Answer:
eeeeeeeeeeeeeeeeeeeee
Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up then it is to cool down. C. Is more important to cool down then is to warm up. D. Both warming up and cooling down are not important
Answer:
No. A is correct because both warming up and cooling down are important
Both warming up and cooling down are important.
This is based on aerobics and human body balance regulation as regards exercising.
Warming up and cooling down in exercising are just based on the level of intensity at which the exercise is carried out.
Now, warming up when exercising involves activities like jogging. Warming up is a very vital part of exercising as it helps to get a person's cardiovascular system ready for the subsequent exercises and physical activities to be engaged. This will help in making sure there is enough blood flowinh to your muscles as well as increasing your body temperature.In another sense, cooling down is also very vital in activity because after the blood pressure and heart rates have been raised after exercising, they will need to be restored to their normal levels at which they were before commencement of the exercise. It also helps to regulate the blood flow.Thus, Both warming up and cooling down are important.
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please help
What variables have the most effect on the strength of the gravitational attraction between two
massive objects?
Answer:
Newton's law also states that the strength of gravity between any two objects depends on two factors: the masses of the objects and the distance between them. Objects with greater mass have a stronger force of gravity between them.
The gravitational force depends upon two variables the mass of the objects and the distance between them.
What is gravitational force?In the universe, each body attracts another body towards itself is called Gravitational Force. Thus, gravitation is the study of the interaction forces between two masses.
Newton’s law of gravitation can be defined as every system of matter in the universe attracts every other system with a definite force. The magnitude of this force has a direct relationship with the masses of the systems.
[tex]F \propto M_1M_2[/tex]
The magnitude of this force has an inverse relation with the square of the distance.
[tex]F \propto 1/r^2[/tex]
Therefore, the gravitational attraction between two bodies can be represented as:
[tex]\displaystyle F = G\frac{ M_1M_2}{r^2}[/tex]
The mass of the physical system is directly proportional to the gravitational attraction therefore when the mass increases then the gravitational force will increase.
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A 7.80-g bullet moving at 600 m/s strikes the hand of a superhero, causing the hand to move 5.90 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force that stops the bullet. N (b) Assuming the force is constant, determine how much time elapses between the moment the bullet strikes the hand and the moment it stops moving. s
Answer:
a) F = 2.3797 10⁴ N, b) t = 1.97 10⁻⁴ s
Explanation:
a) For this exercise let's use the relationship between work and scientific energy
W = ΔK
W = F .d
ΔK = ½ m v² - ½ m v₀²
the bold are vectors. in this case the force of the bullet on the hand and the displacement of the bullet has the opposite direction, therefore the angle between the two is 180. The velocity when the bullet stops is zero.
We substitute
-F x = - ½ m vo2
F = [tex]\frac{m v_o^2}{2 x}[/tex]
let's calculate
F = [tex]\frac{7.80 \ 10^{-3} 600^{2} }{2 \ 5.90 \ 10^{-2} }[/tex]
F = 2.3797 10⁴ N
b) let's find the acceleration
F = ma
a = F / m
a = 2.3797 10⁴ / 7.80 10⁻³
a = 3.05 10⁶ m / s²
now we can use the kinematics relations
v = v₀ - a t
the final velocity is zero v = 0
t = v₀ / a
t = 600 / 3.05 10⁶
t = 1.97 10⁻⁴ s
Four point charges each having charge Q are located at the corners of a square having sides of length a. Find symbolic expressions for the work required to bring a fifth charge q from infinity to the center of the square.
Answer:
Explanation:
We shall first calculate electric potential at the centre of square due to four charges of Q .
potential = k Q / r , k = 9 x 10⁹ , Q is charge which is placed at distance r from the point where potential is measured .
Distance of corner of square to centre = √ 2 a / 2 = a / √2 = .707 a
potential due to one charge
= k x Q / .707 a
= 9 x 10⁹ x Q / .707a
= 12.73 x 10⁹ Q/a
Potential due to 4 charges
= 4 x 12.73 x 10⁹ Q/a
= 50.92 x 10⁹ Q/a .
