For which scientific method do you use experimental and control groups?
A. Forming a hypothesis
B. Analyzing data
C. Drawing conclusions
D. Conducting an experiment

Changing the green sum of forces arrow involves manipulating applied forces, frictional forces, and normal forces, allowing for adjustments in magnitude and direction through various means.

To change the green sum of forces arrow, you can employ the following three strategies:

Adjusting Applied Forces: By modifying the magnitude or direction of the applied forces, you can alter the green sum of forces arrow. If the applied forces are increased or directed in a different way, the green sum of forces arrow will change accordingly. For instance, increasing the magnitude of a pushing force will result in a larger green sum of forces arrow in that direction.

Modifying Frictional Forces: Frictional forces play a crucial role in determining the green sum of forces. By changing the coefficient of friction or applying lubricants, you can affect the magnitude of frictional forces acting on an object. Reducing friction will decrease the green sum of forces arrow, while increasing friction will have the opposite effect.

Varying Normal Forces: The green sum of forces arrow can be influenced by adjusting the normal forces acting on an object. Normal forces are perpendicular to the surface and counteract the weight of an object. By changing the angle or surface on which an object rests, you can modify the normal forces. This alteration will subsequently impact the green sum of forces arrow.

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A 66.1-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.59 m, and ends with a speed of 8.51 m/s. The nonconservative work done on the boy is approximately -42.7 kilojoules.

To find the nonconservative work done on the boy, we need to consider the change in the boy's mechanical energy during the process. Mechanical energy is the sum of the boy's kinetic energy (KE) and gravitational potential energy (PE).

The initial mechanical energy of the boy is given by the sum of his kinetic energy and potential energy when he catches the wave:

E_initial = KE_initial + PE_initial

The final mechanical energy of the boy is given by the sum of his kinetic energy and potential energy after he drops through the height:

E_final = KE_final + PE_final

The nonconservative work done on the boy is equal to the change in mechanical energy:

Work_nonconservative = E_final - E_initial

Let's calculate each term:

KE_initial = (1/2) * m * v_initial^2

= (1/2) * 66.1 kg * (1.60 m/s)^2

PE_initial = m * g * h_initial

= 66.1 kg * 9.8 m/s^2 * 1.59 m

KE_final = (1/2) * m * v_final^2

= (1/2) * 66.1 kg * (8.51 m/s)^2

PE_final = m * g * h_final

= 66.1 kg * 9.8 m/s^2 * 0

Since the boy ends at ground level, the final potential energy is zero.

Substituting the values into the equation for nonconservative work:

Work_nonconservative = (KE_final + PE_final) - (KE_initial + PE_initial)

Simplifying:

Work_nonconservative = KE_final - KE_initial - PE_initial

Calculating the values:

KE_initial = (1/2) * 66.1 kg * (1.60 m/s)^2

PE_initial = 66.1 kg * 9.8 m/s^2 * 1.59 m

KE_final = (1/2) * 66.1 kg * (8.51 m/s)^2

Substituting the values:

Work_nonconservative = [(1/2) * 66.1 kg * (8.51 m/s)^2] - [(1/2) * 66.1 kg * (1.60 m/s)^2 - 66.1 kg * 9.8 m/s^2 * 1.59 m]

Calculating the result:

Work_nonconservative ≈ -42.7 kJ

Therefore, the nonconservative work done on the boy is approximately -42.7 kilojoules. The negative sign indicates that work is done on the boy, meaning that energy is transferred away from the boy during the process.

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The bullet is in the barrel for 4.5610^-5 s, experiences a force of 16713 N, an impulse of 0.763 Ns, and has a momentum of mv = 34.5 kg*m/s as it leaves the barrel.

