Radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 million years. Assuming a constant rate, seafloor spreading in this area occurs at a rate of __________.
A. 5 cm per year
B. 1,012.5 km per year
C. 20,000 cm per year
D. 50 km per year

Answer:

A) [tex] Z = 0.577 C +112.931[/tex]

[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]

B) [tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]

C) [tex] K = C +273.15[/tex]

[tex] K = -22.41 +273.15 =250.739 K[/tex]

Explanation:

For this case we want to create a function like this:

[tex] Z = a C + b[/tex]

Where Z represent the degrees for the Z scale C the Celsius grades and  tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

[tex] a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}[/tex]

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

[tex] 0 = 0.577 (-195.8) + b[/tex]

And if we solve for b we got:

[tex] b = 0.577*195.8 =112.931 Z[/tex]

So then our lineal model would be:

[tex] Z = 0.577 C +112.931[/tex]

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

[tex] Z-112.931 = 0.577 C[/tex]

[tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]

Part C

For this case we know that [tex] K = C +273.15[/tex]

And we can use the result from part B to solve for K like this:

[tex] K = -22.41 +273.15 =250.739 K[/tex]

Final answer:

The boiling point of water on the Z scale is 500°Z. To convert 100°Z to the Celsius scale, the equivalent temperature is 10°C. To convert 100°Z to the Kelvin scale, the equivalent temperature is 283.15 K.

Explanation:

The Z scale is a new temperature scale created by the demented scientist. According to the Z scale, the boiling point of nitrogen is 0°Z and the melting point of iron is 1000°Z.

A) To find the boiling point of water on the Z scale, we need to look at the temperature range between the boiling point of nitrogen and the melting point of iron. Since the boiling point of nitrogen is 0°Z and the melting point of iron is 1000°Z, the range between them is 1000-0 = 1000°Z. Therefore, the boiling point of water on the Z scale is 500°Z, which is halfway between 0°Z and 1000°Z.

B) To convert 100°Z to the Celsius scale, we can use the formula: Celsius = (Z - 0) * (100 - 0) / (1000 - 0). Plugging in the values, we get Celsius = (100 - 0) * (100 - 0) / (1000 - 0) = 10000 / 1000 = 10°C.

C) To convert 100°Z to the Kelvin scale, we use the formula: Kelvin = Celsius + 273.15. Plugging in the value, we get Kelvin = 10 + 273.15 = 283.15 K.

Answer: have "cis C=C double bonds" and "liquid" at room temperature.

Explanation:

The unsaturated fatty acids have one or more C=C double bonds in the cis formation. Thus, this results in the molecules not been as stable as the saturated fats. They have weaker intermolecular bonds thus resulting in lower melting point . The consequently results in it being liquid at room temperature.

Unsaturated fats have one or more double bonds in their carbon chain and are usually liquid at room temperature. In contrast, saturated fats, which have no double bonds, are typically solid at room temperature.

When comparing saturated and naturally occurring unsaturated fats, the unsaturated fats have one or more double bonds in their carbon chain and are typically liquid at room temperature. Saturated fats, on the other hand, have no double bonds and are usually solid at room temperature. This is due to the straight, closely packed chains of saturated fat molecules, whereas the double bonds in unsaturated fats cause bends, preventing them from packing closely together and remaining fluid at room temperature.

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Answer:

-1.03 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².

Mathematically, acceleration is expressed as

a = (v-u)/t ........................ Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t  = time.

Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.

Substituting into equation 2

a = (7.20-13.60)/6.2

a = -6.4/6.2

a = -1.03 m/s²

Note: a is negative because, the hockey puck is decelerating.

Hence the average acceleration = -1.03 m/s²

When the altimeter is set to 29.92 inHg, it will read 4,000 feet MSL at a field elevation of 2,000 feet MSL.

To determine the altimeter reading when the field elevation is 2,000 feet MSL (Mean Sea Level) and the altimeter is set to 29.92 inches of mercury (inHg), we need to consider the difference in pressure settings.

