determined using the formula for the angle of the first minimum in a single-slit diffraction pattern, which is given by θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. In this case, the angle is 45.0 degrees and the wavelength is 470 nm. Converting the angle to radians and rearranging the formula, we can solve for the width of the slit:
d = λ / (2 * tan(θ))
Substituting the given values, we have:
d = (470 nm) / (2 * tan(45.0 degrees))
Calculating this, we find that the width of the slit is approximately 244 nm.
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What is the acceleration of a 32-kg object if a 6.75-N force is applied to it?
A projectile of mass m is launched with an initial velocity vector v i making an angle θ with the horizontal as shown below. The projectile moves in the gravitational field of the Earth. Find the angular momentum of the projectile about the origin when the particle is at the following locations. (Use the following as necessary: vi, θ, m, and g for the acceleration due to gravity.)
(a) at the origin
L with arrow =
rmvcos(θ)
Incorrect: Your answer is incorrect.
(b) at the highest point of its trajectory
L with arrow =
(c) just before it hits the ground
L with arrow =
The angular momentum of the projectile about the origin when the particle is at the following locations is zero
Explanation:A projectile of mass m is launched with an initial velocity making an angle θ with the horizontal as shown below (attached). The projectile moves in the gravitational field of the Earth. The angular momentum of the projectile about the origin when the particle is at the following locations. (Use the following as necessary: vi, θ, m, and g for the acceleration due to gravity.)
[tex]L=r x p[/tex]
Angular momentum is zero because there is no r. Since the angular momentum depends on then cross product of the position vector and momentum vector which is in the same direction as velocity, therefore the formula for the angular momentum can be written as [tex]mvr_{perpendicular}[/tex], where the cross product component of r vector along/parallel to velocity/momentum is 0.
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A 10.00 g sample of a compound containing only carbon, hydrogen, and oxygen forms 23.98 g CO2 and 4.91 g H2O upon complete combustion. What is the empirical formula of the compound?
What is the approximate weight of a 20-kg cannonball on earth?
a. 2 n
b. 20 n
c. 200 n
d. 2000 n?
Answer: The weight of the cannonball on Earth is 200 N
Explanation:
Weight is defined as the force exerted by the body on any surface. It is also defined as the product of mass of the body multiplied by the acceleration due to gravity.
Mathematically,
[tex]W=mg[/tex]
where,
W = weight of the cannonball
m = mass of the cannonball = 20 kg
g = acceleration due to gravity on Earth = [tex]10m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=20kg\times 10m/s^2=200N[/tex]
Hence, the weight of the cannonball on Earth is 200 N
What is an expression for the difference in energy between state c and state d?
A 2.10-kg block starts from rest at the top of a 30.0° incline and slides a distance of 1.90 m down the incline in 1.40 s.
(a) Find the magnitude of the acceleration of the block.
(b) Find the coefficient of kinetic friction between block and plane
(c) Find the friction force acting on the block.
magnitude
N
direction
---Select--- up the incline down the incline normal to the incline and upward normal to the incline and downward
d) Find the speed of the block after it has slid 1.90 m.
Final answer:
To find the acceleration, the kinematic equation is used. The kinetic friction coefficient is derived from force equations, and the friction force can be calculated with this coefficient and the normal force. Finally, the block's final speed after sliding down the incline is determined using another kinematic equation.
Explanation:
Finding the Magnitude of Acceleration, Coefficient of Kinetic Friction, Friction Force, and Final Speed
To solve for the acceleration of a 2.10-kg block sliding down a 30.0° incline, we can use the kinematic equation s = ut + (1/2)at², where s is the distance slid which is 1.90 m, u is the initial velocity which is 0 m/s, t is the time which is 1.40 s, and a is the acceleration. Solving for acceleration, a = 2s/t², we find:
a = (2 * 1.90 m) / (1.40 s)²
a = 3.80 m / 1.96 s²
a = 1.9388 m/s²
The coefficient of kinetic friction μ can be determined by using the equation f_k = μ * N, where f_k is the kinetic friction force and N is the normal force. The gravitational component parallel to the incline is mg sin(θ) and perpendicular to the incline is mg cos(θ). Knowing the total force equation, mg sin(θ) - μ*mg cos(θ) = ma, we can isolate μ and solve for it.
The friction force then can be calculated using f_k = μ * mg cos(θ), and the direction of this force is up the incline since it opposes the motion of the block sliding down.
Finally, to find the final speed v of the block after sliding 1.90 m, we use the kinematic equation v² = u² + 2as. Since the initial speed u is 0, this simplifies to v = √(2as), and plugging in the known values for a and s gives us the final speed.
