PHYSICS COLLEGE

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer 1

Answer:

Answer:

Explanation:

Mass of the gate, $m_1 = 4.5 kg$

Mass of the raven, $m_2 = 1.2 kg$

Initial speed of raven, $v_1 = 4.5 m/s$

Final speed of raven, $v_2 = - 1.5 m/s$

Moment of Inertia of the gate about the axis passing through one end:

$I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2$

Angular momentum of the gate, $L = I \omega$

$L = 5.33 \omega$

Using the law of conservation of angular momentum:

$m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n$

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An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.

Answers

Answer:

Explanation:

Using Conservation of momentum (total final momentum of system is)

m1•v1f + m2•v2 f + m3•v3 f=0

and it must be zero to equal the original momentum( since the original body is at rest).

Given that

original mass M=1.82×10^-26

First disintegrate mass m1=5.18×10^-27kg

In y direction V1f=6×10^6 I'm/s

Second disintegrate mass m2=8.5×10^-27kg

In x direction V2f=4×10^6 im/s

Then the third disintegrate will be

m3=M-m1-m2

m3=1.82×10^-26-5.18×10^-27-8.5×10^-27

m3=4.52×10^-27

And the velocity is unknown

Now using the formula above

m1•v1f + m2•v2 f + m3•v3 f=0

m3•V3f= - m1•v1f - m2•v2 f

4.52E-27V3f=-5.18E-27×6E6j - 8.5E-27×4E6 i

Divide thorough by 4.52E-27

V3f= - 6.88×10^6j - 7.52×10^6i

V3f= - 7.52×10^6i - 6.88×10^6j

The final velocity of the third mass disintegrate is 6.88×10^6j - 7.52×10^6i m/s

A hollow sphere of radius 0.25 m is rotating at 13 rad/s about an axis that passes through its center. the mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?

Answers

The work required to bring the sphere to stop is equal to the kinetic energy possessed by the sphere.

Kinetic energy of a rotating body is given by,

K.E = $(1)/(2)Iw^(2)$

Here, I= Moment of inertia of hollow sphere,

Since, the hollow sphere is rotating about the axis passing through its center, I = $(2)/(3)MR^(2)$

M= Mass of the sphere= 3.8 kg,

R= Radius of gyration= Radius of the sphere= 0.25 m

w= Angular speed of the sphere = 13 rad/s

Substituting the values,

Kinetic energy = $(1)/(2) *(2)/(3) (3.8)(0.25)^(2)(13.0)^(2)$

= 13.4 J

∴ Work required to bring the sphere to stop is 13.4 J.

What displacement do I have if I travel at 10 m/s E for 10 s? A. 1 m E B. 1 m C. 100 m D. 100 m E Scalar quantities include what 2 things? A. Number and direction B. Numbers and units C. Units and directions D. Size and direction What measures distance in a car? A. Odometer B. Pressure gauge C. Speedometer D. Steering wheel What displacement do I have if I travel 10 m E, then 6 m W, then 12 m E? A. 28 m E B. 16 m E C. 16 m D. 28 m

Answers

Hope this will help you

Final answer:

The displacement is 100 m to the east.

Explanation:

The displacement can be calculated using the formula:

Displacement = Velocity × Time

In this case, the velocity is 10 m/s to the east and the time is 10 seconds.

So, Displacement = 10 m/s × 10 s = 100 m to the east.

Learn more about Displacement here:

brainly.com/question/33459975

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.

Answers

maximum speed of cheetah is

$v_1 = v_(max)$

speed of gazelle is given as

$v_2 = v_(g)$

Now the relative speed of Cheetah with respect to Gazelle

$v_(12) = v_1 - v_2$

$v_(12) = v_(max) - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_(12)* t$

so by rearranging the terms we can say

$t = (d)/(v_(12))$

$t = (d)/(v_(max) - v_g)$

so above is the relation between all given variable

Why does an astronaut in a spacecraft orbiting Earthexperience
a feeling of weightlessness?

Answers

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)