A diver goes under water and measures the pressure. At some point his instruments read a pressure of 50,000 Pa. How deep did the diver go in meters? The density of water is 1000 kg/m^3. (Write the number only with 1 significant figure)

Answer: [tex]v \approx 18.37 \frac{m}{s}[/tex]

Explanation:

Let assume that system is conservative. From application of the Principle of Energy Conservation, it is noticed that initial linear kinetic energy must be equal to the gravitational energy at the top of the circle. That is to say:

[tex]K_{1} = U_{2}\\[/tex]

[tex]\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (2\cdot R) \\v = \sqrt{4 \cdot g \cdot R}[/tex]

Where [tex]g = 9.81 \frac{m}{s^{2}}[/tex].

[tex]v = \sqrt{4 \cdot (9.81 \frac{m}{s^2})\cdot(8.6 m)} \\v \approx 18.37 \frac{m}{s}[/tex]

Answer:

P = 128.6 T / V

Explanation:

The ideal gas equation is

         P V = n R T

Where the pressure is

P = 36 pounds / in²

V = 2500 in³

T = 700 K

    PV = k T

    k = PV / T

    k = 36 2500/700

    k = 128.6

    P = 128.6 T / V

Answer:

a) The statement is true, b)  [tex]K_{tr.atom} = 6.073 \times 10^{-21} \frac{J}{atom}[/tex],[tex]v_{rms} \approx 477.904 \frac{m}{s}[/tex]

Explanation:

a) The standard conditions are 273.15 K and 1 atm. Let consider that gas behaves ideally, whose equation of state is:

[tex]P \cdot V = n \cdot R_{u} \cdot T[/tex]

The volume is cleared out:

[tex]V = \frac{n \cdot R_{u} \cdot T}{P}[/tex]

By replacing terms:

[tex]V = \frac{(1 mole)\cdot (0.082 \frac{L \cdot atm}{mole \cdot K} )\cdot (273.15 K)}{1 atm}\\V = 22.399 L[/tex]

Therefore, the statement is true.

b) The average kinetic energy per atom is:

[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{R_{u}\cdot T}{N_{A}}[/tex]

Where [tex]N_{A}[/tex] is the Avogadro constante and is equal to [tex]6.022 \times 10^{23} \frac{atoms}{mole}[/tex].

[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{(8.317 \frac{J}{mole\cdot K} )\cdot (293.15 K)}{6.022 \times 10^{23} \frac{atoms}{mole} } \\K_{tr,atom} = 6.073\times 10^{-21} \frac{J}{atom}[/tex]

The rms speed is determined by the following formula:

[tex]v_{rms} = \sqrt{\frac{3 \cdot k \cdot T}{n \cdot M_{O} \cdot \frac{1}{N_{A}} } }\\v_{rms} = \sqrt{\frac{3\cdot(1.38 \times 10^{-23} \frac{J}{K} )\cdot(293.15 K)}{(2 moles) \cdot (16 \frac{g}{mole} )(\frac{1 kg}{1000 g} ) \cdot (\frac{1 mole}{6.022 \times 10^{23} atoms} )}} \\v_{rms} \approx 477.904 \frac{m}{s}[/tex]

Answer:

the magnitude and direction of the magnetic field in the region through which the current passes is 0.008 T and +z direction.

Explanation:

given information:

current, I = 20 A

magnetic force per unit length, F/L = 0.16 N/m

the conductor in the negative y-direction

θ = 90° (perpendicular)

as we know the formula to calculate magnetic force is

F = B I L sin θ

B = F/(I L sin θ)

   = (F/L) (1/I sin θ)

   = 0.16 (1/15 sin 90)

   = 0.008 T

since F is in the negative y direction, based of the right hand rule the magnetic field is in positive z direction

Answer:

Explanation:

Given:

current, I = 20 A

Magnetic force per unit length, F/L

= 0.16 N/m

Conductor in the negative y-direction, therefore θ = 90° (perpendicular)

For a magnetic field,

F = B I L sin θ

B = F/(I L sin θ)

= 0.16 × (1/15 sin 90)

= 0.008 T

The field is in the +ve z - direction.

Answer:

Check attachment for complete question

Question

Find a unit vector in the direction in which

f increases most rapidly at P and give the rate of change of f

in that direction; Find a unit vector in the direction in which f

decreases most rapidly at P and give the rate of change of f in

that direction.

f (x, y, z) = x²z e^y + xz²; P(1, ln 2, 2).

