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Verdich [7]
3 years ago
8

Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi

n, vertical rod. A constant force of magnitude F is being exerted on the rod. The rod remains perpendicular to the direction of the applied force so that the springs are extended by the same amount. Find the effective spring constant k of the two-spring system. Give your answer for the effective spring constant in terms of k1 and k2.
Physics
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

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A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

Mass of the climber = 69.38 kg

6 0
3 years ago
In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one sid
mojhsa [17]

Answer:

Net pull = 110 N to the left

Explanation:

Group the different pulls according to the direction (right or left)

2 pull 196 N each to the right

4 pull 98 N each to the left

5 pull 62 N each to the left

3 pull 150 N each to the right

1 pull 250 N to the left

Since positive direction is to the right, the pulls to the left will have a minus (-)

Net Force= 2(196)+4(-98)+5(-62)+3(150)+1(-250) \\Net Force = -110

The resulting force is negative, meaning the direction is to the left

6 0
3 years ago
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
What does the slope on a velocity-time graph measure?
Lostsunrise [7]

Answer:The slope in the velocity-time graph represents the acceleration. The slope is defined as the ratio of change in y-axis to change in the x-axis. The slope is represented by the letter m and following is the general formula used for determining the slope:

Explanation:

6 0
3 years ago
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A 1400-kg car is moving at a speed of 25m/s. how much kinetic energy does the car have?
Luba_88 [7]

Answer:

437.5Kjoules

Explanation:

K.E=half multiply by mass multiply by square of velocity

=437.5Kjoules

3 0
3 years ago
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