A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:

Answers

Answer 1

Explanation:

Charge on proton, q₁ = e

Charge on alpha particles, q₂ = 2e

The magnetic force is given by :

[tex]F=qvB\ sin\theta[/tex]

Here, [tex]\theta=90=sin(90) = 1[/tex]

For proton, [tex]F_p=ev_pB[/tex]..........(1)

For alpha particle, [tex]F_a=2ev_aB[/tex]..........(2)

Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,

[tex]ev_pB=2ev_aB[/tex]

[tex]\dfrac{v_p}{v_a}=\dfrac{2}{1}[/tex]

So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.

Answer 2
Final answer:

If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.

Explanation:

The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.

The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.

This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.

Therefore, the proton is moving twice as fast as the alpha particle.

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Related Questions

A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to determine that the Slinky has an inductance 130 µH, when it has been stretched to a length of 3 m. The permeability of free space is 4π × 10−7 N/A 2 . If a slinky has a radius of 4 cm, what is the total number of turns in the Slinky?

Answers

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

[tex]l[/tex] = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

[tex]L = \frac{\mu _{o}N^{2}A}{l}[/tex]

[tex]130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}[/tex]

N = 248

What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.214855 meters D. 0.001386 meter

Answers

Answer:

Option B is the correct answer.

Explanation:

Thermal expansion

            [tex]\Delta L=L\alpha \Delta T[/tex]

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

          [tex]\Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m[/tex]

New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m

Option B is the correct answer.

Answer:

B. 1.201386 meters

Explanation:

The length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters is 1.201386 meters.

On an aircraft carrier, a jet can be catapulted from 0 to 140 mi/h in 1.95 s . If the average force exerted by the catapult is 9.15Ã10^5 N , what is the mass of the jet?

Answers

Answer:

Mass of the jet, m = 28511.5 kg

Explanation:

It is given that,

Initial velocity if the jet, u = 0

Final velocity of the jet, v = 140 mi/h = 62.58 m/s

Time, t = 1.95 s

Average force, [tex]F=9.15\times 10^5\ N[/tex]

We need to find the mass of the jet. According to second law of motion as :

F = m × a

[tex]F=m\times \dfrac{v-u}{t}[/tex]

[tex]m=\dfrac{F.t}{v-u}[/tex]

[tex]m=\dfrac{9.15\times 10^5\ N\times 1.95\ s}{62.58\ m/s-0}[/tex]

[tex]m=28511.5\ kg[/tex]

So, the mass of the jet is 28511.5 kg. Hence, this is the required solution.

he density of copper is 8.96g/cm^3 and the density of water is 1 g/cm^3. When a copper is submerged in a cylindrical beaker whose bottom has surface area 10 cm^2 the water level rises by 2 cm. The volume of the cylinder is the area of its base times its height. a) What is the specific gravity of copper?
b) What is the buoyant force on the copper object?
c) What is the buoyant force on the copper object?
d) What is the mass of the copper object?

Answers

Answer:

(a) 8.96

(b) 19600 dyne

(c) 19600 dyne

(d) 20 g

Explanation:

dcu = 8.96 g/cm^3, dw = 1 g/cm^3, A = 10 cm^2

Water level rises by 2 cm.

(a) The specific gravity of copper = density of copper / density of water

                                                       8.96 / 1 = 8.96

(b) According to the Archimedes's principle, the buoyant force acting on the body is equal to the weight of liquid displaced by the body.

Weight of water displaced by the copper = Area of beaker x rise in water level                          

                                                                      x density of water x gravity

                          = 10 x 2 x 1 x 980 = 19600 dyne

(c) Same as part b

(d) Let mass of copper be m.

For the equilibrium condition,

the true weight of copper = Buoyant force acting on copper

m x g =  19600

m = 19600 / 980 = 20 g

The rate of change of the moment of angular momentum of the fluid flowing through the produced impeller of a turbine is equal to the a. Work b. Power c. Torque d. Thrust

Answers

Answer:

Option (c)

Explanation:

The rate of change of linear momentum is called force.

As the linear motion terms are analogous to the terms in rotational motion.

So, the rate of change of angular momentum is called torque.

When a tennis ball is thrown against a wall it appears to bounce back with exactly the same speed as it struck the wall. Is momentum conserved for this collision? Explain!

Answers

Explanation:

When a tennis ball is thrown against a wall it appears to bounce back with exactly the same speed as it struck the wall. The momentum will remain conserved in this case. The law of conservation of momentum states that when no external force is acting on a system, the initial momentum is equal to the final momentum.

Here, this is a case of inelastic collision. The kinetic energy is not conserved in this case. Some of the energy is lost in the form of heat, sound etc.

Momentum conservation in a tennis ball-wall collision implies that the slight loss of the ball's speed is balanced by the wall slightly flexing or moving, ensuring total system momentum is conserved. The ball's kinetic energy, however, is partly lost in inelastic forms such as heat, sound, or deformation.

When a tennis ball is thrown against a wall and bounces back with a slight reduction in speed, from 15 m/s to 14 m/s, the momentum is not completely conserved in terms of the ball alone because some of it is transferred to the wall. Despite the wall being much more massive and thus not visibly moving, the wall does exert a force on the ball and so it must move very slightly or flex in response, showing that the momentum of the entire system, including the wall, is conserved.

