Carbon-14 is the typical radioisotope used to date materials; however, it has a limitation to 40,000 years. A scientist who wants to date materials older than 40,000 years would most likely use which radioisotope?

Answer:

[tex]p =-1.03[/tex]

Explanation:

From the question we are told that:

Focal length of lens [tex]f=-0.304 m[/tex]

Image distance [tex]q=0.237 m[/tex]

Height of image [tex]H_i=0.343[/tex]  

Generally the lens equation is mathematically given by

[tex]\frac{1}{f} = \frac{1}{q} - \frac{1}{p}[/tex]

Where [tex]p[/tex] is Subject

[tex]p = \frac{(qf) }{(f - q)}[/tex]

[tex]p = \frac{(-0.237)(-0.304)) }{((-0.304) - (0.237))}[/tex]

[tex]p =-1.03[/tex]

Therefore the distance between  the person and the car is

[tex]p =-1.03[/tex]

Answer:

Yes, I agree with rich

Explanation:

Yes I agree with rich.

Thus is because;

The electromagnetic process is what will determine how electromagnetic radiation will be emitted. Now, the mode of travel of electromagnetic energy is in the form of waves and then the wavelength of these waves is what will enable us know the form of energy. This includes light, x-rays, gamma rays e.t.c. This means that light and the others are just parts of the spectrum emitted and are called photons.

Thus, light can be said to be electromagnetic radiation.

I can agree with Rich that light is the same as electromagnetic radiation since the two terms can conveniently be used interchangeably.

Light has become a generic term that is used to describe all electromagnetic waves. All the waves in the electromagnetic spectrum have different frequencies and wavelengths but they are travel at the same speed which has been designated as "speed of light".

Therefore, i can agree with Rich that light is the same as electromagnetic radiation since the two terms can conveniently be used interchangeably.

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Answer:

0.706

Explanation:

Since the other astronaut measures a longer time, this is a time dilation problem. So, our equation for time dilation is given by

T = T₀/√(1 - β²) where T = period on passing space ship = 31.87 s, T₀ = period on other space vehicle = proper time = 22.58 s and β = relative velocity of between the two observers.

T = T₀/√(1 - β²)

√(1 - β²) = T₀/T

squaring both sides, we have

[√(1 - β²)]² = (T₀/T)²

1 - β² = (T₀/T)²

β² = 1 - (T₀/T)²

taking square root of both sides, we have

√β² = √[1 - (T₀/T)²]

β = √[1 - (T₀/T)²]

substituting the values of the variables into the equation, we have

β = √[1 - (22.58 s/31.87 s)²]

β = √[1 - (0.7085)²]

β = √[1 - 0.502]

β = √0.498

β = 0.706

Answer:

b

Explanation:

f1=-f2 that could be thank u

Answer:

Explanation:

From the given information:

Since both stars are in the same cluster, the magnitude and luminosity relationship can be calculated as:

[tex]m_1 - m_2 = -2.5 log _{10} (\dfrac{L_1}{L_2})[/tex]

Given that;

m_1 = 1 and

m_2 = 4

Therefore,

[tex]1 - 4 = -2.5 log _{10} ( \dfrac{L_1}{L_2})[/tex]

[tex]3 = -2.5 log _{10} ( \dfrac{L_1}{L_2})[/tex]

Making [tex]\dfrac{L_1}{L_2}[/tex] the subject of the formula:

[tex]\implies \dfrac{L_1}{L_2}= 10^{(\dfrac{3}{2.5})}[/tex]

=15.84

≅ 16

Hence, we can conclude that star X is more luminous by a factor of 16

Answer:

The 9th harmonic is 1,010.67 Hz.

Explanation:

Given;

length of the pipe, L = 1.23 m

speed of sound at 0°C = 331.5 m/s

A pipe closed at one end is known as a closed pipe;

The wavelength of the sound for the first harmonic is calculated as;

L = Node ------ > Antinode

[tex]L = \frac{\lambda }{4} \\\\\lambda = 4L[/tex]

First harmonic:  [tex]F_0 = \frac{V}{\lambda} = \frac{V}{4L}[/tex]

The wavelength of the sound for the first harmonic is calculated;

L = Node ----> Node + Node ------> Antinode

[tex]L = \frac{\lambda}{2} + \frac{\lambda}{4} = \frac{2\lambda+\lambda}{4} = \frac{3\lambda}{4} \\\\\lambda = \frac{4 L}{3}[/tex]

Second harmonic: [tex]F_1 = \frac{V}{\lambda} = \frac{3V}{4L}[/tex]

F₁ = 3F₀

The increment from F₀ to F₁ is 1 to 3; (odd number).

