Answer:
B
Explanation:
ballistic stretching is bouncing, and if done cold or too aggressively can pull a muscle or cause it to tighten. ballistic stretching is NOT the most recommended
Answer:
B
Explanation:
Ballistic stretching is not the most recommended.
Is the mass of an object measured with scale, yes or no
Answer:
No.
Explanation:
The scale actually measures the force that the object does against it, and that force is called the weight.
Such that if we have an object with mass M and we are on Earth, where the gravitational acceleration is g, the weight is:
W = M*g
Now, there is a unit called "kilogram-force"
Such that on Earth, an object that has a mass of 10 kilograms, weighs 10 kilograms-force.
Then from the weight measure, we can instantly know the mass of the object, but the thing that is being measured is the weight, not the mass.
Which of these properties of light is a constant?
speed in a vacuum
amplitude
wavelength
frequency
Answer:
speed in vacuum
Explanation:
lets say we are in an empty universe and you are moving 10% the speed of light you wont slow down or speed up.
A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyond the edge hanging freely in air. how far beyond the edge of the wall can a 100kg woman walk before the plank began to rotate about the edge of the wall
Answer:
x₂ = 1.33 m
Explanation:
For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall
Στ = 0
W₁ x₁ - W₂ x₂ = 0
where W₁ is the weight of the woman, W₂ the weight of the table.
Let's find the distances.
Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.
x₁ = 2.5 -1.5 = 1 m
The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative
x₂ = [tex]\frac{M_1g }{m_2 g} \ x_1[/tex]
let's calculate
x₂ = [tex]\frac{100}{75} \ 1[/tex]
x₂ = 1.33 m
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 805 N and stands 1.00 m from the left end. Two meters from the left end is the 500 N washing equipment. Joe is 0.500 m from the right end and weighs 820 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable
Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ (805 N)( (3-1) m) + ( 395 N )( [tex]\frac{3}{2} m[/tex]) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ 1610 + 592.5 + 500 + 410 ]
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ 3112.5 ]
[tex]T_{left[/tex] = 1037.5 N
[tex]T_{left[/tex] = 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
[tex]T_{Right[/tex] = ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
[tex]T_{Right[/tex] = 2520 N - 1037.5 N
[tex]T_{Right[/tex] = 1482.5 N
[tex]T_{Right[/tex] = 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
Which statement is FALSE? *
explosives that deflagrate contain explosive materials that react (burn) slower than the speed of sound
an example of a deflagration explosive is C4
some deflagration explosives have shock waves that move faster than the speed of sound
explosives that deflagrate are also called low explosives
Answer:
Some deflagration explosives have shock waves faster than the speed of sound.
Explanation:
A example of a deflagration explosives.
Answer:
an example of a deflagration explosive is C4, is false.
Explanation:
Explosive materials may be categorized by the speed at which they expand. Materials that detonate (the front of the chemical reaction moves faster through the material than the speed of sound) are said to be "high explosives" and materials that deflagrate are said to be "low explosives".
C4 meaning:
C-4 or Composition C-4 is a common variety of the plastic explosive family known as Composition C, which uses RDX as its explosive agent. C-4 is composed of explosives, plastic binder, plasticizer to make it malleable, and usually a marker or odorizing taggant chemical.
Conservation equations are powerful equations, because of how versatile and flexible they are to different physical situations. Conservation of mass is a simple one that can even be applied to fluids with care. The conservation of momentum also applies to any closed system, such as collisions. We've examined the nuances of energy conservation in our labs with total mechanical energy. Does kinetic energy remain conserved in all collisions too?
a) true
b) false
Answer:
b) false
Explanation:
Since in the given situation it is mentioned that the mass conversation would be simple and same applied to the fluids with care also the conservation of momentum would be applied to any type of closed system like collisions
But as we know that
In the inelastic collisions, the total kinetic energy would not be remain conserved
So the given statement is false
A child holds one end of a 33.0-meter long rope in her hand and moves it up-and-down to produce a sinusoidal wave by moving her hand from 8.00 cm above her shoulder to 8.00 cm below her shoulder at a frequency of 2.00 Hz and a wavelength of 75.0 cm. If the child doubles the amplitude of her hand's motion on that same rope with the same tension in the rope, then what will the wavelength of the wave now be
Answer:
Wavelength=75 cm.