(a) Even though the tennis ball changes its momentum upon collision with the wall, the conservation of momentum principle tells us that the rest of the system must account for this change. Since the ball loses a bit of speed, resulting in a decrease in momentum, the wall gains this tiny amount of momentum. Because the wall has a vast mass compared to the tennis ball, its movement is negligible and often imperceptible. (b) The slight decrease in the rebound speed indicates that not all of the ball's kinetic energy is recovered post-collision. Kinetic energy is dissipated due to factors such as sound generation, deformation of the ball, or heat production, leading to an inelastic collision.

What is the change in electric potential energy in moving an electron from a location 3 × 10-10 m from a proton to a location 7 × 10-10 m from the proton? The result should be in joules, with an appropriate sign.

Answers

Answer:

4.39 x 10^-19 J

Explanation:

q1 = 1.6 x 10^-19 C

q2 = - 1.6 x 10^-19 C

r1 = 3 x 10^-10 m

r2 = 7 x 10^-10 m

The formula for the potential energy is given by

U1 = k q1 q2 / r1 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (3 x 10^-10)

U1 = - 7.68 x 10^-19 J

U2 = k q1 q2 / r2 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (7 x 10^-10)

U2 = - 3.29 x 10^-19 J

Change in potential energy is

U2 - U1 = - 3.29 x 10^-19 + 7.68 x 10^-19 = 4.39 x 10^-19 J

Final answer:

The change in electric potential energy is calculated using the formula ΔU = kq1q2/r, with the distances and charges plugged into the formula, resulting in a negative value, indicating a decrease in potential energy.

Explanation:

The change in electric potential energy when moving an electron from 3 × 10-10 m to 7 × 10-10 m from a proton can be calculated using the formula for the potential energy between two charges: ΔU = kq1q2/r. Plugging in the constants and distances, we have ΔU = k(-e)(+e)(1/7 × 10-10 - 1/3 × 10-10) where k is the Coulomb's constant (8.987 × 109 Nm2/C2) and e is the elementary charge (1.602 × 10-19 C).

The result, calculated in joules, will be negative, signifying that the electron is moving to a region of lower potential energy as it moves away from the proton.

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz (M = 106 ). Calculate the energy for a photon having this frequency. For credit show all your work including correct units.

Answers

Answer:

Energy, [tex]E=2.65\times 10^{-25}\ J[/tex]

Explanation:

It is given that,

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz,

[tex]\nu=400\ MHz=400\times 10^6\ Hz=4\times 10^8\ Hz[/tex]

We need to find the energy for a photon having this frequency. The energy of a photon is given by :

[tex]E=h\nu[/tex]

[tex]E=6.63\times 10^{-34}\ J-s\times 4\times 10^8\ Hz[/tex]

[tex]E=2.65\times 10^{-25}\ J[/tex]

So, the energy of the photon is [tex]2.65\times 10^{-25}\ J[/tex]. Hence, this is the required solution.

A 40 g block of ice is cooled to -69°C and is then added to 590 g of water in an 80 g copper calorimeter at a temperature of 22°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g·°C = 2090 J/kg°C.

Answers

The final temperature of the system consisting of the ice, water, and calorimeter is approximately 6.8°C.

Here's how you can calculate it step by step:

1. Calculate the heat absorbed by the ice to reach 0°C:

[tex]\[ Q_1 = mc\Delta T \]\[ Q_1 = 40 g \times 2090 J/kg^\circ C \times (0^\circ C - (-69^\circ C)) \]\[ Q_1 = 40 g \times 2090 J/kg^\circ C \times 69^\circC \]\[ Q_1 = 571560 J \][/tex]

2. Calculate the heat absorbed by the ice to melt completely:

[tex]\[ Q_2 = mL_f \]\[ Q_2 = 40 g \times 334 J/g \]\[ Q_2 = 13360 J \][/tex]

3. Calculate the heat absorbed by the melted ice to reach the final temperature (assuming the final temperature is T):

[tex]\[ Q_3 = mc\Delta T \]\[ Q_3 = 40 g \times 4186 J/kg°C \times (T - 0°C) \]\[ Q_3 = 167440T J \][/tex]

4. Calculate the heat absorbed by the water and calorimeter to reach the final temperature (T):

[tex]\[ Q_4 = mc\Delta T \]\[ Q_4 = (590 g + 80 g) \times 4186 J/kg°C \times (T - 22°C) \]\[ Q_4 = 270104T - 114208 \][/tex]

Now, since heat gained = heat lost (assuming no heat loss to the surroundings):

[tex]\[ Q_1 + Q_2 + Q_3 = Q_4 \]\[ 571560 + 13360 + 167440T = 270104T - 114208 \]\[ 571560 + 13360 + 114208 = 270104T - 167440T \]\[ 699128 = 102664T \]\[ T \aprrox 6.8°C \][/tex]

The final temperature of the system is approximately 6.8°C.

Complete Question:
A 40 g block of ice is cooled to -69°C and is then added to 590 g of water in an 80 g copper calorimeter at a temperature of 22°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g·°C = 2090 J/kg°C.

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The current is: A) 2.0 A B) 0.63 A C) 5.9A D) 300 A E) 26000 A

Answers

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current density[tex]J=2.8\times10^{7}\ A/m^2[/tex]

We need to calculate the current

Using formula of current density

[tex]J = \dfrac{I}{A}[/tex]

[tex]I=J\timesA[/tex]

Where, J = current density

A = area

I = current

Put the value into the formula

[tex]I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2[/tex]

[tex]I=1.97=2.0\ A[/tex]

Hence, The current is 2.0 A.

