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1 month ago

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You are asked to design a spring that will give a 1160-kg satellite a speed of 2.50 m>s relative to an orbiting space shuttle

. your spring is to give the satellite a maximum acceleration of 5.00g. the spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible. (a) what must the force constant of the spring be? (b) what distance must the spring be compressed?

1 answer:

Maru [3.3K]1 month ago

7 0

Given the values: m = 1160 kg g = 9.81 m/s² v = 2.5 m/s Unknowns include: k (spring constant) x (spring compression) 1. To address this, use Newton's second law and Hook's law: 2. The total energy in the spring must correspond to the satellite's energy: By combining these two equations

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A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for____ rays.

Ostrovityanka [3204]

~Greetings! ^_^

Your inquiry: A plant produces flowers with a violet hue. The flowers possess a violet appearance due to their ability to absorb all light waves barring____ rays.

Your reply: A plant produces flowers with a violet hue. The flowers possess a violet appearance because they absorb all light rays except for violet rays.

Hope this is beneficial~

5 0

1 month ago

Read 2 more answers

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

serg [3582]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

8 0

2 months ago

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

kicyunya [3294]

Answer:

a) Q = C*emf

b) Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let $V_{R}$ represent the Voltage across the Resistor

$V_{c}$ and

represent the Voltage across the capacitor

Implementing KVL;

$emf - V_{R} - V_{c} = 0\\$

.........................(1)

Since this is a series connection, the same current traverses through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Integrating $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Substituting the values of t and I₀ into the aforementioned formula for I

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

1 month ago

Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a

Ostrovityanka [3204]

Answer:

Power output, P = 924.15 watts

Explanation:

We have the following parameters:

Length of the ramp, l = 12 m

Weight of the individual, m = 55.8 kg

Incline angle with respect to the horizontal, $\theta=25^{\circ}$

Elapsed time, t = 3 s

Let h represent the vertical height of the hill:

$h=l\ sin\theta$

$h=12\times \ sin(25)$

h = 5.07 m

Power P required for a person to ascend the hill can be expressed as:

$P=\dfrac{E}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{55.8\times 9.8\times 5.07}{3}$

P = 924.15 watts

This indicates that a minimum average power output of 924.15 watts is essential for an individual to ascend this elevation. Thus, this is the answer sought.

3 0

2 months ago

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g

Maru [3345]

Answer:

a) $X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b) D does not influence the long-term results.

Explanation:

Given that

$\dfrac{dx}{dt}=-k(x-A)$

A = A0 cos(ωt)

$\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))$

$\dfrac{dx}{dt}+kx=kA_o cos(\omega t)$ This is a linear equation hence the integration factor, I

$I=e^{\int kdt}$

$I=e^{kt}$ Now using the characteristics of linear equations

$e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C$

$e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C$

b) At t= 0

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

Thus, the initial condition

does not affect the long-term outcome.

$X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C$

5 0

2 months ago