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2 years ago

Marvin Martian is standing on planet Potatoine.

Physics

1 answer:

prisoha [69]2 years ago

4 0

Answer:

W = 113.98 N

Explanation:

Given that,

Radius of Potatoine $R=6.4\times 10^6\ m$

Mass of Potatoine, $M=2\times 10^{24}\ kg$

Mass of Marvin, m = 35 kg

We need to find his weight on Potatoine. Weight of an object is given by :

W = mg

g is acceleration due to gravity, $g=\dfrac{GM}{R^2}$

So,

$W=\dfrac{GMm}{R^2}\\\\W=\dfrac{6.67\times 10^{-11}\times 35\times 2\times 10^{24}}{(6.4\times 10^6)^2}\\\\W=113.98\ N$

So, his weight on Potatoine is 113.98 N.

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A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

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Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

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A 50-cm-long spring is suspended from the ceiling. A 230g mass is connected to the end and held at rest with the spring unstretc

alukav5142 [94]

Answer:

k = 25.07 N/m

Amplitude = 9 cm

f = 1.66 Hz

Explanation:

Given:

- The original length of the spring L_o = 50 cm

- The mass hanged m = 230 g

- The amount of stretch given 2x = 18 cm @lowest point.

Find:

a. What is the spring constant? (K=)

b. What is the amplitude of the oscillation?

c. What is the frequency of the oscillation?

Solution:

- Make a FBD of the hanging mass, There are two external forces acting on it that is the force of gravity due to its weight and the springs restoring force when its stretched to its lowest point. After hanging the mass on the spring a new equilibrium position is achieved which also causes the spring to stretch. We can apply the Equilibrium conditions at this point in vertical direction as:

k*x - m*g = 0

k = m*g / x

Where, x is the extension of the spring or mean stretch. which 0.5*amplitude (Lowest point). x = 9 cm

k = 0.23*9.81 / 0.09

k = 25.07 N/m

Answer: For part a we have the stiffness of the spring k = 25.07 N/m

- The amplitude of the oscillating motion is the half the amount of total stretch or the amount the spring extends above or below the mean position.

Amplitude = x = 9 cm

- The frequency of any oscillatory motion which can be modeled by SHM can be expressed as:

f = 1 / 2*p* sqrt ( k / m )

- Plug the values in:

f = 1 / 2*pi* sqrt (25.07 / 0.23 )

f = 1.66 Hz

Answer: For part c the frequency of oscillation is f = 1.66 Hz

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