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iragen [17]
3 years ago
11

Which statements are false? Money comes in different forms. Money is indivisible. M1 includes notes, coins, and funds in checkab

le deposits. Money is a unit of account. Credit cards are used primarily to withdraw money from a checking account. NextReset
Physics
2 answers:
Naily [24]3 years ago
7 0

Answer:

Wrong statements are:-

1. Money is indivisible.

2. credit cards are used primarily to withdraw money from a checking account.

Explanation:

Some characteristics of money are:-

Money is divisible and it can be present in the form of coins, notes funds, checks, government securities etc. Anything which have a store of value is a form of money. Money is measure of value(fiat) money. so the statement that money is indivisible is a false statement.

2. Credit cards are primarily used to withdraw the money from account. so given statement is also wrong.

Ainat [17]3 years ago
3 0

Answer: Money is indivisible and Credit cards are used primarily to withdraw money from a checking account are two false statements.

Explanation:

Since we know the characteristics of money as follows:

1) Money is divisible.

2) Money has different forms - coins, notes, demand deposits, cheques, tokens, funds in checkable deposits etc.

3) Money is a unit of account.

4) Money is measure of value.

5) Money is store of value.

So, Money is indivisible is false.

And credit cards are used primarily to withdraw money from a checking account is false , as debit cards are used for withdrawing money.

Hence, Money is indivisible and Credit cards are used primarily to withdraw money from a checking account are two false statements.

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Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to
stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

N=\frac{(1962N)(1.5m)}{3m}

N=981N

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

-F_{By}+N_{c}=0

F_{By}=N_{c}

Next, the forces in x.

\sum F_{x}=0

-f_{sB}-f_{sC}+P_{x}=0

We can find the x-component of force P like this:

P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

f_{sB}(1.75m)-P_{x}(0.75m)=0

f_{sB}=\frac{288N(0.75m)}{1.75m}

f_{sB}=123.43N

With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

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3 years ago
Which is the formula to calculate the slope of this line? A mass divided by volume B mass minus volume C volume divided by mass
Hatshy [7]

A- mass divided by volume

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A sprinter reaches his maximum speed in 2.6 seconds from rest with constant acceleration. He then maintains that speed and finis
Delvig [45]

Answer: maximum speed vmax = 11.42m/s

Explanation:

Given that the sprinter maintained constant acceleration during the first 2.6 seconds.

a = vmax/ta .......1

The distance covered during the acceleration period is;

da = 0.5a(ta)^2 .....2

Substituting equation 1 to 2

da = 0.5(vmax/ta)(ta)^2 = 0.5vmax(ta) .....3

The distance covered during the period of constant speed vmax is;

dv = vmax (tv) ......4

The total distance travelled is

d = da + dv = 100 (Given)

da + dv = 100 ......5

Substituting equation 3 and 4 into 5

0.5vmax(ta) + vmax(tv) = 100

vmax ( 0.5ta +tv) = 100

vmax = 100/(0.5ta + tv) ....6

But,

t = ta + tv

tv = t - ta .......7

Substituting equation 6 into equation 7

vmax = 100/(0.5ta + t - ta)

vmax = 100/(t-0.5ta)

t = 10.06 s

ta = 2.6 s

Substituting the values;

vmax = 100/(10.06 -0.5(2.6))

vmax = 11.42m/s

Note:

ta = acceleration time

tv = constant velocity vmax time

t = overall time

da , dv and d = acceleration, constant velocity and overall distance covered respectively.

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