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2 years ago

Assume that Parker Company will receive SF200,000 in 360 days. Assume the following interest rates:

Physics

1 answer:

Anna35 [415]2 years ago

7 0

Answer:

b. $96,914

Explanation:

360-day borrowing rate = 5%

spot rate = 0.48

360-day deposit rate = 6%

Borrow at the rate of 5% to get

SF200,000/1.05 = $190,476.19

Convert at the spot rate of $0.48 to get

190,476.19*0.48 = $91,428.57

Invest at the interest rate of 6% to get

91,428.57/1.06 = 96,914.28

Therefore, Parker Company will receive $96,914 in 360 days.

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How to find the magnitude and direction of a magnetic field? What is the equation?

anzhelika [568]

The magnetic field lines around a long wire which carries an electric current form concentric circles around the wire. The direction of the magnetic field is perpendicular to the wire and is in the direction the fingers of your right hand would curl if you wrapped them around the wire with your thumb in the direction of the current.
hope it helps you

3 0

2 years ago

Read 2 more answers

A force of 6 N will stretch a rubber band 4 cm (0.04 m). Assuming that Hooke's law applies, how far will an 18-N force stre

Bogdan [553]

Explanation:

First, we need to calculate the constant force, that is, the ratio between the applied force and the rubber stretch due to the application of the force:

$k=\frac{F}{d}(1)\\k=\frac{6N}{0.04m}\\k=150\frac{N}{m}$

Now, we can know how far will an 18N force stretch the rubber. From (1):

$d=\frac{F}{k}\\d=\frac{18N}{150\frac{N}{m}}\\d=0.12m=12cm$

The work done by the external force on the rubber is equal to its elastic potential energy:

$W=\frac{kd^2}{2}\\W=\frac{150\frac{N}{m}(0.12m)^2}{2}\\W=1.08J$

3 0

2 years ago

When riding a 10-speed bicycle up a hill, a cyclist shifts the chain to a larger-diameter gear attached to the back wheel. Why i

dimaraw [331]

To solve this problem we will use the concepts of the moment of rotational inertia, angular acceleration and the expression of angular velocity.

The rotational inertia is expressed as follows:

$I = \sum mr^2$

Here,

m = Mass of the object

r = Distance from the rotational axis

The rotational acceleration in terms of translational acceleration is

$\alpha = \frac{a}{R}$

Here,

a = Acceleration

R = Radius of the circular path of the object

The expression for the rotational speed of the object is

$\omega = \frac{\Delta \theta}{\Delta t}$

Here,

$\Delta \theta$ is the angular displacement of the object

The explanation by which when climbing a mountain uphill is changed to a larger pinion, is because it produces a greater torque but it is necessary to make more pedaling to be able to travel the same distance. Basically every turn results in less rotations of the rear wheel. Said energy that was previously used to move the rotation of the wheel is now distributed in more turns of the pedal. Therefore option a and c are correct.

This would indicate that the correct option is D.

8 0

3 years ago

A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

liberstina [14]

Answer:

x = 7.62 m

Explanation:

First we need to calculate the weight of the rocket:

W = mg

we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:

W = 0.5 * 9.8 = 4.9 N

We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:

F - W = m*a

a = F - W/m

Solving for a:

a = (8 - 4.9) / 0.5

a = 6.2 m/s²

As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:

y = vo*t + 1/2 at²

y = 1/2at²

Solving for t:

t = √2y/a

t = √2 * 20 / 6.2

t = √6.45 = 2.54 s

Now that we have the time, we can calculate the horizontal distance:

x = V*t

Solving for x:

x = 3 * 2.54 = 7.62 m

4 0

3 years ago

Is a fault or igneous intrusion that cuts across rock layers younger or older than the rocks it disrupts?

Ksju [112]

It is younger because there had to be rock layers already present for that to be possible.

4 0

2 years ago