A Physicist is testing a brand new projectile launcher in her lab that is designed to launch human skulls in a vacuum. She can adjust the angle to whatever she wants. She tests it by launching a skull at an angle of 70 degrees above the horizontal at a velocity of 18.0m/s at ground level.

The relative velocity with respect to my friend is 32.7 m/s.

We need to know about relative velocity to solve this problem. The relative velocity depends on the object and observer velocity. It can be written as

v = vo ± v'

where v is relative velocity, vo is object velocity and v' is observer velocity.

From the question above, we know that

vo = 29 m/s

v' = 3.7 m/s

The velocity will be added each other because the direction of the observer is different from the object.

v = vo + v'

v = 29 + 3.7

v = 32.7 m/s

Hence, the relative velocity with respect to my friend is 32.7 m/s.

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Answer:

The statements are not given, so I will answer in a general way.

We know that when one object is moving with a velocity V, and it wants to come to stop, it needs to accelerate in the opposite direction to the initial velocity (Or decelerate).

As larger this acceleration is, the faster the object will come to a full stop.

Now we have a car (with a mass m) and a truck (with a mass M) are moving with a velocity V.

We can assume that the mass of the truck is larger than the mass of the car, then:

M > m.

Now also remember Newton's second law:

F = m*a

Force equals mass times acceleration.

And we know that both vehicles stop with the same stopping force.

Then the acceleration (deceleration actually) that experiences the car is:

a = F/m

While the acceleration that experiences the truck is:

a' = F/M

Because M is larger than m, we will have:

a' < a

Then the deceleration of the truck is smaller than the one of the car, which means that the car will come to a full stop faster than the truck (or the truck needs more time to stop)

Answer:

The stone thrown upward lands farther away from the building.

Explanation:

Given that,

Angle = 30° above the horizontal

Speed = 18.0 m/s

Angle = 30° below the horizontal

Suppose, Which stone lands farther away from the building ?

We need to find which stone lands farther away from the building

Using formula of range

$R=v_{x}t$

The horizontal velocity for the both stones is equal but the time will be different.

The time of stone thrown upward will be greater than that of downward.

Hence, The stone thrown upward lands farther away from the building.

Answer:

(a) 0 (b) $F=1.67\times {10}^{9}N$

Explanation:

Given that,

Mass of a supertanker, $m=1.7\times {10}^{8}kg$

The engine of a generate a forward thrust of, $F=7.4\times {10}^{5}N$

(a) As the supertanker is moving with a constant velocity. We need to find the magnitude of the resistive force exerted on the tanker by the water. It is given by :

F = ma, a is the acceleration

For constant velocity, a = 0

So, F = 0

(b) The magnitude of the upward buoyant force exerted on the tanker by the water is equal to the weight of the ship.

F = mg

$$\begin{gathered} F=1.7\times {10}^8\times 9.8 \\ F=1.67\times {10}^{9}N \end{gathered}$$

Hence, this is the required solution.

Answer:

2450 J

Explanation:

Given that,

A 50Kg girl jumps off a 5-meter-high diving board.

We need to find the kinetic energy of the girl before she hits the water. At this point the kinetic energy becomes equal to the potential energy such that,

$$\begin{gathered} K=mgh \\ K=50\times 9.8\times 5 \\ K=2450J \end{gathered}$$

So, her kinetic energy right before she hits the water is equal to 2450 J.

When the powerlifter deadlifts a  167 kg barbell to a height of 0. 85 m, the work done by the powerlifter is mathematically given as

W=-1057.11J

Question Parameters:

A powerlifter deadlifts a 365 lb (167 kg) barbell to a height of 0. 85 m,

The bar reaches the ground with a speed of 2 m/s

Generally, the equation for the work done  is mathematically given as

W=0.5mvf^2+mghf)-(0.5mv1^2+mghi)

Therefore

W=0.5*16*167*2^2-(167*9.8*0.85)

W=-1057.11J

In conclusion, work done is

W=-1057.11J

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A battery rating given in ampere-hours measures the total amount of charge the battery can deliver over time.