Explanation:

The function, z = f(x, y,z), increases most rapidly at (a, b,c) in the

direction of the gradient and decreases

most rapidly in the opposite direction

Given that

F=x²ze^y+xz² at P(1, In2, 2)

1. F increases most rapidly in the positive direction of ∇f

∇f= df/dx i + df/dy j +df/dz k

∇f=(2xze^y+z²)i + (x²ze^y) j + (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= (2×1×2×e^In2+2²)i +(1²×2×e^In2)j +(1²e^In2+2×1×2)

∇f=12i + 4j + 6k

Then, unit vector

V= ∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=(12i+4j+6k)/14

V=6/7 i + 2/7 j + 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

2. F increases most rapidly in the positive direction of -∇f

∇f=- (df/dx i + df/dy j +df/dz k)

∇f=-(2xze^y+z²)i - (x²ze^y) j - (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= -(2×1×2×e^In2+2²)i -(1²×2×e^In2)j -(1²e^In2+2×1×2)

∇f=-12i -4j - 6k

Then, unit vector

V= -∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=-(12i+4j+6k)/14

V= - 6/7 i - 2/7 j - 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

There's a part of the question missing and it is:

f(x, y) = 4{x(^3)}{y^(2)} ; P(-1,1)

Answer:

A) Unit vector = 4(3i - 2j)/ (√13)

B) The rate of change;

|Δf(1, - 1)|= 4/(√13)

Explanation:

First of all, f increases rapidly in the positive direction of Δf(x, y)

Now;

[differentiation of the x item alone] to get;

fx(x, y) = 12{x(^2)}{y^(2)}

So at (1,-1), fx(x, y) = 12

Similarly, [differentiation of the y item alone] to get; fy(x, y) =

8{x(^3)}{y}

At (1,-1), fy(x, y) = - 8

Therefore, Δf(1, - 1) = 12i - 8j

Simplifying this, vector along gradient = 4(3i - 2j)

Unit vector = 4(3i - 2j)/ (√(3^2) + (-2^2) = 4(3i - 2j)/ (√13)

Therefore, the rate of change;

|Δf(1, - 1)|= 4/(√13)

The wavelength of green light in meters using exponential notation is 5.5 × 10-7 m.

The green light in the visible portion of the electromagnetic radiation spectrum has a wavelength of around 550 nm (nanometers).

To express this wavelength in meters using exponential notation, we can convert nanometers to meters by dividing by 109. So, the wavelength of green light is 5.5 × 10-7 m (meters).

Answer:

0.080625 m/s

Explanation:

Horizontally, the total momentum must be conserved, meaning the momentum before and after the throw must be the same.

The total momentum before the throw is 0, and so is after the throw:

mv + MV = 0

where m, M are the masses of the ball and the quarterback, respectively. v and V are the velocities of the ball and the quarterback, respectively.

0.43*15 + 80V = 0

80V = -6.45

V = -6.45 / 80 = -0.080625 m/s

So he will be moving in the opposite direction with the ball at the rate of 0.080625 m/s

Final answer:

To determine the quarterback's movement after throwing the football, one would need to use conservation of momentum principles. Initial velocity, time in the air, and the maximum height of the football once thrown can be calculated using the equations of projectile motion.

Explanation:

The scenario presented in the question can be analyzed using the principles of conservation of momentum. Since the situation described involves horizontal motion and the exchange of momentum between the quarterback and the football, we will apply the concepts of momentum conservation and the horizontal aspect of projectile motion.

Conservation of Momentum

For part of the scenario where we determine the quarterback's movement after throwing the ball, we use the conservation of momentum which states that the total momentum of a system is conserved if there are no external forces acting on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity. Just before the quarterback throws the ball, both he and the ball are part of the same system, and their combined momentum is zero if he jumps straight up in the air. After throwing the ball, the quarterback will have a backward velocity to conserve the total momentum of the system; we calculate this by equating the momentum of the thrown ball and the momentum of the moving quarterback.

Projectile Motion

In a situation where the football is thrown in a certain manner, the motion of the football can be separated into horizontal and vertical components. The initial velocity of the football can be calculated based on the given range, height, and angle of throw using kinematic equations. The time it takes for the ball to reach the receiver and its maximum height can also be determined through these equations.

Final answer:

The maximum height a wrench can be thrown on the Moon with the same starting speed as on Earth is 6 times higher.

Explanation:

To determine how many times higher an astronaut could jump on the Moon compared to Earth, we need to compare the gravitational accelerations of both locations. The gravitational acceleration on the Moon is about 1/6 of the acceleration on Earth, denoted as g.

When an object is thrown vertically upward, its maximum height is determined by the initial velocity and the gravitational acceleration. Since the astronaut gives the wrench the same starting speed in both places, the maximum height it can reach on the Moon will be 6 times higher than on Earth. This is because the weaker gravitational acceleration on the Moon allows the object to reach a greater height before being pulled back down.

Therefore, if the astronaut can throw the wrench 15.0 m vertically upward on Earth, they could throw it about 90.0 m (15.0 m * 6) on the Moon with the same starting speed.