The correct answer is A) 2.0 A.

To find the current carried by the wire, we need to calculate the cross-sectional area of the wire and then multiply it by the current density.

First, let's calculate the cross-sectional area (A) of the wire. The wire has a radius (r) of 0.15 mm, which we need to convert to meters to be consistent with the SI units used in the current density (J). Since 1 mm = 10^-3 m, we have:

[tex]\[ r = 0.15 \text{ mm} = 0.15 \times 10^{-3} \text{ m} \][/tex]

The cross-sectional area (A) of a circle is given by the formula:

[tex]\[ A = \pi r^2 \][/tex]

Substituting the value of r in meters, we get:

[tex]\[ A = \pi (0.15 \times 10^{-3} \text{ m})^2 \][/tex]

[tex]\[ A = \pi (0.15^2 \times 10^{-6} \text{ m}^2) \][/tex]

[tex]\[ A = \pi (0.0225 \times 10^{-6} \text{ m}^2) \][/tex]

[tex]\[ A = \pi (2.25 \times 10^{-8} \text{ m}^2) \][/tex]

[tex]\[ A \approx 3.1416 \times 2.25 \times 10^{-8} \text{ m}^2 \][/tex]

[tex]\[ A \approx 7.0686 \times 10^{-8} \text{ m}^2 \][/tex]

Now, we have the current density (J) given as:

[tex]\[ J = 2.8 \times 10^7 \text{ A/m}^2 \][/tex]

The current (I) is the product of the current density (J) and the cross-sectional area (A):

[tex]\[ I = J \times A \][/tex]

Substituting the values we have:

[tex]\[ I = (2.8 \times 10^7 \text{ A/m}^2) \times (7.0686 \times 10^{-8} \text{ m}^2) \][/tex]

[tex]\[ I = 1.98 \times 10^1 \text{ A} \][/tex]

[tex]\[ I \approx 2.0 \text{ A} \][/tex]

Therefore, the current carried by the wire is approximately 2.0 A, which corresponds to option A.

A ball is released at a velocity of 400 m/s at an angle of 85 degrees from the horizontal. How far does it travel horizontally before the ball lands on the ground? (use a-10m/s2 if needed, round answer to a whole number-no decimals). Do not put the units in the answer. QUESTION 5 angle of 85 degrees from the horizontal. How long will it take for the ball to get if needed, round answer to a whole number-no decimals). Do not put the units in the answer. A ball is released at a velocity of 400 m/s at an to its apex? (use a-10m/s QUESTION 6 A ball is released at a velocity of 400 m/s at an angle of 45 degrees from the horizontal. What is the vertical component of the velocity? (use a--10m/s if needed, round answer to a whole number-no decimals), Do not put the units in the answer.

Answers

Answer:

8000

Explanation:

V²=U²+2as

V=400m/s

U=0

a=10

s=?

(400)²=(0)²+2*10*a

160,000=20s

s=160,000/20

s=8000

A proton moves at 4.50 × 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 × 103 N/C. Ignoring any gravitational effects, find

(a) the time interval required for the proton to travel 5.00 cm horizontally,

(b) its vertical displacement during the time interval in which it travels 5.00 cm horizontally,

(c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.

Answers

(a) The time interval required for the proton to travel 5.00 cm horizontally is: 1.11 x 10⁻⁷ s.(b) Its vertical displacement during the time interval in which it travels 5.00 cm horizontally is: 5.66 mm.(c) The horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally is: 1.021 x 10⁵ m /s.

Here, given that,

Horizontal velocity V_x = 4.5 x 10⁵ m /s

vertical electric field E_y = 9.6 x 10³ N/C

so, we get,

acceleration in vertical direction a_y = force on proton / mass

= 9.6 x 10³ x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷

= 9.2 x 10¹¹ m /s²

a ) In horizontal direction it will move with uniform velocity

time required = distance / velocity

= 5 x 10⁻² / 4.5 x 10⁵

= 1.11 x 10⁻⁷ s

b ) vertical displacement in time 1.11 x 10⁻⁷ s

h = 1/2 x at² , initial vertical velocity is zero.

= .5 x 9.2 x 10¹¹ x ( 1.11 x 10⁻⁷ )²

= 5.66 x 10⁻³ m

=5.66 mm

c )Horizontal velocity will be unchanged ie 4.5 x 10⁵ m /s

vertical velocity will change due to acceleration

= u + at

0 + 9.2 x 10¹¹ x1.11 x 10⁻⁷

= 10.21 x 10⁴ m / s

= 1.021 x 10⁵ m /s

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Final answer:

The time interval required for the proton to travel 5.00 cm horizontally is 1.11 × 10^-7 s. The vertical displacement of the proton during this time interval can be calculated using the formula displacement = 0.5 * (qE/m) * (time^2). The horizontal component of the proton's velocity remains unchanged, while the vertical component changes due to the electric field force.

Explanation:

(a) To find the time interval required for the proton to travel 5.00 cm horizontally, we need to use the formula: time = distance/speed. In this case, the distance is 5.00 cm, which is 0.05 m, and the speed is 4.50 × 10^5 m/s. So, the time interval is 0.05 m / (4.50 × 10^5 m/s) = 1.11 × 10^-7 s.