The wavelength well remain unchanged which is 75 cm.
Explanation:
The formula which will help us to answer the question is:
V=f*λ
Where:
V is the velocity
f is the frequency of wave
λ is the wave length
Now:
λ=V/f Eq (1)
The equation show's that wavelength is independent of the amplitude but it depends on the frequency and the velocity with which wave is moving.
The wavelength well remain unchanged which is 75 cm.
If a wave has a frequency of 6 waves per second and a wavelength of 10 m, what is the wave's speed?
0.6 m/sec
1.6 m/sec.
60 m/sec.
none of the above
helpp me plzzz
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?
Answer:
λ = 5.734 x 10⁻⁷ m = 573.4 nm
Explanation:
The formula of the Young's Double Slit experiment is given as follows:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x d}{L}[/tex]
where,
λ = wavelength = ?
L = distance between screen and slits = 8.61 m
d = slit spacing = 1.09 mm = 0.00109 m
Δx = distance between consecutive bright fringes = [tex]\frac{4.53\ cm}{10}[/tex] = 0.00453 m
Therefore,
[tex]\lambda = \frac{(0.00453\ m)(0.00109\ m)}{8.61\ m}[/tex]
λ = 5.734 x 10⁻⁷ m = 573.4 nm
The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
Calculate the decay constant of the radioisotope.
Answer: [tex]0.0353\ s^{-1}[/tex]
Explanation:
Given
Radioactive material is found to decrease 40% of its original value in [tex]2.59\times 10\ s[/tex]
Sample at any time is given by
[tex]N=N_oe^{-\lambda t}[/tex]
where, [tex]\lambda=\text{decay constant}[/tex]
Put values
[tex]\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10[/tex]
Taking natural logarithm both side
[tex]\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}[/tex]
What is the frequency of a wave traveling at 300,000,000 m/s with a wavelength of .0025 m/cycle? Please help me !!
Answer:
its 00.0035474
Explanation:
..... .. . . .. . . . . .. .
A sound wave has a wavelength of 2M and a frequency of 100 Hz. The speed of the wave is
Answer:
200 meters per second
Explanation:
wave speed = wavelength x frequency
2 meters x 100 hertz = 200 meters per second
Choose the correct statement regarding the sign conventions for lenses.
a. The focal length f is positive for diverging lenses.
b. Virtual images appear on same side of the lens as the object and have a negative value for the image distance.
c. Real images appear on the opposite side of the lens from the object and have a negative value for the image distance.
d. The focal length f is negative for converging lenses.
e. Virtual images appear on same side of the lens as the object and have a positive value for the image distance.
Answer:
a) false
b) True
c) True
d) False
e) False
Explanation:
a) False
For a diverging lens, the focal length is negative while it is positive for a converging lens
b) True
Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged
c) True
d) False
For a diverging lens, the focal length is negative while it is positive for a converging lens
e) False
Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 1.61THz. Calculate the wavelength of the infrared radiation. Round your answer to 3 significant digits.
Answer:
λ = 1.86 x 10⁻⁴ m = 186 μm
Explanation:
The relationship between the wavelength and the frequency of a wave is given by the following equation:
[tex]c = f\lambda\\\\\lambda = \frac{c}{f}[/tex]
where,
λ = wavelength of infrared radiation = ?
c = speed of infrared radiation = speed of light = 3 x 10⁸ m/s
f = frequency of infrared radiation = 1.61 THz = 1.61 x 10¹² Hz
Therefore,
[tex]\lambda = \frac{3\ x\ 10^8\ m/s}{1.61\ x\ 10^{12}\ Hz}[/tex]
λ = 1.86 x 10⁻⁴ m = 186 μm
PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
a - is the amount of matter in this object
Ohm's Law is a relationship between which variables?
