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3 years ago

Problem 4:

Physics

1 answer:

Olegator [25]3 years ago

6 0

Answer:

v2^2 - v1^2 = 2 g s fundamental formula

v2 = v1 + 2 g = v1 + 19.8 increase in velocity in 2 sec

v1^2 + 39.6 v1 + 392 - v1^2 = 2 * 9.8 * 123.1 = 2412.76

v1 = (2412.76 - 392) / 39.6 = 51.03

v2 = 51.03 + 19.6 = 70.63

T = 70.63 / .8 = 7.207 sec time to fall height of tower

S = 1/2 g T^2 = 4.9 * 7.207^2 = 254.5 m

(Note v2^2 - v1^2 = 70.63^2 - 51.03^2 = 2385 m

2385 / (2 * 9.8) = 122 m (close to 123.1 as was given

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3 years ago

A ball thrown vertically upward is caught by the thrower after 2.93 s. Find the initial velocity of the ball. The acceleration o

almond37 [142]

Answer:

The initial velocity of the ball is 28.714 m/s

Explanation:

Given;

time of flight of the ball, t = 2.93 s

acceleration due to gravity, g = 9.8 m/s²

initial velocity of the ball, u = ?

The initial velocity of the ball is given by;

v = u + (-g)t

where;

v is the final speed of the ball at the given time, = 0

g is negative because of upward motion

0 = u -gt

u = gt

u = (9.8 x 2.93)

u = 28.714 m/s

Therefore, the initial velocity of the ball is 28.714 m/s

7 0

3 years ago

A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi

Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

$\Delta x_1=0.50m-0.35m=0.15m$

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

$k\Delta x_1=m_1g$

Solving for the spring constant k, we get:

$k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}$

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

$k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m$

Finally, we sum this to the unstretched length to obtain the length of the spring:

$x_2=0.35m+0.088m=0.44m$

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0

4 years ago

In deep space (no gravity), the bolt (arrow)

salantis [7]

t = 0.527 s

<u>It accelerates for 0.527 s.</u>

<u>Explanation:</u>

We use the formula:

v = u+at

Given:

v = 106 m/s

u = 0 (since no gravity)

$a=201 \mathrm{m} / \mathrm{s}^{2}$

So applying the formula,

v = u+at

106 = 0 + 201t

t = 106/201

t = 0.527 s

4 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

3 years ago