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2 years ago

15

If the Sun were to mysteriously turn into a black hole (don't worry . . . it won't!) and retain its current mass, what would hap

pen to Earth?

1 answer:

kap26 [50]2 years ago

8 0

The Earth would continue to orbit it
as usual.(But it would be very cold and dark around here.)

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GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

$v^2-u^2=2ah$ , h is the maximum height and a = -g

$0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m$

Hence, it will go up to a height of 1.8 m.

4 0

2 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

2 years ago

A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti

ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = $6.31\times 10^{4} T$

Solution:

To find the number of turns, N, we use the given formula:

$B = \frac{\mu_{o}NI}{2R}$

Therefore,

$N = \frac{2BR}{\mu_{o}I}$

$N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}$

N = 22.74 = 23 turns (approx)

8 0

2 years ago

Which of the following is the best example of uniform circular motion? A. A child riding on a constantly spinning merry-go-round

Ulleksa [173]

A. because as the merry-go-round spins the child accelerates towards the center of the merry-go-round at a uniform rate.

5 0

2 years ago

The speed of light is 3.0 x 108 m/s. How far does light travel in one day? Remember that you should first convert the day into s

Anna35 [415]

Answer:

$2.59\times 10^{13}\ m$

Explanation:

The speed of light is 3.0 x 10⁸ m/s

We need to find the distance traveled by it in one day.

1 day = 24 hours

1 day = 24 × 3600 seconds

= 86400 s

Now, using the formula of speed to find the distance covered as follows :

$d=v\times t$

Putting all the values, we get :

$d=3\times 10^8\ m/s\times 86400\ s\\\\d=2.59\times 10^{13}\ m$

Hence, light will cover $2.59\times 10^{13}\ m$ of distance in one day.

6 0

2 years ago