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aleksandrvk [35]

2 years ago

13

Block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is p

arallel to the incline and passes over amassless frictionless pulley at the top. block B with a massof 8.0kg is attached to the dangling end of the string. theacceleration of B is:
a. 0.69 up
b. 0.69 down
c. 2.6 up
d. 2.6 down
e. 0

1 answer:

Tomtit [17]2 years ago

6 0

Answer:

Please find attached

Explanation:

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you are hiking along a river and see a tall tree on thhe opposite bank. You measure the angle of elevation of the top of the tre

Sidana [21]

Answer:

Explanation:

Let the height of the tree is y and the distance of tree from point B is x.

According to the diagram

$tan61 = \frac{y}{x}$

x = 0.55 y ..... (1)

$tan49.5 = \frac{y}{50+x}$

(50 + 0.55y) 1.17 = y ..... from equation (1)

58.5 + 0.644 y = y

0.356 y = 58.5

y = 164.3 ft

3 0

2 years ago

An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d

just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

3 0

2 years ago

. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the

vichka [17]

Transverse wave as the wave is going up and down no compressions

3 0

2 years ago

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

7 0

2 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

6 0

2 years ago