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creativ13 [48]
3 years ago
9

Which of the following type of personal debts is usually considered wise and justified? A. credit card purchases of luxuries tha

t you want immediately B. a loan for the car you have always dreamed about C. money to pay off other debts and interest D. home mortgage
Physics
2 answers:
timama [110]3 years ago
8 0

Money to pay off other debts and interest are usually considered wise and justified.

Answer:Option c

<u>Explanation: </u>

Personal loans give persons an access to funds that can be used as their wish and they are unsecured. This means that they do not need you to put down collateral for obtaining the loan.  

A personal loan usually has higher rate of interests than that of secured loan. They offer lower rate of interest than credit cards. It is advisable to use personal loans to pay off debts form credit cards will support you in saving interests and help you to get out of the debt faster period of time.

NeTakaya3 years ago
3 0

Answer:

C. Money to pay off other debts and interest

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A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque
labwork [276]

Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

Explanation:

given data

rectangular block above water  = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block  that is

specific gravity = block vol below ÷ total block vol × specific gravity  water   ..............1

put here value we get

specific gravity =  \frac{1.60}{1.60+0.400}  × 1

specific gravity = 0.8

and now we get here specific gravity of  solution  that is express as

specific gravity of  solution  = total block vol ÷ block vol below × specific gravity  block   ........................2

put here value we get

specific gravity of  solution  = \frac{1.60+0.400}{0.800} × 0.8

specific gravity of  solution  = 2

6 0
3 years ago
A particle of unit mass moves in a potential V(x)= ax^2+b/x^2 . Where a and b are constnts. Calcuate the angular frequency of sm
Sholpan [36]

Answer:

Explanation:

Given

Potential Energy is given by

U(x)=ax^2+\frac{b}{x^2}

And Force is given by

F=-\frac{\mathrm{d} U}{\mathrm{d} x}

Particle will be at equilibrium when Potential Energy is either minimum or maximum

F=-\left ( 2ax-\frac{2b}{x^3}\right )

i.e.ax=\frac{b}{x^3}

x_0=(\frac{b}{a})^{0.25}

So angular Frequency of small oscillation is given by

\omega =\sqrt{\frac{U''(x)}{m}}

for m=1

we get    \omega =\sqrt{\frac{U''(x_0)}{1}}

U''(x_0)=2a+6a= 8a

\omega =\sqrt{8a}

4 0
3 years ago
A block slides down a frictionless plane having an inclination of 15.0°. The block starts from rest at the top, and the length o
nalin [4]

Answer: check the pic

Explanation:

8 0
3 years ago
One student made the incomplete diagram shown below to represent the relationship between igneous rocks, sediments, sedimentary
Kryger [21]

Answer: Option (C)

Explanation: Rock cycle plays an important role in the alteration of rocks from one form to another.

  1. Igneous rocks when undergoes high temperature and pressure condition, it transforms into a metamorphic rock.
  2. Sedimentary rocks are formed from the sedimentation and consolidation of sediments
  3. Igneous rocks are formed due to the crystallization of magma.

Hence the correct answer is option (C)

3 0
3 years ago
A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object
Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    Em_{f} = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   Em_{f}

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      p_{f} = M v1f + m v

   p₀ =   p_{f}

   M v₁₀ = M v_{1f}+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = K_{f}

    ½ M v₁₀² = ½ M v_{1f}² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  v_{1f} + m v

    M v1₁₀² = M v_{1f}² + m v²

We cleared v1f in the first we replaced in the second

   v_{1f} = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

     2 m v₁₀ - v (m + 1) m/ M = 0

     v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s

5 0
3 years ago
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