(1, 3, 5, 7, 9, 11, 13, 15, 17, 19..... n+2, where n is odd number)

The 9th harmonic = F₈

F₈ = 15F₀

[tex]F_8= 17(F_0) = 15 (\frac{V}{4L} )\\\\F_8 = 15(\frac{331.5}{4\times 1.23} )\\\\F_8 = 1,010.67 \ Hz[/tex]

Therefore, the 9th harmonic is 1,010.67 Hz.

The frequency of the 9th harmonic if the pipe is closed at one end is 605.48 Hz.

Based on the given information,

• The length (l) of the shortest pipe given is 1.23 m.

The speed of sound at 0 degree C is,

[tex]V = 331\sqrt{1+\frac{T}{273} } \\V = 331\sqrt{1+\frac{0}{273} } \\V = 331 m/s[/tex]

Now the condition is that the pipe is closed at one end. The formula to use in the given case is,

[tex]fn = \frac{nv}{4l} \\[/tex]

Here l is the length of the organ pipe.

Now the frequency of the 9th harmonic will be,

[tex]f9 = \frac{9v}{4l} \\\\f9 = \frac{9*331}{4*1.23} \\f9 = 605.48 Hz[/tex]

Thus, the frequency of the 9th harmonic if the pipe is closed at one end is 605.48 Hz.

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Answer:

K.E. = ½ × mv²

= ½ × 6 × (2)²

= ½ × 6 × 4

= 3 × 4

= 12 J

Answer:

A

Explanation:

Answer:

In a way it does, but overall, there are many factors that affect your rank. In general, and talking about the average Platinum II, they are pretty decent according to casual player standards.

Explanation:

Answer:

[tex]4.12\times 10^{-5}\ J[/tex].

Explanation:

Given that,

Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]

Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

[tex]E=\dfrac{Q^2}{2C}[/tex]

Put all the values,

[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]

So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].

Answer:

Earth has a much greater mass than objects on its surface

Answer:

D)

The mass of the reactants is greater than the mass of the products.

Answer:

Truck [tex]\dfrac{g}{10}[/tex]

Road [tex]-\dfrac{g}{10}[/tex]

Explanation:

[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]

[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]

Frictional force is given by

[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]

Net acceleration is given by

[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]

The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.

Answer:

Explanation:

The classification will be made into 3 categories, which are

Ones that shortens wavelengths

Ones that lengthens wavelengths

Ones that has no effect on wavelengths

Shortens wavelengths -> Increase frequency

Lengthens wavelengths -> Decrease frequency

No effect -> Increase amplitude, decrease amplitude, increase damping, decrease damping.

Answer:

a= 4.4×10 m/s^2

Explanation:

pressure P  = E/c

Where, E = 100 W/m^2 intensity of light

c= speed of light  = 3×10^8 m/s

P = 1000/ 3×10^8

P = 3.33×10^(-6) Pa

Force F = P×A

F = 3.33×10^{-6}×6.65×10(-29)

= 2.22×10^{-6}

acceleration a  = F/m = 2.22×10^{-6}/ 5.10×10^{-27}

a= 4.4×10 m/s^2

Answer:

80m/s2

Explanation:

a = v-u/ t

=>a = 0- 400/ 5

= -400/5

= -80m/s2

v- sign is negligible

Answer:

statement 1 is correct

2 not sure

Explanation:

Answer:

Bacteria: Unicellular organisms that can cause disease in living things

Plant: A multicellular organism that makes it's own food

Animal: A multicellular organism that does not make it's own food

Answer:

heavy molecules are pumped much more efficiently than light molecules. Most gases are heavy enough to be well pumped but it is difficult to pump hydrogen and helium efficiently.

Answer:

B) It accelerated for 0.25s

Explanation:

Given the following data;

Acceleration = 20 m/s²

Final velocity, v = 5 m/s

Initial velocity, u = 0 m/s

To find the time, we would use the first equation of motion;

V = u + at

Substituting into the formula, we have;

5 = 0 + 20t

5 = 20t

Time, t = 5/20

Time, t = 0.25 seconds

Answer:

I think it's C. Long root systems to reach water deep underground.

Answer:

C

stilt roots provide extra support

Answer:

carbon dioxide plus water in the presence of energy from light, results in glucose plus oxygen

The process is like carbon dioxide plus water in the presence of sunlight result in glucose plus oxygen.

Photosynthesis is a process that plants and other organisms use to convert light energy into chemical energy that can then be released to fuel the organism's activities via cellular respiration.

Photosynthesis is important because it is the process by which plants convert carbon dioxide into oxygen using sunlight. Animals require oxygen to survive, and plants serve as nature's air filter, filtering out harmful carbon dioxide.

