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3 years ago

If you know that the rock fell 4.9 m in 1 s , how far did it drop in the first 0.5 s after you dropped it?

Physics

2 answers:

iren [92.7K]3 years ago

8 0

<u>Answer</u>

1.225 m

<u>Explanation</u>

The formula for getting the distance cover in this case is;

s = ut + 1/2 gt²

Where u⇒intial velocity

t⇒time taken

g ⇒ acceleration due to gravity

But in this case u = 0, t = 1 and s =4.9.

So, the first thin is to calculate the value of g.

S = 1/2 gt²

4.9 = 1/2 g × 1²

g = 4.9 × 2

= 9.8 m/s².

Now calculate the distance after 0.5 s.

S = 1/2 × 9.8 × 0.5²

= 1.225 m

klio [65]3 years ago

7 0

When an object is free-falling, no other force is acting upon it but the gravitational force. Because of this, the equations of motion are simplified. We can determine first the initial velocity:

v = √2gy = √2(9.81)(4.9) = 9.805 m/s

Then, we use this to the equation below:
y = vt + 1/2*at²
y = (9.805)(0.5) + 1/2(9.81)(0.5)²
y = 6.13 m

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The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull

Talja [164]

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>

Mass of block 1 = m1
Mass of block 2, = m2
Height of block 1 above the ground, = h1
Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

$m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0$

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

$m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 ) - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 ) = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2$

Thus, we can conclude the following before the block m2 reaches the ground;

For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

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3 years ago

A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin

Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

$Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa$

The shearing strain is defined as the tangent of the displacement that the block over its length:

$SS=tan\theta =\frac{Displacement}{L} =\frac{10}{40} =0.25$

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

$Sm=\frac{10208.3333}{0.25} =40833.3332Pa$

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3 years ago

If a freely falling rock were equipped with a speedometer, by how much would its speed readings increase with each second of fal

Klio2033 [76]

Its speed reading would increase to 10 m/s every second

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3 years ago

ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly

Igoryamba

The period T of a pendulum is given by:
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where L is the length of the pendulum while $g=9.81 m/s^2$ is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is $T=0.200 s$ . Using this data, we can solve the previous formula to find L:
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