Group of answer choices
Voltage and Resistance
Current and Voltage
Current and Resistance
Voltage, Current, and Resistance
Answer:
Voltage and resistance
A string is wrapped several times around the rim of a small hoop. The hoop has radius 0.08 m and mass 1.2 kg. If the free end of the string is held in place and the hoop is released from rest, calculate a) the tension in the string while the hoop is descending; 5.88 N b) the time it takes the hoop to descend 0.5 m; 0.452 s c) the angular velocity of the rotating hoop after it has descended 0.5 m
Answer:
Hello From your question you already have the answers for ( a ) and ( b ) hence I will provide the solution to option ( c )
answer :
a) 5.88 N
b) 0.452 s
c) 27.68 rad/s
Explanation:
Hoop radius ( R ) = 0.08 m
mass of hoop = 1.2 kg
height hoop descended ( h ) = 0.5 m
Calculate The angular velocity of the rotating hoop after it descends 0.5 m
since it is released from rest we will apply the law of conservation
mgh = 1/2 mv^2 + 1/2 * Ic * ω^2 ----- ( 1 )
where : Ic (moment of inertia ) = m*R^2 , ω = angular velocity , v = Rω
Back to equation 1 above
mgh = m*R^2*ω^2 ------- ( 2 )
therefore angular velocity ( ω ) = [tex]\sqrt{\frac{gh}{R^2} }[/tex] , h = 0.5 m, g = 9.81, R = 0.08 m
∴ ω = 27.68 rad/s
Drag each tile to the correct box.
Adam takes a bus on a school field trip. The bus route is split into the five legs listed in the table. Find the average velocity for each leg of the trip. Then arrange the legs of the trip from highest velocity to lowest.
Leg Distance (km) Time (min)
A 18 9
B 25 15
C 24 8
D 48 12
E 15 7
Answer:
E
A
C
B
D
Explanation:
Which of the following types of electromagnetic radiation has waves with the highest frequency ? A. Infrared radiation B. Visible light c. Microwaves D. X-rays
Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted ray?
A.
The refracted ray is unpolarized whereas the reflected ray is horizontally polarized.
B.
The refracted ray is unpolarized whereas the reflected ray is vertically polarized.
C.
The refracted ray is horizontally polarized whereas the reflected ray is unpolarized.
D.
The refracted ray is vertically polarized whereas the reflected ray is unpolarized.
E.
The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.
Answer:
E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.
Explanation:
#PLATOLIVESMATTER
A cell has an emf of 1.5 V and an internal resistance of 0.65 Q.
The cell is connected to a resistor R.
[0[8].[4] State what is meant by an emf of 1.5 V.
[2 marks]
I
[0] 8]-[2] The current in the circuit is 0.31 A.
Show that the total power output of the cell is approximately 0.47 W.
[1 mark]
[o[s).[3] Calcutate the energy dissipated per second in resistor R.
[2 marks]
energy dissipated per second = Ist
[o[si.[4] The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.
Calculate the maximum time during which the fully-charged cell can deliver energy to
resistor R.
[2 marks)
maximum time = 5
1. A cell has an emf of 1.5 V means a battery with an Emf rating of 1.5v can supply a maximum of 1.5v of electrical energy in an open circuit. Once it is connected to a closed circuit, it will have a potential difference of 1.5v(considering it has negligible resistance).
2. Voltage(V) in the circuit is 1.5 V and The current (I) in the circuit is 0.31 A.
So we have to find out power (p) in the circuit.
So P = VI
P= 1.5(0.31)
P = 0.465
P= 0.47 W
the total power output of the cell is approximately 0.47 W.
3. We have to obtain the energy dissipated per second in resistor R.
P = 0.465 - I²R
P = 0.465 - 0.31²(0.65)
P = 0.402535 J/s
P = 0.40 J
Total 0.40 Joule per second energy has been dissipated from the resistor R.
4. The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.
The energy dissipated per second(e) in resistor R is .40
Cell have total (E) 14 kJ (14000 J) of energy.
We have to calculate the maximum time (t) during which the fully-charged cell can deliver energy to resistor R.