A 15.0-lb block rests on the floor. (a) What force does
the floor exert on the block? (b) A rope is tied to the
block and is run vertically over a pulley. The other end
is attached to a free-hanging 10.0-lb object. What now
is the force exerted by the floor on the 15.0-lb block?
(c) If the 10.0-lb object in part (b) is replaced with a
20.0-lb object, what is the force exerted by the floor on
the 15.0-lb block?
A 15.0 lbs block rests on the floor. Then, the force exerted on the block is in the upward direction to counterbalance the Earth's force on the block. If a rope is tied to it, then the force from the floor and the rope together balance the weight.
What is Force?A force is an external agent which is capable of changing an object's state of rest or motion. Force has both the magnitude and the direction, hence it is a vector quantity. The direction towards which this force is applied is known as the direction of the force, and the application of force is the point at which the force is being applied.
The force exerted on the block by the floor is 15.0lb upwards, to counterbalance the Earth’s force on the block. If a rope is tied to the block, and is running vertically over a pulley, and other object of 10lb is attached to it, then a force of 5.00lb towards upward direction is exerted by the floor on the block. The Earth pulling it down with 15.0lb and the rope pulling it up with a force of 10.0lb. The forces from the floor and rope together balance the weight of objects.
If the 10 lb object is replaced with a 20lb object, then the force exerted by the floor on the 15lb block is 0, because the block now accelerates up away from the floor.
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Which best describes why winds at the beach change direction between day and night
Is newspaper turning yellow over time a physical change or a chemical change
A circular plate of 500-mm diameter is maintained at T1 = 600 K and is positioned coaxial to a conical shape. The back side of the cone is well insulated. The plate and the cone, whose surfaces are black, are located in a large, evacuated enclosure whose walls are at 300 K.
(a) What is the temperature of the conical surface, T2?
(b) What is the electrical power that would be required to maintain the circular plate at 600K?
The student’s questions involve the principles of thermal radiation and heat conduction in Physics, relating to the determination of the equilibrium temperature in a radiative system and the electrical power needed to maintain a temperature. Calculations require additional data on materials and geometry to provide accurate answers.
Explanation:The student's questions pertain to the topics of thermal radiation and heat conduction in Physics, which are within the realm of thermodynamics. Specifically, they are asking about (a) the equilibrium temperature of a conical surface and (b) the power required to maintain a certain temperature on a circular plate. These calculations involve understanding Stefan-Boltzmann law, conduction, and power calculations.
Unfortunately, without further information or input parameters for materials and geometries, providing a definitive answer for a system like the one described (circular plate and conical surface in an enclosure) isn't possible. To determine the electrical power required, one would need to know properties like thermal conductivity, surface area, and possibly the geometry of the objects in question to find out the heat transfer rates.
The refracting telescope at the Yerkes Observatory has a 1.00-m diameter objective lens of focal length 20.0 m. Assume it is used with an eyepiece of focal length 1.90 cm. (a) Determine the magnification of Mars as seen through this telescope.
A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0700 m, an angular speed of 88.0 rad/s, and a moment of inertia of 0.850 kg · m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 4.40 s.(a) Find the magnitude of the angular deceleration of the cylinder. rad/s 2 (b) Find the magnitude of the force of friction applied by the brake shoe. N
The magnitude of the angular deceleration of the cylinder is -10.0 rad/s^2, and the magnitude of the force of friction applied by the brake shoe is -8.50 N.
Explanation:(a) To find the magnitude of the angular deceleration of the cylinder, we can use the formula:
angular deceleration = (final angular velocity - initial angular velocity) / time
Given that the initial angular speed is 88.0 rad/s, the final angular speed is half of that, and the time is 4.40 s, we can calculate:
angular deceleration = (44.0 rad/s - 88.0 rad/s) / 4.40 s = -10.0 rad/s^2
(b) To find the magnitude of the force of friction applied by the brake shoe, we can use the formula:
force of friction = moment of inertia * angular deceleration
Given that the moment of inertia is 0.850 kg·m², and the angular deceleration is -10.0 rad/s², we can calculate:
force of friction = 0.850 kg·m² * -10.0 rad/s² = -8.50 N
You pull upward on a stuffed suitcase with a force of 105N , and it accelerates upward at 0.710m/s2, what is the wieght and mass of the suitcase?
Suppose the suitcase is on a rocket. The rocket launches with an acceleration of 29.0m/s2, What is the apparent weight of an 86-kg astronaut aboard this rocket?
1. The weight of the suitcase is approximately 1448.32 N.
2. The apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.
To find the weight and mass of the suitcase, we'll use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Given:
Force (F) = 105 N
Acceleration (a) = 0.710 m/[tex]s^2[/tex]
We can use the equation F = ma to solve for mass (m):
105 N = m * 0.710 m/s^2
Rearranging the equation, we have:
m = F / a
m = 105 N / 0.710 m/[tex]s^2[/tex]
m ≈ 147.89 kg
So, the mass of the suitcase is approximately 147.89 kg.
