Weight of a person's muscles, bones, tendons, and ligaments.
A. flexibility
B. lean mass
C. aerobic
What type of research based on approach that used to describe variables rather than to test a predicted relationship between variables?
Answer:
Correlational research can be used to see if two variables are related and to make predictions based on this relationship.
g (a) Calculate the block's final speed when it reaches the bottom of the frictionless inline. Keep 2 decimal places.
Answer:
Explanation:
The question is incomplete. Here is the complete question with attachment below.
In the above figure on the LEFT side, a block (mass = 2.8 kg) starts from rest at the top of a frictionless inline and slides to the bottom. The height of the incline is h=6.5 m, angle
A sleigh is being pulled horizontally by a train of horses at a constant speed of 6.38 m/s. The magnitude of the normal force exerted by the snow-covered ground on the sleigh is 7.50 ✕ 103 N.
(a) If the coefficient of kinetic friction between the sleigh and the ground is 0.26, what is the magnitude of the kinetic friction force experienced by the sleigh?
N
(b) If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh, what must be the magnitude of this force?
N
Answer:
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
Explanation:
Given that,
Constant speed = 6.38 m/s
Force [tex]F=7.50\times10^{3}\ N[/tex]
Kinetic friction = 0.26
(a). We need to calculate the friction force
Using formula of friction force
[tex]f_{k}=\mu F_{N}[/tex]
Put the value into the formula
[tex]f_{k}=0.26\times7.50\times10^{3}[/tex]
[tex]f_{k}=1950\ N[/tex]
(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,
We need to calculate the magnitude of this force
According to given data,
The same force will be applied to keep constant velocity.
Hence, (a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force.
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
The calculation is as follows;a. The magnitude of the kinetic friction force experienced by the sleigh is
[tex]= 0.76 \times 7.50 \times 10^3[/tex]
= 1950 N
b. It should be equivalent to the friction force.
Learn more: https://brainly.com/question/24908711?referrer=searchResults
One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?
Answer:
The one with the faster velocity is the one with a velocity of -10m/s
What's the difference between an open cluster and a globular cluster
An open cluster is a group of up to a few thousand stars that were formed from the same giant molecular cloud, and are still loosely gravitationally bound to each other. In contrast, globular clusters are very tightly bound by gravity. ... Open clusters are very important objects in the study of stellar evolution.
what phase changes take place when you are adding energy to the substance
Answer:
During a phase change, a substance undergoes transition to a higher energy state when heat is added, or to a lower energy state when heat is removed. Heat is added to a substance during melting and vaporization. Latent heat is released by a substance during condensation and freezing. Explanation:
A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1.3 dB. What is the power output P
Answer:
The value of the power is [tex]P_c = 38.55 \ W [/tex]
Explanation:
From the question we are told that
The power rating [tex]P_{1000} =P_b= 52 \ W[/tex]
The frequency is [tex]f = 1000 \ Hz[/tex]
The frequency at which the sound intensity decreases [tex]f_k = 20 \ Hz[/tex]
The decrease in intensity is by [tex]\beta = 1.3 dB[/tex]
Generally the initial intensity of the speaker is mathematically represented as
[tex]\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ][/tex]
Generally the intensity of the speaker after it has been decreased is
[tex]\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ][/tex]
So
[tex]\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ][/tex]
=> [tex]\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3[/tex]
=> [tex]\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3[/tex]
=> [tex]\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3[/tex]
=> [tex]10log_{10} [\frac{P_b}{P_c} ] = 1.3[/tex]
=> [tex]log_{10} [\frac{P_b}{P_c} ] = 0.13[/tex]
taking atilog of both sides
[tex][\frac{P_b}{P_c} ] = 10^{0.13}[/tex]
=>[tex][\frac{52}{P_c} ] = 10^{0.13}[/tex]
=> [tex]P_c = \frac{52}{1.34896}[/tex]
=> [tex]P_c = 38.55 \ W [/tex]
3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?
Answer:
[tex]E_s = 277.13V[/tex]
Explanation:
Given
[tex]Load\ Voltage = 480V[/tex]
Required
Determine the voltage dropped in each stage.
The relation between the load voltage and the voltage dropped in each stage is
[tex]E_l = E_s * \sqrt3[/tex]
Where
[tex]E_l = 480[/tex]
So, we have:
[tex]480 = E_s * \sqrt3[/tex]
Solve for [tex]E_s[/tex]
[tex]E_s = \frac{480}{\sqrt3}[/tex]
[tex]E_s = \frac{480}{1.73205080757}[/tex]
[tex]E_s = 277.128129211[/tex]
[tex]E_s = 277.13V[/tex]
Hence;
The voltage dropped at each phase is approximately 277.13V
Define reflection of sound?
Explanation:
When sound travels in a given "medium", it would touch the surface of another "medium" and will bounce back in some other direction, this occurrence is called the reflection of sound.
How much would a 15.0 kg object weigh on that planet? Round the answer to the nearest whole number.
Answer:
168
Explanation:
Answer: a 15 kg object would weigh the most on Neptune
168 N
the radius of the earth social
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef
Answer:
(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the bowling ball is 113.272 joules.