The altimeter measures altitude based on the atmospheric pressure. When the altimeter setting (also known as the barometric pressure setting) is changed, it adjusts the reference point for sea level pressure. The standard atmospheric pressure at sea level is approximately 29.92 inHg.

Given:

Field elevation = 2,000 feet MSL

Altimeter setting = 29.92 inHg

We need to find the altimeter reading when the altimeter is set to 29.92 inHg.

To calculate the altimeter reading, we need to determine the difference in pressure between the current pressure (altimeter setting) and the pressure at the field elevation (2,000 feet above sea level).

The standard atmospheric pressure decreases as we go higher in altitude. A typical lapse rate for pressure is around 1 inHg per 1,000 feet of altitude gain. Since the field elevation is 2,000 feet above sea level, the difference in pressure would be approximately 2 inHg.

Therefore, the corrected altimeter reading would be 2,000 feet + 2,000 feet (for the 2 inHg difference) = 4,000 feet MSL.

Hence, when the altimeter is set to 29.92 inHg, it will read 4,000 feet MSL at a field elevation of 2,000 feet MSL.

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Answer:

When a converging lens is placed under water then its focal length increases.

Explanation:

When any lens, be it converging or diverging is immersed into water then the speed of light before and after the refraction through the lens has less speed and due to less difference in speed it shows less deviations as a result the focal length is increased for the lenses.

Lesser the difference between the refractive index of the medium and the lens glass greater becomes its focal length.

The focal length of the lens immersed in a medium is given by:

[tex]\frac{1}{f} =\frac{n_l-n_m}{n_m} \times (\frac{1}{R_1} +\frac{1}{R_2} )[/tex]

where:

[tex]n_l=[/tex] refractive index of the glass material with resp. to air

[tex]n_m=[/tex] refractive index of the medium with resp. to air

[tex]R_1\ \&\ R_2=[/tex] radii of curvature of the two surfaces of the lens

Answer:

Explanation:

According to the lens maker's formula

The focal length of the converging lens in air is given by

[tex]\frac{1}{f_{a}}=\left ( \mu _{a}-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]       .... (1)

Where, R1 and R2 be the radius of curvature of the lens and μa is the refractive index of glass in air.

When the lens is placed in water. the focal length is given by

[tex]\frac{1}{f_{w}}=\left ( \frac{\mu _{a}}{\mu _{w}}-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]        .... (2)

Where, R1 and R2 be the radius of curvature of the lens and μa is the refractive index of lens in air and μw is the refractive index of glass in water.

Dividing equation (1) by (2), we observe that the value of focal length,

the focal length of lens in water increases.

Answer:

[tex]t=25.6446\ s[/tex]

is the time after which Jill will be able to catch the cart.

Explanation:

Given:

Now the component of gravity acting on the cart along the inclined plane:

[tex]g'=g.sin\ \theta[/tex]

[tex]g'=9.8\times sin\ 3^{\circ}[/tex]

[tex]g'=0.5129\ m.s^{-2}[/tex]

Time taken by Jill to reach this speed:

[tex]v=u+a.t[/tex]

where:

t = time taken

u = initial velocity = 0

[tex]51.289=0+2\times t[/tex]

[tex]t=25.6446\ s[/tex] is the time after which Jill will be able to catch the cart.

Answer:

Explanation:

Gravitational Potential Energy at earth surface [tex]U_1=\frac{GM_em}{R_e}[/tex]

Gravitational Potential Energy at height h is [tex]U_2=\frac{GM_em}{R_e+h}[/tex]

Energy required to lift the satellite [tex]E_1=U_1-U_2[/tex]

[tex]E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}[/tex]

Now Energy required to orbit around the earth

[tex]E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}[/tex]

[tex]\Delta E=E_1-E_2[/tex]

[tex]\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}[/tex]

[tex]E_1=E_2[/tex]  (given)

[tex]\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0[/tex]

[tex]\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0[/tex]

[tex]h=\frac{R_e}{2}[/tex]

[tex]h=3.19\times 10^6\ m[/tex]