To find the weight of the suitcase, we can use the equation W = mg, where W is the weight and g is the acceleration due to gravity (approximately 9.8 m/[tex]s^2[/tex]).
W = 147.89 kg * 9.8 m/[tex]s^2[/tex]
W ≈ 1448.32 N
Therefore, the weight of the suitcase is approximately 1448.32 N.
Now let's calculate the apparent weight of the astronaut aboard the rocket.
Given:
Mass (m) = 86 kg
Acceleration (a) = 29.0 m/[tex]s^2[/tex]
Using the same formula W = mg, we can find the apparent weight (W) of the astronaut:
W = m * a
W = 86 kg * 29.0 m/[tex]s^2[/tex]
W ≈ 2504 N
Therefore, the apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.
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What value resistor will discharge a 2.80 μf capacitor to 10.0% of its initial charge in 2.50 ms ?
Examine the statement.
Two balls of the same mass are thrown at different velocities. Ball A has a higher velocity than ball B.
Which statement best describes the kinetic energy of the balls?
Hint: KE = 1/2mv2
The kinetic energy of Ball A and Ball B is the same.
Ball B has a higher kinetic energy than Ball A.
Ball A has a higher kinetic energy than Ball B.
Ball A and Ball B do not have a kinetic energy.
A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.30mmcos[(6.98rad/m)x + (742 rad/s)t]. Being more practical, you measure the rope to have a length of 1.35 m and a mass of 0.00338 kg. You are then asked to determine the following:
A) Amplitude
B) Frequency
C) Wavelength
D) Wave Speed
E) Direction the wave is traveling
F) Tension in the rope
G) Average power transmitted by the wave
The amplitude is 2.30mm, frequency is 118 Hz, wavelength is 0.899 m, wave speed is 106 m/s, the wave is moving in the direction of decreasing x, tension in the rope is 28 N and the average power transmitted by the wave is approximately 0.38W.
Explanation:The wave function of a travelling wave on a thin rope is given by y(x,t) = 2.30mm * cos[(6.98rad/m) * x + (742 rad/s) * t]. From this equation, we can derive the following information:
Amplitude - This is the maximum displacement of a point on the wave and is determined by the coefficient before the cosine in the equation. Therefore, the amplitude is 2.30mm.
Frequency - This can be obtained by dividing the wave speed by the wavelength. The coefficient in front of the 't' in the equation is equivalent to 2π times the frequency (since there are 2π rad in one wave cycle). Therefore, the frequency f = (742 rad/s)/(2π) = 118 Hz.
Wavelength - This is represented by the reciprocal of the coefficient in front of 'x' in the equation. Therefore, the wavelength λ = 2π/(6.98 rad/m) = 0.899 m.
Wave Speed - This can be found by multiplying frequency and wavelength together. Hence, wave speed v = f * λ = 118 Hz * 0.899 m = 106 m/s.
Direction the wave is travelling - The symbol '+' in front of 't' in the equation tells us that the wave is moving in the direction of decreasing x.
Tension in the rope - This can be found using the wave speed that we have already calculated and the rope's mass per unit length µ (mass/length = 0.00338 kg / 1.35 m = 0.00250 kg/m). The tension T can be calculated using the formula v = sqrt(T/µ), which gives T = µ * v² = 0.00250 kg/m * (106 m/s)² = 28 N.
Average power transmitted by the wave - The power P carried by wave is given by the formula P = 0.5 * µ * v * A² * ω², where A is the amplitude, ω is the angular frequency (which is the coefficient in front of 't' in the equation). Substituting the values we find: P = 0.5 * 0.00250 kg/m * 106 m/s * (2.30 mm)² * (742 rad/s)² ≈ 0.38 W.
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A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is the acceleration of the ride?
4.0 m/s2
0.67 m/s2
0.075 m/s2
54 m/s2
should an electrician wear gloves made of a metal or a nonmetal element when working around electrical wires explain your answer
Light propagating in the glass n1= 1.65 wall of an aquarium tank strikes the interior edge of the wall with incidence angle 19.0 degrees . What is the angle of refraction in the water?
To find the angle of refraction in water when light propagates in a glass wall, we can use Snell's law.
Explanation:To find the angle of refraction in water, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the indices of refraction of the two media:
sin(θ1) / sin(θ2) = n2 / n1
Given that the incidence angle in glass is 19.0 degrees and the index of refraction of glass is 1.50, we can substitute these values into the equation:
sin(19.0) / sin(θ2) = 1.333 / 1.50
Solving for θ2, the angle of refraction in water, we find:
θ2 ≈ 13.14 degrees
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The angle of refraction in the water is approximately 24.21 degrees.
