if the force acting on a body of mass 40 k.g is doubled by how much will the acceleration change​

Answer:

Explanation:

2.Big bang, 3.contraction of the solar nebula, 5.stellar ignition in our sun, 4.outgassing of earths secondary atmosphere, 1.appearance of the first ocean on earth ,7.evolution of photosynthesis, 6.build- up of oxygen in earths atmosphere.

Explanation:

PE=mgh

5=m(9.8)(2)

m=5/19.6

m=0.2251 kg

m=225.1 grams

The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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Answer:

Basketball

Explanation:

Im asuming basketball because a basketball has the least movement out of the rest of the answers.

Answer:

3 m/s²

Explanation:

The acceleration is equal to the change in velocity divided by the change in time. Since the velocity increases by 0.3 m/s for every tenth of a second, we can say that the change in velocity is 0.3 m/s and the change in time is 0.1 s, then the acceleration is equal to:

Therefore, the acceleration of the object is 3 m/s²

The current flowing in the 2Ω and 1Ω is 1.14 A and the current flowing in the 3Ω and 4Ω is 0.286 A.

The value of the current in each resistor is calculated by applying Kirchoff voltage law as follows;

The total voltage in loop 1 is calculated as;

2 + 4 - I₁R₁ - (I₁ - I₂)R₂ - I₁R₃ = 0

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

The current flowing in loop 2 is calculated as;

I = V/R

I₂ = ( 6 V - 4 V ) / (3 + 4)

I₂ = 0.286 A

The value of the current flowing in loop 1 is calculated as;

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

6 - 2I₁ - 3(I₁ - 0.286) - 1₁ = 0

6 - 3I₁ - 3₁ + 0.858 = 0

-6I₁ = -6.858

I₁ = 6.858 / 6

I₁ = 1.14 A

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Frequency is defined as number of waves or vibration per unit time.

Therefore, option (a), frequency is the number of waves passing a given point each second, is the correct choice.

Complete Question

X-ray diffraction is used to study the structure of crystallized proteins, nucleic acids, and other biological macromolecules. It was actually the technique used to study the structure of DNA in the early 1950s. X-rays of wavelength 0.22 nm are to used to study the structure of a protein. One of the maxima in intensity is located at 0.8 rad from the crystal planes responsible for this maximum.

1. Which wavelength of x-ray listed below would give you a maximum at the same location?

a. 0.55 nm

b. 0.33 nm

c. 0.77 nm

d. 0.44 nm

2. If this is the first maximum for the X-rays of wavelength 0.22 nm, what is the crystal plane separation of the protein that is responsible?

a. 0.16 nm

b. 0.31 nm

c. 0.14 nm

d. 0.22 nm

Answer:

1

  The correct option is D

2

  The correct option is A

Explanation:

From the question we are told that

   The wavelength of X-ray used is  \(\lambda = 0.22 \ nm = 0.22 *10^{-9} \ m\)

   The  angular displacement is  \(\theta = 0.8 \ rad = 0.8 *\frac{180}{3.142} = 45 ^o\)

Generally from  Brags equation we have that

      \(2d sin \theta = n \lambda\)

Here d is the distance of separation between that planes of the protein

and  2d is constant given that the separation does not change

So

      \(2d sin (45.84) = n \lambda\)    

Given that 2d is constant in order for the angle for \(\lambda\) to be the same for \(\lambda '\) then  

      \(\lambda'\) must be an integral multiple of  \(\lambda\)

So  

         \(\lambda' = 2 * \lambda\)

=>      \(\lambda' = 2 *0.22\)

=>      \(\lambda' = 0.44 \ nm\)

Considering question 2

From the question we are told that

    The wavelength is  \(\lambda = 0.22 \ nm = 0.22 *10^{-9} \ m\)

Generally from the above equation we have that

         \(d = \frac{\lambda }{2 * sin (\theta )}\)

=>      \(d = \frac{0.22 * 10^{-9} }{2 * sin (45.84 )}\)

=>       \(d = 1.6 *10^{-10 } \ m\)

=>       \(d = 0.16 *10^{-9 } \ m\)

=>       \(d = 0.16 \ nm\)

Answer:

how close a true value a measurement is

Answer:

The correct answer is "\(\frac{4E}{3}\)".