Photosynthesis is the conversion of light energy by phototrophs into chemical energy, which is then used to power cellular activities. Sugars, which are formed from water and carbon dioxide, are used to store chemical energy.

The process of photosynthesis includes:

[tex]6CO_2+6H_2O \frac{Sunlight}{chlorophyll} > C_6H_1_2O_6+6O_2[/tex]

Thus, this is the basic process of photosynthesis.

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#SPJ2

Answer:

F = -319.2 N

Explanation:

Given that,

The mass of a bicyclist, m = 70 kg

Mass of the bicycle = 9.8 kg

The speed of a bicycle, v = 16 m/s

We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.

The braking force is given by :

[tex]F=ma\\\\=\dfrac{m(v-u)}{t}\\\\=\dfrac{(70+9.8)(0-16)}{4}\\\\=-319.2\ N[/tex]

So, the required force is 319.2 N.

Answer:

300

Explanation:

the momentum is 300

p=mv

p=5×60

5×60 =300

Answer:

1.35 A

Explanation:

Applying,

V = IR

I = V/R..................... Equation 1

I = Current, V = Voltage, R = Resistance.

But,

R = Lρ/A............... Equation 2

Where L = Length of the wire, ρ = resistivity, A = Cross-sectional area of the wire.

Sustitute equation 2 into equation 1

V = AV/Lρ............... Equation 3

From the question,

Given: V = 0.7 V, A = 0.290 mm² = 2.9×10⁻⁷ m², L = 1.5 m, ρ = 10×10⁻⁸ Ω.m

Substitute these values into equation 3

I = (0.7× 2.9×10⁻⁷)/(1.5× 10×10⁻⁸ )

I = (2.03×10⁻⁷)/(15×10⁻⁸)

I = 1.35 A

This is my alt account just grinding points

Answer:

what the heck is sakurfa

Explanation:

Answer:

No. Because it would correspond to zero Instantaneous acceleration.

Explanation:

hope this helps

Answer:

Because the earth's axis is tilted. ... It is all about the tilt of the Earth's axis. Many people believe that the temperature changes because the Earth is closer to the sun in summer and farther from the sun in winter. In fact, the Earth is farthest from the sun in July and is closest to the sun in January

[tex]V_{f2} =0.29 \dfrac{m}{s}[/tex] :Speed of the block immediately after the collision.

From elastic collision theory.

Momentum of a body is given as

[tex]P=m\times v[/tex]

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For elastic collision the change in momentum remains constant

[tex]P_{i} =P_{f}[/tex]

Nomenclature and data

m₁: ball mass= 5.21 g= 5.21*10⁻³kg

V₀₁: initial ball speed,  =412 m/s

[tex]V_{f1}[/tex]: final ball speed

m₂: block mass =  14.8 kg  

V₀₂: initial block speed, = 0

[tex]V_{f2}[/tex]: final block speed

We apply furmula

[tex]P_{i} =P_{f}[/tex]

m₁[tex]\times[/tex]V₀₁+m₂[tex]\times[/tex]V₀₂=m₁[tex]\times[/tex]Vf₁+m₂[tex]\times[/tex]Vf₂

5.21[tex]\times[/tex]10⁻³[tex]\times[/tex]412+14.8[tex]\times[/tex]0= 5.21[tex]\times[/tex]10⁻³[tex]\times[/tex]Vf₁+14.8[tex]\times[/tex]Vf₂

2.15= 5.21*10⁻³[tex]\times[/tex]Vf₁+14.8[tex]\times[/tex]Vf₂ Equation (1)

For perfectly elastic collision the coefficient of elastic restitution (e) is equal to 1, and e is defined like this:

[tex]e=\dfrac{v_{f2-V_{f1} } }{v_{01-V_{ 02} } }[/tex]

1*(V₀₁-V₀₂) =Vf₂-Vf₁  , V₀₂=0, V₀₁ =412 m/s

412=Vf₂-Vf₁

Vf₁=Vf₂-412 Equation (2)

We replace Equation (2) in Equation (1)

2.15= 5.21[tex]\times[/tex]10⁻³(Vf₂-412)+14.8[tex]\times[/tex]Vf₂

2.15= 5.21[tex]\times[/tex]10⁻³*Vf₂-2.15+14.8*Vf₂

4.3=14.805Vf₂

[tex]Vf_{2} =\dfrac{4.3}{14.805} =0.29 \dfrac{m}{s}[/tex]: (+) ,with equal direction  of the movement of the ball before the collision.

Hence    [tex]V_{f2} =0.29 \dfrac{m}{s}[/tex]   Speed of the block immediately after the collision.

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