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2 years ago

Sodium-24 has a half-life of 15 hours. After 45 hours, how much sodium-24 will remain of an original 50.0-g sample?

Physics

2 answers:

Ivahew [28]2 years ago

6 0

Answer:

6.25 g

Explanation:

Svetlanka [38]2 years ago

3 0

45 hours/15 hours = 3 Half lifes
50/2 = 25
25/2 = 12.5
12.5/2 = 6.25

6.25g is the answer

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What is the inverse square law and how does it relate to gravity?

Nesterboy [21]

Answer:

Inverse Square Law Newton proposed the Inverse Square Law. The effect of gravity (and also on forces such as sunlight) works like this. If say we have a half-mass Earth, it would produce a gravity of not half but a quarter (the square of 2).

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3 years ago

The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co

motikmotik

Explanation:

The position vector r:

$\overrightarrow{r(t)}=lcos\theta\hat{i}+lsin\theta\hat{j}$

The velocity vector v:

$\overrightarrow{v(t)}=\overrightarrow{\frac{dr}{dt}}=\dot{l}cos\theta-lsin\theta\dot{\theta}\hat{i}+\dot{l}sin\theta+lcos\theta\dot{\theta}\hat{j}$

The acceleration vector a:

$\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}$

$\overrightarrow{v(t)}=0.13\hat{i}+0.18\hat{j}$

$\overrightarrow{a(t)}}=-0.3\hat{i}-0.1\hat{j}$

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2 years ago

Which of the following statements describes nucleons?

Yuliya22 [10]

The correct answer is D.

A nucleon is one of either of the two types of subatomic particles (neutrons and protons) which are located in the nucleus of atoms.

The total number of nucleon in the nucleus of an atom gives you an idea about the mass of that atom. In fact, one may refer mass number as nucleon number.

Simply put, nucleons are the particles that make nucleus of an atom and are held up together inside the nucleus due to nuclear force.

7 0

2 years ago

Read 2 more answers

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

2 years ago

An electron has a charge of 1.602 X 10-19.coulomb. When two electrons are separated by 1.2 X 10-9m, what force will they exert o

amid [387]

Answer:

The force they will exert on each other is 1.6*10⁻¹⁰ N

Explanation:

The electromagnetic force is the interaction that occurs between bodies that have an electric charge. When the charges are at rest, the interaction between them is called the electrostatic force. Depending on the sign of the interacting charges, the electrostatic force can be attractive or repulsive. The electrostatic interaction between charges of the same sign is repulsive, while the interaction between charges of the opposite sign is attractive.

Coulomb's law is used to calculate the electric force acting between two charges at rest. This force depends on the distance "r" between the electrons and the charge of both.

Coulomb's law is represented by:

$F=k*\frac{q1*q2}{r^{2} }$

where:

F = electric force of attraction or repulsion in Newtons (N). Like charges repel and opposite charges attract.
k = is the Coulomb constant or electrical constant of proportionality.
q = value of the electric charges measured in Coulomb (C).
r = distance that separates the charges and that is measured in meters (m).

In this case:

k= 9*10⁹ $\frac{N*m^{2} }{C^{2} }$
q1= 1.602*10⁻¹⁹ C
q2= 1.602*10⁻¹⁹ C
r= 1.2*10⁻⁹ m

Replacing:

$F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{1.602*10^{-19} C*1.602*10^{-19} C}{(1.2*10^{-9} )^{2} }$

and solving you get:

F=1.6*10⁻¹⁰ N

The force they will exert on each other is 1.6*10⁻¹⁰ N

3 0

2 years ago