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3 years ago

Suppose a spring compresses by 5cm if you put a 250 gram weight on top of it. How

Physics

1 answer:

Leokris [45]3 years ago

8 0

Answer:

Option D. 0.061 J

Explanation:

We'll begin by converting 5 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

5 cm = 5 cm × 1 m / 100 cm

5 cm = 0.05 m

Next, we shall convert 250 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

250 g = 250 × 1 Kg / 1000

250 g = 0.25 Kg

Next, we shall determine the weight of the object. This can be obtained as follow:

Mass (m) = 0.25 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

W = m × g

W = 0.25 × 9.8

W = 2.45 N

Next, we shall determine the spring constant. This can be obtained as follow:

Compression (e) = 0.05 m

Weight = Force (F) = 2.45 N

Spring constant (K) =?

F = Ke

2.45 = K × 0.05

Divide both side by 0.05

K = 2.45 / 0.05

K = 49 N/m

Finally, we shall determine the potential energy contained in the compressed spring. This can be obtained as follow:

Compression (e) = 0.05 m

Spring constant (K) = 49 N/m

Energy (E) =?

E = ½Ke²

E = ½ × 49 × (0.05)²

E = 24.5 × 0.0025

E = 0.061 J

Therefore, the potential energy contained in the compressed spring is 0.061 J

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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2m/s2 . A watermelon has a weight of 56.0 NN at

brilliants [131]

Answer: 5.71kg and 10.33N

Explanation:

Recall,

Weight =mass × g

Where w = 56.0 N , g = 9.8 m/s2

56= m × 9.8

M = 56/9.8

M = 5.71 kg

At the surface of the moon, the mass still remains 5.71kg because it is a fundamental quantity it does not change irrespective of position.

Weight at the moon surface will change as a result of change in gravity. Hence, weight = mg

Where g=1.81m/s2, m= 5.71

W = 5.71×1.81

W = 10.33N

5 0

3 years ago

Read 2 more answers

What is the energy of a single photon of infrared light with a wavelength of 1.00 µm? 1 µm = 10^-6m

notka56 [123]

Answer:

Energy is found as:

E = 1.987·10⁻¹⁹ J

Explanation:

Energy of a single photon of infrared light can be found by using the following formula:

$E=hf$

where

E = in Joules

h = Planck's constant = 6.627×10 ⁻³⁴ J

f = frequency in hertz

It can also be written as:

$E = \frac{hc}{\lambda}$

where

c = 2.998×10⁸ ms⁻¹

λ = wavelength

Wavelength is given in the question which is:

λ = 1×10⁻⁶m

Substitute all the values in the Energy formula

$E = \frac{hc}{\lambda}\\E = \frac{(6.627\cdot10^{-34})(2.998\cdot10^8)}{1\cdot10^{-6}}\\E = 1.987\cdot10^{-19}J$

3 0

3 years ago

Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?

Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

$x_{a}(t) = at+bt^2$

$x_{b}(t) = ct^2-dt^3$

Constants are:

$a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3$

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both $x_a(t)$ and $x_b(t)$ depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

$at+bt^2 = ct^2-dt^3$

$2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0$

$t_1 = 4 - \sqrt{3} = 2.27$ s, $t_1 = 4 + \sqrt{3} = 5.73$ s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

$v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0$

$t_1 = 1$ s, $t_2 = 4.33$ s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

$a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67$ s