(b) The vertical displacement of the proton during the time interval can be found using the formula: displacement = 0.5 * acceleration * time^2. Since the proton is not affected by gravity, the only force acting on it is the electric field force, which is given by: F = qE, where q is the charge of the proton and E is the electric field magnitude. The acceleration is equal to F/m, where m is the mass of the proton. The vertical displacement is then given by: displacement = 0.5 * (qE/m) * (time^2). Substituting q = +e (the charge of a proton) and m = mass of a proton, we can calculate the vertical displacement.
(c) After traveling 5.00 cm horizontally, the horizontal component of the proton's velocity remains unchanged since there are no forces acting on it in the horizontal direction. The vertical component of the velocity, on the other hand, changes due to the electric field force. To find the new vertical velocity, we can use the formula: final velocity = initial velocity + acceleration * time, where the acceleration is given by F/m and the time is the time interval we calculated in part (a).

Heated lithium atoms emit photons of light with an energy of 2.961 × 10−19 J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?

Answers

Answer:

4.5 x 10¹⁴ Hz

666.7 nm

1.8 x 10⁵ J

The color of the emitted light is red

Explanation:

E = energy of photons of light = 2.961 x 10⁻¹⁹ J

f = frequency of the photon

Energy of photons is given as

E = h f

2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f

f = 4.5 x 10¹⁴ Hz

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of photon

Using the equation

c = f λ

3 x 10⁸ = (4.5 x 10¹⁴) λ

λ = 0.6667 x 10⁻⁶ m

λ = 666.7 x 10⁻⁹ m

λ = 666.7 nm

n = number of photons in 1 mole = 6.023 x 10²³

U = energy of 1 mole of photons

Energy of 1 mole of photons is given as

U = n E

U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)

U = 1.8 x 10⁵ J

The color of the emitted light is red

Final answer:

The frequency and wavelength of a photon with energy of 2.961 × 10-19 J are approximately 4.468 x 10¹4 s¯¹ and 656.3 nm, respectively. The energy of one mole of these photons is approximately 1.823 x 10⁵ J mol-¹. The color of the light emitted by these photons is red.

Explanation:

The energy of a photon of light can be calculated into both a frequency and wavelength using the Planck-Einstein relation E = hf, and the speed of light relation c = λf, where h is Planck's constant, c is the speed of light, and λ is wavelength.

Let's first find the frequency. Given that the energy E is 2.961 × 10-19 J and the value of Planck's constant h is 6.63 × 10-³4 J·s (approximately), you can solve for the frequency f = E/h, which results in approximately 4.468 × 10¹4 s¯¹.

To find the wavelength, you can use the light speed equation c = fλ. Given the speed of light c is approximately 3.00 × 10⁸ m/s and using the frequency calculated above, solve for λ = c/f, which results in approximately 656.3 nm.

 

To find the energy of one mole of these photons, use Avogadro's number (6.022 x 10²³ photons/mole) and the provided energy of a single photon. The total energy is then calculated as E = (2.961 × 10-19 J/photon) x (6.022 x 10²³ /mole) = 1.823 × 10⁵ J mol-¹.

Lastly, the color of the photon can be inferred from its wavelength. As the wavelength is 656.3 nm, this falls within the range of the visible light spectrum for the color red.

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What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Answers

Answer:

The velocity at discharge is 100.46 ft/s

Explanation:

Given that,

Pressure = 68 psi

We need to calculate the pressure in pascal

[tex]P=68\times6894.74\ Pa[/tex]

[tex]P=468842.32\ Pa[/tex]

We need to calculate the velocity

Let the velocity is v.

Using Bernoulli equation

[tex]P=\dfrac{1}{2}\rho v^2[/tex]

[tex]468842.32=0.5\times1000\times v^2[/tex]

[tex]v=\sqrt{\dfrac{468842.32}{0.5\times1000}}[/tex]

[tex]v=30.62\ m/s[/tex]

Now, We will convert m/s to ft/s

[tex]v =30.62\times3.281[/tex]

[tex]v=100.46\ ft/s[/tex]

Hence, The velocity at discharge is 100.46 ft/s

Final answer:

The speed of water discharged from a hose depends on the nozzle pressure and the constriction of the flow, but the specific speed cannot be determined from pressure alone without additional parameters.

Explanation:

The question is asking about the velocity or speed achieved by water when it is forced out of a hose with a nozzle pressure of 68 psi. To understand this, we need to know that the pressure within the hose is directly correlated with the speed of the water's exit. This is due to the constriction of the water flow by the nozzle, causing speed to increase.

However, the specific velocity at discharge can't be straightforwardly calculated from pressure alone without knowing more details, such as the dimensions of the hose and nozzle, and the properties of the fluid. Therefore, based on the provided information, a specific answer in ft/sec can't be given.

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The mass of a proton at rest is 1.67 10 2kg . Calculate the value of e/m for a proton.

Answers

Answer:

[tex]\frac{e}{m} = 9.58 \times 10^7 [/tex]

Explanation:

mass of the proton is given is

[tex]m_P = 1.67 \times 10^{-27} kg[/tex]

also we know that charge of proton is same as charge of electron

[tex]e = 1.6 \times 10^{-19} C[/tex]

now we need to find the charge mass ration of proton

so here we have

[tex]\frac{q}{m} = \frac{1.6 \times 10^{-19}}{1.67 \times 10^{-27}}[/tex]

[tex]\frac{e}{m} = 9.58 \times 10^7 [/tex]

So above is the charge mass ratio of proton

A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \mu_kμ k ​  = 0.30. Use the work–energy theorem to find how far the text-book moves from its initial position before coming to rest.

