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svlad2 [7]
3 years ago
5

What is the mechanical energy of the cart on the right at position A. The

Mathematics
2 answers:
drek231 [11]3 years ago
6 0

Answer:

1800 J

Step-by-step explanation:

At point A when the cart is not in motion, it only has potential energy.

We know that potential energy is given by mgh where m is the mass of the cart, g is acceleration due to gravity and h is the height.

Substituting 12 kg for m, 10 m/s2 for g and 15 m for h then

PE= 12*10*15=1800 J

QveST [7]3 years ago
5 0

Answer:

1800

Step-by-step explanation:

P.E. =m.g.h = 12x10x15 = 1800 Joules

K.E/ = 1/2 mv2 = 1/2x12x02 = 0

P.E = 1800 | K.E. 0

M.E. = 1800 + 0 = 1800 Joules

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The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 37 o
just olya [345]

Answer:

a) 68 % will lie between 30 and 44 ounces

b) between 23 and 44 ounces will lie 81,5 %

Step-by-step explanation: See Annex

a) Normal Distribution  N ( 37, 7)

The Empirical Rule establishes that 68 % of values will be at

μ  ±  1 σ      where μ is the mean and σ the standard deviation

Then:   37 - 7  = 30    and 37 + 7 = 44

68 % of the values will lie between  30 and 44 ounces

b) And between  23 and 44 we should find:

We have 95 %  of values between

μ  ± 2*σ  

In our case to the left of the value (mean 0 ) and up to the value 23 we get 95/2 = 47,5 % and to the right of the value (mean 0 ) and up to 44 we half 68/2 = 34 %

Then between 23 and 44, we have 47,5 + 34 = 81,5 %

8 0
3 years ago
3. How much interest does $5300 earn at a rate of 2.8% interest compounded quarterly after 6
azamat

Answer:

The interest is I=\$74.46

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=6/12=0.5\ years\\ P=\$5,300\\ r=0.028\\n=4  

substitute in the formula above  

A=5,300(1+\frac{0.028{4})^{4*0.5}

A=5,300(1.007)^{2}  

A=\$5,374.46  

<em>Find the interest</em>

I=A-P

substitute

I=\$5,374.46-\$5,300=\$74.46

5 0
2 years ago
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