Explanation:

According to the question,

Energy of satellite,

⇒ \(E_s=-\frac{GM_sM_E}{2r}\)

For the very 1st case:

\(r = R_E+R_E\)

  \(=2R_E\)

or,

⇒ \(E=-\frac{GM_sM_E}{4R_E}\)...(1)

For the new case:

\(r = R_E+\frac{R_E}{2}\)

  \(=\frac{3R_E}{2}\)

then,

⇒ \(E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }\)

        \(=-\frac{GM_sM_E}{3R_E}\)...(2)

From equation (1) and (2), we get

⇒ \(E'=\frac{1}{3}(4E)\)

        \(=\frac{4E}{3}\)  

Answer:

a. i) The pressure drop per unit length is 52,151.89 Pa

ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta

b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts

ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta

c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel

ii) The velocity at the center is approximately 2.04 m/s

Explanation:

The given diameter of the aorta, D = 2.5 cm = 0.025 m

The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s

Assumptions;

The blood flow is laminar

The blood is a Newtonian fluid

The viscosity of water ≈ 0.01 poise = 1 cp

a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;

\(Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)\)

Where;

Q = The flow rate = A·v

A = The cross sectional area

R = The radius = D/2

Δp/L = The pressure drop per unit length of the pipe

Therefore, we have;

\(\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89\)

The pressure drop per unit length ΔP/L = 52,151.89 Pa

ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;

∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa

Given that the atmospheric pressure, \(P_{atm}\) = 10⁵ Pa, we have;

\(P_{atm}\)/ΔP = 10⁵/5,215.189 ≈ 19.175

Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta

b. i) The power, P = Q × ΔP

Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts

ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;

P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35

The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length

c. i) The velocity profile across the aorta is given as follows;

\(v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2\)

Where;

\(v_m\) = The velocity at the center

We get;

\(v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04\)

The velocity at the center, \(v_m\) ≈ 2.04 m/s

ii) The velocity profile, v(r), is given by the following formula;

\(v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right]\)

Therefore, we have;

\(v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2\)

The velocity profile of the pipe is created with Microsoft Excel

If A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and the frictional force between the box and surface is 100 n, Then the distance covered by the box is 10.89 meters.

To calculate the distance covered by the box, we need to analyze the forces acting on it and apply the work-energy principle.

Given:

Mass of the box, m = 210 kg

Tension in the string, T = 1300 N

The angle of inclination, θ = 35°

Frictional force, f = 100 N

Initial speed, u = 0 m/s

Final speed, v = 10 m/s

First, let's resolve the tension force into components parallel and perpendicular to the incline. The parallel component of the tension force can be calculated as:

T_parallel = T * cos(θ)

Next, let's calculate the net force acting on the box along the incline. The net force is given by:

Net force = T_parallel - f

Now, using Newton's second law, we can calculate the acceleration (a) of the box:

Net force = m * a

From the given information, we have the final velocity (v), initial velocity (u), and acceleration (a). We can use the following kinematic equation to calculate the distance covered (s):

v^2 = u^2 + 2as

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Now, let's plug in the given values and calculate the distance covered:

T_parallel = 1300 N * cos(35°) ≈ 1067.35 N

Net force = 1067.35 N - 100 N = 967.35 N

a = (967.35 N) / (210 kg) ≈ 4.61 m/s^2

s = (10 m/s)^2 - (0 m/s)^2 / (2 * 4.61 m/s^2) ≈ 10.89 m

Therefore, the distance covered by the box is approximately 10.89 meters.

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Answer:

F/L =  8*10^-4 N/m

Explanation:

To calculate the magnitude of the force per meter in the central wire, you take into account the contribution to the force of the others two wires:

\(F_N=F_{1,2}+F_{2,3}\)   (1)

F1,2 : force between first and second wire

F2,3 : force between second and third wire

The force per meter between two wires of the same length is given by:

\(\frac{F}{L}=\frac{\mu_oI_1I_2}{2\pi r}\)

μo: magnetic permeability of vacuum =  4pi*10^-7 T/A

r: distance between wires

Then, you have in the equation (1):

\(\frac{F_N}{L}=\frac{\mu_oI_1I_2}{2\pi r}+\frac{\mu_oI_2I_3}{2\pi r}\\\\\frac{F_N}{L}=\frac{\mu_oI_1}{2\pi r}[I_2+I_3]\)

But

I1 = I2 = I3 = 10A

r = 10cm = 0.1m

You replace the values of the currents and the distance r and you obtain:

\(\frac{F_N}{L}=\frac{\mu_oI^2}{\pi r}\\\\\frac{F_N}{L}=\frac{(4\pi*10^{-7}T/A)(20A)^2}{2\pi (0.1m)}=8*10^{-4}\frac{N}{m}\)

hence, the net force per meter is 8*10^-4 N/m

The x and y components of the plane's velocity is:

1. How do I determine the horizontal component (Vx) of the velocity?

We can obtain the horizontal component (Vx) of the velocity as illustrated below:

Vx = u × Cosθ

Vx = 40 × Cos 15

Vx = 38.6 m/s

Thus, the horizontal component (Vx) of velocity is 38.6 m/s

2. How do I determine the vertical component (Vy) of the velocity?

The vertical component (Vy) of the velocity can be obtained as

Vy = u × Sine θ

Vy = 40 × Sine 15

Vy = 10.4 m/s

Thus, the vertical component (Vy) of velocity is 10.4 m/s

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By supplying the fundamental knowledge required to create new instruments and techniques for medical use, physics enhances our quality of life

From can openers, light bulbs, and mobile phones to muscles, lungs, and brains; from paintings, piccolos, and pirouettes to cameras, vehicles, and cathedrals; from earthquakes, tsunamis, and storms to quarks, DNA, and black holes, physics aids us in understanding the workings of the world around us.

The science of physics is the most fundamental and has many applications in contemporary technology. Because it makes it possible for smartphones, computers, televisions, watches, and many other modern technologies to function automatically, physics is crucial to modern technology.

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Considering the definition of redox reactions, the correct answer is the last option: Oxygen is reduced, and iron is oxidized.

Oxidation is a reaction where an atom, ion or molecule loses electrons while reduction corresponds to the gain of electrons from an atom, ion or molecule.

The oxidation and reduction reactions always occur simultaneously for what are generally known as oxidation-reduction reactions or redox reactions.

That is, a redox reaction is a chemical reaction that occurs between an oxidizing substance and a reducing substance. During the reaction, the oxidizing substance loses electrons and the reducing substance gains electrons.

That is, the oxidizing agent is the one that traps the electrons while the reducing agent is the one that provides the electrons.

In this case, when iron pipes come in contact with water and oxygen, iron transfers electrons to the oxygen molecules. So, iron loses electrons while oxygen gains electrons. So oxygen is reduced, and iron is oxidized.

Then, the correct answer is the last option: Oxygen is reduced, and iron is oxidized.

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Answer:

The correct answer is: Oxygen is reduced, and iron is oxidized.

Answer:

 (C) Mechanical Force

Explanation:

Answer:

The speed will be "0.144 rad/s".

Explanation:

Given that,

Diameter,

d = 7.50 km

Radius,

R = \(\frac{7.5}{2} \ Km\)

Acceleration on inner curve,

= 8 times

Now,

As we know,

⇒ \(\omega^2R=8g\)

or,

⇒ \(\omega=\sqrt{\frac{8g}{R} }\)

On substituting the values, we get

⇒     \(=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }\)

⇒     \(=\sqrt{\frac{78.4}{3750} }\)

⇒     \(=\sqrt{0.0209}\)

⇒     \(=0.144 \ rad/s\)

Answer:

Charged

Explanation:

When an atoms loses or gains electrons they become charged(ion)

When an atom loses electrons it becomes positively charged and it is called a Cation

When an atom gains electrons it becomes negatively charged and it is called an Anion

Answer:

It takes approximately 0.625 seconds for the person to be stopped by the airbag.

Explanation:

We can use the formula for average acceleration to solve this problem:

a = (v_f - v_i) / t

where:

a = average acceleration

v_f = final velocity (0 m/s)

v_i = initial velocity (15 m/s)

t = time

We know that the airbag provides a stopping force of 1200 N, which is equal to the net force on the person-car system. We can use Newton's second law of motion to find the acceleration of the person:

F_net = ma

where:

F_net = net force (1200 N)

m = mass of person (50 kg)

a = acceleration

Solving for a, we get:

a = F_net / m

a = 1200 N / 50 kg

a = 24 m/s^2

Now we can substitute the values of a, v_f, and v_i into the formula for average acceleration and solve for t:

a = (v_f - v_i) / t

24 m/s^2 = (0 m/s - 15 m/s) / t

t = -15 m/s / 24 m/s^2

t = 0.625 s

Therefore, it takes approximately 0.625 seconds for the person to be stopped by the airbag.

Answer:

PE = 137.2931 J

Explanation:

PE = 137.2931 J

Answer:

When the cold weather fills hard surfaces with water, it causes it to freeze and expand, causing cracks in the surface.

Answer:

C

Explanation:

Sled A has more potential energy because it's mass is 100 kg, and it is higher up than Sled B. The more high up the sled is and the lighter it is, the faster it gets, it creates more and more potential energy.

Hope this helps!

Answer: just say no and never talk to that person again.

Explanation:

because that is what i would do to prevent drugs and/or nicotine to enter my body.

The density of water is approximately 1 g/cm^3. Therefore, the volume of 2800 g of water would be 2800 cm^3 because density is mass/volume, and so volume is mass/density.