Answers

Answer:

L = 1.06 m

Explanation:

As per work energy theorem we know that work done by all forces must be equal to change in kinetic energy

so here we will have

[tex]W_{spring} + W_{friction} = KE_f - KE_i[/tex]

now we know that

[tex]W_{spring} = \frac{1}{2}kx^2[/tex]

[tex]W_{friction} = -\mu mg L[/tex]

initial and final speed of the book is zero so initial and final kinetic energy will be zero

[tex]\frac{1}{2}kx^2 - \mu mg L= 0 - 0[/tex]

here we know that

k = 250 N/m

x = 0.250 m

m = 2.50 kg

now plug in all data in it

[tex]\frac{1}{2}(250)(0.250)^2 = 0.30(2.50)(9.81)L[/tex]

now we have

[tex]7.8125 = 7.3575L[/tex]

[tex]L = 1.06 m[/tex]

The distance that the textbook moves from its initial position before coming to rest is 1.0629m

According to the  work-energy theorem, the work done on the spring is equal to the work done by friction

Work done by the spring = [tex]\frac{1}{2}kx^2[/tex]

Work done by friction = [tex]\mu mgL[/tex]

k is the spring constant = 250 N/m

x is the distance moved by the spring = 0.250m

[tex]\mu[/tex] is the coefficient of friction = 0.30

m is the mass of the textbook = 2.50kg

g is the acceleration due to gravity = 9.8m/s²

L is the distance from its initial position before coming to rest

Using the formula

[tex]\frac{1}{2} kx^2=\mu mgL[/tex]

Substitute the given parameters into the formula as shown;

[tex]\frac{1}{2} (250)(0.25)^2=0.3 (2.5)(9.8)L\\7.8125=7.35L\\L=\frac{7.8125}{7.35}\\L= 1.0629m[/tex]

Hence the distance that the textbook moves from its initial position before coming to rest is 1.0629m

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What is the kinetic energy of a 1450 kg sports car traveling down the road with a speed of 30 m/s?

Answers

[tex]E_{k}[/tex] = 652500J

The easiest way to solve this problem is using the kinetic energy equation:

[tex]E_{k}[/tex] = mv²/2

A 1450kg sports car traveling down the road at a speed of 30m/s is kinectic energy will be:

[tex]E_{k}[/tex] = [(1450kg)(30m/s)²]/2 = 1305000kg-m²/s²/2

[tex]E_{k}[/tex] = 652500J

Answer:

652500 J

Explanation:

A potter's wheel, with rotational inertia 49 kg·m2, is spinning freely at 40 rpm. The potter drops a lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm what is the mass of the clay?

Answers

Answer:

8.5 kg

Explanation:

Angular momentum is conserved.

I₁ ω₁ = I₂ ω₂

I ω₀ = (I + mr²) ω

(49 kg m²) (40 rpm) = (49 kg m² + m (1.2 m)²) (32 rpm)

61.25 kg m² = 49 kg m² + (1.44 m²) m

m = 8.5 kg

The mass of the clay is 8.5 kg.

Which of the following statements are true about Electric Force, Electric Field, Electric Potential and Electric Potential Energy?
(i) Electric Force, Electric Field, and Electric Potential are vectors; Electric Potential Energy is a scalar value.
(ii) The Electric Field reports the Electric Force on a 1C test charge placed at a particular location.
(iii) The Electric Potential reports the Electric Potential energy of a 1C test charge place at a particular location.
(iv) The spatial derivative of the electric potential at a particular location reports the electric field at that location.

Answers

Answer:

(ii) The Electric Field reports the Electric Force on a 1C test charge placed at a particular location.

(iii) The Electric Potential reports the Electric Potential energy of a 1C test charge place at a particular location.

(iv) The spatial derivative of the electric potential at a particular location reports the electric field at that location.

ii, iii and iv are true.

Explanation:

Electric force per unit charge  is the electric field. It is a vector. E = F/q.

Electric potential is the electric potential energy per unit charge and is a scalar quantity, like the electric potential. V = U/q where U is the electric potential energy and q the charge.

Electric field exists due to a difference of potential, between two different points.E =dV/dx .

Option (ii),(iii) & (iv) is correct.

The question posed deals with the concepts of Electric Force, Electric Field, Electric Potential, and Electric Potential Energy in the context of physics. Let's analyze each statement:

Electric Force, Electric Field, and Electric Potential are vectors; Electric Potential Energy is a scalar value. This statement is partially incorrect. Electric Force and Electric Field are indeed vectors because they have both magnitude and direction. However, Electric Potential is a scalar quantity, not a vector, because it only has magnitude without direction. Electric Potential Energy is correctly identified as a scalar value.

The Electric Field reports the Electric Force on a 1C test charge placed at a particular location. This statement is true. The Electric Field is defined as the Electric Force per unit charge, and it indeed represents the force that a 1 coulomb test charge would experience at that point.

The Electric Potential reports the Electric Potential energy of a 1C test charge placed at a particular location. This statement is true. Electric Potential is the amount of Electric Potential Energy per unit charge that a test charge would have in a particular location in the electric field.

The spatial derivative of the electric potential at a particular location reports the electric field at that location. This statement is true. According to electrostatics, the electric field at a point is the negative gradient (spatial derivative) of the electric potential at that point.

