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]
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.
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an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.
Answer:
0.5 m/s 2 is the acceleration
Explanation:
hope it helped!!!
Another word for kinetic energy
could be
energy.
A. Safe
B. Moving
C. Stored
D. Potential
Answer:
moving
Explanation:
hope it helped!!!
Answer:
B
Explanation:
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!
A transparent oil with index of refraction 1.15 spills on the surface of water (index of refraction 1.33), producing a maximum of reflection with normally incident violet light (wavelength 400 nm in air). Assuming the maximum occurs in the first order, determine the thickness of the oil slick.
Answer:
The thickness of the oil slick. t = 173.91 nm
Explanation:
Oil film thickness t is given by the equation
[tex]t = \frac{\lambda}{2n}[/tex]
where λ = wavelength of incident light in air = 400 nm
and n = index of refraction of oil
therefore,
[tex]t =\frac{400}{2\times 1.15}\\t= 173.91 nm[/tex]
The thickness of the oil slick. t = 173.91 nm
Two vectors have magnitudes 3 and 4 . how are the directions of the two vectors related if: a/the sum has magnitude 7.0
please help me with this! i will give brainliest if you get it right! NO LINKS OR IM REPORTING
Determine the angle between the directions of vector A with rightwards arrow on top = 3.00i + 1.00j and vector B with rightwards arrow on top = -3.00i + 3.00j.
A) 117°
B) 88.1°
C) 26.6°
D) 30.0°
E) 45.2°
Answer:
C) 26.6
Explanation:
I don't know how to calculate vector
The angle between the two vectors is 117⁰
The given parameters;
vector A = 3.00i + 1.00j
vector B = -3.00i + 3.00j
The angle between the two vectors is calculated as follows;
[tex]cos \ \theta = \frac{A\ . \ B}{|A| \ . \ |B|}[/tex]
The dot product of vector A and B is calculated as;
[tex]A \ . \ B = (3i \ + j) \ . \ (-3i \ + 3j) = (3\times -3) + (1 \times 3) = -9 + 3 =- 6[/tex]
The magnitude of vector A and B is calculated as;
[tex]|A| = \sqrt{3^2 + 1^2} = \sqrt{10} \\\\|B| = \sqrt{(-3)^2 + (3)^2} = \sqrt{18}[/tex]
The angle between the two vectors is calculated as;
[tex]cos \ \theta = \frac{-6}{\sqrt{10} \ . \sqrt{18} } \\\\cos \ \theta = \frac{-6}{\sqrt{180} } \\\\cos \ \theta = -0.4472\\\\\theta = cos \ ^{-1} (-0.4472) \\\\\theta = 116.6^0 \approx 117^0[/tex]
Thus, the angle between the two vectors is 117⁰
Learn more here: https://brainly.com/question/15006306
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
In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- vestigated occurs when the external field Bext has magnitude 0.78 T. Assume that Bint and Bext are in the same direction and take the pro- ton magnetic moment component u, to be 1.41 X 10-26 J/T. What is the magnitude of Bint?
Answer:
[tex]B_{int}=-0.015T[/tex]
Explanation:
From the question we are told that:
RF source oscillation speed [tex]\sigma= 34 MHz[/tex]
The external field [tex]Bext =0.78 T[/tex].
Pro- ton magnetic moment component [tex]\mu=1.41 X 10-26 J/T[/tex]
Generally the equation for magnitude of [tex]B_{int}[/tex] is mathematically given by
[tex]B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}[/tex]
[tex]B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}[/tex]
[tex]B_{int}=0.78-0.7957[/tex]
[tex]B_{int}=-0.015T[/tex]
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
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.
Types of telescope
for Space
observation
Answer:The three main types are reflecting telescopes, refracting telescopes, and catadioptric telescopes. Radio telescopes collect and focus radio waves from distant objects. Space telescopes orbit Earth, collecting wavelengths of light that are normally blocked by the atmosphere.
Answer:
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