(b)For greater height [tex]E_1[/tex]  is greater than [tex]E_2[/tex]

thus energy to lift the satellite is more than orbiting around earth

Answer:

axial stress in the diagonal bar =36,000 psi

Explanation:

Assuming we have to find axial stress

Given:

width of steel bar: 1.25 in.

height of the steel bar: 3 in

Length of the diagonal member = 20ft

modulus of elasticity E= 30,000,000 psi

strain in the diagonal member ε = 0.001200 in/in

Therefore, axial stress in the diagonal bar σ = E×ε

=  30,000,000 psi×  0.001200 in/in =36,000 psi

The stress in the bridge truss diagonal tension member is computed to be 36000 psi, and the elongation of the member under this stress is calculated to be approximately 0.384 inches.

The subject of the question is in the field of Engineering, particularly dealing with the stress and strain effects in a steel structural member of a bridge.

The stress in the member can be calculated using stress and strain relationship along with Young's modulus, given by the formula Stress = Strain x Young's Modulus. Given that strain is 0.001200 in./in. and Young's Modulus (E) is 30,000,000 psi, stress (σ) in the member is σ = E x Strain = 30,000,000 psi x 0.001200 in./in. = 36000 psi.

The steel bar's cross-sectional area can also be calculated as Area = Width x Depth = 1.25 in. x 3 in. = 3.75 in^2. The change in length (∆L) of the member under this stress can be calculated using the formula ∆L = (Stress x Original length) / (Young's Modulus x Area) = (36000 psi x 20 ft) / (30,000,000 psi x 3.75 in^2) = 0.032 ft or 0.384 in.

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Answer:

zero

Explanation:

Work done = - 400 J

In an isothermal process, the temperature remains constant. So, according to the first law of thermodynamics

dQ = dU + dW

As the temperature remains constant and the change in internal energy is the function of change in temperature, here change is temperature is zero so the change in internal energy is also zero.

Final answer:

The change in internal energy of an ideal gas during an isothermal process is 0 J. This is because in an isothermal process for an ideal gas, any work done by the gas is invariably compensated by an equivalent amount of heat absorbed, resulting in no net change in internal energy.

Explanation:

The question pertains to the change in internal energy of an ideal gas during an isothermal process in the context of thermodynamics, specifically according to the first law of thermodynamics. The first law states that the change in internal energy of a system (Triangle U) is equal to the heat added to the system (Q) minus the work done by the system (W). In an isothermal process, the temperature remains constant, and if the process is performed on an ideal gas, the internal energy also remains constant since it depends only on temperature for an ideal gas. Therefore, the change in internal energy ( triangle U) is zero.

In this specific scenario where the work done on the system is -400 J (-W), which means that the system has done 400 J of work on the surroundings. Since no heat transfer information is given, we can infer that in an isothermal process, whatever work is done by the gas is compensated by the heat received from the surroundings, keeping the internal energy unchanged. Thus, the change in internal energy of the gas is 0 J, as the energy spent as work would have been absorbed as heat from the surroundings.

Answer:

Explanation:

Check attachment for answer.

The acceleration of the center of the cylinder and the tension force in the tape is mathematically given as

a=2g/3

T=ug/3

Generally, the equation for the Liner motion  is mathematically given as

Mg-T=Ma

Therefore

Mg.R=(0.5MR^2+MR^2)\alpha

[tex]\alpha=2g/3R[/tex]

b)

Where

a=\alpha R

Hence

a=2g/3

c)

For the tension T

ug-T=Ma

T=u(g-a)

T=ug/3

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Answer:

Option C. 3 m/s

Explanation:

From the question it is stated that we can neglect the force of air resistance,  therefore there is no force acting on the ball horizontally. From Newton's first law of motion an object will remain at rest or in uniform motion unless acted upon by an external force. thus the ball velocity will remain 3m/s.

