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Margarita [4]
4 years ago
6

Two pendulum bobs have equal masses and lengths (8.100 m). bob a is initially held horizontally while bob b hangs vertically at

rest. bob a is released and collides elastically with bob
b. how fast is bob b moving immediately after the collision?
Physics
2 answers:
Karo-lina-s [1.5K]4 years ago
5 0
Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.

Speed will be = squreroot ( 2*g*L)

L is length of pendulum
Ira Lisetskai [31]4 years ago
5 0

bob is the bobbiest of them all

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Which type of electromagnetic radiation cannot be focused?
Irina18 [472]
The answer is A. Hope this helps. :)
7 0
3 years ago
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A(n) 69.8 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 60.4 m away from the shuttle
alukav5142 [94]

Answer:

The time taken is 6.7  min

Explanation:

Using the linear momentum conservation theorem, we have:

m_1*v_{o1}+m_2*v_{o2}=m_1*v_{f1}+m_2*v_{f2}

when she was 60.4m from the shuttle, she has zero speed, so the initial velocity is zero.

m_1*0+m_2*0=m_1*v_{f1}+m_2*v_{f2}\\m_1*_{f1}=-m_2*v_{f2}\\v_{f1}=-\frac{m_2*v_{f2}}{m_1}\\\\v_{f1}=-\frac{0.886kg*12m/s}{69.8kg}\\\\V_{f1}=-0.15m/s

That is 0.15m/s in the opposite direction of the camera.

the time taken to get to the shuttle is given by:

t=\frac{d}{v_{f1}}\\\\t=\frac{60.4m}{0.15m/s}\\\\t=403s\\t_{min}=403s*\frac{1min}{60s}=6.7min

6 0
3 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

5 0
3 years ago
Meeta used an elastic tape to measure the length of her window to stitch a curtain. Do you think she will be able to stitch a cu
Yuki888 [10]

Answer:

No

Explanation:

She will not be able to measure the length of her window accurately due to instrumental error from her choice of instrument. The elastic nature of her tape would alter the measurement because it will stretch as she is taking her readings, thus reducing the true measurement of the length of her window.

To measure the length of her window, she could use an inelastic tape rule or a metre rule. These instruments would eliminate instrumental error.

3 0
3 years ago
Bailey was driving down the street on her motorcycle and reduced her velocity from 25
Sav [38]

Answer:

a = - 25 m/s²

Explanation:

The magnitude of acceleration or deceleration of an object can be found by using the third equation of motion as follows:

2as = Vf² - Vi²

where,

a = magnitude of acceleration = ?

Vf = Final Velocity = 15 m/s

Vi = Initial Velocity = 25 m/s

s = distance traveled = 8 m

Therefore,

2a(8 m) = (15 m/s)² - (25 m/s)²

a = (- 400 m²/s²)/(16 m)

<u>a = - 25 m/s²</u>

<u>Here negative sign indicates deceleration</u>

5 0
3 years ago
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