A battery rating given in ampere-hours (Ah) quantifies the total amount of charge a battery can deliver over a specific period of time. It represents the product of the battery's current (in amperes) and the duration of discharge (in hours) until the battery is depleted.

For example, a 5 Ah battery can supply a current of 1 ampere for 5 hours or 0.5 amperes for 10 hours before it is fully discharged. Ampere-hours provide a measure of a battery's capacity or energy storage capability.

This rating is commonly used for rechargeable batteries in various applications, such as automotive batteries, portable electronics, and renewable energy systems, allowing users to estimate the battery's performance and plan their usage accordingly.

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The pendulum bob B of mass M is released from rest when $\theta =0^0$. the tension in the cord at the instant the bob reaches point D, $\theta =\theta 1={45}^0$ is 41.72 N.

The motion of a simple pendulum is idealized. It consists of a point mass attached to an inextensible, massless cable or rod suspended from a pivot point. The pendulum oscillates about the pivot point in a plane that is orthogonal to the rod or cable and has a period that depends solely on the pendulum's length and gravitational acceleration.

As a result, the time period of a pendulum is calculated. The time it takes for a pendulum to swing back and forth is referred to as the time period. It is denoted by T and measured in seconds.

A simple pendulum's time period is calculated using the formula:

$T=2\pi (L/g{)}^{(1/2)}$

where T = time period, L = length of the pendulum, and g = acceleration due to gravity = $9.81m/s^2$.

Tension in the string at the time of release $\theta =0^0$in the pendulum. It implies that the pendulum is vertical. At this point, the tension in the cable is equal to the weight of the bob.

Mg = Tension = 3 x 9.81 = 29.43 N

When the bob reaches point D, $\theta =\theta 1={45}^0$, the tension in the cord is determined. At point D, the velocity of the bob is

$v=L\sqrt{2}gL=\sqrt{(}2\ast 9.81\ast 2)=6.26m/s$

Kinetic Energy of bob KE = $(1/2)m{v}^{2}KE=(1/2)\ast 3\ast (6.26{)}^2=58.46J$

Potential energy of the bob at D = mghU = mghU = 3 × 9.81 × (2 - 2cos45) = 39.22 J

Total mechanical energy at D = KE + U58.46 + 39.22 = 97.68 J

The total mechanical energy at D is equal to the initial mechanical energy since no energy is lost in the absence of any external forces.

WE = KE + UE = PE = mgh = 3 x 9.81 x 2 = 58.86 J

From the formula of total mechanical energy,

WE = Tension x L cosθTension

= (WE / L cosθ)Tension = (58.86 / (2cos45))

= 41.72 N

Therefore, the tension in the cord at the instant the bob reaches point D, $\theta =\theta 1={45}^0$ is 41.72 N.

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Both energy and work do have the same units which is the Joule. The formula is P = E÷ t, where P = power in watts (W),

Work is defined as the transfer of energy that results from a force acting on an object as it moves through a distance. In other words, work is done when a force causes an object to move. The formula for work is W = Fd, where W is work, F is the force applied, and d is the distance over which the force is applied.

Power, on the other hand, is a measure of the rate at which work is done, or the amount of work done per unit time. Power is expressed in units of watts, and the formula for power is P = W/t, where P is power, W is work, and t is time.

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Answer:

Group D

Explanation:

Sound waves through gases travel slowly because of how spread out they are.

Answer:

by paying attention in class, studying and using notes on everything

Explanation:

without specific information about the duration of each repetition and individual factors, we can estimate that approximately 23.3 repetitions or more may be required to burn off 140 calories while lifting a 13 lb barbell through a distance of 1.6 ft.

The energy expenditure of an exercise depends on various factors such as body weight, intensity, and duration of the exercise. Without specific information about these factors, we cannot provide an accurate calculation.

