a rifle is fired horizontally and travels 400.0 m. the rifle barrel is 1.90 m from the ground. what speed must the bullet have been traveling at? ignore friction.

The mass flow rate of water is 0.08757 kg/s

mCp (T2 - T1) = Q =w

m x 1.015 (360-315) = 4000

m = 4000/ 1.015 x 10^3 x (360-315)

m = 0.08757 kg/s

The density of water at 68°F / 20°C has a flow rate in our example of about 998 kg/m3. However, as the density of water varies with temperature, salinity, and pressure, use our water density calculator if you want to be extremely accurate. 30.58 lbs/s of mass flow were shown by the gadget.

By dividing the volume flow rate by the fluid’s mass density, one can determine the mass flow rate. The cross-sectional vector area, A, is multiplied by the flow velocity of the mass elements, v, to determine the volume flow rate.

The mass of a substance that moves per unit of time is known as the mass flow rate in physics and engineering.

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Answer:

Potential energy is the energy stored within an object, due to the object's position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy.

Explanation:

Answer:

The answer for Work done is 60J or 60Nm

Explanation:

Work done=Force×distance

W=15×4

W=60J or 60Nm

Neglecting air resistance the direction of acceleration of a rock thrown straight upward is upward.

The acceleration of the rock as it is rising is upward. This is due to the force of gravity acting on the rock, which is constantly pulling it down toward the ground. As the rock moves upward, the force of gravity is still pulling it down, so the acceleration of the rock is also in the downward direction. However, since the rock is moving upwards, the acceleration is in the opposite direction of the force of gravity, so the acceleration of the rock is actually upward.

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Conventional method of heat transfer.​

How it works?

For example:

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The additional charge that flows to the positive plate from the 8 Volts battery is  12 x 10⁻³ Coulomb's.

We have an uncharged capacitor which is charged in two different ways using battery of two different voltages. Firstly, it is connected to the terminals of a 4.0 V battery, and 12 μc of charge flows to the positive plate. Then, the 4.0 V battery is disconnected and replaced with a 8.0 V battery

We have to determine the Additional charge that flows to the Positive plate from this 8 V battery.

Capacitance is the ability of an object to collect and store energy in the form of an electrical charge. Mathematically -

C = Q/V

According to the question, we have -

Electric potential of battery 1 = V[1] = 4 V

Charge flow from battery 1 to the positive plate = Q[1] = 12 x 10⁻³ C

Electric potential of battery 2 = V[2] = 8 V

Now, using the following relation, we will find Q[2] {Charge flow from battery 2 to the positive plate} →

V[1] / V[2] = Q[1] / Q[2]

Substituting the values -

Q[2] = (12 x 10⁻³ x 8)/4 = 24 x 10⁻³ C

Additional Charge that flowed to the positive plate →

Q[2] - Q[1] = 12 x 10⁻³ Coulomb's

Hence, the additional charge that flows to the positive plate is  \(12\times 10^{-3}\) Coulomb's.

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[ The question given is incomplete, the complete question is -

"An uncharged capacitor is connected to the terminals of a 4.0 v battery,  and 12 μc flows to the positive plate. the 4.0 v battery is then disconnected and replaced with a 8.0 v battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate" ]

Answer:

See below ↓↓↓

Explanation:

1. C. Gas

2. B. Speed

3. A. Pascal

Work is equal to energy. Since your braking distance is 10.8 m. you will not stop before you hit the oil spill

Energy is the ability to do work. The work done on an object is tantamount to energy used on the object.

If you are first place in Mario Kart at the moment. Your 5.0 kg Mario Kart is moving at 5.5 m/s when you see an oil spill 10 m ahead. If your maximum braking force is 7.0 N, will you stop before you hit the oil spill? What is your braking distance?

By using the concepts of work and energy, that is,

1/2mv² = F × s

Where

Substitute all the parameters into the formula

1/2 × 5 × 5.5² = 7 × s

75.625 = 7s

s = 75.625/7

s = 10.8 m

Since the calculated distance is greater than the given distance (10 m), you will not stop before you hit the oil spill. And your braking distance will be 10.8 m

Therefore,  you will not stop before you hit the oil spill.

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option b.  The starts visible from that location would be very different from the stars visible from Earth at that time of year. it's important to understand that the Moon is tidally locked to the Earth.

which means that it rotates on its axis at the same rate that it orbits the Earth. This results in the same side of the Moon always facing the Earth. The side facing away from the Earth is known as the far side of the Moon. Because the far side of the Moon always faces away from the Earth, you would not be able to see Earth from that location. The Sun would also be invisible as it would be hidden by the Moon itself. However, because the Moon rotates on its axis, over the course of a lunar day (which lasts about 29.5 Earth days), you would see different stars rise and set. So the stars visible from the far side of the Moon would indeed be very different from the stars visible from Earth at any given time of year.

