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3 years ago

What is the leading hypothesis for the origin of short gamma-ray bursts?

Physics

1 answer:

Arturiano [62]3 years ago

6 0

Answer:

Gamma rays are the most energetic form of rays. A gamma ray burst can emit a large amount of energy in a very short interval of time. In fact, a short gamma-ray burst occurs in less than 2 seconds. A leading hypothesis states that when two neutron stars merge or when a neutron star merges with a black hole, short gamma-ray burst originates. Such merger are expected to produce kilonovoe.

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What is electricity

irina [24]

D I know this is right

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2 years ago

Which use of iron is due to its chemical properties?

AfilCa [17]

Colored sparks and rusting.

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2 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

In a science fiction story, a microscopic black hole is given an enormous positive charge by firing an un-neutralized ion drive

Olegator [25]

Answer: distance d = 4.73e10m

Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.

Using electric potential V formula:

V = kq / d

Where K = 9.05×10^9Nm^2/C

And e = 1.6×10^-19C

But you don't need to substitute it.

1090 V = 8.99e9N·m²/C² * 5740C /d

Make d the subject of formula

d = 4.73e10 m

6 0

3 years ago

How much kinetic energy does a 50 kg object have if its moving at a velocity of 2 m/s?

bulgar [2K]

Answer:

C) 100 joules

Explanation:

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where m is the mass of the object and v its speed.

In this problem, we have an object of mass m = 50 kg and v = 2 m/s, so by using the formula we can find its kinetic energy:

$K=\frac{1}{2}(50 kg)(2 m/s)^2=100 J$

3 0

2 years ago