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Simora [160]
3 years ago
14

In September 2016 you purchased a 125 shares of Grainger stock for $224.84. It is now September 2019 and Grainger stock is curre

ntly priced at $295.56. If you were to sell today at this price, what would be your capital gain (loss) on the sale, ignoring commissions
Physics
1 answer:
Temka [501]3 years ago
6 0

Answer:

Gain in capital = $ 70.72

Explanation:

Given:

- The price of stocks when purchased P_o = $ 224.84

- The price of stocks when sold P_s = $ 295.56

Find:

what would be your capital gain (loss) on the sale, ignoring commissions

Solution:

- The capital gain or loss on the selling of stocks stems from the difference of buying and selling value of stocks. The original price of stock was P_o and the selling price would be P_s. The difference would be:

                             capital gain = P_s - P_o

                             capital gain =  $295.56 - $224.84

                             capital gain = $ 70.72

- Hence, there would be a gain in capital if sold today for about $ 70.72.

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The figure shows a positive charge placed in a uniform electric field. Is the force exerted on the +1 C charge directed up or do
raketka [301]

Answer:

the answer is down on e2020

Explanation:

3 0
3 years ago
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An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul
PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight  due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ =  400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

= 100 N .

weight at height = 100 N

5 0
3 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w
Vesna [10]

Answer:

\ \text{m/s}

Explanation:

u_1 = Velocity of one lump = 3x+3y-3z

u_2 = Velocity of the other lump = -4x+0y-4z

m = Mass of each lump = 30\ \text{g}

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

4 0
3 years ago
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

6 0
3 years ago
A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri
fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0
3 years ago
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