The section of a longitudinal wave in which the particles are densely packed together

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

Anna35 [415]

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.

7 0

2 years ago

I need help by today guys plz help

NISA [10]

radio waves bc they have the longest wave lenthgs in a magnetic spectrum

5 0

2 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

2 years ago

A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0

2 years ago

Read 2 more answers

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

2 years ago