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2 years ago

11

A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3

meters.
What is the SPE?

1 answer:

vaieri [72.5K]2 years ago

3 0

22.5 J

Explanation:

Given:

x = 3 m

$k = 5\:\text{N/m}$

The spring potential energy $PE_s$ is

$PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2$

$\:\:\:\:\:\:\:=22.5\:\text{J}$

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The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

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3 years ago

A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2 m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0

2 years ago

2. A 50 kg diver is standing on the edge of a 15 m high cliff. What is his potential energy?

Lera25 [3.4K]

Answer:

3675 J

Explanation:

Gravitational Potential Energy = $\frac{1}{2}$ × mass × g × height

( g is the gravitation field strength )

Mass = 50 kg

G = 9.8 N/kg ( this is always the same )

Height = 15 m

Gravitational Potential Energy = $\frac{1}{2}$ × 50 ×9.8 × 15

= 3675 J

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2 years ago

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

Pepsi [2]

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a force of 12.344 N

at distance r

and we know Electrostatic force is given

$F=\frac{kq_1q_2}{r^2}$

$F=\frac{kQ\cdot Q}{r^2}$

$F=\frac{kQ^2}{r^2}$

Now the magnitude of charge is 2Q and is at a distance of $\frac{r}{2}$

$F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}$

F'=16F

F'=197.504 N

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3 years ago

The image formed by a plane mirror is _____.

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The image formed by a plane mirror is virtual, upright and the same size with the actual object. The upright image of an object in a plane mirror is can be found on the other side of the mirror which is why it is also virtual.

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3 years ago