[tex]\(E=\into{{e}_}}{t}\\\)[/tex]
14000 = 0.40 × t
t =14000 / 0.40
t = 35 000 s or about 9.75 hrs.
Fully-charged cell can deliver energy to resistor R up to 9.75 hrs.
What is Electromotive Force?
Electromotive force, abbreviation E or emf, energy per unit electric charge that is imparted by an energy source, such as an electric generator or a battery. Energy is converted from one form to another in the generator or battery as the device does work on the electric charge being transferred within itself.
Learn more about Electromotive force:
https://brainly.com/question/820393
Object X of mass 4 kg travels with a speed of 3 ms toward object Y of mass 2 kg that is initially at rest. Object X then collides with and sticks to object Y . After the collision, object X and object Y remain stuck together. How much mechanical energy is converted into nonmechanical energy during the collision?
The nonmechanical energy or the loss of energy after the collision will be equal to [tex]E=6\ J[/tex]
What is the conservation of momentum?The conservation of the momentum is defined as when two bodies collide with each other then the total energy of the masses will remain constant.
It is given in the question that
Mass of the body X is [tex]M_X=4\ kg[/tex]
The velocity of the body X is =[tex]V_x=3 \ \frac{m}{s}[/tex]
Mass of the body Y is [tex]M_Y= 2\ kg[/tex]
The velocity of body Y is [tex]V_y= 0[/tex]
Now to find out the energy converted after the collision we will first find the final velocity of both the bodies.
Now from the conservation of the momentum
[tex]M_X V_X +M_YV_Y=(M_X+M_Y)V_F[/tex]
[tex](4\times 3)+(2\times0)=(4+2)\times V_F[/tex]
[tex]V_F= \dfrac{12}{6}[/tex]
[tex]V_F = 2 \frac{m}{s}[/tex]
Now to find the Change in the energy of the body
[tex]\Delta E=E_i-E_f[/tex]
[tex]\Delta E=\dfrac{1}{2} M_XV_X^2+\dfrac{1}{2} M_YV_Y^2-\dfrac{1}{2} (M_X+M_Y)V_F^2[/tex]
[tex]\Delta E=\dfrac{1}{2} (4)(3^2)+\dfrac{1}{2} (2)(0^2)-\dfrac{1}{2} ( 4+2) 2^2[/tex]
[tex]\Delta E= 6\ J[/tex]
Thus the nonmechanical energy or the loss of energy after the collision will be equal to [tex]E=6\ J[/tex]
To know more about the collision follow
https://brainly.com/question/7694106
A football is thrown down off of a building with a force of 70
Newtons. Gravity is pulling down the football with a force of
19 Newtons. What is the net force on the football?
Answer:
89 N downward.
Explanation:
Both forces are working in the same direction -- down.
Therefore you can add them together.
F = F1 + F2
F is the total force
F1 is the gravitational force
F2 is the force the person provides
F1 = 19
F2 = 70
F = ?
F = 19 + 70
F = 89 Newtons downward.
Answer:
wE dONt taLk aBt brunooooOoooOoOo ;')
Explanation:
Which of the following should a warm up NOT include?
Select one:
A. A few minutes of cardiovascular exercise
B. A light version of the exercise you will be performing
C. Intense stretching
D. All of the above should be included in a warm up
One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the 12 C and 13 C isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 km/s, and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the 12 C. The measured masses of these isotopes are 1.99×10−26kg(12C) and 2.16×10−26kg(13C).
(a) What strength of magnetic field is required?
(b) What is the diameter of the 13 C semicircle?
(c) What is the separation of the 12 C and 13 C ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?