Answer:

[tex]v_{avg}=29.499\ m.s^{-1}[/tex]

Explanation:

Given:

distance as a function of time is:

[tex]s=e^t+2-sin\ t[/tex]

Average velocity of the particle from time t=1s to t=5s can be given by the total distance covered divided by the total time consumed.

Position at t=1s:

[tex]s_1=e^1+2-sin\ 1[/tex]

[tex]s_1=3.8768\ m[/tex]

Position at t=5s:

[tex]s_5=e^5+2-sin\ 5[/tex]

[tex]s_5=151.3721\ m[/tex]

Therefore the distance covered during this time is:

[tex]\Delta s=s_5-s_1[/tex]

[tex]\Delta s=151.3721-3.8768[/tex]

[tex]\Delta s =147.4952\ m[/tex]

Now the average speed for the duration:

[tex]v_{avg}=\frac{\Delta s}{\Delta t}[/tex]

[tex]v_{avg}=\frac{147.4952}{5}[/tex]

[tex]v_{avg}=29.499\ m.s^{-1}[/tex]

Answer:

If the truck were lifted twice as high, it would have twice as much potential energy

Explanation:

Potential energy = mgh

where;

m is mass of the truck (kg)

g is acceleration due to gravity (m/s²)

h is the height above the floor.

Initial Potential Energy, E₁ = mgh₁

When the truck was lifted twice as high, New height (h₂) = 2h₁

New potential Energy, E₂ =mgh₂ = mg(2h₁)

E₂ = 2mgh₁

Recall, E₁ = mgh₁, then substitute in E₁  into E₂

E₂ = 2(mgh₁)

E₂ = 2E₁

Therefore, If the truck were lifted twice as high, it would have twice as much potential energy

Final answer:

The speed of the ram at the instant of contact is 8.413 m/s.

Explanation:

To find the speed of the ram at the instant of contact, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is converted into the kinetic energy of the ram as it hits the staple. Since the mass of the spring is ignored, we can consider the system as consisting of only the ram.

Using the equation for the potential energy of the spring (PE = 0.5kx^2), we can calculate the initial potential energy. The compressed length of the spring is 3.20 x 10^-2 m, so the initial potential energy is 0.5(37107 N/m)(3.20 x 10^-2 m)^2 = 192.128 J.

At the instant of contact, all of the potential energy has been converted into kinetic energy. Therefore, we can equate the initial potential energy to the final kinetic energy of the ram. The kinetic energy is given by KE = 0.5mv^2, where m is the mass of the ram (0.179 kg) and v is the speed of the ram at the instant of contact.

Solving for v, we have: 192.128 J = 0.5(0.179 kg)v^2. Rearranging the equation and solving for v, we find that v = 8.413 m/s. Therefore, the speed of the ram at the instant of contact is 8.413 m/s.

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

[tex]f'=f(\dfrac{v+v_{p}}{v-v_{s}})[/tex]

Where, f = frequency

v = speed of sound

[tex]v_{p}[/tex] = speed of passenger

[tex]v_{s}[/tex] = speed of source

Put the value into the formula

[tex]f'=262\times(\dfrac{344+18}{344-30})[/tex]

[tex]f'=302.05\ Hz[/tex]

Hence, The frequency is 302.05 Hz.

Answer:

Time takes the player to read the entire CD is 94.9 minutes.

Explanation:

compact disc (CD) contains 783.216 megabytes of digital information.

Each byte consists of exactly 8 bits.

constant rate of 1.1 megabits per second

How many minutes does it take the player to read the entire CD?

1 Megabites = 1000000 bites

783.216 megabytes = 783216000 bites = 783216000 x 8 = 6265728000 bits

1.1 megabits/s = 1100000 bits/s

thus

t =  6265728000 bits/1100000 bits/s

 = 5696 s

 = 94.9 min

Answer:

c. 8.67 cm/s

Explanation:

From the law of conservation of momentum,

Total momentum before the thread was burned = Total momentum after was burned

mu + m'u' = mv + m'v'...................... Equation

Where m = mass of the heavier cart, m' = mass of the lighter cart, u = initial velocity of the bigger cart, u' = initial velocity of the smaller cart, v = final velocity of the bigger cart, v' = final velocity of the smaller cart.