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(last choice)

The speed increases from X to W because a component of the gravitational force acts in the direction of the velocity.

D)The speed increases from X to W because a component of the gravitational force acts in the direction of the velocity.

The gravitational force is inversely proportional to the square of the separation distance between the 2 interacting gadgets, extra separation distance will result in weaker gravitational forces. so as items are separated from every other, the pressure of gravitational attraction between them also decreases.

The force of gravity relies immediately upon the masses of the 2 gadgets, and inversely on the square of the gap between them. this means that the pressure of gravity will increase with mass, but decreases with growing distance among items.

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The work done by the gas during the isothermal expansion is approximately -4125.40 J.  The calculation involves considering the gas constant, temperature, and initial and final volumes.

To calculate the work done by the gas during an isothermal expansion, we can use the formula:

W = -nRT ln(Vf/Vi)

Where:

W is the work done

n is the number of moles of the gas

R is the gas constant (8.314 J/(mol K))

T is the temperature in Kelvin

Vf is the final volume

Vi is the initial volume

Given:

n = 1 mole

R = 8.314 J/(mol K)

T = 20°C

= 293.15 K

Vi = 0.8 m³

Vf = 2.1 m³

Substituting the values into the formula:

W = -1 * 8.314 J/(mol K) * 293.15 K * ln(2.1 m³ / 0.8 m³)

≈ -4125.40 J

Therefore, the work done by the gas during the isothermal expansion is approximately -4125.40 J. The negative sign indicates work done on the gas.

By using the formula for work done during an isothermal expansion and substituting the given values, we calculated that the work done by the gas is approximately -4125.40 J. The negative sign indicates that work is done on the gas during the expansion. The calculation involves considering the gas constant, temperature, and initial and final volumes.

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\(\\ \bull\tt\longmapsto Volume=LBH\)

\(\\ \bull\tt\longmapsto Volume=8(4)(3)\)

\(\\ \bull\tt\longmapsto Volume=96m^3\)

Answer:

Formular:

\({ \boxed{ \bf{v = length \times width \times height}}}\)

» length, l = 8m

» width, w = 4m

» height, h = 300cm = 3 m

\({ \tt{• Therefore :→}} \: { \sf{ volume = (8 \times 4 \times 3)}} \\ → { \sf{ \underline{ \: \: volume = 96 \: {m}^{3} \: \: }}}\)

The following statements have italicized word, and has 2 correct  statement and 1 incorrect statement.

a. Taylor's weight is 65 kilograms. : Its correct.

b. I went on a diet and my mass decreased. : Its correct.

c. I went to the Moon and I lost mass, even though I didn't lose any of my matter. : Its incorrect.

The weight is a term shows the mass of the object and the gravity working on it. Example is that a person who feel heavy in earth that same person feel very light on moon cause the gravity will work on that object verry small in comparison to earth.

a. Taylor's weight is 65 kilograms. It is a complete sentence. So its a correct statement.

b. I went on a diet and my mass decreased. : Its correct. Its also a complete sentence. So its a correct statement.

But, c. I went to the Moon and I lost mass, even though I didn't lose any of my matter. This sentence is incorrect because the mass of the person remains same in the moon but the gravity become lesser in the moon so he felt lighter.

So, according to the theory the statement is incorrect.

The correct statement would be:

c. I went to the Moon and I lost weight, even though I didn't lose any of my matter.

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Answer:

Musicians use frequency is pitch.

The shorter the wavelength, the higher the frequency, and the higher the pitch, of the sound. The shorter the wavelength, the higher the frequency, and the higher the pitch, of the sound. Hope it helps!

You can change the volume, pressure, or temperature of a gas.

Change the volume of a gas to raise or lower the temperature. As the temperature increases, the gas expands and increases in volume. Conversely, as the temperature decreases, the volume of the gas decreases. As the temperature increases, the volume of the gas increases. Conversely increasing the pressure will decrease the volume of the gas.

Reducing the gas volume increases the gas pressure. An example of this is when gas is trapped inside a cylinder by a piston. When the piston is pushed in, the volume occupied by the gas decreases, so there is less room for gas particles to move. The relationship between pressure and volume Boyle's Law. As the pressure of a gas increases the gas particles are pushed together thus decreasing the volume of the gas.

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Answer:

the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.