Answer:
[tex]0.0084575\ \text{T}[/tex]
[tex]0.272\ \text{m}[/tex]
2.2 cm easily observable
Explanation:
[tex]m_1[/tex] = Mass of 12 C = [tex]1.99\times 10^{-26}\ \text{kg}[/tex]
[tex]m_2[/tex] = Mass of 13 C = [tex]2.16\times 10^{-26}\ \text{kg}[/tex]
[tex]r_1[/tex] = Radius of 12 C = [tex]\dfrac{25}{2}=12.5\ \text{cm}[/tex]
B = Magnetic field
v = Velocity of atom = 8.5 km/s
[tex]r_2[/tex] = Radius of 13 C
The force balance of the system is
[tex]qvB=\dfrac{m_1v^2}{r}\\\Rightarrow B=\dfrac{m_1v}{rq}\\\Rightarrow B=\dfrac{1.99\times 10^{-26}\times 8500}{12.5\times 10^{-2}\times 1.6\times 10^{-19}}\\\Rightarrow B=0.0084575\ \text{T}[/tex]
The required magnetic field is [tex]0.0084575\ \text{T}[/tex]
Radius is given by
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r\propto m[/tex]
So
[tex]\dfrac{r_2}{r_1}=\dfrac{m_2}{m_1}\\\Rightarrow r_2=\dfrac{m_2}{m_1}r_1\\\Rightarrow r_2=\dfrac{2.16\times 10^{-26}}{1.99\times 10^{-26}}\times 12.5\times 10^{-2}\\\Rightarrow r_2=0.136\ \text{m}[/tex]
The required diameter is [tex]2\times 0.136=0.272\ \text{m}[/tex]
Separation is given by
[tex]2(r_2-r_1)=2(0.136-0.125)=0.022\ \text{m}[/tex]
The distance of separation is 2.2 cm which is easily observable.
Two vectors have magnitudes 3 and 4 . how are the directions of the two vectors related if: a/the sum has magnitude 7.0
One way to probe the nucleus is to bombard a sample with high-energy electrons. To learn about the nuclear structures in a sample, the de Broglie wavelengths of these electrons would need to be a little smaller than a nuclear radius. Estimate the energy of such electrons. Give your answer in electron-volts. (Assume that the wavelength used is about 9.0 fm.) eV
Answer:
E = 1.38 x 10⁸ eV = 138 MeV
Explanation:
The energy associated with the given wavelength can be found from the following formula:
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = Energy of electron = ?
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
c = Speed of Light = 3 x 10⁸ m/s
λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m
Therefore,
[tex]E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})[/tex]
E = 1.38 x 10⁸ eV = 138 MeV
What would the current be for a circuit that has a voltage of 0.8 V and a resistance of 0.01 Q?
0 1 = 0.01 A
0 1 = 0.8 A
0 1 = 80 A
O I = 0.08 A
Answer:
80 A
Explanation:
Hi there!
Ohm's law states that [tex]V=IR[/tex] where V is the voltage, I is the current and R is the resistance.
Plug the given information into Ohm's law (V=0.8, R=0.01) and solve for I
[tex]V=IR\\0.8=I(0.01)[/tex]
Divide both sides by 0.01 to isolate I
[tex]0.8=I(0.01)\\\frac{0.8}{0.01}= \frac{I(0.01)}{0.01} \\80=I[/tex]
Therefore, the current for this circuit would be 80 A.
I hope this helps!
Two carts, cart AA (mass 4.00 kgkg) and cart BB (mass 6.50 kgkg) move on a frictionless, horizontal track. Initially, cart BB is at rest and cart AA is moving toward it at 4.00 m/sm/s. The carts are equipped with ideal spring bumpers, meaning they conserve energy. The collision is head-on, so all motion before and after the collision is along a straight line. Let x x be the direction of the initial motion of cart AA.
Answer:
[tex]V{_a}'=-0.95m/s[/tex]
Explanation:
From the question we are told that:
Mass of cart A [tex]m_a=4kg[/tex]
Mass of cart B [tex]m_a=6.50kg[/tex]
Speed of cart AA [tex]V-{a}=4.00[/tex]
Generally the equation for velocity of A after collision [tex]V{_a}'[/tex] is mathematically given by
[tex]V_{a}'=\frac{M_a-M_b}{M_a+M+b} V_a[/tex]
[tex]V{_a}'=\frac{4-6.5}{4+M6.5} *4[/tex]
[tex]V{_a}'=\frac{4-6.5}{4+M6.5} *4[/tex]
[tex]V{_a}'=-0.95m/s[/tex]
Help me please please help me