Ross training for a half marathon which is 13.1 miles long. Within four months he has progressed to an 11 mile practice run if Ross takes two months off from training what will likely occur?
If Ross takes two months off from training, his fitness level will reduce in comparison to what it was two months ago.
In as little as 3–4 weeks after beginning strength training, Ross will probably experience weight increase, energy loss, diminished balance, diminished strength (making it tougher to carry out daily tasks), and overall fewer fitness levels.Many people mistakenly believe they lose muscle mass far more quickly than they actually do because their muscles' ability to store water and glycogen is declining.A decrease in strength and muscle mass, with beginners experiencing a smaller decline in strength than experienced lifters.Ross will experience Increased VO2 Max from exercise. VO2 Max is almost completely lost in people who train at lower intensities.learn more about fitness here: https://brainly.com/question/13490156
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Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and constant acceleration Of 1m/s². Q Starts from N at the Same time with velocity 6m/s and at a Constant acceleration of 3m/s². find the time when the
a) Particles are 30m apart, 13
b) Particles meet
c) velocity of P is ¾ of the velocity of Q
The time required by the particles are as follows:
a. t = 1.5 seconds
b. t = 3 seconds
c. t = 0.4 seconds
What is the time required?The time required for the particles to be at several distances apart is calculated using the equation of motion given below:
[tex]S = ut + \frac{1}{2}at^{2}[/tex]
a) Time required to be 30 m apart:
Assuming the distance covered by P is S1 and distance covered by Q is S2.
S1 + S2 = 51 - 30
S1 + S2 = 21
Substituting the values of velocity and acceleration in the equation of motion above:
5t + 1/2t^2 + 6t + 3t^2 = 21
2t^2 + 11t - 21 = 0
Solving for time, t by factorization, t = 1.5 seconds
b) Time required to meet:
Assuming the distance covered by P is S1 and distance covered by Q is S2.
S1 + S2 = 51
Substituting the values of velocity and acceleration in the equation of motion above:
5t + 1/2t^2 + 6t + 3t^2 = 51
2t^2 + 11t - 51 = 0
Solving for time, t by factorization, t = 3 seconds
c) Time required for velocity of P is ¾ of the velocity of Q:
Using the equation of motion: V = u + at
Vp = 3/4 Vq
4Vp= 3Vq
Substituting the values:
4(5 + t) = 3(6 + 3t)
5t = 2
t = 0.4 seconds
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The practice of science can answer only scientific questions. And scientific
questions guide the design of investigations. What must be true of the
possible answers to a scientific question?
OA. They could be shown to be false by evidence.
B. They are at odds with established scientific theories.
C. They support an established scientific theory.
OD. They do not require the use of physical measurements.
SUBMIT
Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific topic.Option A is correct.
What is a Scientific hypothesis?A scientific hypothesis is the main element in the scientific method and experiment.
It is said that a hypothesis is a scientific hypothesis. The majority must agree on it, or it must be done after thorough experimentation.
Only scientific inquiries can have their answers provided by scientific endeavor. Scientific concerns have an impact on the research's design as well.
Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific question.
Hence option A is correct.
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Resolve the vector shown below into its components.
Answer:
D. s=3x+4y
Explanation:
The line is at the 3rd column in the 4th row.
For which length of wire are the reading of resistance most precise
Answer:
Explanation:
There are different ways to investigate the factors that affect resistance. In this practical activity, it is important to:record the length of the wire accuratelymeasure and observe the potential difference and currentuse appropriate apparatus and methods to measure current and potential difference to work out the resistance
(The question and setup are below)
The gauge pressure at bottom of vaccine solution will be 16 kPa
Positive pressure is another name for gauge pressure. When a system's internal pressure exceeds that of its surroundings, it is said to be under positive pressure. Any leak that develops in the positively pressured system will therefore escape into the outside world. In contrast, a negative pressure chamber draws air into it.
Given As seen in the illustration, a syringe is held vertically. The container carries a 3 cm tall column of vaccine solution and has an open inner diameter of 1 cm. The needle contains a 2 cm column of vaccine solution and has an open inner diameter of 0.5 mm. At the needle's open end, the solution is exposed to the air. The vaccination solution has a density of 1200 kg/m3.
