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meriva
3 years ago
10

It takes 500 J of work to compress a spring 10 cm. What is the force constant of the spring? (b) When a 3.0 kg block is pushed a

gainst a massless spring of force constant 4.5. 10' N/m, the spring is compressed 8.0 cm. The block is released and it slides 2.0 m from the point at which it is released across a horizontal surface before friction stops it. What is the coefficient of kinetic friction between the block and the surface?
Physics
1 answer:
NARA [144]3 years ago
8 0

Answer:

Explanation:

a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²

500 = 5 x 10⁻³ k ,

k = (500/5) x 10³ = 10⁵ N/m

b )

k = 4.5 x 10¹ = 45 N/m

Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J

This energy gets dissipated by friction .

work done by friction = μ mg d

d is the distance traveled under friction

so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2

μ = 245 x 10⁻⁴  or 0.00245 which appears to be very small. .

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A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl
Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

(1). \:F = (m_1+m_2)a,

where a =0.265m/s^2 is the upward acceleration, and F = 688N is the net propelling force (counts the gravitational force).

Also, the tension T in the rope is 79.8 N more than the basket's weight; therefore,

(2). \:T = m_2g+79.8

and this tension must equal

T -m_2g =m_2a

(3). \:T = m_2g +m_2a

Combining equations (2) and (3) we get:

m_2a = 79.8

since a =0.265m/s^2, we have

\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

688N = (m_1+301.13kg)(0.265m/s^2)

\boxed{m_1 = 2295 kg}

Thus, the mass of the balloon and the basket is  2295 kg and 301 kg respectively.

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3 years ago
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3 years ago
A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you
soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s

The interaction time to avoid that the water balloon breaks is 0.029s

5 0
3 years ago
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