However, we can provide a general estimate based on average values. On average, weightlifting burns about 5-8 calories per minute for a person weighing around 150-180 lbs. Assuming an average energy expenditure of 6 calories per minute, we can calculate the approximate number of minutes required to burn 140 calories:

140 calories / 6 calories per minute ≈ 23.3 minutes

Since we don't have information about the duration of each repetition, we cannot provide an exact number of repetitions required. However, if we assume that each repetition takes approximately 1 minute (including rest periods), the estimated number of repetitions needed to burn off 140 calories would be around 23.3 repetitions. This calculation is a rough estimate and can vary based on individual factors and workout intensity.

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Answer:

Average speed = 541.67 mph

Explanation:

The value of resistor R that gives the circuit in the figure a time constant of 22 μs is 220 Ω.

The circuit that is in the figure is shown below:Given that time constant (RC) = 22 μs. To find the value of resistor R, we need to use the formula for the time constant:

RC = τ, where R is the resistance and C is the capacitance of the circuit.

Rearranging the above formula, we get:R = τ / C

Where τ is the time constant and C is the capacitance of the circuit.

From the figure, the capacitance is given as 0.1 μF

.Substituting the values of τ and C in the above formula, we get:

R = (22 × 10⁻⁶ s) / (0.1 × 10⁻⁶ F)

R = 220 Ω

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The ball that is projected 125 meters straight upward has a total time in the air of: 10.1 s

The formulas for the vertical launch upward and the procedures we will use are:

Where:

Information about the problem:

Applying the maximum height formula and clearing the initial velocity we get:

y max = v₀²/(2*g)

v₀ = √(y max * (2*g))

v₀ = √( 125 m * (2 * 9.8 m/s²))

v₀ = √( 125 m * 19.6 m/s²)

v₀ = √2450 m²/s²

v₀ = 49.497 m/s

Applying the maximum time formula we get:

t max= v₀ / g

t max= 49.497 m/s / 9.8 m/s²

t max = 5.050 s

Applying  the time of flight formula, we get:

t(of) =2 * t max

t(of) =2 * 5.050 s

t(of) = 10.1 s

In physics vertical launch upwards is the motion described by an object that has been launched vertically upwards in which the height and the effect of the earth's gravitational force on the launched object are taken into account.

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The time spent in the air by the ball is 10.01 seconds.

The total time spent in the air by the ball is the sum of the time it spends moving up and the time it spends falling down.

The time spent by the ball moving up - time spent by the ball travelling down.

Time spent by the ball traveling down a distance of 125 meters is calculated using the formula given as follows:

t = √2h/g

Total time spent = 2 * √2h/g

where:

h = 125 m

g = 9.81 m/s²

T = 2 * √(2 * 125/9.81)

T = 10.01 s

Therefore, the time spent in the air by the ball is the sum of the time spent moving up and the time it spends falling down which are both equal.

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The force exerted on the electron has the magnitude of  $8\times {10}^{-14}N$. The force on an electric charge q in an electric field E is given by the formula F = qE.

Here, the charge on the electron is $q=1.60\times {10}^{-19}C$, and the electric field is $E=5\times {10}^{5}N/C$.

the electron's Magnitude of force is given by :$F=(1.60\times {10}^{-19}C)(5\times {10}^{5}N/C)=8\times {10}^{-14}N$

This force is directed opposite to the direction of the electric field because the electron has a negative charge. Since all other forces acting on the electron are negligible in comparison to the electric field force, we can assume that the electron moves with a constant acceleration given by F = ma. The acceleration of the electron can be determined by using the formula a = F/m.

Substituting the values, we get

$a=\frac{8\times {10}^{-14}N)}{(9.11\times {10}^{-31}kg)}=8.78\times {10}^{16}m/s^2.$

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The flow of blood from into the heart to out of the heart is given as Right atrium→tricuspid valve→right ventricle→pulmonary semilunar valve→pulmonary trunk→pulmonary arteries→lung capillaries→pulmonary veins→left atrium→bicuspid valve→left ventricle→aortic semilunar valve→aorta

The right atrium receives deoxygenated blood from the body through two large veins, the posterior (inferior) and anterior (superior) vena cava.