What are the time, force, impulse, and momentum of a bullet fired from a gun with given mass, barrel length, and muzzle velocity

(a) The time the bullet spends in the barrel can be found using the equation for distance traveled by an object with constant acceleration, where the initial velocity is zero and the distance is the barrel length:

d = 1/2 at^2

Solving for time t, we get:

t = √(2d/a)

where a is the acceleration of the bullet as it is propelled out of the barrel. To find the acceleration, we can use the equation for force:

F = ma

where F is the force exerted on the bullet, m is the mass of the bullet, and a is its acceleration. Since the bullet is in the barrel, the force is the force of the expanding gases from the gunpowder. We can find this force using the ideal gas law:

PV = nRT

where P is the pressure of the gases, V is the volume of the barrel, n is the number of moles of gas produced by the gunpowder, R is the ideal gas constant, and T is the temperature of the gases. Solving for P, we get:

P = nRT/V

We can assume that the volume of the barrel remains constant during the firing of the bullet, so we can write:

P = k nT

where k is a constant that depends on the volume of the barrel and the gas constant R. Since the bullet is fired in a fraction of a second, we can assume that the temperature of the gases remains constant, so we have:

P = k n

The force exerted on the bullet is then:

F = PA = k nA

where A is the cross-sectional area of the barrel.

a = F/m = k nA/m = k P A/m = (kRT/V)(πr^2)/m

where r is the radius of the barrel. we get:

a = (k)(8.31 J/molK)(293 K)/(0.00066 m^3)(π(0.0033 m)^2)/(0.075 kg) = 2.4110^7 m/s^2

Plugging this into the equation for time, we get:

t = √(2d/a) = √(20.66 m/2.4110^7 m/s^2) = 4.56*10^-5 s

So the bullet is in the barrel for 4.56*10^-5 s.

(b) The force on the bullet while it is in the barrel is the force exerted by the expanding gases from the gunpowder, which we found to be:

F = k nA

Substituting in the given values, we get:

F = (k)(8.31 J/mol*K)(293 K)/(0.00066 m^3)(π(0.0033 m)^2) = 16713 N

So the force on the bullet while it is in the barrel is 16713 N.

(c) The impulse exerted on the bullet while it is in the barrel is the product of the force and the time:

J = FΔt = (16713 N)(4.5610^-5 s) = 0.763 Ns

So the impulse exerted on the bullet while it is in the barrel is 0.763 N*s.

(d) The momentum of the bullet as it leaves the barrel can be found using the equation:

p = mv

where p is the momentum, m is the mass of the bullet, and v is its velocity. We can find the velocity using conservation of energy, since the gun and bullet

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An electromagnet consists of a conductive wire wrapped around a magnetic core, creating a magnetic field when an electrical current is passed through it.

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Explaining the concept:

To explain this, we first need to know what density actually is:

Density is the mass of a given substance per unit volume, which means, that is the amount of mass present in unit volume (1 cm³ , 1m³)

Impure water is water, but with impurities added to it

Why is the density of Impure water is greater:

When we compare the mass of equal volumes of Water and Impure water, we will find that the mass of Impure water will be more, that is because the extra mass is of the impurities

Since impure water has impurities uniformly spread in it, we will find more matter in unit volume of Impure water than in Pure water

That is why the density of Impure water is greater

Explanation:

They observe gravitational lensing bending the light of other objects in space.

Answer:

a) ω = 4.49 rad/s

b) v = 2.69 m/s

Explanation:

a) The angular speed can be found by using the rotational rate:

\( \omega = \frac{1 rev}{1.4 s}*\frac{2\pi rad}{1 rev} = 4.49 rad/s \)

Hence, the angular speed is 4.49 rad/s.

b) The tangential speed is given by:

\( v = \omega*r = 4.49 rad/s*0.6 m = 2.69 m/s \)

Therefore, the tangential speed is 2.69 m/s.

I hope it helps you!            

Answer:

0.75 m³/s

Explanation:

Applying,

Q = vA.................... Equation 1

Where Q = flow rate of the water, v = velocity of the water, A = area of the pipe.

From the question,

Given: v = 2.5 m/s, A = 0.3 m²

Substitute these values into equation 1

Q = 2.5(0.3)

Q = 0.75 m³/s

Hence the flow rate of water in the pipe is 0.75 m³/s

The displacement of the object after 1 second, is 3 m.

The displacement of the object after 3 seconds, is 9 m.

The displacement of the object after 5 seconds, is 15 m.

The displacement of an object is the change in the position of the object.

The displacement of the object after 1 second, x = 3 m/s x 1 second = 3 m

The displacement of the object after 1 second, x = 3 m/s x 3 s = 9 m

The displacement of the object after 1 second, x = 3 m/s x 5 s = 15 m

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It is not recommended to fire a gun straight up into the air.

When a bullet is fired into the air, it will eventually come down and can pose a danger to people and property below. The bullet can still be lethal when it reaches the ground, especially if it lands on a hard surface or hits someone directly.

