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2 years ago

What is the measure of minor arc AB?

Physics

1 answer:

kumpel [21]2 years ago

7 0

The answer is b, you are correct. here’s why.

a is angle p which is 286

b is correct because it looks like it would fit based on what 286 looks like

c is too small

d is way too small

hope this helps brainlist would be much appreciated:)

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A 2.0 kg pendulum has an initial total energy of 20 J. Calculate the energy lost as heat if the pendulum is 0.10 m high and is t

Maurinko [17]

The correct answer is (A) 2.0 J

Total energy of the pendulum is the sum of its kinetic and potential energy. At the instant of time, when the pendulum is at a height h and has a speed v, Its energy is given by,

$E=mgh+\frac{1}{2} mv^2$

Substitute 2.0 kg for m, the mass of the pendulum, 9.81 m/s² for g, the acceleration due to gravity, 0.10 m for h and 4.0 m/s for v.

$E=mgh+\frac{1}{2} mv^2\\ =(2.0kg)(9.81m/s^2)(0.10 m)+\frac{1}{2}(2.0kg)(4.0m/s)^2\\ =17.962J$

The pendulum has an initial energy of 20 J. the energy lost is given by,

$\Delta E=(20J)-(17.962J)\\ =2.038J=2.0J$

Thus, the energy lost by the pendulum is (A) 2.0 J

4 0

2 years ago

One of two 25-year-old identical twins begins a trip on a spaceship traveling at 0.8 c while her twin remains on Earth. The twin

borishaifa [10]

Answer:

This equation is based on twin paradox - a phenomena where one of the twin travels to space at a speed close to speed of light and the other remains on earth. the twin from the space on return discovers that the one on earth age faster.

Solution:

$t_{o}$ = 10 years

v = 0.8c

c = speed of light in vacuum

The problem can be solved by time dilation equation:

$t = \frac{t_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ (1)

where,

t = time observed from a different inertial frame

Now, using eqn (1), we get:

$t = \frac{10}{\sqrt{1 - \frac{(0.8c)^{2}}{c^{2}}}}$

t = 16.67 years

The age of the twin on spaceship according to the one on earth = 25+16.67 =41.66 years

8 0

3 years ago

A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a

Tom [10]

Answer:

The value of new value of angular speed of merry go round. $\omega_{2}$ = 0.96 $\frac{rad}{sec}$

Explanation:

Given data

r = 1.4 m

Moment of inertia $I_{1}$ = 265 kg - $m^{2}$

$N_{1} =$ 11 RPM

$\omega_{1} = \frac{2 \pi N}{60}$

$\omega_{1} = \frac{2 \pi (11)}{60}$

$\omega_{1}$ = 1.15 $\frac{rad}{sec}$

From conservation of momentum principal

$I_{1} \omega_{1} = I_{2} \omega_{2}$ ------- (1)

$I_{2} = m r^{2} + 265$

$I_{2} = 27 (1.4)^{2} + 265$

$I_{2} = 317.92 \ kg m^{2}$

Put all the values in equation (1)

265 × 1.15 = 317.92 × $\omega_{2}$

$\omega_{2}$ = 0.96 $\frac{rad}{sec}$

This is the value of new value of angular speed of merry go round.

6 0

2 years ago

the momentum of a spring coil when the external compressing force is removed b the difference between the final momentum and the

Fed [463]

Answer:

the correct one is b

the difference between the final moment and the initial moment

Explanation:

The momentum is related to the moment

I = ΔP

∫ F dt = p_f - p₀

where p_f and p₀ are the final and initial moments, respectively

When checking the different answers, the correct one is b

the difference between the final moment and the initial moment

7 0

2 years ago

The two major parts of the optical telescope

nydimaria [60]

The "objective" (lens or mirror) is the major major major part of
the optical telescope. It's really the only part you need in order
to make a telescope (besides something to hold the objective).
You can put a piece of film or a CCD right at the focal point of
the objective lens or mirror and capture 'images' (pictures) there.

If you want to use the telescope for looking through and seeing stuff
with your eye, then you need the other major part ... the eyepiece lens.

5 0

2 years ago