Explanation:
The statement is incomplete. The complete question is:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）
(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.
Now we expand the expression by definition of gravitational potential energy:
[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]
[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:
[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]
[tex]K_{2} = 102.974\,J[/tex]
The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]
[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]
[tex]\Delta W = 102.974\,J[/tex]
The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]
Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.
[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]
Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]
[tex]\Delta W = 10.298\,J[/tex]
Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:
[tex]\Delta W' = K_{2}+\Delta W[/tex]
([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])
[tex]\Delta W' = 113.272\,J[/tex]
The work done by the mattress on the bowling ball is 113.272 joules.
2. For a rotating rigid body, which of the following statements is NOT correct?
a. All points along a rotating rigid body move with constant linear speed.
b. Points along a rotating body have same angular velocities.
c. Points along a rotating body move through the same angle in equal time intervals.
d. All points have the same angular acceleration.
Answer:
dasgfwe
Explanation:
what happens to the temperature of water as time elapses? IF YOU ANSWER IT I WILL MARK YOU A BRAINLEST ANSWER
Answer:
I think it will get colder
Explanation:
Answer:
The water molecules go faster as it gets colder they go slower
Explanation:
trust me thats the answer
Which statement accurately describes impulse?
State corrrect ans
Answer:
2nd option on edge2021
Explanation:
Work Done by a Varying Force 06 Work and Energy
w of Energy
100%
2.) The force required to stretch a spring by 1 m from its unstretched length is
150 N. What is the force required to stretch the spring by 3 m?
A. 600 N
B. 450 N
C. 300 N
D. 200 N
Answer:
B. 450 N
Explanation:
Use Hooke's law:
F = kx
150 N = k (1 m)
k = 150 N/m
F = kx
F = (150 N/m) (3 m)
F = 450 N
Which fallacy is committed in the following argument?
Tell your representative that you don't want health care reform. If the health care reform is passed, private insurance companies won't be able to compete with the public options. With the private companies out of the picture, individuals will only be able to turn to the public option, and then it will be up to the government to decide whether we should live or die.
a.
begging the question
b.
slippery slope
c.
hasty generalization
d.
false dilemma
A pendulum can be formed by tying a small object, like a tennis ball, to a string, and then connecting the other end of the string to the ceiling. Suppose the pendulum is pulled to one side and released at t1. At t^2, the pendulum has swung halfway back to a vertical position. At t^3, the pendulum has swung all the way back to a vertical position. Rank the three instants in time by the magnitude of the centripetal acceleration, from greatest to least. Most of the homework activities will be Context-rich Problems.
Answer:
1- t^3
2- t^2
3- t1
Explanation:
The acceleration produced in a body, while travelling in a circular motion, due to change in direction of motion is called centripetal acceleration. The formula of the centripetal acceleration is as follows:
ac = v²/r
where,
ac = centripetal acceleration
v = speed
r = radius
for a constant radius the centripetal acceleration will be directly proportional to the speed of object. The speed of pendulum will be lowest at t1 due to zero speed initially. Then the speed will increase gradually having greater speed at t^2 and the highest speed and centripetal acceleration at t^3. Therefore, the three instants in tie can be written in following order from greatest centripetal acceleration to lowest:
1- t^3
2- t^2
3- t1
Problem 9.23 A uniform-density 8 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 36 N through a distance of 0.9 m. Now what is the angular speed?
Answer:
4.63rad/s
Explanation:
The centripetal force is expressed as;
F = mv²/r
M is the mass of the disk = 8kg
v is the linear speed = wr
r is the radius = 0.21m
Force = 36N
Write the force in term of the angular velocity
F = m(wr)²/r
F = mw²r²/r
F = mw²r
36 = 8w²(0.21)
36 = 1.68w²
w² = 36/1.68
w² = 21.43
w = √21.43
w = 4.63rad/s
Hence the angular speed is now 4.63rad/s
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
Answer:
The two charged objects will exert equal and opposite forces on each other.
Explanation:
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.
This force of attraction or repulsion between the two charged objects is always equal and opposite.
Therefore, the two charged objects will exert equal and opposite forces on each other.
In which medium does the light move faster, water or diamond?
Help me out on this?
Luck walked to a store that is 250m away and it took him 50secs while Layne walked to the mall that is 1000m away and took her 200s to do. What do they have in common?
A. Average speed
B. Acceleration
C. Displacement
D.mass
Answer:
Average speed
Explanation:
250/50=5
1000/20=5
What do mammoths and tigers need energy for
N₂ + H₂
NH3
how do i balance this equation?
Answer:
N2 + 3H2 -----> 2NH3
Explanation:
Reactants side:
2 Nitrogen
5 Hydrogen
Products Side:
2 Nitrogen
5 Hydrogen
A plastic block of dimensions 2.00 cm x 3.00 cm x 4.00 cm has a mass of 30.0 g. What is its density?
Answer:
1.25 g/cm^3
Explanation:
mass-30.0g
volume- 4cm×2cm×3cm=24cm^3
density?
*to find density
Density=Mass/Volume
=30÷24
=1.25g/cm^3
If Mary runs 5 miles in 50 minutes, what is her speed with the correct
label?
Pls help pls pls pls pls
A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m
Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m