It is essential to understand that electric potential is different from an electric field: the former is a scalar that represents potential energy per unit charge, while the latter is a vector representing the force per unit charge. Calculating the scalar electric potential is generally simpler than calculating the vector electric field.

Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00 cm away from the line of charge and released from rest. What kinetic energy in eV will it have when it reaches a position of 5.0 cm away from the initial position? (Note: use the potential difference equation for an infinite line.) a. 350 eV b. 390 ev c. 350k ev d. 6.2 eV e. 170 e

Answers

Answer:

b.  [tex]\Delta KE = 390 eV[/tex]

Explanation:

As we know that the electric field due to infinite line charge is given as

[tex]E =\frac{\lambda}{2\pi \epsilon_0 r}[/tex]

here we can find potential difference between two points using the relation

[tex]\Delta V = \int E.dr[/tex]

now we have

[tex]\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr[/tex]

now we have

[tex]\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})[/tex]

now plug in all values in it

[tex]\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})[/tex]

[tex]\Delta V = 216ln6 = 387 V[/tex]

now we know by energy conservation

[tex]\Delta KE = q\Delta V[/tex]

[tex]\Delta KE = (e)(387V) = 387 eV[/tex]

Upper A 16​-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 ​feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the​ wall?

Answers

Answer:

The ladder is moving at the rate of 0.65 ft/s

Explanation:

A 16​-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 ​feet/second. We need to find the rate at which the top of the ladder moving down when the foot of the ladder is 5 feet from the​ wall.

The attached figure shows whole description such that,

[tex]x^2+y^2=256[/tex].........(1)

[tex]\dfrac{dx}{dt}=2\ ft/s[/tex]

We need to find, [tex]\dfrac{dy}{dt}[/tex] at x = 5 ft

Differentiating equation (1) wrt t as :

[tex]2x.\dfrac{dx}{dt}+2y.\dfrac{dy}{dt}=0[/tex]

[tex]2x+y\dfrac{dy}{dt}=0[/tex]

[tex]\dfrac{dy}{dt}=-\dfrac{2x}{y}[/tex]

Since, [tex]y=\sqrt{256-x^2}[/tex]

[tex]\dfrac{dy}{dt}=-\dfrac{2x}{\sqrt{256-x^2}}[/tex]

At x = 5 ft,

[tex]\dfrac{dy}{dt}=-\dfrac{2\times 5}{\sqrt{256-5^2}}[/tex]

[tex]\dfrac{dy}{dt}=0.65[/tex]

So, the ladder is moving down at the rate of 0.65 ft/s. Hence, this is the required solution.

Final answer:

This is a calculation of a related rates problem in calculus, using the Pythagorean Theorem. The calculation finds the rate of descent (db/dt) for the top of a ladder sliding down a wall. The final answer found is -1.5 feet/second, meaning the top of the ladder is moving downward at a rate of 1.5 feet per second when the ladder is 5 feet away from the building.

Explanation:

The question involves a concept in calculus known as related rates problem. It's a practical extension of the Pythagorean theorem where the sides of the right triangle formed by the ladder, building, and the ground are changing over time. In this case, a 16-feet ladder is sliding down a wall and we are to find the rate at which the top of the ladder is moving down when the foot of the ladder is 5 feet from the wall. First, we label the bottom (horizontal) side of our triangle 'a', the side along the wall (vertical) 'b', and the hypotenuse (ladder) 'c'. The Pythagorean theorem gives us a² + b² = c². Differentiating both sides with respect to time (t) gives us 2a(da/dt) + 2b(db/dt) = 2c(dc/dt). Here, we know that da/dt = 2 feet/sec (rate at which the ladder is moving away from the wall), c = 16 feet (length of the ladder), and a (distance from the wall) = 5 feet. Also, the ladder's length is not changing, so dc/dt = 0. Substituting these values, we can solve for db/dt (rate of descent of the ladder). The resultant rate of descent of the top of the ladder is -1.5 feet/second.

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A model rocket is constructed with a motor that can provide a total impulse of 35.0 N·s. The mass of the rocket is 0.192 kg. What is the speed that this rocket achieves when it is launched from rest? Neglect the effects of gravity and air resistance.

Answers

Answer:

[tex]v_f = 182.3 m/s[/tex]

Explanation:

As we know that total impulse is given as the product of mass and change in velocity

so here we will have

[tex]I = m(\Delta v)[/tex]

here we will have

[tex]I = 35.0 N s[/tex]

also we know that mass of the rocket is

m = 0.192 kg

now we will have

[tex]35 = 0.192(v_f - 0)[/tex]

[tex]v_f = 182.3 m/s[/tex]

A machine part is vibrating along the x-axis in simple harmonic motion with a period of 0.27 s and a range (from the maximum in one direction to the maximum in the other) of 3.0 cm. At time t = 0 it is at its central position and moving in the +x direction. What is its position when t = 55 s?