Additionally, firing a gun in a residential area can be illegal and can result in legal consequences. In general, guns should only be fired in designated shooting ranges or in self-defense situations where there is an immediate threat to life. It is important to handle firearms responsibly and follow all safety guidelines to prevent accidents and injuries.

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Answer:

200 meters per second

Explanation:

wave speed = wavelength x frequency

2 meters x 100 hertz = 200 meters per second

Answer: ITS B

Explanation:

BECAUSE I GOT IT WRONG AND IT WAS ACTUALLYT B F IN THE CHAT FOR ME

IF U GIVE ME HEART FOR HELPING THAT WOULD BE GREAT, THANKS......

Answer: The answer is B. Nuclear fission splits an atom apart, while nuclear fusion fuses two atoms together.

Explanation:

Answer: tennis ball

Explanation:

Answer:

a1 =  v1^2 / R

a2 =  v2^2 / R

a2 = (v2 / v1)^2 = (3 / 2)^2 = 9/4

F2 = 9/4 * F1 = 9/4 * 146 = N     329 N      since F = m * a

Answer:

Ants position at point A = -7 cm

Ants position at point B = 38 cm

Ants position at point C = 26 cm

Ants position at point D = -7 cm

Explanation:

In this question, we are dealing with displacement which is the change in position of an object.

Now, we are told the journey began from A to B and then from C to D.

Now, AB = 31 cm and AO = 7cm

But along AB, we have the origin O.

Since A is on the left hand side of the origin, it means it is negative. Thus, position A = -7cm

Then position B = 31 - (-7) = 31 + 7 = 38 cm

Since BC = 12cm, then position C = 38 - 12 = 26 cm

Position D is same as position A = -7cm

To find the magnitude of the velocity, we can use the law of cosines. The law of cosines states that in a triangle, the square of one side (c) is equal to the sum of the squares of the other two sides (a and b), minus twice the product of the two sides (a and b) multiplied by the cosine of the included angle (θ).

In this case, we have a triangle with sides a, b, and c, and angles θ1 and θ2.

Given:

Sum of velocities (c) = 4.5 m/s

Angle θ1 = 32.4°

Angle θ2 = 22.4°

We can write the equation for the law of cosines as follows:

c² = a² + b² - 2ab * cos(θ)

We want to find the magnitude of the velocity, which is equivalent to the length of side c.

Plugging in the given values, we have:

(4.5 m/s)² = a² + b² - 2ab * cos(θ)

Simplifying and rearranging the equation, we have:

a² + b² - 2ab * cos(θ) = 20.25

Now, we need to find the values of a and b. We can use trigonometric relationships to express a and b in terms of the given angles and the magnitude of the velocity.

a = c * cos(θ1)

b = c * cos(θ2)

Plugging in the given values, we have:

a = 4.5 m/s * cos(32.4°)

b = 4.5 m/s * cos(22.4°)

Now, substitute the values of a and b back into the rearranged equation:

(4.5 m/s * cos(32.4°))² + (4.5 m/s * cos(22.4°))² - 2 * (4.5 m/s * cos(32.4°)) * (4.5 m/s * cos(22.4°)) * cos(θ) = 20.25

Simplify the equation and solve for cos(θ):

cos(θ) = (20.25 - (4.5 m/s * cos(32.4°))² - (4.5 m/s * cos(22.4°))²) / (2 * (4.5 m/s * cos(32.4°)) * (4.5 m/s * cos(22.4°)))

Calculate the value of cos(θ) using a calculator.

Finally, calculate the magnitude of the velocity by taking the square root of both sides of the equation:

velocity = √(a² + b² - 2ab * cos(θ))

Evaluate the expression to find the numeric value of the magnitude of velocity in meters per second.

The Specific Heat Capacity of the liquid is 43.846 KJ/(Kg K).

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Explanation:

electricity is the answer I think

Explanation:i think its nuetrons

The resistance indicated by the platinum thermometer at 200°C is 1.648 times the reference resistance Ro at 0°C.

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Answer:

Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position).