Potential at infinity = 0
Work done in carrying charge q from infinity to centre of square
= potential difference x charge
= ( 50.92 x 10⁹ Q/a - 0 ) x q
= 50.92 x 10⁹ x Qq / a J .
A roller coaster moving along its track rolls into a circular loop of radius r. In the loop, it is only affected by its initial velocity, gravity, and the shape of the track. Let v denote the instantaneous speed and a denote the magnitude of the instantaneous acceleration of the roller coaster in the loop. Which of the following is true in the loop?
a. The roller coaster is not in uniform circular motion, but we still have a=v^2/r everywhere on the loop
b. The roller coaster is not in uniform circular motion, but the tangential acceleration is so small that we can approximate a by v^2/r everywhere on the loop
c. The roller coaster is in uniform circular motion
d. The roller coaster is not in uniform circular motion, and a=v^2/r is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes
Answer:
c. The roller coaster is in uniform circular motion
Explanation:
Since the loop is circular with radius r, and its instantaneous speed, v is always constant, and also, its centripetal acceleration, a' = v²/r.
Since the angular speed, ω = v/r does not change, the magnitude of its tangential acceleration is zero although there is a change in its direction because the direction of its initial velocity changes. That is a" = rα and α = Δω/Δt since Δω = 0, α = 0 and a" = r(0) = 0
So, there is no tangential acceleration. Since there is no tangential acceleration, our instantaneous acceleration which is the vector sum of our centripetal acceleration and tangential acceleration is a = √(a'² + a"²) = √(a'² + 0²) = √a'² = a' = v²/r
So, a is always v²/r.
Since the instantaneous acceleration is always (a = v²/r) constant, the motion is uniform. So, it is uniform circular motion.
The roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.
The given parameters;
radius of the circular path, = r instantaneous speed = v instantaneous acceleration = aThe motion tension on the loop at the lowest point in the circular motion is given as;
[tex]T = mg + \frac{mv^2}{r}[/tex]
The motion tension on the loop at the highest point in the circular motion is given as;
[tex]T = \frac{mv^2}{r} - mg[/tex]
This shows that circular motion is affected by;
acceleration due to gravity, gradius of the circular path, rspeed of the motion, vmass of the object, mThus, we can that the roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.
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What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)
Answer:
30.4 s
Explanation:
A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g
a = 4 x 9.8 = 39.2 m /s²
initial velocity u = 0
Final velocity = 3.60 times speed of sound
= 3.6 x 331 = 1191.6 m /s
v = u + at
Putting the values
1191.6 = 0 + 39.2 t
t = 30.4 s .
Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measure of the electric force on any charged object. It is a measure of the ratio of the charge on an object to its mass. It is a measure of the electric force per unit mass on a test charge. It is a measure of the electric force per unit charge on a test charge. It is a measure of the total charge on the object.
Answer:
It is a measure of the electric force per unit charge on a test charge.
Explanation:
The magnitude of the electric field is defined as the force per charge on the test charge.
Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).
Because from the definition of the electric field intensity, it can be defined as force per unit charge. The correct answer is option D
ELECTRIC FIELDElectric field is the region of space where electric force can be felt. It can also be expressed as electric field intensity E. Mathematically, it can be expressed as;
E = F/q or E = V/d
From the question, the statements that is true concerning the magnitude of the electric field at a point in space is " It is a measure of the electric force per unit charge on a test charge "
Because from the definition of the electric field intensity, is can be defined as force per unit charge.
Therefore, option D is the right answer.
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what is the frequency of a wave having a period equal to 18 seconds
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The shoes are pushed against the surface with a downward force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine only has to pull with a force of 200 N to keep the material moving. What is the coefficient of static friction between the shoe and the material
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
A particle has 37.5 J of kinetic energy and 12.5 J of gravitational potential energy at one point during its fall from a tree to the ground. An instant before striking the ground, how much mechanical energy-rounded to the nearest Joule-does the particle have? Ignore air resistance.