Answers

Answer:

[tex]x = -1.437 cm[/tex]

Explanation:

The general equation for position of Simple harmonic motion is given as:

[tex]x = A sin(\omega t)[/tex]          ........(1)

where,

x = Position of the wave

A = Amplitude of the wave

ω = Angular velocity

t = time

In this case, the amplitude is just half the range,

thus,

[tex]A =\frac{3cm}{2}=1.5cm[/tex]  (Given range = 3cm)

A = 1.5 cm  

Now, The angular velocity is given as:

[tex]\omega=\frac{2\pi}{T}[/tex]

Where, T = time period of the wave =0.27s (given)

[tex]\omega=\frac{2\pi}{0.27s}[/tex]

or

[tex]\omega=23.27s^{-1}[/tex]

so, at time t = 55 s, the equation (1) becomes as:

[tex]x = 1.5 sin(23.27\times 55)[/tex]

on solving the above equation we get,

[tex]x = -1.437 cm[/tex]

here the negative sign depicts the position in the opposite direction of +x

Final answer:

To determine the position of a machine part in simple harmonic motion after 55 seconds, we utilize the period and amplitude to establish the displacement function. The part completes full periods, and at t = 55 seconds, it returns to the central position. The exact position can be calculated using the cosine function for displacement in SHM.

Explanation:

The machine part is undergoing simple harmonic motion (SHM), which is a type of periodic oscillation. Given the period T is 0.27 seconds and the range of 3.0 cm, which is the total distance from the maximum position in one direction to the maximum in the other, we can determine the amplitude (A) is half of this range, which is 1.5 cm or 0.015 meters.

The displacement in simple harmonic motion as a function of time can be modeled using the cosine function:
x(t) = A cos(2πt/T). The part starts at its central position at t = 0 and moves in the +x direction. To find its position at t = 55 s, we can substitute these values into the displacement formula. Since 55 is not a multiple of the period, we need to account for the number of full periods that have passed. After 55 seconds, which is 55/0.27 or approximately 203.7 periods, the part will have completed 203 full periods and be halfway through the 204th cycle. The next step is to find the remainder of the division to obtain the phase of the cycle at t = 55 s, which is 55 mod 0.27. The cosine function is periodic with a period of 2π, hence we consider the fraction of the period completed to model the motion at that specific time.

The exact calculation for the phase at t = 55 leads us to determine that the machine part's displacement will be back at the central position. Since the cosine function repeats every 2π and the part was initially at the central position moving in the +x direction (which corresponds to cos(0) = 1), after an even number of complete cycles at t = 55 s, the part will again be at the central position (x = 0) and about to start moving in the +x direction.

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A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal ( s = 0.900 J/g·°C), find the final temperature of the block and the water

Answers

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature [tex]T_i[/tex] = 93.2°C

We know

[tex]Q = m c \Delta T[/tex]   ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in [tex]J/g^{\circ}C[/tex]

[tex] \Delta T[/tex] = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = [tex]125\times 0.9\times (93.2-T_f)[/tex]

Heat gained by the water = [tex]100\times 4.184\times(T_f -18.3)[/tex]

thus, we have

[tex]125\times 0.9\times (93.2-T_f)[/tex] = [tex]100\times 4.184\times(T_f -18.3)[/tex]

[tex]10485-112.5T_f = 418.4T_f - 7656.72[/tex]

⇒ [tex]T_f = 34.17^oC[/tex]

Therefore, the final temperature will be = 34.17°C

For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency (1.2×1010 Hz ) and wavelength (0.026 m ). Recall that the Avogadro constant is 6.022×1023 mol−1.

Answers

Answer:

4.79 J

Explanation:

The energy of a single photon is given by

[tex]E=hf[/tex]

where

[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant

f is the frequency of the photon

Here we have

[tex]f=1.2\cdot 10^{10} Hz[/tex]

so the energy of one photon is

[tex]E_1=(6.63\cdot 10^{-34})(1.2\cdot 10^{10})=7.96\cdot 10^{-24} J[/tex]

Here we have 1 mol of photons, which contains

[tex]N=6.022\cdot 10^{23}[/tex] photons (Avogadro number). So, the total energy of this mole of photons is:

[tex]E=NE_1 = (6.022\cdot 10^{23})(7.96\cdot 10^{-24})=4.79 J[/tex]

Find the weight required to stretch a steel road by 2 mm, if the length of the road is 2 m and diameter 4 mm. Y= 200 GPa and g -9.8 ms2? (a) 356.4 kg (b) 295.4 kg (c) 365.4 kg (d) 256.4 kg 3.

Answers

Answer:

Option D is the correct answer.

Explanation:

We equation for elongation

   [tex]\Delta L=\frac{PL}{AE}[/tex]      

Here we need to find weight required,

We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m

[tex]A=\frac{\pi d^2}{4}=\frac{\pi ({4\times 10^{-3}})^2}{4}=1.26\times 10^{-5}m^2[/tex]

E = 200GPa = 2 x 10¹¹ N/m²

L=2m

Substituting

[tex]0.002=\frac{P\times 2}{1.26\times 10^{-5}\times 2\times 10^{11}}\\\\P=2520N=256.4kg[/tex]

Option D is the correct answer.

A molecule of roughly spherical shape has a mass of 6.10 × 10-25 kg and a diameter of 0.70 nm. The uncertainty in the measured position of the molecule is equal to the molecular diameter. What is the minimum uncertainty in the speed of this molecule? (h = 6.626 × 10-34 J · s)

Answers

Final answer:

To calculate the minimum uncertainty in the speed of the molecule, we can use the Heisenberg uncertainty principle. The uncertainty in position is equal to the molecular diameter, and the uncertainty in momentum can be calculated using the uncertainty principle equation. Finally, the uncertainty in velocity can be found by dividing the uncertainty in momentum by the mass of the molecule.