Explanation:When light passes from one medium to another, it changes direction. This change in direction is called refraction. Snell's Law describes the relationship between the angles of incidence and refraction. The equation is: n1*sin(θ1) = n2*sin(θ2), where n1 and n2 are the indices of refraction of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.
In this case, light is propagating in glass with an index of refraction of 1.65 and is striking the interface between the glass and water at an angle of 19.0 degrees. The index of refraction of water is 1.333. To find the angle of refraction in the water, we can use Snell's Law:
1.65*sin(19.0) = 1.333*sin(θ2).
Solving for θ2, we find that the angle of refraction in the water is approximately 24.21 degrees.
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Select the answer choices that describe where plasmas can be found.
(Select all that apply.)
in stars
in fireworks
in lightning
in lightbulbs
Answer:
A + C + D
Explanation:
A) in stars
C) in lightning
D) in light-bulbs
(P.S. I just took a test on this and got it correct)
From the small number of impact craters on ____, planetary geologists suspect that virtually all its original surface has been covered over by younger volcanic activity
Planetary geologists suspect that the original surface of a certain celestial body has been covered by younger volcanic activity, based on the small number of impact craters. Evidence of past impact events is gradually erased by the constant renewal of Earth's crust through plate tectonics.
Explanation:Based on the small number of impact craters, planetary geologists suspect that virtually all the original surface of a certain celestial body has been covered by younger volcanic activity.
This can be observed in the case of Earth, where the constant renewal of its crust through plate tectonics gradually erases evidence of past impact events.
Geologists have identified eroded remnants of impact craters on Earth, providing evidence for the influence of these impacts on the planet's evolution over time.
Draw a picture of the bohr atom and label the three particles that comprise it
Below is a picture of the bohr atom and the three particles that comprise it.
When an electron moves from one energy level to another, it either absorbs or emits a photon of electromagnetic radiation.
The Bohr model of the atom is a description of the structure of atoms, especially that of hydrogen, proposed by the physicist Niels Bohr in 1913. The model was a radical departure from earlier, classical descriptions of the atom, and it was the first that incorporated quantum theory.
Therefore, Bohr model of the atom consists of a small, dense nucleus surrounded by orbiting electrons. The nucleus is positively charged, and the electrons are negatively charged. The electrons are attracted to the nucleus by the electromagnetic force, but they are also moving around the nucleus, so they are not pulled in.
When you throw a ball the work you do to accelerate it equals the kinetic energy the ball gains?
A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.
a) How much work is done by the worker on the crate?
b) How much work is fone by the floor on the crate?
c) What is the net work done on the crate?
(a) The work done by the worker on the crate is 8,280 J.
(b) The work done by the floor on the crate is 7,920 J.
(c) The net work done on the crate is 360 J.
Work done by the workerThe total work done by the worker is calculated as follows;
[tex]W = Fd\\\\W = 345 \times 24\\\\W = 8,280 \ J[/tex]
Work done by the floor on the crateThe work done by the floor on the crate is calculated as follows;
[tex]W = F_f d\\\\W = \mu F_n d\\\\W = 0.22 \times 1.5 \times 10^3 \times 24\\\\W = 7,920 \ J[/tex]
Net work done on the crateThe net work done on the crate is calculated as follows;
[tex]W = 8,280 - 7,920\\\\W = 360 \ J[/tex]
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The work done by the worker is 8280 Joules. The work done by the floor (accounting for friction) is -7920 Joules. This makes the net work done on the crate 360 Joules.
Explanation:The question is asking about the concept of work and net work in physics, specifically in the context of friction and forces. To start, we need to understand that work is defined as the force applied on an object times the distance that object moves in the direction of the force (Work = Force x Distance).
a) The work is done by the worker on the crate can be calculated using the formula Work = Force x Distance. Hence, Work = 345 N x 24.0 m = 8280 Joules.
b) The work done by the floor on the crate is the frictional force times the distance. The frictional force can be calculated using the formula Force = Coefficient of friction x Normal force. In this case, the normal force is the weight of the crate (1.50 x 10^3 N). Hence, Frictional force = 0.220 x (1.50 x 10^3 N). Work done by the floor is Frictional force x Distance, so the work done by the floor = Frictional force x 24.0 m = -7920 Joules (It's negative because the force of friction acts in the opposite direction to motion).
c) The net work done on the crate is the sum of the work done by the worker and the work done by the floor. Hence, Net work = 8280 Joules - 7920 Joules = 360 Joules.
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