Explanation:

Given;

length of the wire, = 34.9 m

length of solenoid, L = 0.24 m

radius of the solenoid, r = 0.051 m

current in the solenoid, I = 11.0 A

The number of turns of the wire is calculated as follow;

\(N = \frac{34.9}{2\pi \times 0.051} = 109 \ turns\)

The strength of the magnetic field inside the solenoid is calculated as follows;

\(B = \mu_0 (\frac{N}{L} )I\\\\B = 4\pi \times 10^{-7} \times (\frac{109}{0.24} )\times 11.0 \\\\B = 6.278 \times 10^{-3} \ T\)

Therefore, the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.

Mast cell stabilizer is part of a treatment program for laboratory animal allergies.

What are animal allergies? Animal allergies occur when a person's immune system overreacts to an allergen in an animal's body. Pet allergies are more common, but anyone who works with or around laboratory animals may develop allergies to them. When a person is exposed to animal allergens, their immune system produces a response that causes allergic symptoms in the body.

The following are some of the treatment options for animal allergies:

Antihistamines: Antihistamines are used to alleviate allergic symptoms such as itching, sneezing, and runny nose. Antihistamines work by blocking histamine, a chemical produced by the body in response to an allergen.

Mast cell stabilizers: Mast cell stabilizers prevent the release of histamine and other chemicals that cause allergy symptoms. These drugs can take a few weeks to become fully effective.

Nasal corticosteroids: These nasal sprays relieve inflammation in the nasal passages, which can reduce the severity of allergy symptoms.

Allergy shots: Allergy shots, or immunotherapy, can be used to treat severe animal allergies. Immunotherapy works by exposing the patient to small amounts of the allergen over time, which can reduce the body's immune response to the allergen. Dilantin is an anticonvulsant medication that is used to treat seizures. It is not used in the treatment of animal allergies.

Therefore, Mast cell stabilizer is part of a treatment program for laboratory animal allergies.

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Given the initial distance between the two vehicles as 3200 ft and their speeds as v_A = 65 mi/h and v_B = 40 mi/h, we can use the formula for distance and time to calculate their accelerations.

Let's call the acceleration of vehicle A as a_A and that of vehicle B as a_B.

Using the first equation of motion:

v_A = v_0 + a_A * t

v_B = v_0 + a_B * t

Where v_0 is the initial velocity and t is the time taken to reach point P and Q.

t = 40 s for A and t = 42 s for B.

We can substitute the values and solve for a_A and a_B.

When the two vehicles pass each other, their relative velocity will be zero.

Using the formula for relative velocity:

v_rel = v_A - v_B

0 = v_A - v_B

v_A = v_B

At this point, we can use either of the two vehicles' equation of motion to find the time when they meet.

Using the equation of motion for vehicle A:

x = x_0 + v_0 * t + 0.5 * a_A * t^2

Substituting the values, we can solve for t to find the time when the vehicles meet.

The speed of the vehicles at that time will be equal to v_A or v_B, whichever we have used in our calculations.

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Answer:

team A - team B = 450 N - 415 N

                                               = 35 N

Explanation:

net force on the rope is 35 N and the direction is towards A.since the two teams are pulling away from each other with the given forces, the net force can be calculated by subtracting the forces.

The net force on the rope is 35N in right direction

As We know that ,

\(F_{net} = F_1+F_2\)

Now as The direction of the Team A is right so the Force applied will be positive, that is 450N

Simultaneously , The direction of Team B is left so the Force applied will be negative , that is -415N

Here,

F1 = 450N

F2 = -415N

\(F_{net} = F_1+F_2\)

= 450+ (-415N)

= 450-415

=35N

Hence the net force applied is 35N and as the net force is positive the direction will be right

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Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, remains invariant under Galilean transformations. This principle implies that the form of the equation remains the same regardless of the observer's reference frame.

The Galilean transformations describe the relationship between different reference frames in classical mechanics. When an observer changes from one inertial reference frame to another, these transformations ensure that the laws of physics remain consistent.

To demonstrate the invariance of Newton's second law under Galilean transformations, we consider two inertial reference frames: one in which the equation is initially stated (the "primed" frame) and another that is moving at a constant velocity relative to the first frame (the "unprimed" frame).

In the unprimed frame, the position, velocity, and acceleration of an object may be different than in the primed frame due to the relative motion between the two frames. However, upon applying the Galilean transformations, it can be shown that the form of Newton's second law remains unchanged. The mass of the object, the acceleration it experiences, and the force acting on it all transform in a consistent manner, such that the equation retains its original structure.

Therefore, the invariance of Newton's second law under Galilean transformations ensures that the fundamental relationship between force, mass, and acceleration remains consistent across different inertial reference frames.

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Answer:

The mass will stay the same throughout time

The total distance traveled in one full cycle of simple harmonic motion is twice the amplitude (A) of the oscillation.

Hence, the correct option is C.