Note: Both cart where momentarily at rest, as  such u = u' = 0. i.e the total momentum before the thread was burn = 0

And assuming the left is positive,

We can rewrite equation 1 as

mv + m'v' = 0............................................ Equation 2

Given: m = 4.5 kg, m' = 1.5 kg, v' = 26 cm/s

Substitute into equation 2,

4.5v + 1.5(26) = 0

4.5v + 39 = 0

4.5v = -39

v = -39/4.5

v = -8.67 cm/s.

Note: v is negative because it moves to right.

Hence the velocity of the 4.5 kg cart = 8.67 cm/s.

The right option is c. 8.67 cm/s

The velocity of the 4.5 kg cart after the thread is burned is 10 cm/s.

When the thread is burned, the compressed spring pushes the carts apart. Since the 1.5 kg cart moves to the left with a speed of 26 cm/s, the 4.5 kg cart would also move to the left but with a slower speed. To determine the velocity of the 4.5 kg cart, we can use the principle of conservation of momentum:

Momentum before = Momentum after

(Mass of 1.5 kg cart) x (Initial velocity) + (Mass of 4.5 kg cart) x (Initial velocity) = (Mass of 1.5 kg cart + Mass of 4.5 kg cart) x (Final velocity)

Solving for the final velocity of the 4.5 kg cart:

(1.5 kg x 26 cm/s + 4.5 kg x 0 cm/s) / (1.5 kg + 4.5 kg) = (6.9 kg) x (Final velocity)

Final velocity = 10 cm/s

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Answer:

Susie has strabismus.

Explanation:

Strabismus is a medical condition in which the patient can not keep his/her eyes aligned under the normal conditions. In strabismus one or both of eyes rolls into up, down, in or out direction. Physicians usually recommend the exercises for eyes or placing a patch on the unaffected eye or they also do surgery where it's required like in severe cases.  

Answer:

a. [tex]v'=8.925\ m.s^{-1}[/tex]

b. [tex]I=1585397.5872\ N.s[/tex]

c. [tex]s=265.4125\ m[/tex]

Explanation:

a.

Velocity of the two cars just after the collision:

Using the law of conservation of

[tex]m_m.v+m_s\times 0=(m_m+m_s)\times v'[/tex]

[tex]2100\times 17+ 0=(2100+1900)\times v'[/tex]

[tex]v'=8.925\ m.s^{-1}[/tex]

b.

Frictional force acting on the combined mass of cars:

[tex]f=\mu_k\times N[/tex]

where:

N = reaction normal to the contact surface by the ground

[tex]f=0.68\times (2100+1900)\times 9.8[/tex]

[tex]f=26656\ N[/tex] is the resistance force to the motion of the wheels

Now, by Newton's second law of motion:

[tex]f=\frac{dp}{t}[/tex]

[tex]f=\frac{(m_m+m_s)\times v'}{t}[/tex]

[tex]26656=\frac{(2100+1900)\times 8.925}{t}[/tex]

[tex]t=59.4762\ s[/tex]

Now impulse is the impact load acting acting for certain time:

[tex]I=f\times t[/tex]

[tex]I=26656\times 59.4762[/tex]

[tex]I=1585397.5872\ N.s[/tex]

c.

The distance the the cars skid before coming to rest:

using equation of motion:

[tex]v'_f=v'+a.t[/tex]

we've [tex]v'_f=0\ m.s^{-1}[/tex] since final velocity is zero

[tex]0=8.925+a\times 59.4762[/tex]

[tex]a=0.1501\ m.s^{-2}[/tex]

now since we have:

[tex]s=v'.t+\frac{1}{2} a.t^2[/tex]

[tex]s=8.925\times 59.4762-0.5\times 0.1501\times 59.4762^2[/tex]

[tex]s=265.4125\ m[/tex]

