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3 years ago

8

The tent meteorite, found in 1897 near cape York, on the west coast of Greenland, is the largest meteorite exhibited by any muse

um. It has a mass of 30 883kg. How much is the mass of this meteorite in milligrams? Show work please.

1 answer:

sweet-ann [11.9K]3 years ago

6 0

Answer:
3.0883 x 10^10mg

Explanation:
1 kilogram = 1000 000 milligrams
So, 30 883 x 1000 000 = 30 883 000 000mg

You might be interested in

How dose an atom change if all of its electrons are removed

cluponka [151]

Answer:

The atom becomes a positively charged ion.

Explanation:

The building blocks of an atom are protons, neutrons, and electrons.
The protons and neutrons present in the core of the atom are called nucleus.
The electrons are scattered in an ordered way around the nucleus.
The protons are positively charged and the electrons are negatively charged particles. The neutrons do not possess any charges.
Binding energy is supplied to the atom to remove an electron.
It is possible to remove the electrons of the lighter elements.
When an electron is removed from the hydrogen atom. It becomes positively charged ion or simply proton.
When all of the electrons are removed from the helium atom, it becomes a positively charged α particle.
It is practically very difficult to remove all of the electrons from the heavier elements.
When all of the electrons are removed from an atom it becomes an unstable positively charged ion.

7 0

3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0

3 years ago

You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type

alexdok [17]

Answer:

$30 N \cdot m$

Explanation:

The torque applied by a force can be calculated as

$\tau = F d sin \theta$

where

F is the magnitude of the force

d is the length of the arm

$\theta$ is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

d = 2.0 m

$\theta=90^{\circ}$

Substituting into the equation, we find

$\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m$

7 0

3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

so

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago