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3 years ago

14

Please help im really confused

1 answer:

Travka [436]3 years ago

8 0

Answer:

1,8 and 2,13 and 3,2 and 4,7 and 5,18

Explanation:

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Sam’s father bought him a new car for $26,304.00. He expects to pay for it in equal monthly payments for 60 months. How much wil

Alex_Xolod [135]

Answer:

438.4$ each month

Explanation:

you get this answer by dividing the total payment by the months

8 0

3 years ago

Read 2 more answers

A closed, 0.4-m-diameter cylindrical tank is completely filled with oil (SG 0.9)and rotates about its vertical longitudinal axis

egoroff_w [7]

Answer: $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa

Explanation: To determine the pressure of a liquid in a rotating tank,it is used:

p = $\frac{p_{fluid}.w^{2}.r^{2} }{2}$ - γfluid . z + c

where:

$p_{fluid}$ is the liquid's density

w is the angular velocity

r is the radius

γfluid.z is the pressure variation due to centrifugal force.

For this question, the difference between a point on the circumference and a point on the axis will be:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}.r_{B} ^{2} }{2}$ - γfluid. $z_{B}$ - ( $\frac{p_{fluid}.w^{2}.r_{A} ^{2} }{2}$ - γfluid. $z_{A}$ )

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$ - γfluid( $z_{B}$ - $z_{A}$ )

Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$

$p_{B} - p_{A} = \frac{0.9.10^3.40^2}{2}(0.2^2 - 0)$

$p_{B} - p_{A}$ = 28800

The difference in pressure between two points, one on the circumference and the other on the axis is $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa

8 0

3 years ago

Which of the following are true about algorithms? (You may select more than one)

r-ruslan [8.4K]

Have a orderf. Fr

R
F
F
D
Djeiiekee

7 0

4 years ago

Read 2 more answers

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

Ymorist [56]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

the half life of the given circuit is given by

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

$t'_{1\2} =R'Cln2$

Divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

putting all value we get new half life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

7 0

3 years ago

In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the l

Juli2301 [7.4K]

Answer:

a) 0.26

b) 1077 MPa

Explanation:

a) The following equation can be used to determine the volume fraction:

$\frac{F_f}{F_m} =\frac{E_fV_f}{E_m(1-V_f)}$

$\frac{0.97}{1-0.97} =\frac{260V_f}{2.8(1-V_f)}$

$32.3 = \frac{260V_f}{2.8-2.8V_f}$

$V_f = 0.26$

b) Tensile strength can be found by using the following equation:

$\sigma_{cl} = \sigma_m(1-V_f)+\sigma_fV_f = 50*(1-0.26)+4000*0.26 = 1077$ MPa

5 0

4 years ago