The total distance traveled by the block in one full cycle of its motion can be determined by analyzing the nature of simple harmonic motion.

In simple harmonic motion, the block oscillates back and forth around its equilibrium position. During one complete cycle, it moves from its initial position to one extreme position, then returns to the equilibrium position, and finally moves to the other extreme position before returning back to the equilibrium position again.

The total distance traveled in one full cycle of simple harmonic motion is twice the amplitude (A) of the oscillation. This is because the block travels a distance of A in one direction to reach the first extreme position, and then it travels the same distance of A in the opposite direction to reach the other extreme position.

Therefore, the correct option is (c) 2A.

Hence, the correct option is C.

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The gauge pressure at the second point is 3.6x10^4 Pa.

The gauge pressure in a horizontal pipeline can be determined using Bernoulli's equation, which relates the speed of the fluid to the pressure it exerts. At the first point in the pipeline, the water's speed is given as 2.50 m/s, and the gauge pressure is 1.8x10^4 Pa.

To find the gauge pressure at the second point, where the cross-sectional area is twice that at the first point, we can apply the principle of continuity, which states that the product of the fluid's speed and the cross-sectional area is constant along the pipeline.

Let's denote the cross-sectional area at the first point as A1 and the speed as v1, and similarly, the cross-sectional area at the second point as A2 and the speed as v2. Since A2 is twice A1, we can write A2 = 2A1.

According to the principle of continuity, we have A1v1 = A2v2. Since A2 = 2A1, we can substitute this into the equation to get A1v1 = 2A1v2.

By canceling out A1 from both sides of the equation, we have v1 = 2v2. This means that the speed at the second point is half the speed at the first point.

Now, let's apply Bernoulli's equation to find the gauge pressure at the second point. Bernoulli's equation states that the sum of the pressure, the kinetic energy per unit volume, and the potential energy per unit volume is constant along a streamline.

At both points in the pipeline, the potential energy per unit volume is the same since the pipeline is horizontal. Therefore, we only need to consider the gauge pressure and the kinetic energy per unit volume.

At the first point, we have P1 + 1/2ρv1^2 = 1.8x10^4 Pa + 1/2ρ(2.50 m/s)^2, where ρ is the density of water.

At the second point, since the speed is half of that at the first point, we have P2 + 1/2ρ(v2^2) = P2 + 1/2ρ((1/2)(2.50 m/s))^2.

Comparing the two equations, we can see that P2 = P1 + 1/2ρ(v1^2 - (1/2v1)^2).

Substituting the known values, we have P2 = 1.8x10^4 Pa + 1/2ρ(2.50 m/s)^2 - 1/2ρ((1/2)(2.50 m/s))^2.

By simplifying the equation and substituting the density of water, we find that P2 is approximately 3.6x10^4 Pa.

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Answer:

The gravitational pull at a distance of 19320 km above the surface of the Earth would be much weaker than it is at the surface. The strength of the gravitational field decreases with distance from the center of the Earth, so at a distance of 19320 km above the surface, the gravitational pull would be much weaker than it is at the surface.

To calculate the gravitational pull at a distance of 19320 km above the surface of the Earth, you can use the following formula:

g = G * M / r^2

where g is the gravitational acceleration, G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth.

Using this formula, we can calculate that the gravitational acceleration at a distance of 19320 km above the surface of the Earth would be approximately 0.0012 m/s^2. This is much weaker than the gravitational acceleration at the surface of the Earth, which is approximately 9.81 m/s^2.

It's important to note that this calculation assumes that the Earth is a perfect sphere with a uniform mass distribution. In reality, the Earth is not a perfect sphere and its mass is not distributed uniformly, so the gravitational acceleration at a given distance from the surface may be slightly different than what is calculated using this formula.

Explanation:

Answer:

Acceleration, \(a=4\ m/s^2\)

Explanation:

Given that,

Initial velocity, u = 20 m/s

Final velocity, v = 60 m/s

Time, t = 10 s

We need to find the acceleration of the car. It is equal to the change in velocity divided by time.

\(a=\dfrac{v-u}{t}\\\\a=\dfrac{60-20}{10}\\\\a=\dfrac{40}{10}\\\\a=4\ m/s^2\)

So, the acceleration of the car is \(4\ m/s^2\).

The car's acceleration while assuming no resistance. 4000 N of net force are required.

Speed and direction changes in velocity with time. When a point or object moves faster or slower along a straight line, it is said to be accelerated. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.

The pace at which displacement changes is known as velocity. The rate at which velocity changes is known as acceleration. Because it includes both magnitude and direction, velocity is a vector quantity. As acceleration is merely the rate at which velocity changes, acceleration is likewise a vector quantity.

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