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2 years ago

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identical springs are placed side-by-side (in parallel), and connected to a large massive block. The stiffness of the 43-spring

combination is 16770 N/m. What is the stiffness of one of the individual springs? ks= N/m

1 answer:

allsm [11]2 years ago

6 0

Answer:

The stiffness of one of the individual spring is 390 N/m.

Explanation:

It is given that 43 identical springs are placed side-by-side and connected to a large massive block.

The stiffness of the 43 spring combination is 16770 N/m

We need to find the stiffness of one of the individual springs. Let k is the stiffness of one spring. The effective spring stiffness of this width spring is given by :

$K=N\times k$

$k=\dfrac{K}{N}$

$k=\dfrac{16770\ N/m}{43}$

k = 390 N/m

So, the stiffness of one of the individual spring is 390 N/m. Hence, this is the required solution.

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The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

2 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

2 years ago

The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver

luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

Stage 2:

Time = 2 hours

Speed = 200km

Stage 3:

Time = 4 hours

Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0$

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100$

Lets find the speed at stage 1

Speed = $\frac{Distance }{Time}$

Speed = $\frac{80}{1}$

Speed 1= 80 km/hr

The speed at stage 2

Speed = $\frac{Distance }{Time}$

Speed = $\frac{200}{2}$

Speed 2 = 100 km/hr

The speed at stage 3

Speed = $\frac{Distance }{Time}$

Speed = $\frac{x}{4}$

Speed 3 = $\frac{x}{4}$

we kow that average is ,

$\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100$

$\frac{ 80 + 100+ \frac{x}{4} }{3} = 100$

$\frac{ 180 + \frac{x}{4} }{3} = 100$

$\frac{ \frac{720 +x}{4} }{3} = 100$

$\frac{720 +x}{4} \times \frac{1}{3} = 100$

$\frac{720 +x}{12} = 100$

$720 +x = 100 \times 12$

$720 +x = 1200$

$x = 1200- 720$

x = 480

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