Classical physics is a good approximation to modern physics under certain circumstances. What are they?

Answers

Answer 1

Answer:

Answer and Explanation:

In classical physics we study about macroscopic objects and in modern physics we study about microscopic objects but under some circumstances classical physics is a approximation to modern physics such as

The object must be microscopic but they should be large enough so that we can see that through a microscope
The speed of the object should be very less it should be about 1 % of the speed of the light
They should be involved with very week gravitational field

Answer 2

Answer:

Answer:

Classical physics is a good approximation to modern physics under these certain circumstances: Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields, such as the field generated by the Earth, can be involved.

Explanation:

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Which of the Following is NOT a layer of the atmosphere?A) Mesosphere
B) Thermosphere
C) Exosphere
D) Lithosphere

Answers

lithosphere is not a layer of the atmosphere

The lithosphere is not a layer of the atmosphere. It is a layer of the earth-in the mantle

The volume occupied by three milliliters of water is the same as three cubic centimeters of water

Answers

The volume occupied by three milliliters of water is the same as three cubic centimeters of water is true. The conversion is as follows.

3mL (1 L/1000 mL) (1 m3/1000 L) (100 cm/1 m)^3 = 3 cm3

Think about multiplying the mass of each student by a factor to calculate each student’s kinetic energy. Is there a common factor that works for every student? If so, what’s this factor?

Answers

Answer:

KE= (1/2)mv^2

given m, the remaining components of the equation are=

1/2v^2?

Explanation:

KE= (1/2)mv^2

plug in...

=(1/2)(27 kg)(3 m/s)^2= 121.5

KE= 121.5 J

Answer:

Multiplying the mass of any student by a factor of 4.5 gives the kinetic energy of the student.

A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the table of height h2, landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h1 is increased, h2 must decrease by the same amount so that the sum h1+h2 remains equal to H. The student wants to adjust h1 and h2 to make d as large as possible.A) 1) Without using equations, explain why making h1 very small would cause d to be small, even though

h2 would be very large?
2) Without using equations, explain why making h2 very small would cause d to be small, even though
h1 would be large

B) Derive an equation for d in terms of h1, h2, m, and physical constants as appropriate.

Answers

(A)

(1) The reason for making $h_(1)$ very small is due to smaller value of horizontal component of launch velocity.

(2) The reason for making $h_(2)$ very small is due to smaller value of time of flight.

(B) The distance (d) covered by the block is $2\sqrt{h_(1)h_(2)}$ .

Given data:

The mass of block is, m.

The height of table from the top of slide is, $h_(1)$ .

The height of table at the end of slide is, $h_(2)$ .

The height of room is, H.

(A)

(1)

If the launch velocity of the block is v, then its horizontal component is very small, due to which adjusting the height $h_(1)$ to be very small will cause the d to be small.

(2)

The height $h_(2)$ is dependent on the time of flight, and since the time of flight taken by the block to get to the floor is very less, therefore the block will not get sufficient time to accomplish its horizontal motion. That is why making $h_(2)$ very small will cause d to be smaller.

(B)

The expression for the distance covered by the block is,

$v=(d)/(t)\nd = v * t$ ..............................(1)

Here, v is the launch speed of block and t is the time of flight.

The launch speed is,

$v^(2)=u^2+2gh_(1)\nv=\sqrt{u^2+2gh_(1)}\nv=\sqrt{0^2+2gh_(1)}\nv=\sqrt{2gh_(1)}$

And the time of flight is,

$h_(2)=ut+(1)/(2)gt^(2)\nh_(2)=0 * t+(1)/(2)gt^(2)\nh_(2)=0+(1)/(2)gt^(2)\nt=\sqrt(2h_(2))/(g)$

Substituting the values in equation (1) as,

$d = v * t\nd = \sqrt{2gh_(1)}* \sqrt(2 h_(2))/(g)}\nd=2\sqrt{h_(1)h_(2)}$

Thus, the distance (d) covered by the block is $2\sqrt{h_(1)h_(2)}$ .

Learn more about the time of flight here:

brainly.com/question/17054420?referrer=searchResults

A1) The reason why making h₁ very small would cause d to be small is; Because the horizontal component of the launch velocity would be very small.

A2) The reason why making h₂very small would cause d to be small is;

Because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) The equation for d in terms of h₁ and h₂ is;

d = 2√(h₁ × h₂)

A) 1) The reason why making h₁ very small would cause d to be small is because the horizontal component of the launch velocity would be very small.

A) 2) The reason why making h₂ very small would cause d to be small is because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) Formula for Launch Velocity is;

V = √(2gh₁)

h₁ was used because the top of the slide from where the student released the block has a height of h₁.

Also, the time it takes to fall which is time of flight is given by the formula;

t = √(2h₂/g)

h₂ was used because the height of the table the object is on before falling is h₂.

Now, we know that d is distance from edge of the table and formula for distance with respect to speed and time is;

distance = speed × time

Thus;

d = √(2gh₁) × √(2h₂/g)

g will cancel out and this simplifies to give;

d = 2√(h₁ × h₂)

Read more at; brainly.com/question/20427663

a 64kg skateboarder on a 2.0kg skateboard is on top of a ramp with a vertical height of 5.0 m what is the skateboarders maximum kinetic energy at the bottom of the ramp

Answers

The maximum kinetic energy of the skateboarders at the bottom of the ramp is 3234 J.

What is kinetic energy?

Kinetic energy is the energy of a body due to motion.

To calculate the maximum kinetic energy of the skateboarders at the bottom of the ramp, we use the formula below.

Formula:

K.E = Mgh............ Equation 1

Where:

M = mass of the skateboarder and the skateboard

K.E = maximum kinetic energy of the skateboarders at the bottom of the ramp

h = Height of the ramp
g = acceleration due to gravity.

From the question,

Given:

M = (2+64)M = 66 kgh = 5 mg = 9.8 m/s²

Substitute these values into equation 2

K.E = 66×5×9.8K.E = 3234 J

Hence, the maximum kinetic energy of the skateboarders at the bottom of the ramp is 3234 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

The formula for a kinetic energy KE of a falling body is
KE = mgh
where m = mass, g = acceleration due to gravity (9.8 m/s^2, constant), h = height.

The total mass of a skateboader and a skateboard is 64 + 2.0 = 66 kg.

Finally,
KE = 66*9.8*5.0 = 32340 J

If you have a vector which is 12 Newtons at 75 degrees south of west, what other vector is equivalent to this vector? a) 12 N at 15 degrees west of south b) 12 N at 15 degrees south of east c) 12 N at 15 degrees north of east d) 12 N at 15 degrees south of west

Answers

To find the vector that is equivalent to a vector of 12 Newtons at 75 degrees south of west, we need to determine the direction and magnitude of the equivalent vector.

The equivalent vector will have the same magnitude of 12 Newtons but a direction that is 180 degrees opposite to the given vector.

The given vector is 75 degrees south of west, which means it is pointing towards the south but slightly to the west.

Therefore, the direction of the equivalent vector will be 75 degrees north of east (opposite direction).

Among the given options:

a) 12 N at 15 degrees west of south - This is not the correct direction.

b) 12 N at 15 degrees south of east - This is not the correct direction.

c) 12 N at 15 degrees north of east - This is the correct direction.

d) 12 N at 15 degrees south of west - This is not the correct direction.

Therefore, the vector that is equivalent to the given vector is option c) 12 N at 15 degrees north of east.

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