A foot is 12 inches and a mile is 5280I ft exactly. A centimeter is exactly 0.01m or mm. Sammy is 5 feet and 5.3tall. what is Sammy's height in inches?

Answers

Answer 1

Answer:

Answer:

65.3 Inches tall

Explanation:

If Sammy is 5 feet and 5.3 inches tall, we simply need to convert the feet to inches, and sum the remaining inches from his height to determine his overall height in inches.

So, 5 feet = (12 inches/1foot) * (5 feet) = 60 inches

And 60 inches + 5.3 inches = 65.3 inches.

Hence, Sammy is 65.3 inches tall.

Cheers.

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5. The order is to give 600 mg of Ampicillin IM q8h. The directions for dilution on the 2 gm vial reads: Reconstitute with 4.8 mL of sterile water to obtain a concentration of 400 mg per mL. How many mL will you administer per dose?

Answers

The volume of Ampicillin IM q8h to be administered per dose is 1.5 mL when the order is to give 600 mg of it from a concentration of 400 mg per mL prepared by the dilution of 2 g in 4.8 mL of sterile water.

1. The information we know

The order is to give 600 mg of Ampicillin.

From the dilution of 2 g in 4.8 mL was obtained a solution of 400 mg per mL of Ampicillin.

2. We need to find:

The volume of Ampicillin in mL per dose

3. Calculation of the Ampicillin's volume to be administered

We can calculate the volume of Ampicillin as follows:

$V = 600 mg*(1 \:mL)/(400 \:mg) = 1.5 \:mL$

Therefore, we need to administer 1.5 mL of Ampicillin per dose.

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Answer:

1.5 ml

Explanation:

The nurse is to administer 600 mg of Ampicillin IM q8h

the reconstitute yield 400 mg per mL

400 mg is in 1 ml

600 mg will be (600 mg × 1 ml) / 400 mg = 1.5 ml

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Answers

The work ( $W_f$ ) done by the friction force between the ramp and the skateboarder is given by $-\mu_k \cdot m_s \cdot g \cdot h_y$ .

The workdone by the friction force ( $W_f$ ) can be calculated using the formula for work, which is the product of the force applied ( $F_f$ ) and the displacement (d) over which the force is applied:

$W_f = F_f \cdot d$

In this scenario, the frictionforce works against the skateboarder's momentum down the ramp, therefore it does no good.

Given:

Mass of skateboarder ( $m_s$ ) = 54 kg

Height of the ramp ( $h_y$ ) = 3.3 m

Final velocity ( $v_f$ ) = 6.2 m/s

Coefficient of kineticfriction ( $\mu_k$ ) between skateboarder and ramp

Acceleration due to gravity (g) = $9.81 \, \text{m/s}^2$

The normal force ( $F_{\text{normal}}$ ) is equal to the weight of the skateboarder:

$F_{\text{normal}} = m_s \cdot g$

The displacement (d) is the vertical distance ( $h_y$ ) that the skateboarder descends down the ramp.

Now we can write the expression for the work done by the friction force ( $W_f$ ):

$W_f = -\mu_k \cdot F_{\text{normal}} \cdot d$

Substitute the expression for the normal force:

$W_f = -\mu_k \cdot (m_s \cdot g) \cdot h_y$

Thus, this expression represents the work done by the friction force between the ramp and the skateboarder in terms of the given variables.

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Your question seems incomplete, the probable complete question is:

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

Final answer:

The momentum of the box with respect to the floor can be found by multiplying its mass by its velocity. When the box is put down on the frictionless skating surface, its velocity becomes zero and its momentum with respect to the floor is also zero.

Explanation:

To find the momentum of the box, we can use the formula:

Momentum = mass x velocity

a. The momentum of the box with respect to the floor is: 5 kg x 5 m/s = 25 kg·m/s

b. When the box is put down on the frictionless skating surface, its velocity becomes zero. So, the momentum of the box with respect to the floor is also zero.

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How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.00×10−15 m (a typical nuclear distance)?

Answers

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field. The work would it take to push two protons will be 7.7×10⁻¹⁴.

What is electric potential?

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electricfield.

The given data in the problem is;

q is the charge= 1.6 ×10⁻¹⁹ C

V is the electric potential

r₁ is the first separation distance= 2.00×10−10 m

r₂ is the second separation distance= 3.00×10−15 m

The electric potential generated by the proton at rest at the two points, using the formula:

Firstly the electric potential at loction 1

$\rm V=(Kq)/(r) \n\n v_i= 9* 10^9 * (1.6*10^(-19))/(2.0*10^(-10))$

The electric potential at loction 2

$V_f = 9 * 10^9 (1.6 * 10^(-19))/(3.0*10^(-15)) \n\n \rm v_f= 4.8 *10^5 \ V$

The product of difference of electric potential and charge is defined as the workdone.