Explanation:

To calculate the minimum uncertainty in the speed of the molecule, we can use the Heisenberg uncertainty principle. The uncertainty in position (Ax) is equal to the molecular diameter, which is given as 0.70 nm. The uncertainty in momentum (Ap) can be calculated using the equation AxAp ≥ h/4. Once the uncertainty in momentum is found, the uncertainty in velocity can be found from Ap = mΔv, where m is the mass of the molecule.

Using the given values, the uncertainty in position (Ax) is 0.70 nm. Plugging this into the uncertainty principle equation and solving for Ap, we get Ap ≥ h/4Ax. Substituting the values for h and Ax, we have Ap ≥ (6.626 × 10-34 J·s)/(4 × 0.70 × 10-9 m).

Next, we can find the uncertainty in velocity (Δv) by using Ap = mΔv. Rearranging the equation, we have Δv = Ap/m. Plugging in the values for Ap (obtained from the previous calculation) and the mass of the molecule, we can calculate the minimum uncertainty in the speed of the molecule.

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At rest, a car’s horn sounds the note A (440 Hz). The horn is sounded while the car is moving down the street. A bicyclist moving in the same direction with one-third the car’s speed hears a frequency of 415 Hz. (a) Is the cyclist ahead of or behind the car? (b) What is the speed of the car?

Answers

Final answer:

The cyclist is behind the car, and the speed of the car is approximately 13.3 m/s.

Explanation:

The question you have asked is related to the Doppler effect. When a car's horn sounds the note A (440 Hz) at rest, the frequency heard by a stationary observer is 440 Hz. However, when the car is moving and approaching a bicyclist, the frequency heard by the bicyclist is lower (415 Hz). From this information, we can determine the answers to the two parts of your question:

(a) The cyclist is behind the car: The fact that the frequency heard by the cyclist is lower than the frequency produced by the car suggests that the cyclist is behind the car and moving in the same direction.(b) The speed of the car is approximately 13.3 m/s: By using the formula for the Doppler effect, we can find the speed of the car. The formula is:

(v - vc)/(vs - vc) = fs/fr

Where v is the speed of sound, vc is the speed of the car, vs is the speed of the cyclist, fs is the frequency heard by the cyclist, and fr is the frequency produced by the car. By substituting the known values (v = 343 m/s, fs = 415 Hz, and fr = 440 Hz) and solving for vc, we can find the speed of the car.

By plugging in the values, we get:
(343 - vc)/(0 - vc) = 415/440

Simplifying the equation further, we get:
vc = 343(415 - 440) / (415)

After substituting the values and solving the equation, we find that:
vc ≈ 13.3 m/s

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Final answer:

The sound heard by the bicyclist is lower than the frequency of the car's horn due to the Doppler effect. The bicyclist is behind the car and the speed of the car can be calculated using the formula for the Doppler effect.

Explanation:

The phenomenon described in the question is an example of the Doppler effect. The Doppler effect is the apparent change in frequency of a sound wave due to the relative motion between the source of the sound and the observer. In this case, the frequency of the horn sounds lower to the bicyclist due to their relative motion with the car.

(a) The frequency heard by the bicyclist is lower than the frequency of the car's horn, indicating that the bicyclist is behind the car.

(b) To find the speed of the car, we can use the formula for the Doppler effect:

fo = fs (V + Vo) / (V - Vs)

Where fo is the observed frequency, fs is the source frequency, V is the speed of sound, Vo is the speed of the observer, and Vs is the speed of the source.

In this case, fo is 415 Hz, fs is 440 Hz, and Vo is one-third the speed of the car. We can rearrange the formula and solve for V (the speed of the car):

V = Vs (fo / fs) - Vo

Plugging in the values, we find that the speed of the car is approximately 40 m/s.

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Which of the following statements is representative of the second law of thermodynamics? Heat represents a form of energy that can be used by most organisms to do work. Conversion of energy from one form to another is always accompanied by some gain of free energy. Cells require a constant input of energy to maintain their high level of organization. Without an input of energy, organisms would tend toward decreasing entropy. Every energy transformation by a cell decreases the entropy of the universe.

Answers

Answer: Cells require a constant input of energy to maintain their high level of organization.

Explanation:

According to the second principle of thermodynamics:

"The amount of entropy in the universe tends to increase over time ".

That is, in any cyclic process, entropy will increase, or remain the same.

So, in this context, entropy is a thermodynamic quantity defined as a criterion to predict the evolution or transformation of thermodynamic systems. In addition, it is used to measure the degree of organization of a system.  

In other words: Entropy is the measure of the disorder of a system and is a function of state.

Now, in the specific case of cells, in order to maintain their high level of organization, which goes against the natural tendency to disorder, a constant input of energy is necessary to maintain that level.

The Second Law of Thermodynamics explains how energy transfers result in some energy being lost in unusable forms, such as heat energy.

The Second Law of Thermodynamics states that in every energy transfer, some amount of energy is lost in a form that is unusable, often in the form of heat energy. This principle explains why energy transfers are never completely efficient, leading to increased disorder or entropy in the universe.

The magnetic force which a particle feels is: A) inversely proportional to the velocity B) inversely proportional to the magnetic field C) directly proportional to the electric field D) inversely proportional to the charge E) directly proportional to the charge

Answers

Answer:

Magnetic force is directly proportional to the charge.

Explanation:

The magnetic force of a charged particle is given by :

[tex]F=qvB[/tex]..... (1)

Where

q = charged particle

v = velocity of particle

B = magnetic field

From equation (1) it is clear that the magnetic force is directly proportional to the charge. Hence, the correct option is (E).

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