Answer: 50J
Explanation:
Mechanical energy follows the same principles of kinetic energy and potential energy, it is conserved. So Ei = Ef.
Mechanical energy is the sum of ALL energy's. There is no friction, so its just kinetic plus potential.
37.5 + 12.5 = 50J
Since the particle has not touched the ground, it has not transferred any energy to the ground yet, therefore the mechanical energy must still be 50J; mostly in kinetic energy with a very small amount of potential because of the low height relative to the ground.
Which of the following statements are TRUE about nonmetals?
SELECT ALL THAT APPLY
a They can exist as solids, liquids, or gases
b
They have a wide variety of properties, most of
which are the opposite of metals.
С
They tend to have higher melting points than
metals.
d
They make up the majority of the periodic table of
elements
Answer:
c
Explanation:
The statement that are true regarding nonmetals are that they can exist as solids, liquids, or gases, have a wide variety of properties, most of which are the opposite of metals. The correct options are a and b.
What are nonmetals?Non-metals are naturally occurring materials that do not generate heat or electricity and are structurally brittle.
Non-metallic elements in the periodic table include hydrogen, carbon, nitrogen, oxygen, phosphorus, arsenic, and selenium.
Metals make up the majority of the elements on the periodic table. Depending on how the artificially synthesized elements are classified, approximately 93 elements are metals. Only about 18 elements are considered nonmetals.
Because most metals have high melting points, they are solid at room temperature. At room temperature, most nonmetals have low melting points because they are not solid.
Nonmetals can exist as solids, liquids, or gases and have a wide range of properties, the majority of which are diametrically opposed to metals.
Thus, the correct options are a and b.
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There are two metallic spheres and a positive-charge generator. Throughout the entire problem, one sphere is always closest to the generator (call it C), and one is always farther away (call it F). Both spheres start electrically neutral and are in contact with each other. These spheres are now brought near the positivecharge generator (call it G), but do not touch it. The two metallic spheres C and F are then separated from each other, and then individually moved away from the positive-charge generator.
Immediately after the two spheres are moved away from the positive-charge generator, the sphere farther away from the charge generator is charged_______
a. positive
b. negative
c. neutral
d. cannot be determined
Answer:
The correct answer is C
Explanation:
In this problem we must remember that charges of the same sign repel and of the same sign attract.
Generator G induces a charge in the closest spheres C, otherwise the charge is negative, therefore in the furthest sphere F, with the ac that is in contact, a positive charge is formed, they do not indicate that the spheres It is in contact with Earth, so the charges remain on them.
As the spheres move away from the generator, the load is redistributed, they return to their initial load, that is, they remain Neutral since no load left or entered the spheres, there was only a redistribution due to the proximity of the generator.
The correct answer is C
Calculate the heat energy required to increase the temperature of 3 kg of water by 20ºC. The specific heat capacity of water is 4200 J/ kg °C
Answer:
Heat energy required = 252000J or 252KJ.
Explanation:
Given the following data;
Mass = 3kg
Temperature = 20ºC
Specific heat capacity of water = 4200 J/kg°C
To find the heat energy required;
Heat capacity is given by the formula;
[tex] Q = mct [/tex]
Where;
Q represents the heat capacity or quantity of heat. m represents the mass of an object. c represents the specific heat capacity of water. t represents the temperature.Substituting into the equation, we have;
[tex] Q = 3*4200*20 [/tex]
Q = 252000 Joules or 252 Kilojoules.
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 520 lines/mm , and the light is observed on a screen 1.4 m behind the grating.
Required:
What is the distance between the first-order red and blue fringes?