Answer:

1.7 m

Explanation:

Draw a free body diagram of the ladder.  There are 5 forces:

Normal force N pushing up at the base of the ladder.

Friction force Nμ pushing right at the base of the ladder.

Weight force mg pushing down a distance d up the ladder.

Weight force Mg pushing down a distance L/2 up the ladder.

Reaction force R pushing left at the top of the ladder.

Sum of forces in the x direction:

∑F = ma

Nμ − R = 0

Sum of forces in the y direction:

∑F = ma

N − mg − Mg = 0

Sum of moments about the base of the ladder:

∑τ = Iα

mg (d cos θ) + Mg (L/2 cos θ) − R (L sin θ) = 0

Use the first equation to substitute for R:

mg (d cos θ) + Mg (L/2 cos θ) − Nμ (L sin θ) = 0

Use the second equation to substitute for N:

mg (d cos θ) + Mg (L/2 cos θ) − (mg + Mg) μ (L sin θ) = 0

Simplify and solve for d:

m (d cos θ) + M (L/2 cos θ) − (m + M) μ (L sin θ) = 0

m (d cos θ) = (m + M) μ (L sin θ) − M (L/2 cos θ)

d = [ (m + M) μ (L sin θ) − M (L/2 cos θ) ] / (m cos θ)

Plug in values and solve:

d = [ (67 kg + 12 kg) (0.39) (5 m sin 43°) − (12 kg) (2.5 m cos 43°) ] / (67 kg cos 43°)

d = 1.70 m

Rounded to two significant figures, the maximum distance is 1.7 m.

The maximal distance,  [tex]d_{max[/tex], the person will reach before the ladder slips will be 3.41 m.

Here, we need to consider the forces acting and torque equilibrium. Let's start by noting down the given data:

The forces acting on the ladder are:

Using Newton’s second law for horizontal and vertical equilibrium and setting torques about the base of the ladder:

1. Horizontal forces:

2. Vertical forces:

For torque equilibrium about the base:

Substitute: [tex]\[N = (m_l + m_p)g\][/tex]

And frictional force: f = μN

We solve for d:

Simplify:

Substituting values:

Answer:

Option D is correct.

Explanation:

Bmax = Emax / c

The general form for electromagnetic wave equation is

E = jEmax ×cos(kx-wt)

We were given

(360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ].

So from the equation above

Emax = 360V/m

Bmax = 360/(3×10⁸) = 1.2 ×10‐⁶ T.

Answer

Option D

Amplitude of Magnetic field = B = 1.2×10⁻⁶ T

Explanation:

The relationship between electric field and magnetic field of an electromagnetic wave is given by

B = E/c

Where B is the amplitude of magnetic field and E is the amplitude of electric field and c is the speed of light

The amplitude of electric field is given as 360 V/m

B = (360 V/m)/(3×10⁸ m/s)

B = 1.2×10⁻⁶ V.s/m²

Since 1 Tesla is equal to 1 V.s/m²

B = 1.2×10⁻⁶ T

Therefore, option D is correct

Answer:

The balloon falls to the ground before the professor gets there. The student is DEFINITELY in for some TROUBLE!

Explanation:

The balloon picks up speed due to gravity and we can calculate the time taken for it to fall to the ground as follows:

Gravity (g) = 9.81 m/s^2

Height or distance (s) = 11 meters

Initial Speed (u) = 0 m/s

[tex]s = u*t + 0.5 * (a*t^2)[/tex]

[tex]11 = 0*t + 0.5 (9.81*t^2)[/tex]

[tex]t= 1.4975 s[/tex]

So we can see that the balloon takes 1.4975 seconds to fall to the ground, and since the professor takes 2 seconds to get to that place, the balloon hits the ground right before the professor gets there.

Answer:

77.88 lbm/ft³

Explanation:

Given,

Specific gravity, SG = 1.25

Density of water, ρ = 62.30 lbm/ft³

density of the fluid =

   = S.G x ρ_{water}

   = 62.30 x 1.25

   = 77.88 lbm/ft³

Density of the fluid is equal to 77.88 lbm/ft³

D. Torque

Explanation:

Torque is a pattern of the force that can make a victim to revolve on an axis. The torque's direction vector based on the force's direction on the axis. The SI unit for torque is the Newton-meter. Torque can be both static or dynamic.

A static is one that does not generate an angular acceleration.The  Torque's magnitude  vector ζ for a torque generated by a given force F is

ζ = F .r sin(θ)

where,

r is the width of the moment arm

θ is the angle within the force vector and the moment arm.

In rotational kinematics, torque is measured as

ζ = Iα

Where,

α is the angular acceleration

I is the rotational inertia

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