Through the tricuspid valve, blood is transferred from the right atrium to the right ventricle. Tricuspid valve closure occurs when the ventricle is full to stop blood from returning to the atrium.

Through the pulmonary vein, blood that has been oxygenated by the lungs enters the left atrium.

Blood can go from the left atrium into the left ventricle when the mitral valve is open. When the ventricle is full, the mitral valve closes to prevent blood from returning to the atrium.

Blood may move from the heart to the aorta and from there to the rest of the body thanks to the aortic valve.

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Answer:

b 8W

Explanation:

i dont know u find out

Answer:

B. 8 I believe that this is the answer

The tension in the string must be changed to 508 N to raise the wave speed to 180 m/s.

The wave speed on a string is given by the equation:

v = sqrt(T/μ)

where v is the wave speed, T is the tension in the string, and μ is the linear density of the string.

To find the new tension required to achieve a wave speed of 180 m/s, we can rearrange the equation as:

T = μv^2

We can use the given information to find the initial value of μ:

130 m/s = sqrt(116 N / μ)

Solving for μ, we get:

μ = 0.002938 kg/m

Now we can use this value of μ to calculate the new tension required to achieve a wave speed of 180 m/s:

T = (0.002938 kg/m) x (180 m/s)^2

T = 508 N

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The power series representation for the function f(x) = ln(9 - x) is:
ln(9 - x) = -x/9 - (x²)/162 - (x³)/2916 - (x⁴)/52488 - (x⁵)/944784 -
with -9 < x < 9.

To find a power series representation for the function f(x) = ln(9 - x), we can use the Taylor series expansion of the natural logarithm function centered at x = 0. The Taylor series expansion of ln(1 + x) is given by:

ln(1 + x) = x - (x²)/2 + (x³)/3 - (x⁴)/4 + (x⁵)/5 - ...

To apply this formula to our function, we need to make a substitution. Let's substitute x with -x:

ln(9 - x) = -ln(1 - (x/9))

Now, we can substitute (x/9) into the Taylor series expansion of ln(1 + x):

-ln(1 - (x/9)) = -(x/9) - ((x/9)²)/2 - ((x/9)³)/3 - ((x/9)⁴)/4 - ((x/9)⁵)/5 - ...

Simplifying, we have:

ln(9 - x) = -x/9 - (x²)/162 - (x³)/2916 - (x⁴)/52488 - (x⁵)/944784 - ...

This power series representation holds true for values of x within the interval of convergence, which in this case is -9 < x < 9.

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A power series representation for the function f(x) = ln(9 − x) is as ; f(x) = -ln(1 - x/9)f(x) = -[-x/9 + x²/2(9²) - x³/3(9³) + ...]f(x) = x/9 - x²/2(9²) + x³/3(9³) - ..

To find a power series representation for the function f(x) = ln(9 − x), first, we need to recall the formula for the Maclaurin series of ln(1+x).The formula for the Maclaurin series of ln(1+x) is given by;

ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + ... (-1 < x ≤ 1)This is valid for x ∈ (-1, 1].

Hence, we can find a power series representation of f(x) by substituting x/9 in place of x in the formula above.

We have ;f(x) = ln(9-x)f(x) = ln[1 - (-x/9)]f(x) = -ln(1 - x/9).

The expression -x/9 is less than 1, hence we can use the formula above to write ln(1 - x/9) as a power series expansion.

We have; f(x) = -ln(1 - x/9)f(x) = -[-x/9 + x²/2(9²) - x³/3(9³) + ...]f(x) = x/9 - x²/2(9²) + x³/3(9³) - ..

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Answer:

The difference is that water is an incompressible fluid — its density is almost constant as the pressure changes — while air is a compressible fluid — its density changes with pressure. ... Atmospheric pressure is the pressure exerted on a surface by the weight of the atmosphere (a compressible fluid) above it

Explanation:

change some words

Answer:

Wavelength!

Explanation:

At least I think? Or wavelength might be crest to crest! Sorry if I'm incorrect. Let me know how I did!