A. The velocity of the two cars just after the collision is 8.925 m/s

B. The impulse that acts on the skidding cars just after the collision is 35700 Ns

C. The distance travelled by the skidding cars before coming to rest is 5.98 m

A. Determination of the velocity the two cars

Mass of moving car (m₁) = 2100 Kg

Initial velocity of moving car (u₁) = 17 m/s

Mass of stationary car (m₂) = 1900 Kg

Initial velocity of stationary car (u₂) = 0 m/s

[tex]v = \frac{u_1m_1 + u_2m_2}{m_1m_2} \\ \\ v = \frac{(2100 \times 17) + (1900 \times 0)}{2100 + 1900} \\ \\ v = 8.925 \: m/s[/tex]

Thus, the velocity of the two cars is 8.925 m/s

B. Determination of the impulse

Mass of moving car (m₁) = 2100 Kg

Mass of stationary car (m₂) = 1900 Kg

Velocity of the two cars (v) = 8.925 m/s

I = Ft = mv

I = (2100 + 1900) × 8.925

Thus, the impulse on the skidding cars is 35700 Ns

C. Determination of the distance travelled by the skidding cars.

Mass of moving car (m₁) = 2100 Kg

Mass of stationary car (m₂) = 1900 Kg

Total mass = 2100 + 1900 = 4000 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (N) = mg = 4000 × 9.8 = 39200 N

Coefficient of kinetic friction (μ) = 0.68

F = μN

F = 0.68 × 39200

Impulse (I) = 35700 Ns

Force (F) = 26656 N

I = Ft

35700 = 26656 × t

Divide both side by 26656

t = 35700 / 26656

Initial velocity of the two cars (u) = 8.925 m/s

Final velocity of the two cars (v) = 0 m/s

Time (t) = 1.34 s

[tex]s = \frac{(v + u)t}{2} \\ \\ s = \frac{(8.925 + 0)1.34}{2} \\ \\ s = \frac{8.925 \times 1.34}{2} \\ \\ s = 5.98 \: m[/tex]

Therefore, the distance travelled by the skidding cars before coming to rest is 5.98 m

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Answer:

F=G[tex]\frac{M1XM2}{R^{2} }[/tex]

Explanation:

F=F=G[tex]\frac{M1XM2}{R^{2} }[/tex]

Newton law define gravity as the product of two mass and the mean squired distance between the two bodies.

Where F is the gravitational pull and G is the gravitational constant and M1 and M2 are masses of two bodies while R is their mean distance

Replacing the value in the equation it becomes:

3.2x10N=G[tex]\frac{M1X55}{2.1X10^{2} }[/tex] Where G is the gravitation constant,[tex]6.67x10^{-11}[/tex]

M1=[tex]\frac{3.2X10X(2.1X10)^{2} }{55}[/tex]

M1=[tex]3.8468x10^{-11}[/tex]Kg

Answer:

A. 3.1m/s²

Explanation:

Since the body is sliding down the plane, the frictional force (Ff) being a force of opposition will be acting upwards.

The forces acting on the body in the vertical direction will be the weight (W) and the normal reaction(R) acting in the opposite direction.

Taking the sum of the forces along the plane,

Fm - Ff = mass × acceleration (newton's law)

Since Fm = Wsin(theta)

Ff = nR where n is the coefficient of friction we will have;

Wsin(theta) - nR = ma... (1)

W = mg = 6×10 = 60N

n = 0.40

R = Wcos (theta) {resolving weight to the vertical)

R = 60cos39°

R = 46.6N

Substituting this values into eqn 1 to the get the acceleration,

60sin39° - 0.4(46.6) = 6a

37.8 - 18.64 = 6a

19.16 = 6a

a = 19.16/6

a = 3.19m/s²

The acceleration of the body down the plane will be 3.1m/s²

Wavelength of the most strongly reflected is approximately 707 nm, which corresponds to red light in the visible spectrum.

To determine the wavelength most strongly reflected by a soapy water film, we can use the principle of thin film interference. Here,

For constructive interference to occur in a thin film illuminated perpendicularly, the condition we use is:

2nt = mλ

where n is the refractive index of the film,

Since we need the most strongly reflected wavelength, we'll consider

m = 1 (the first order of constructive interference).

