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3 years ago

Can someone please explain the quantum theory in simple terms

1 answer:

6 0

Quantum theory is the theoretical basis of modern physics that explains the nature and behavior of matter and energy on the atomic and subatomic level. The nature and behavior of matter and energy at that level is sometimes referred to as quantum physics and quantum mechanics.

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Distance meters Time (seconds) Trial Tria 2 Trials Average 3 053 059 0.52 055 6 106 719 104 Uning the values trom the data table

kumpel [21]

Jsisisusuagahannananana this is worth 22 points g that’s crazy 25

5 0

2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0

3 years ago

types of plate boundaries where two plates separate or move apart?? does anyone know please help ??????

Shtirlitz [24]

Erosion i belive it is called

7 0

3 years ago

Read 2 more answers

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$