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3 years ago

What is sir Richard Branson's personal dilemma ?

Physics

1 answer:

kakasveta [241]3 years ago

4 0

Answer:

Sir Richard Branson's personal dilemma is that he is concerned about the environment and climate change, but he has made his fortune with an airline industry that contributes to the greenhouse gases.

brainliest plz

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A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g

dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

$f = \frac{4.04}{60} =0.067 rev/s$

here we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(0.067)$

$\omega = 0.42 rad/s$

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

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3 years ago

Is an imaginary "line" from the earth to the moon a line, a ray, or a segment? Why?

igomit [66]

Answer;

- Line segment

Explanation;

"from earth to moon" implies endpoints at both locations and it is thus a line segment

A line extends forever in both directions, a line segment is just part of a line. It has two endpoints, and a ray starts at one point and continues on forever in one direction.

5 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

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3 years ago

The acceleration of gravity is -9.8 m/s2. A sling shot fires a rock straight up into the air with a speed of +39.2 m/s. 1. what

Vanyuwa [196]

Given that,

The acceleration of gravity is -9.8 m/s²

Initial velocity, u = 39.2 m/s

Time, t = 2 s

To find,

The final velocity of the shot.

Solution,

Let v is the final velocity of sling shot. Using first equation of motion to find it.

v = u +at

Here, a = -g

v = u-gt

v = (39.2)-(9.8)(2)

v = 19.6 m/s

So, its velocity after 2 seconds is 19.6 m/s.

4 0

3 years ago

A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph

Nina [5.8K]

Think about how each of noise, heat, visible light, radiation and atmospheric shock waves travel. Which ones require air particles to travel?

A vacuum has no air particles within it, it is completely empty.

Therefore, any of the above that requiring particles to travel will not be able to cross it and the observer will not experience it.

8 0

3 years ago

What is sir Richard Branson's personal dilemma ?​

What is sir Richard Branson's personal dilemma ?