Substituting the values into the equation:

2 x 1.33 x 266 nm = 1 x λ

This simplifies to:

λ = 2 x 1.33 x 266 nm

λ ≈ 707 nm

Therefore, the most strongly reflected wavelength is approximately 707 nm, which corresponds to red light in the visible spectrum.

Answer: The astronomer who, at the turn of the century, measured the spectra of hundreds of thousands of stars (leaving a catalog that astronomers used for the rest of the century) was: Annie Jump Cannon.

Annie Jump Cannon was born on 11th December 1863 in Dover, Delaware. She was an american astronomer whose work contributes in the expansion of current stellar classification. She made the first accurate attempt to organize and classify stars on the basis of their temperature and spectral form. She suffered from hearing loss during her childhood. She was almost deaf during her career. She was member of National women's party and by her nature, she was a extreme suffragist.

Annie Jump was the eldest of the three daughters of Wilson Cannon (shipbuilder and state senator) and Mary Jump (second wife of Wilson Cannon). It was Annie's mother who encouraged her to peruse her education and career in her interests. She suggested her to study in Wellesley college, perusing in mathematics, biology and chemistry. Cannon's mother teaching like household economics, helped her to organize her research later on.

She took her early education and was brilliant student of mathematics, then went to Wellesley College of Massachusetts in 1880 and studied physics and astronomy, later she went to work at Harvard Observatory.

Astronomer Annie Jump was the pioneer, who develop a simple spectral classification system. She categorized nearly 400,000 stars in total. She for the very first time found a double star, three hundred variable stars and 5 novas. She worked as astronomer for more then forty years till 1940, until she got retired. During her career she also work as suffragist, and helped many women to attain respect and reputation in scientific society. She was a hard working woman with a calm nature who build the ways for upcoming women in the field of scientific research and astronomy.

Annie Jump Canon died on 13th April, 1941 in Cambridge, Massachusetts. she died at the age of 72. She was hospitalized prior to her death due extreme illness. After her death American Astronomical Society announce an award named after her (The Annie Jump Cannon Award), yearly to those female astronomers, who worked remarkably in astronomy.

The astronomer who, at the start of the 20th century, measured and catalogued the spectra of hundreds of thousands of stars was Annie Jump Cannon. She worked under Edward C. Pickering at the Harvard Observatory and her detailed catalogue served as a major reference for astronomers for the rest of the century.

The astronomer you're referring to is Annie Jump Cannon who worked at the Harvard Observatory under Edward C. Pickering. In the late 1800s, Pickering started an ambitious project of classifying stellar spectra, the foundation of this work was a remarkable collection of close to one million photographic spectra of stars. Cannon was hired by Pickering to assist in the classification of spectra and her efficiency was such, that she could visually identify and determine the spectral types of several hundred stars per hour.

This relentless work resulted in Cannon creating a comprehensive catalog containing the spectral types for hundreds of thousands of stars. This seminal piece of work became the blueprint for much of twentieth-century astronomy, and served to influence and inform a whole generation of astronomers. Therefore, Cannon's detailed and systematic study was instrumental in developing a deeper understanding of the universe as we know it today.

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Answer: The wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.

Explanation:

To calculate the wavelength of light, we use the equation:

[tex]\lambda=\frac{c}{\nu}[/tex]

where,

= wavelength of the light  = ?

c = speed of light = 300000 km/s = [tex]3\times 10^8m/s[/tex]   (1km=1000m)

[tex]\nu[/tex] = frequency of light = 100kHz = 100000Hz or    (1kHz=1000Hz)

[tex]\lambda=\frac{3\times 10^8m/s}{100000s^{-1}}[/tex]

[tex]\lambda=3000m[/tex]

Thus the wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.

Answer:

The correct options are b. This scale rates an earth quake according to how much damage it causes. and d. This scale uses roman numerals to rank the damage caused by an earth quake.