Hope this helps! :)

Explanation:

1. The substance equation for water is \(H_2O\).

2. The compound name for table salt is sodium chloride \((NaCl)\).

3. The synthetic equation for carbon dioxide is \(CO_2\).

4. The substance name for baking soft drink is sodium bicarbonate \((NaHCO_3)\).

5.  The substance recipe for methane is \(CH_4\).

6. The compound name for vinegar is acidic corrosive \((CH_3COOH)\)

7.  The compound equation for smelling salts is \(NH_3\).

8. The compound name for blanch is sodium hypochlorite \((NaClO)\).

9.  The synthetic equation for hydrogen peroxide is \(H_2O_2\).

10. The synthetic name for rust is iron oxide \((Fe_2O_3)\).

11.  The compound recipe for sodium hydroxide is \(NaOH\).

12. The substance name for lye is sodium hydroxide \((NaOH)\).

13. The synthetic equation for sulfuric corrosive is \(H_2SO_4\).

14. The compound name for gypsum is calcium sulfate dihydrate \((CaSO_4.2H_2O)\).

15. The substance equation for table sugar is \(C_1_2H_2_2O_1_1\).

16. The synthetic name for anti-inflamatory medicine is acetylsalicylic corrosive \((C_9H_8O_4)\).

17. The compound recipe for calcium carbonate is \(CaCO_3\).

18. The substance name for baking powder is a combination of sodium bicarbonate \((NaHCO_3)\) and an acidic fixing like cream of tartar \((KC_4H_5O_6)\).

19. The synthetic equation for sodium chloride is \(NaCl\).

20. The substance name for hydrogen gas is \(H_2\).

The questions are:

1. What is the synthetic recipe for water?

2. What is the substance name for table salt?

3. What is the compound equation for carbon dioxide?

4. What is the synthetic name for baking pop?

5. What is the substance equation for methane?

6. What is the synthetic name for vinegar?

7. What is the substance equation for alkali?

8. What is the substance name for blanch?

9. What is the compound recipe for hydrogen peroxide?

10. What is the synthetic name for rust?

11. What is the synthetic equation for sodium hydroxide?

12. What is the synthetic name for lye?

13. What is the substance recipe for sulfuric corrosive?

14. What is the substance name for gypsum?

15. What is the synthetic recipe for table sugar?

16. What is the substance name for anti-inflamatory medicine?

17. What is the synthetic equation for calcium carbonate?

18. What is the synthetic name for baking powder?

19. What is the substance equation for sodium chloride?

20. What is the substance name for hydrogen gas?

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Answer:

Explanation:

When a physical change occurs, the mass of the substance is conserved. This means that the total mass of the substance remains the same from beginning to end. The physical properties of the substance, such as size and shape, may change, but the amount of matter in the substance does not change.

( sample response )

Answer:

Its C on edge

Explanation:

trust me

Answer:

Yes.

Explanation:

A negative power would just represent a loss of power. So in your case it lost -1252.16 W

Answer:

the car have travelled 0.31 mile during that​ time

Explanation:

Applying the Equation of motion;

s = 0.5(u+v)t

Where;

s = distance travelled

u = initial speed = 0 mph

v = Final speed = 50 mph

t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour

Substituting the given values into the equation;

s = 0.5(0+50)×(1/80)

s = 0.3125 miles

s ~= 0.31 mile

the car have travelled 0.31 mile during that​ time

An angle '\(\Theta\)' on the circle Corresponds to an angle 2\(\Theta\) on the part.

On the Mohr, the circle has a positive counterclockwise rotation, similar to the physical space convention for shear stress. is the clockwise shear stress and both are plotted. Mohr's circle is a graphical representation of the general state of stress at a point.

This is a graphical method for evaluating principal and maximum shear stresses. Normal and tangential stress in any plane. Soil mechanics use Mohr's circles to visualize the relationship between normal and shear stresses, and to estimate the maximum stress based on three or more soil samples taken from a site.

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The quantity with units of \(kgm^3/s^1\) is mass flow rate or mass flow density.

Mass flow, is a physical quantity that represents the amount of mass flowing through a unit area per unit time. It is the mass flow per unit area and is typically denoted by the symbol ṁ/A. It is a measure of how much mass is flowing per unit area in a given system.

For example, it is often used to describe the rate of mass transfer in a chemical reaction, where a mass flow density is used to measure the amount of reactant or product flowing per unit area per unit time. Mass flow density is also used to describe the rate of mass flow in a fluid through a particular area, such as in a pipe or channel.

The quantity with units of \(kgm^3/s^1\) is mass flow rate or mass flux density.

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