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2 years ago

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A spring is used to launch a coffee mug. The 20cm long spring can be compressed by a maximum of 8cm. The mug has a mass of 350 g

. What kind of spring do you need to launch the mug at least 2 stories high (7 m) - as measured from the relaxed spring position - using the maximum compression?

2 answers:

Zepler [3.9K]2 years ago

8 0

Answer:

7503.13 N/m

Explanation:

Use principle of conservation of energy.

Here, energy stored in the spring due to compression shall be utilized in attaining the potential energy of the mug.

Given that,

Length of the spring = 20 cm = 0.20 m

Compression, x = 8 cm = 0.08 m

mass of the mug, m = 350 g = 0.35 kg

h = 7 m

use the expression for energy balance -

(1/2)*k*x^2 = m*g*h

=> k = (2*m*g*h) / x^2

input the values

k = (2*0.35*9.8*7) / 0.08^2

= 7503.13 N/m

Nina [5.8K]2 years ago

7 0

Answer:

The kind of spring needed to launch the 0.35 kg mug to 7 m high from start is 7510.78 N/m.

Explanation:

To solve the question, we note that the

Mass m of the mug = 350 g

Height to which mug is to be launched to = 7 m

Length of spring = 20 cm

Therefore we look for the governing equations of motion as follows

Energy required to raise the mug to 7 m = Gravitational Potential of the mug at 7 m and

Gravitational Potential = m·g·h

Where m = mass = 350 g = 0.35 kg

h = height = 7 m

g = Acceleration due to gravity = 9.81 m/s²

Energy of spring to raise the height of the mug to 7 m = $\frac{1}{2}\times k \times x^2$

Where:

K = Spring constant

x = Length of compression or expansion

Equating the Energy supplied to the energy gained we get

m·g·h = $\frac{1}{2}\times k \times x^2$

Making k the subject of the formula gives

$K = \frac{m \times g \times h}{\frac{1}{2} \times x^2}$ = $K = \frac{0.35 \times 9.81 \times 7}{\frac{1}{2} \times 0.08^2}$ = 7510.78 N/m.

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A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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Answer:

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Due to the unusual nature of water; at about 4°c, the behavior of the density of water in relation to its temperature reverses. This means that water becomes less dense as it becomes colder below 4°c. The colder parts therefore floats to the top of the water body while the warmer part sinks allowing the top to freeze and the remaining body below to remain in its liquid state.

The freezing of the top of the lake alone protects the remaining depth of water from freezing by acting as an insulator and preventing further heat loss from the water to the ambient space. If this had not been the case, and water froze all through, marine lives will freeze to death and it will be more difficult to melt the ice come the next summer.

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storchak [24]

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In the exercise they tell us

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KATRIN_1 [288]

Answer:

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Explanation:

The formula for thermal conductivity states that the rate of heat loss H is as follows:

H= $\frac{KA (T_{1}-T_{2})}{x}$ where,

K=Coefficient of thermal conductivity

A=Area of Cross Section

H=Rate of heat loss

x=Thickness of the material

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$T_{1}$ =37°C ie. Body Temperature

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For first case, $T_{1}-T_{2}$ =17°C or 17°F

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H=714 Watts of energy is given out.

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2570400 J=614.3403442 KCal

Therefore, 614.3403442 KCal energy is burnt in an hour.

3) Only the temperature difference becomes 37°C from 17°C , rest everything remains the same, therefore the energy will also vary due to the temperature factor and more energy will be spent as in freezing climate $T_{2}=0$ °C

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