Explanation:

According to the definition of the Mercalli scale " it is a scale used for the measurement of the intensity of the earthquakes" and was first invented in 1884 by the Italian volcanologist Giuseppe Mercalli. This scale ranges from I to XII.

Answer:

b d

Explanation:

did the quiz

Answer:

35.7 kg

Explanation:

We are given that

Left scale reads =200N

Right scale reads=150 N

We have to find the mass in kg.

We know that

Weight  of child= Sum of two scales=200+150=350 N

Acceleration due to gravity=[tex]9.8 m/s^2[/tex]

We know that

Weight= mg

Using the formula

[tex]350=9.8 m[/tex]

[tex]m=\frac{350}{9.8}=35.7 kg[/tex]

Hence, the mass of child=35.7 kg

Answer:

total magnification = 400 X

Explanation:

given data

ocular lens = 10 X

objective lens = 40 X

to find out

total magnification

solution

we know that total magnification is express as

total magnification = Objective magnification ×  ocular magnification  .................1

put here value we get

total magnification =  10 × 40

total magnification = 400 X

The total magnification for a microscope with a 10X ocular lens and a 40X objective lens is 400X.

When using a microscope, the total magnification can be found by multiplying the magnification of the ocular lens (also known as the eyepiece) by the magnification of the objective lens. Thus, in your case, the total magnification of the specimen being viewed with a 10X ocular lens and a 40X objective lens would be 400X. This is because 10 times 40 equals 400. Therefore, you are observing the specimen at 400 times its actual size.

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Answer:

the magnet or electric current produces a magnetic field

Explanation:

this magnetic field can be visualized as a pattern of a circular field lines surrounding a wire. hall probes can determine the magnitude of the field

hope it helps you

The direction of a magnetic field produced by an electric current in a straight wire can be determined using the right-hand rule. The Biot-Savart law is used to calculate the magnetic field strength at a distance from the wire.

To describe and determine the direction of the magnetic field produced by an electric current, one commonly uses the right-hand rule. For a straight, current-carrying wire, point your right-hand thumb in the direction of the current; then, the direction in which your fingers curl around the wire represents the direction of the magnetic field lines, which form concentric circles around the wire.

For example, if the current flows from right to left in a wire, and you are looking at the wire end-on from the left end, your thumb would point to the left, indicating the direction of the current. Your fingers would naturally curl in a counterclockwise direction, suggesting the magnetic field direction at that point.

For calculating the magnetic field strength, B, produced by a long straight wire carrying current I, the Biot-Savart law can be used. It states that the strength of the magnetic field is directly proportional to the magnitude of the current and inversely proportional to the distance from the wire (assuming the wire is very long compared to the distance).

Answer: The final speed of the balls will be 8.12m/s

Explanation:

For least collision of bodies, both momentum and kinetic energy is conserved.

According to Law of conservation of energy,

The sum of momentum of the objects before collision is equal to the sum of their momentum after collision. Note that the objects will move with a common velocity after impact.

Let m1 and m2 be the masses of the objects

u1 and u2 be their velocities before impact

v be their common velocity after impact

Since momentum is mass × velocity

Mathematically,

m1u1 + m2u2 = (m1+m2)v

Given m1= 3.4kg m2 = 5.7kg u1 = 10.5m/s u2 = 6.7m/s v = ?

Substituting this values in the formula, we have

3.4(10.5)+5.7(6.7) = (3.4+5.7)v

35.7+38.19 = 9.1v

73.89 = 9.1v

V = 73.89/9.1

V= 8.12m/s

Answer:

392307.6923 m/s²

Explanation:

t = Time taken

u = Initial velocity = 560 m/s

v = Final velocity = 460 m/s

s = Displacement = 13 cm

a = Acceleration

From the equation of motion we have

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{460^2-560^2}{2\times 13\times 10^{-2}}\\\Rightarrow a=-392307.6923\ m/s^2[/tex]

The acceleration of the bullet as it passes through the board is -392307.6923 m/s²