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Marina CMI [18]

2 years ago

6

Some gamma ray bursts are hypothesized to come from mergers of neutron stars or black holes. if this hypothesis is correct, what

else should we in principle be able to detect from such mergers?

1 answer:

Umnica [9.8K]2 years ago

6 0

We should see (and have now detected with LIGO) gravitational waves

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If a train travels 500 kilometers from Stockholm to

o-na [289]

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

$\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}$

Therefore, the average speed is 100 km/h. Please let me know if you have any questions!

4 0

2 months ago

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

$\Sigma \tau = 0$

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
$\tau = r \times F$

Doing the summation using their respective lever arms:

$0 = L Tsin\theta - dF_g$

$dF_g = LTsin\theta$

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

$tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o$

Now, let's solve for 'T'.

$T = \frac{dMg}{Lsin\theta}$

Plugging in the values:
$T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}$

3 0

11 months ago

2. An alternating current is represented by the equation I=20sin 100mt.

Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

$I=I_o sin\omega t$

Where

I₀ is the peak value of current

$\omega$ is angular frequency

$As\ \omega=2\pi f$

So,

$f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{100\pi}{2\pi}\\\\f=50\ Hz$

We know that,

$I_{rms}=\dfrac{I_0}{\sqrt{2}}\\\\=\dfrac{20}{\sqrt{2}}\\\\I_{rms}=14.14\ A$

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.

3 0

2 years ago

On a hot summer day several swimmers decide to jump into a river. They step off the bridge and strike the water at 1.5s later. I

xxTIMURxx [149]

That is right because base*length is how you get the height

6 0

2 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

2 years ago