Answer:
0.143 m
Explanation:
Since
d = 1/N = 1/520 = 1.92 * 10^-3 mm
For red light;
θ = sin^-1 (1 * λred/d) = sin^-1 (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°
L = 1.4 * (tan 19.98) = 0.509 m
For blue light;
θ = sin^-1 (1 * λblue/d) = sin^-1 (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°
L = 1.4 * (tan 14.66°) = 0.366 m
Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m
A lake has a surface area of 410 m2 and a volume of 1140 m3 . Suppose that during the day, sunlight with a power averaging 820 W/m2 shines on the lake, and that about 10% of this power is absorbed in the lake, producing heat. Assume that the temperature of the lake water stays constant because the absorbed solar power is exactly balanced by heat lost due to evaporation of water from the lake surface. What is the evaporation rate, in g/s
Answer:
Explanation:
Total solar energy falling on total surface per second
= 410 x 820 W
= 336200 W
10 % of 336200 = 33620 J is converted into heat which is absorbed by lake water . But its temperature does not rise because heat is used up in evaporating water in the form of vapor .
Total heat released during evaporation = 33620 J
Let evaporation rate be m gram /s
heat absorbed by m gram water = m x latent heat of evaporation
= m x 2260 J .
Given ,
m x 2260 = 33620
m = 14.87 g /s .
You want to paddle a boat straight across the Mississippi River. You are able to paddle the boat with a speed of 2.2 m/s relative to the water, and the current flows at a speed of 1.1 m/s. In what direction (angle), relative to the river bank, do you need to point the boat to cross the river in the shortest time? The width of the river is 95.0 m.
A) 60 degrees
B) 80 degrees
C) 90 degrees
D) 70 degrees
E) 50 degrees
Answer:
C) 90 degrees
Explanation:
If we could have an imaginative illustration about what the question is all about.
We could think of the velocity of the river (V_r cosθ) as a horizontal component and the velocity of the boat (V_b sinθ) as the vertical component.
So the velocity of the boat V_b = 2.2 m/s; &
the velocity of the river V_r = 1.1 m/s
∴
The minimum time (which is the shortest time) required for the boat to cross the river is θ = 90°
the time require is:
[tex]t = \dfrac{d}{v_b \ sin \theta}[/tex]
[tex]t = \dfrac{95}{2.2}[/tex]
t = 43.18 sec
A locomotive is accelerating at 1.60 m/s2. It passes through a 20.0-m-wide crossing in a time of 2.40 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 31.6 m/s
Answer:
The time to reach 31.6 m/s is 14.54 s.
Explanation:
Given;
acceleration of the locomotive, a = 1.60 m/s²
distance traveled by the locomotive, s = 20 m
time of motion, t = 2.4 s
final velocity of the locomotive, v = 31.6 m/s
The speed of the of the locomotive as it leaves the 20 m wide crossing;
[tex]v = \frac{d}{t} \\\\v = \frac{20}{2.4} \\\\v = 8.333 \ m/s[/tex]
The time to reach 31.6 m/s is calculated from the equation below;
v = u + at
31.6 = 8.333 + 1.6t
1.6t = 31.6 - 8.333
1.6t = 23.267
t = 23.267 / 1.6
t = 14.54 s
Therefore, the time to reach 31.6 m/s is 14.54 s.
How much horizontal force FFF must a sprinter of mass 48 kgkg exert on the starting blocks to produce this acceleration
Answer:
720N
Explanation:
According to Newtons second law;
Force = Mass* Acceleration
Let the acceleration given be 15m/s²
Mass = 48kg
Substitute
Force = 48 * 15\
Force = 720N
Hence the required horizontal force is 720N
Note that the value of the acceleration was assumed
What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
Answer:
the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
Explanation:
Given;
mass of the particle, m = 1.43 g = 0.00143 kg
electric field experienced by the particle, E = 700 N/C
The force experienced by the particle is calculated as;
F = mg = EQ
Where;
Q is the magnitude of the charge
[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]
The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.
Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.
The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
What is electric charge?The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.
Given information-
The mass of the particle is 1.43 g or 0.00143 kg.
The magnitude of the downward-directed electric field is 700 N/C.
The magnitude of the free-fall acceleration is 9.80 meter per second squared.
The electric field is defined as the electric force per unit charge. It can be given as,
[tex]E=\dfrac{F}{q}[/tex]
Rewrite the equation for the charge,
[tex]q=\dfrac{F}{E}[/tex]
Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,
[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]
Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.
Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
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Newton's second law says that when an___force is applied to a___ it causes it to____
Answer:
Newton's second law says that when an unbalanced force is applied to a mass it causes it to accelerate
Explanation:
A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises
to 46°C. What is the new pressure of the gas relative to its initial pressure (Pi)?
Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture
Explanation:
The new pressure of the gas relative to its initial pressure P₁ is P₁ × 1.09
From the question given above, the following data were obtained:
Initial temperature (T₁) = 20 °C = 20 + 273 = 293 K
Initial pressure (P₁) = P₁
New temperature (T₂) = 46 °C = 46 + 273 = 319 K
New pressure (P₂) =?The new pressure of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
P₁ / 293 = P₂ / 319
Cross multiply
293 × P₂ = P₁ × 319
Divide both side by 293
P₂ = (P₁ × 319) / 293
P₂ = P₁ × 1.09Thus, the new pressure relative to it's initial pressure is P₁ × 1.09
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1)
Which of the following objects has the greatest inertia?
A)
Pencil
B)
Ping pong ball
C)
Baseball
Bowling ball
Answer:
the answer is C baseball and bowling ball
dose contact or noncontact force weaken with distance
Answer:
The more massive an object is, the greater the gravitational force. Since gravitational force is inversely proportional to the distance between two interacting objects, more separation distance will result in weaker gravitational forces.
Explanation:
hope this help:)
A spring has a spring constant of 25 Newtons per meter. The minimum force required to
stretch the spring 0.20 meter from its equilibrium position is approximately
Answer:
6.3N
Explanation:
Guessed it right on castle learning
Answer:
6.3 N
Explanation:
F=kx
F=(25N/m)(0.25m)
6.3 N
Calculate the net force on the crate if the magnitude of the forces in the diagram are Fg=10N, Fapp= 7.5N, Ff=2.5N and Fn=10N
Answer:
5 units to the left
Explanation:
I had the same question :)
before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv
Answer:
22
Explanation:
before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv
The equilibrant is the equal to the resultant magnitude but opposite in direction.
True
False
Answer and I will give you brainiliest
Answer:
The answer is False......
Answer:
true
it is equal but opposite
A point charge of -11 [Coulombs] is placed inside a spherical conducting shell with net charge of 5 [Coulombs]. Calculate the net charge on the outer surface of the conducting shell. Enter your answer without units (example 100 for 100 [Coulombs] or -100 for -100 [Coulombs] ).
Answer:
20 C
Explanation:
To do this, is pretty easy, we just need to do a little reasoning of what is happening.
When any charge called q is placed inside this metallic shell which is spheric, all the opposite and even equal charges are induced on the inner and outer surface of the shell. Hence, we can say that if in the inner shell we have +q, in the outer will be -q.
Now, here we have the shell with 5 C, and when the charge of -11 C is placed inside the shell we can have the following changes on the inner surface and the outer surface:
Inner surface: +11 C
Outer surface: 9 + 11 = 20 C
Net charge on the outer surface: 20 CHope this helps
A car of mass 150 kg is riding down at constant velocity. What is the normal force?
The formula is ( Fg=mg ; Fnet =ma ; Fx = Fcos theta ; Fy = Fsin theta)
Answer and I will give you brainiliest
Answer:
1500N
Explanation:
Normal force = mg - F sin theta
constant velocity means acceleration = 0
F= ma = 150× 0 = 0N
thus;
normal force = mg = 150 × 10 = 1500N
Answer:
864N
Explanation:
An aircraft flies 800km due East and then 600km due north. Determine the magnitude of it's displacement. please it's urgent
Answer:
Magnitude of it's displacement = 1,000 km
Explanation:
Given:
Distance towards east = 800 km
Distance towards north = 600 km
Find:
Magnitude of it's displacement
Computation:
Magnitude of it's displacement = √800² + 600²
Magnitude of it's displacement = √640000 + 360000
Magnitude of it's displacement = √10000000
Magnitude of it's displacement = 1,000 km