We have to find the gauge pressure at bottom of vaccine solution
Since the 5N force is applied to vaccine solution the pressure exerted will be much more
Hence the gauge pressure at bottom of vaccine solution will be 16 kPa
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An object is released from an aeroplane which is diving at an angle of 30° from the horizontal with a speed of 50m/s. If the plane is at a height of
4000m from the ground when the object is released, find
(a) the velocity of the object when it hits the ground.
(b) the time taken for the object to reach the ground.
please I need the solution urgently.
a)The velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec
b) The time taken for the object to reach the ground will be 11.4 × 10³ sec.
What is projectile motion?The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.
The velocity in the x-direction;
[tex]\rm v_x = (vcos \theta)^2 +2gh \\\\ v_x = (50 cos 30)^2+ 2 \times 9.81 \times 4000 \\\\ v_x = 80355 m/sec[/tex]
Velocity in y-direction;
[tex]\rm v_ y = (vsin \theta)^2 + 2gh \\\\ v_ y = (25)^2+2 \times 9.81 \times 4000 \\\\ v_y =79105 \ m/sec[/tex]
The resultant velocity is found as 1.12 × 10⁵ m/sec.
The time taken to reach the ground is found as;
[tex]\rm v = u+gt \\\\ 1.12 \times 10^5 \ m/sec = 50 \ m/sec + 9.81 \times t \\\\ t = 11.4 \times 10^3 \ sec[/tex]
Hence, the velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec, and the time taken for the object to reach the ground will be 11.4 × 10³ sec.
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A uniform, 255 N rod that is 1.80 m long carries a 225 N weight at its right end and an unknown weight W toward the left end. When W is placed 60.0 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 75.0 cm from the right end.
Part a) Find W.
Part b) If W is now moved 20.0 cm to the right, how far must the fulcrum be moved to restore balance? Δx = ?
Part c) If W is now moved 20.0 cm to the right, in what direction must the fulcrum be moved to restore balance?
(a)The value of W for the given condition will be 237.3 N.
(b)The distance must the fulcrum be moved to restore balance will be 0.070 m.
What is the center of mass?A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.
Given data;
Weight is applied to the right end of the rod, W₁= 225 N
Rod's weight,W₂ = 255
W₃ is the Weight at "x" distance from the left end.
x1 is the location of W₁ with respect to the left end, where L = 1.80 m
m is the location of the rod's center of mass from the left end.
The position of W from the left end; x₃ = 50 cm = 0.50 m
Xcm is the position of the center of gravity of the rod at the fulcrum from the left end;
Given that the fulcrum is balanced in this situation, the rod's center of gravity will be there.
Xcm= L - 75 cm
Xcm=1.80 m - 0.75 m
Xcm= 1.05 m
The value of x₂
x₂= 1.80 m/2 = 0.90 m
Currently, the horizontal center of gravity is determined by
[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]
Substitute the values we get;
[tex]\rm 1.05 = \fraca{ (225 \times 1.80 + 255\times 0.90 + W\times 0.50)}{(225 + 255 + W)}\\\\ W = \frac{ (225\times 1.80 + 255\times 0.90 - 480 \times 1.05)}{(1.05 - 0.50)}\\\\ W = 237.3 \ N[/tex]
After moving the fulcrum 20.0 cm to the right,
The location of the W from the left end;
x₃ = .50 cm + 20 cm
x₃ = 70 cm
x₃ = 0.70 m
W₁, W₂, W, x₁, and x₂ have fixed values. Therefore, based on these values, the new center of gravity will be:
[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]
[tex]\rm X_{cm} = \frac{(225\times 1.80 + 255\times 0.90 + 237.3\times 0.70)}{(225 + 255 + 237.3)}\\\\ X_{cm} = 1.12 m[/tex]
Fulcrum must be shifted by;
Y = (1.12 - 1.05)
Y= 0.070 meters.
W has been moved in the appropriate direction, thus the fulcrum should likewise be moved in the appropriate direction.
Hence, the value of W for the given condition will be 237.3 N. and the distance must the fulcrum be moved to restore balance will be 0.070 m.