Answer:

1. the robber carrying the metal safe will hurt more because the metal safe is heavier, so the heavier will be the force on the opposite direction from the one he was running when he hit the wall

2. ???

Explanation:

1. 3rd newton law: "If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction"

2. I don't know

To determine the minimum film thickness for a layer of MgF2 that acts as an infrared filter to prevent 781.1-nm light from entering a lens, we can use the equation for the optical path difference (OPD) for a thin film:

OPD = 2nt

where n is the refractive index of the film, t is the thickness of the film, and the factor of 2 accounts for the two reflections at the air-film and film-air interfaces.

To prevent 781.1-nm light from passing through the filter, the OPD of the film must be equal to an odd multiple of half the wavelength of the light, which is given by:

OPD = (2m + 1) * λ/2

where m is an integer.

Therefore, we can set the OPD equal to the above equation and solve for the minimum film thickness:

2nt = (2m + 1) * λ/2

t = [(2m + 1) * λ/4n]

Substituting the given values of λ = 781.1 nm and n = 1.38 for MgF2, we get:

t = [(2m + 1) * 781.1 nm / (4 * 1.38)]

Simplifying this expression, we get:

t = (m + 0.5) * 283.2 nm

where 283.2 nm is the optical path difference for one half-wavelength of 781.1-nm light in vacuum.

Therefore, the minimum film thickness for a layer of MgF2 that acts as an infrared filter to prevent 781.1-nm light from entering a lens is given by the equation:

t = (m + 0.5) * 283.2 nm

where m is an integer. The value of m depends on the desired level of filtering and can be chosen accordingly. For example, for a first-order filter, m = 0, and the minimum film thickness would be 141.6 nm.

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When a bead is sliding along a wire that is bent into a circular shape and rotating about its vertical axis, it experiences a radial acceleration towards the axis of rotation. At a certain point along the wire, the bead will be in vertical equilibrium, meaning that it experiences no net force in the vertical direction.

This occurs when the gravitational force acting on the bead is balanced by the centrifugal force resulting from the circular motion of the wire. The centrifugal force is directed radially outwards from the axis of rotation and is proportional to the square of the angular velocity of the wire.

At the point of vertical equilibrium, the magnitude of the centrifugal force is equal to the magnitude of the gravitational force acting on the bead. This allows the bead to remain in a stable position on the wire, without moving up or down.

Therefore, the point of vertical equilibrium can be determined by balancing the gravitational force and the centrifugal force acting on the bead, which occurs at a specific distance from the axis of rotation and at a specific angular velocity of the wire.

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Answer:

1. 0 J

2. 7500 J

3. 7500 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) of car = 600 Kg

Initial velocity (v₁) of car = 0 m/s

Final velocity (v₂) of car = 5 m/s

Original kinetic energy (KE₁) =?

Final kinetic energy (KE₂) =?

Work used =?

1. Determination of the original kinetic energy.

Mass (m) of car = 600 Kg

Initial velocity (v₁) of car = 0 m/s

Original kinetic energy (KE₁) =?

KE₁ = ½mv₁²

KE₁ = ½ × 600 × 0²

KE₁ = 0 J

Thus, the original kinetic energy of the car is 0 J.

2. Determination of the final kinetic energy.

Mass (m) of car = 600 Kg

Final velocity (v₂) of car = 5 m/s

Final kinetic energy (KE₂) =?

KE₂ = ½mv₂²

KE₂ = ½ × 600 × 5²

KE₂ = 300 × 25

KE₂ = 7500 J

Thus, the final kinetic energy of the car is 7500 J

3. Determination of the work used.

Original kinetic energy (KE₁) = 0

Final kinetic energy (KE₂) = 7500 J

Work used =?

Work used = KE₂ – KE₁

Work used = 7500 – 0

Work used = 7500 J

Answer:

because we naturally use energy everyday in everyway , energy is also a bill for instance you electric bill used by energy

Explanation:

Answer:

Explanation:

That’s dangerous who would do that.

Me

Explanation:

so I think it B hope that I helped you can good lick