$\rm W= q \triangle V \n\n \rm W= 1.6 * 10^-19 *( 4.8*10^5 -7.2) \n\n \rm W= 7.7 * 10^(-14)$

Hence the work would it take to push two protons will be 7.7×10⁻¹⁴.

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We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

$W=q \Delta V$

where $q=1.6 \cdot 10^(-19)C$ is the charge of the proton and $\Delta V$ is the potential difference between the final position and the initial position of the proton. To calculate this $\Delta V$ , we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

$V=k(Q)/(r)$

where $k=9.0 \cdot 10^9 N m^2 C^(-2)$ is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

$V_i = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(2.0 \cdot 10^(-10))=7.2 V$

$V_f = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(3.0 \cdot 10^(-15))=4.8 \cdot 10^5 V$

Therefore, the work done is

$W=q \Delta V=(1.6 \cdot 10^(-19)C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^(-14) J$

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. Part A How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Answers

Final answer:

The total work done in moving an alpha particle from the center to the side of a square with electrons at its corners involves finding the potential energy change, which can be calculated using the charges, distances, and Coulomb's constant.

Explanation:

The question deals with the fundamental concepts of electrostatics and the energy associated with moving charges in an electric field. Given the aforementioned question, we are required to find the work done moving an alpha particle (a helium nucleus, having a charge of +2e) from the center of a square to one of its sides, with electrons (each having a charge of -e) being situated at its corners.

To determine the work done, we must consider the potential energy changes resulting from moving the alpha particle. The potential energy associated with two point charges is given by the formula: U = k*q1*q2/r, where q1 and q2 are charges, r is the distance between them, and k is Coulomb's constant.

First, we calculate the potential energy at the center due to all four electrons then find the potential energy at the midpoint of the side. The work done is the difference between these two potential energies. As the electrons are all at an equal distance from the alpha particle (in the center and on the side), the calculations would involve plugging in the values for the charge of an electron, the charge of an alpha particle, the given distance values, and Coulomb's constant into the aforementioned formula.

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Final answer:

The work required to move the alpha particle from the midpoint to the midpoint of one of the side of the square with four electrons at its corners would be zero as the net electric field at the midpoint due to the electrons is zero.

Explanation:

The subject of this question pertains to the concept of electrostatics and potential energy in physics. In this scenario, the alpha particle is initially at the midpoint of a square with four electrons at its corners. As per Coulomb's Law, the electrostatic force between two charges is inversely proportional to the square of the distance between them.

Since the alpha particle placed in the center of the square and four electrons at the corners form a symmetrical system, the net force and hence the net electric field at the midpoint due to the electrons is zero. Thus, no work would be required to move the alpha particle to the midpoint of one of the sides of the square as work done is calculated by the formula W = F x d x cos(θ), where F is force, d is the displacement, and θ is the angle between the force and displacement. Since F is equal to zero, the work done will also be zero.

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Two long ideal solenoids (with radii 20 mm and 30 mm, respectively) have the same number of turns of wire per unit length. The solenoid is mounted inside the larger, along a common axis. The magnetic field with in the inner solenoid is zero. The current in the inner solenoid must be: a. two-thirds the current in the outer solenoid
b. one-third the current in the outer solenoid
c. twice the current in the outer solenoid
d. half of the current in the outer solenoid
e. the same as the current in the outer solenoid

Answers

Answer: The current in the inner solenoid is the same as the current in the outer solenoid.

The correct option is e

Explanation: Please see the attachment below

If gear a rotates with a constant angular acceleration of aa = 90 rad>s2, starting from rest, determine the time required for gear d to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear d to attain this angular velocity. Gears a, b, c, and d have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

Answers

solution:

$Given: (let a = alpha)aA = 90 rad/s^2(wd) = 600 rpm = 62.831 rad/s(w0) = 0 rpmra = .015 mrb = .05 mrc = .025 mrd = .075aB = aA*(ra/rb) = 90 * (.015/.05)aB = 27 rad/s^2aC = aB, Therefore aC = 27 rad/s^2aD = aC*(rc/rb) = 27 * (.025/.075)aD = 9 rad/s^2$

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