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Two small plastic spheres are given positive electric charges. When they are 15 cm apart the repulsive force between them has magnitude 0.22 N. What is the charge on each sphere?
The charge on each sphere is 742 n
What is Charge?In physics, Charge, also known as electric charge is a electrical charge, or electrostatic charge and symbolized q, is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.
According to the question,
Distance = 15cm or 0.15m
Force = 0.22N
[tex]K=9*10^{9} kgm^2/c^2[/tex] ( this is a constant value of K)
By using the formula,
F = [tex]\frac{kQq}{d^{2} }[/tex]
0.220 N = 8.99
N·m²/C² * Q² / (0.15m)²
Q = 7.42e-7
C = 742 n
C is the charge of each sphere .
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An athlete is running at a constant velocity with a javelin held in his right hand. The force he is applying on the javelin to carry it is 5.0 newtons. If he covers a distance of 10 meters, what work has he done on the javelin?
Answer:
50J
Explanation:
Work done=Fd
=5×10
=50J
Question 12 of 25
Two identical satellites circle the Earth in orbits. The distance between the
Earth and satellite B is twice the distance between the Earth and satellite A.
What is the relationship between the gravitational force acting on satellite A
(FA) and the force on B (FB)?
Earth
OA. FB = FA
OB. Cannot be determined
OC. FB< FA
OD. Fa> FA
Satellite A
Satellite B
Answer:
F = G M m / R^2 gravitational force on satellite of mass m
Fma / Fmb = (ma / Ra^2) \ (mb / Rb^2) = (ma / mb) * (Rb / Ra)^2
Fma / Fmb = 1 * 4 = 4
The force on A is 4 times the force on B
If the atmospheric pressure is 15 lb/in, what is the corresponding downward force on the top of a horizontal square area 5 inches on each side ?
The corresponding downward force on the top of a horizontal square is 375 lb.in
What is pressure?The pressure is the amount of force applied per unit area. It is represented as
Pressure p = Force/Area
If the atmospheric pressure is 15 lb/in and area 5 inches on each side is
A = 5² = 25 in²
The force applied is
15 lb/in = F /25 in²
F = 375 lb.in
Hence, the corresponding downward force on the top of a horizontal square is 375 lb.in
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A 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal. How much work is done in pulling the sled? What is the acceleration in the x direction, assuming that friction is negligible? Assuming the sled started at rest, how long does it take to pull the sled 185 m?
The work done in pulling the sled is 445,930.2 Joules.
The acceleration in the x direction, assuming that friction is negligible, is 20.85 m/s².
Time taken to pull the sled 185 m is 4.21s.
What is work done?Work done is equal to product of force applied and distance moved.
Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.
Work = Force x Distance x cos(angle)
W= 1225 x 385 x cos 19°
W = 445,930.2 Joules
Thus, the work done in pulling the sled is 445,930.2 Joules
From the Newton's second law of motion, we have'
F = ma
acceleration, a = 1225cos19° / ( 545 /9.81)
a = 20.85 m/s²
Thus, the acceleration in the x direction is 20.85 m/s²
Using the second equation of motion, we get
s = ut+ 1/2 at²
Substitute the values, we have
185 m = 0 +1/2 x 20.85 x t²
t = 4.21 s
Thus, the time taken to pull the sled 185 is 4.21s.
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What is the energy of a photon of blue light with a frequency of 6.53 × 10¹4 Hz?
Answer:
V = 6.53 × 10^14
h = 6.626 × 10^-34
E = hV
E = (6.626 × 10^-34) (6.53 × 10^14)
E = 4.33 × 10^-19
What type of power plant burns material to make electricity?
O A. Geothermal
OB. Hydroelectric
OC. Fossil fuel
OD. Radiant
The type of power plant burns material to make electricity is Fossil fuel. The correct option is C.
What is energy?Energy is the ability to do work.
There are different kind of energies like kinetic energy, potential energy, gravitational energy and heat energy.
The energy obtained from burning different type of materials like Coal, gas and oil and produces heat.
Fossil fuel power plants burn coal or oil to create heat and generate steam to run turbines which generate electricity.
Thus, the correct option is C.
