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RSB [31]
2 years ago
12

A startled armadillo leaps upward, rising 0.537 m in the first 0.220 s. (a) what is its initial speed as it leaves the ground? (

b) what is its speed at the height of 0.537 m? (c) how much higher does it go? use g=9.81 m/s2.
Physics
1 answer:
Elena L [17]2 years ago
3 0
Let u = initial vertical velocity.

Assume that
g = 9.81 m/s²,
Wind resistance is ignored.

When t = 0.220 s, the height is h = 0.537 m. Therefore
0.537 m = (u m/s)*(0.220 s) - (1/2)*(9.81 m/s²)*(0.220 s)²
0.537 = 0.22u - 0.2372
u = 3.519 m/s

The upward velocity after 0.220 s is
v = 3.519 - 9.81*0.22 = 1.363 m/s

At maximum height, the upward velocity is zero. The maximum height, H, is given by
(3.519 m/s)² - 2*(9.81 m/s²)*(H m) = 0
12.3834 - 19.6H = 0
H = 0.632 m
It goes higher by 0.632 - 0.537 = 0.095 m

Answers:
(a) The initial speed is 3.519 m/s.
(b) The speed at  0.537 m height is 1.363 m/s.
(c) It goes higher by 0.095 m.

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2 years ago
the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0
3 years ago
A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

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