Since this volume is inside a cubic box, the length of each side of the cube (a, for instance) could be found by taking the cubic root of the volume. This is because the volume of a cube is calculated by a^3 (length of one side cubed). Hence, a = cube root of 2800 cm^3 ≈ 14.1 cm.

Converting centimeters to meters (as 1 meter is equal to 100 centimeters), we get approximately 0.141 meters.

So the filled cubic box has a side length of approximately 0.141 m.

The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:

∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0

(right is positive, left is negative)

∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0

(up is positive, down is negative)

Solve the system of equations. I use elimination here:

• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):

sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0

cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0

T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0

T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)

• Subtract the first equation from the second to eliminate T₁ :

T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

• Solve for T₂ :

T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

T₂ sin(74.0°) = (215 N) cos(29.5°)

… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))

T₂ = (215 N) cos(29.5°) / sin(74.0°)

T₂ ≈ 195 N

• Solve for T₁ :

T₁ cos(29.5°) - T₂ cos(44.5°) = 0

T₁ cos(29.5°) = T₂ cos(44.5°)

T₁ = T₂ cos(44.5°) / cos(29.5°)

T₁ ≈ 160. N

The young's modulus in a cantilever will be 3.92 x 10¹⁰ N/m².

Young's modulus (E) is a material property that indicates how easily it can stretch and deform and is defined as the ratio of tensile stress () to tensile strain (). Where stress denotes the amount of force applied per unit area ( = F/A) and strain denotes the extension per unit length

Given that the length is 1m which is suspended by a load of 150g. The depression is found to be 4cm. The thickness of the beam is 5mm and the breath is 3cm.

The young's modulus will be calculated by the formula below,

Y = (4gl³) / (bd³) x ( M / y )

Y = ( 4 x 9.81 x 1³ x 0.150 ) / ( 0.03 x 0.005³ x 0.04 )

Y = 3.92 x 10¹⁰ N/m².

Therefore, young's modulus will be 3.92 x 10¹⁰ N/m².

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The elasticity of supply of labor varies among occupations. The supply of labor is elastic for occupations (a) and (b) and inelastic for occupations (c).

The elasticity of the supply of labor is an important concept that measures the responsiveness of the quantity of labor supplied to changes in the wage rate. An occupation's supply of labor is said to be elastic if its elasticity is greater than 1, inelastic if the elasticity is less than 1, and unitary elastic if the elasticity of supply equals 1. Let's calculate the elasticity of supply for the following occupations and determine whether the supply of labor is elastic, inelastic, or unitary elastic.

a. %ΔES = 7, %ΔW = 3

The formula for the elasticity of supply is (% change in quantity supplied / % change in wage). Therefore, the elasticity of supply for this occupation would be:

Elasticity of supply = (7% / 3%) = 2.33

Since the elasticity of supply is greater than 1, the supply of labor is elastic.

b. ES = 120, W = $8

E’S = 90, W’ = $6

The formula for the elasticity of supply is ((% change in quantity supplied) / (% change in wage)). Therefore, the elasticity of supply for this occupation would be:

Elasticity of supply = ((90 - 120) / ((90 + 120) / 2)) / ((6 - 8) / ((6 + 8) / 2))

The elasticity of supply = (-30 / 105) / (-2 / 7)

Elasticity of supply = 2.00

Since the elasticity of supply is greater than 1, the supply of labor is elastic.

c. ES = 100, W = $5

E’S = 120, W’= $7

The formula for the elasticity of supply is ((% change in quantity supplied) / (% change in wage)). Therefore, the elasticity of supply for this occupation would be:

Elasticity of supply = ((120 - 100) / ((120 + 100) / 2)) / ((7 - 5) / ((7 + 5) / 2))

The elasticity of supply = (20 / 110) / (2 / 6)

Elasticity of supply = 0.81

Since the elasticity of supply is less than 1, the supply of labor is inelastic. The elasticity of supply is a critical concept in labor economics and helps us understand how changes in wage rates affect the quantity of labor supplied in different occupations.

Complete question:

As with the own-wage elasticity of demand for labor, the elasticity of supply of labor can be similarly classified. The elasticity of the supply of labor is elastic if the elasticity is greater than 1. It is inelastic if the elasticity is less than 1, and it is unitary elastic if the elasticity of supply equals 1. For each of the following occupations, calculate the elasticity of supply, and state whether the supply of labor is elastic, inelastic, or unitary elastic. E’s and W’ are the original supply of workers and wages. and are the new supply of workers and wages.

a %ΔES = 7, %ΔW = 3

b. ES = 120, W = $8

E’S = 90, W’ = $6

c. ES = 100, W = $5

E’S = 120, W’= $7

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