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How to put a new command in the command prompt
Typing in the commands and instructions and pressing enter will put a new command in the command prompt.
What is a Command?This refers to a words or phrase which causes the computer to execute certain tasks or functions.
Typing the exact instruction and entering it will ensure that the chosen tasks are found in the command prompt.
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The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel from 20 °C to 66 °C. Now imagine those same joules were used instead to accelerate the same mass of nickel from rest. What would be the final speed, in m/s?
Hints:
The formula for kinetic energy is: Ek=1/2 mv^2
For joules needed for heating, use the mass in grams, but for kinetic energy, convert the mass to kilograms.
Round your answer to the nearest whole number.
The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.
Final speed of the nickelApply the principle of conservation of energy.
Q = mcΔθ
Q = (18)(0.444)(66 - 20)
Q = 367.63 J
Q = K.E = ¹/₂mv²
2K.E = mv²
v = √(2K.E/m)
where;
v is the final speedv = √(2 x 367.63)/(0.018))
v = 202.1 m/s
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Where are electrons located in an atom?
A. in the nucleus
B. outside the nucleus
C. Either inside or outside the nucleus?
Answer for brainlist
Answer:
B. Outside the nucleus.
Explanation:
Electrons orbit the nucleus of the atom.
Which statement is a true catalyst in a reaction
Answer:
it alters the rate of chemical reaction
hope it is useful
One mole of a substance contains 6.02 × 1023 protons and an equal number of electrons. If the protons could somehow be separated from the electrons and placed in very small, individual containers separated by a million meters, what would be the magnitude of the electrostatic force exerted by one box on the other?
Explanation:
it is almost zero .this is because the distance and the electrostatic force are inversely proportional
In a small village in southern Italy, childbirth is considered?
Answer:
a community event
Explanation:
If the frequency of a FM wave is 8.85 × 107 hertz, what is the period of the FM wave?
Answer:
1.1299 x 10^-8 second
Explanation:
Period = 1 / f = 1 / (8.85 * 10^7) = 1.1299 x 01^-8 sec
A small mass of m = 187 g hangs from a string of length l = 1.21 m.
Calculate the initial speed v the mass m should be given at the bottom, so that the string reaches a maximum angle of θ = 36.5°?
The initial speed v, the mass m should be given at the bottom, so that the string reaches a maximum angle is 2.16 m/s.
What is mechanical energy?The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.
M.E = KE +PE
Given is a small mass of m = 187 g hangs from a string of length l = 1.21 m. The string reaches a maximum angle of θ = 36.5. The acceleration due to gravity, g = 9.8 m/s²
The vertical height of the point mass h = l- lcosθ
h = 1.21 - (1.21cos 36.5°)
h = 0.2373 m
The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.
Kinetic energy = Potential energy
1/2 mv²= mgh
Substituting the values, we have
1/2 x v² = 9.8 x 0.2373
v = sqrt [2 x9.8 x 0.2373]
v= 2.16 m/s
Thus, the initial speed v the mass m should be given at the bottom, so that the string reaches a maximum angle is 2.16 m/s.
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If the velocity of an object doubles and it’s mass triples, the objects momentum increases by how much? A)1.5 times B)2 times C)3 times D)6 times
Answer:
(D)6 times
Explanation:
Hello !
Because when we see the equation of momentum P=mv, where p=momentum
m=mass of the object&
v=velocity of the object
And, p is directly propositional to both mass and velocity.
Thus, p=mv
p=3m.2v....giving on the valuesp=6(mv)Therefore, we can conclude by when the mass is tripled and the velocity is doubled the momentum is increased by 6 units.
A 17.0-m-high and 11.0-m-long wall and its bracing under construction are shown in the figure. Calculate the force, in newtons, exerted by each of the 10 braces if a strong wind exerts a horizontal force of 655 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall.
The force, in newtons, exerted by each of the 10 braces is 2.135 x 10⁴ N.
What is force?The force is defined as the shear stress or pressure applied per unit area.
F = P/A
Given is a 17.0-m-high and 11.0-m-long wall and its bracing under construction are shown in the figure(attached). The force is exerted by each of the 10 braces, if a strong wind exerts a horizontal force of 655 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths.
