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vekshin1
4 years ago
11

Accelerated depreciation has an advantage for profitable firms in that it moves some cash flows forward, thus increasing their p

resent value. On the other hand, using accelerated depreciation generally lowers the reported current year's profits because of the higher depreciation expenses. However, the reported profits problem can be solved by using different depreciation methods for tax and stockholder reporting purposes. a. True b. False
Physics
1 answer:
scZoUnD [109]4 years ago
6 0

Answer: A. True

Explanation:

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A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J. What is the mass of the soccer ball
Savatey [412]
A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J.The mass of the soccer ball is 0.4 kgs. This answer is derived from the formula K=1/2 MV^2.So velocity and kinetic energy are given from that mass of the ball is calculated.By substituting the values 500=1/2*M*50*50 which gives M=0.4 Kgs.<span>
</span>
4 0
4 years ago
Read 2 more answers
A 0.200-kg mass is attached to the end of a spring with a spring constant of 11 N/m. The mass is first examined (t = 0) when the
inn [45]

Answer:

a) x (t) = 0.3187 cos (7.416 t + 1.008) ,  b)  v = -2,363 sin (7,416 t + 1,008)

c)  a = - 17.52 cos (7.416t + 1.008)

Explanation:

The spring mass system creates a harmonic oscillator that is described by the equation

    x = Acos (wt + φ)

Where is the amplitude, w the angular velocity and fi the phase

a) Let's reduce the SI system

    x = 17.0 cm (1 m / 100 cm) = 0.170 m

The angular velocity is given by

      w = √ (k / m)

      w = √ 11 / 0.200

      w = 7.416 rad / s

Let's look for the terms of the equation with the data for time zero (t = 0 s)

      0.170 = A cos  φ

Body speed can be obtained by derivatives

      v = dx / dt

      v = -A w sin (wt + φ)

     2.0 = -A 7.416 sin φ

Let's write the two equations

     0.170 = A cos φ

     2.0 / 7.416 = -A sin φ

Let's divide those equations

    tan φ= 2.0 / (7.416 0.170)

     φ= tan⁻¹ (1,586)

     φ= 1.008 rad

We calculate A

   A = 0.170 / cos φ

   A = 0.170 / cos 1.008

   A = 0.3187 m

With these values ​​we write the equation of motion

    x (t) = 0.3187 cos (7.416 t + 1.008)

b) the speed can be found by derivatives

      v = dx / dt

      v = - 0.3187 7.416 sin (7.416 t +1.008)

      v = -2,363 sin (7,416 t + 1,008)

c) the acceleration we look for conserved

    a = dv / dt

    a = -2,363 7,416 cos (7,416 t + 1,008)

    a = - 17.52 cos (7.416t + 1.008)

6 0
3 years ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².​
Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2}  } =0.0001[m^{2} ]

Now replacing:

P=100/0.0001\\P=1000000[Pa]

3 0
3 years ago
You are holding one end of a horizontal stretched string. Flicking your wrist will send a pulse down the string. Which actions w
kumpel [21]

Answer:

Use a lighter string of the same length, under the same tension.

Stretch the string tighter to increase the tension

Explanation:

The wave speed depends on propertices of the medium, not on how you generate the wave. For a string

Increasing the tension or decreasing the linear density (lighter string) will increase the wave speed.

5 0
3 years ago
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