Considering the pivot at the base of wall.
From the equilibrium of forces, we have
r₁ x Fwind = r₁ x Fbsinθ
Put the values, we get
Fb = Fwind /10sin35°.............(1)
The wind force is also given by
Fwind = Horizontal force or pressure x Area
Fwind = F/A x hw
=655 x 17 x 11
F wind = 122,485 N
From equation (1), we have
Fb = Fwind /10sin35°
Fb = 122,485 /10sin35°
Fb = 21,354.6080 N
0r Fb = 2.135 x 10⁴ N
Thus, the force exerted on each of the 10 brace is 2.135 x 10⁴ N.
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What quantity is represented by a unit called a newton (N)?
A. Inertia
B. Acceleration
C. Force
D. Mass
[tex]\textsf {C. Force}[/tex]
[tex]\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}[/tex]
A ball is projected with initial velocity 50m/s at an angle of elevation of 37 degrees from the top of a cliff 55 m high. Calculate the total time the ball is in the air and the maximum horizontal distance covered.
The total time the ball is in the air is 1.47 s and the maximum horizontal distance covered is 58.7 m.
Time of motion of the ball
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
55 = (50sin37)t + ¹/₂(9.8)t²
55 = 30.1t + 4.9t²
4.9t² + 30.1t - 55 = 0
Solve using formula method;
t = 1.47 s
Horizontal distance of the ballX = vxt
where;
vx is the horizontal velocityt is time of motionX = (50 x cos37) x 1.47
X = 58.7 m
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n Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C.
What is the concentration of H+ ions at a pH = 2?
mol/L
What is the concentration of OH– ions at a pH = 2?
mol/L
What is the ratio of H+ ions to OH– ions at a pH = 2?
: 1
At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².
What is pH?The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.
The concentration of H⁺ ions is calculated as:
pH = -log [H⁺]
2 = -log [H⁺]
[H⁺] = 10⁻²
The concentration of OH⁻ ions is calculated as:
pH + pOH = 14
pOH = 14 - 2
pOH = 12
Solving further:
pOH = -log [OH⁻]
12 =-log [OH⁻]
[OH⁻] = 10⁻¹²
At a pH = 2, the ratio of hydrogen ions to hydroxide ions:
10⁻² ÷ 10⁻¹² = 10¹⁰
Therefore, at pH = 2 the ratio is 10¹⁰.
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A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation. Describe a method the teacher could use (IMPORTANT REPLAY ASAP) I will give Brainly
By using an electric field, it is feasible to differentiate between these different forms of radiation.
What is a radioactive source?A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.
The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.
Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.
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A spring is compressed by 0.0880 m and is used to launch an object horizontally with a speed of 2.76 m/s. If the object were attached
to the spring, at what angular frequency (in rad/s) would it oscillate?
Answer:
Approximately [tex]3.14\; {\rm rad \cdot s^{-1}}[/tex].
Explanation:
Fact: the angular velocity [tex]\omega[/tex] of a simple harmonic oscillator is the ratio between the maximum velocity [tex]v_{\text{max}}[/tex] and the maximum displacement [tex]x_\text{max}[/tex] of this oscillator. In other words:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}[/tex].
Derivation of the previous equation:
Let [tex]A[/tex] denote the amplitude of this oscillation, and let [tex]\omega[/tex] denote the angular velocity.
The displacement of the oscillator at time [tex]t[/tex] would be:
[tex]x(t) = A\, \sin(\omega\, t)[/tex].
The maximum displacement of this oscillator would be [tex]x_\text{max} = A[/tex].
The velocity of this oscillator at time [tex]t[/tex] is the derivative of displacement with respect to time:
[tex]\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}[/tex].
The maximum velocity of this oscillator would be [tex]v_\text{max} = A\, \omega[/tex].
Notice that dividing [tex]v_\text{max} = A\, \omega[/tex] by [tex]x_\text{max} = A[/tex] would give:
[tex]\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega[/tex].
It is given that [tex]v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}[/tex] while [tex]x_\text{max} = 0.0880\; {\rm m}[/tex]. Therefore:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}[/tex].
(Radians per second.)
