PHYSICS
MIDDLE SCHOOL

Answer: In my opinion, calling exercise "work" wouldn't be a good way of describing it, I don't think people would exercise if its called work

HIGH SCHOOL

How long will it take to travel 200 km traveling 100 m/s?

Convert: 1km = 1000 m 200000 m * 0.01 s/m = 2000 s or 33 minutes and 20 seconds.

COLLEGE

A 24.0N picture hangs from a nail in the wall. The diameter of the nail is 2.00 mm. The shear stress on the nail is

Weight tending to shear the pin is

W = 24.0 N

The shear area of the pin is

A = (π/4)*(2.0 x 10⁻³ m)² = 3.1416 x 10⁻⁶ m².

The shear stress is

τ = (24.0 N)/(3.1416 x 10⁻⁶ m²)

= 7.64 x 10⁶ Pa

= 7.64 MPa

Answer: 7.64 Mpa

W = 24.0 N

The shear area of the pin is

A = (π/4)*(2.0 x 10⁻³ m)² = 3.1416 x 10⁻⁶ m².

The shear stress is

τ = (24.0 N)/(3.1416 x 10⁻⁶ m²)

= 7.64 x 10⁶ Pa

= 7.64 MPa

Answer: 7.64 Mpa

COLLEGE

Which of the following scenarios would not decrease the diversity within that environment? (A) A hurricane hits an island hard and wipes out a large number of animals only found on that island (B)Climate change causes the temperature of an ecosystem to drop lower than what many species in that environment can tolerate (C) Trout are released into a lake that already contains trout to increase the population size (D) An invasive species of fish is released into an estuary to control a pest species

Answer:

(C) Trout are released into a lake that already contains trout to increase the population size

Explanation:

Introduction of new trouts into a lake with an already established trout population will not decrease the diversity within that environment.

Biodiversity is the variation of life forms in a particular environment or study area. The more the number of different species, the high its diversity.

- Natural disasters like volcanic eruptions, hurricane, flooding are harmful to life and reduces diversity.
- Climate change and introduction of invasive species are not diversity friendly measures.

HIGH SCHOOL

A student uses an audio oscillator of adjustable frequency to measure the depth of a water well. The student reports hearing two successive resonances at 53.95 Hz and 59.00 Hz. (Assume that the speed of sound in air is 343 m/s.) (a) How deep is the well?

(b) How many antinodes are in the standing wave at 53.95 Hz?

Answer:

a) L = 33.369 m , b) 21

Explanation:

The analysis of the ocean depth can be performed assuming that at the bottom of the ocean there is a node and the surface must have a belly, so the expression for resonance is

λ = 4 L / n

n = 1, 3, 5, ...

The speed of the wave is

v = λ f

v = 4L / n f

L = n v / 4f

Let's write the expression for the two frequencies

L = n₁ 343/4 53.95

L = n₁ 1,589

L = n₂ 343/4 59

L = n₂ 1.4539

Let's solve the two equations

n₁ 1,589 = n₂ 1,459

n₁ / n₂ = 1.4539 / 1.589

n₁ / n2 = 0.91498

Since the two frequencies are very close the whole numbers must be of consecutive resonances, let's test what values give this value

n₁ n₂ n₁ / n₂

1 3 0.3

3 5 0.6

5 7 0.7

7 9 0.77

9 11 0.8

17 19 0.89

19 21 0.905

21 23 0.913

23 25 0.92

Therefore the relation of the nodes is n₁ = 21 and n₂ = 23

Let's calculate

L = n₁ 1,589

L = 21 1,589

L = 33.369 m

b) the number of node and nodes is equal therefore there are 21 antinode

COLLEGE

Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.59 cm. Determine the magnitude of the force that each sphere now experiences.

Answer:

(a):

(b):

Explanation:

Given:

- Charge on one sphere,

- Charge on second sphere,
- Separation between the spheres,

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

where,

k is called the Coulomb's constant, whose value is

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres,

The new charges on both the spheres are equal and given by

The magnitude of the force that each sphere now experiences is given by

COLLEGE

A batter hits a softball over a third baseman's head with speed v0 and at an angle ?from the horizontal. Immediately after the ball is hit, the third baseman turns and runs at a constant velocity v=7.000m/s, for a time t=2.000s. He then catches the ball at the same height at which it left the bat. The third baseman was initiallyl=18.00m from home plate (the location where the ball was hit from). a) Find v0. Use g=9.807m/s2 for the magnitude of the acceleration due to gravity. Assume that there is no air resistance.

b) Find the angle ? in degrees.

c)Find the components vxand vy of the ball’s velocity, v, 0.100 s before the ball is caught.

d)Find the vector components x and y of the ball’s position, r, 0.100 s before the ball is caught.

Answer:

a) The magnitude of the initial velocity is 18.77 m/s.

b) The launching angle is 31.51°.

c) The horizontal component of the velocity at t = 1.900 s is 16.00 m/s.

The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.

d) The horizontal component of the position vector at time t = 1.900 s is 30.40 m.

The vertical component of the position vector at time t = 1.900 s is 0.9375 m

Explanation:

Hi there!

The equations for the velocity and position vector of the ball are the following:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity.

v = velocity vector at time t.

a and b) First, let´s find the range of the ball, i.e. the horizontal distance traveled by the ball.

The distance traveled by the baseman can be calculated with this equation:

x = v · t

Where:

x =traveled distance.

v = velocity.

t = time

Then:

x = 7.000 m/s · 2.000 s

x = 14.00 m

The baseman runs 14.00 m. Since he was located 18.00 from the home plate, the horizontal distance traveled by the ball is (14.00 m + 18.00 m) 32.00 m.

If we locate the origin of the frame of reference at the point where the ball is hit, the initial vertical and horizontal positions (x0 and y0) are zero. Since the ball is caught at the same height at which it left the bat, the vertical position of the ball when it is caught is 0.

So, the position vector of the ball at the time when it is caught (2 s after it is hit), is the following:

r = (32.00 m, 0 m)

Using the equations of the x- and y-components of the position vector, we can obtain the initial velocity and the angle:

rx = x0 + v0 · t · cos α (x0 = 0)

ry = y0 + v0 · t · sin α + 1/2 · g · t² (y0 = 0)

rx = 32.00 m = v0 · 2.000 s · cos α

ry = 0 m = v0 · 2.000 s · sin α - 1/2 · 9.807 m/s² · (2.000 s)²

Solving the first equation for v0:

16.00 m/s / cos α = v0

And replacing v0 in the second equation:

0 m = 32 m · sin α / cos α - 1/2 · 9.807 m/s² · (2.000 s)²

1/2 · 9.807 m/s² · (2.000 s)² = 32 m · tan α

1/2 · 9.807 m/s² · (2.000 s)² / 32 m = tan α

α = 31.51°

b) The launching angle is 31.51°

The initial velocity will be:

16.00 m/s / cos α = v0

16.00 m/s / cos (31.51°) = v0

v0 = 18.77 m/s

a) The magnitude of the initial velocity is 18.77 m/s.

c) Let´s use the equation of the velocity vector:

v = (v0 · cos α, v0 · sin α + g · t)

vx = v0 · cos α

vy = v0 · sin α + g · t

The horizontal component of the velocity does not depend on time (neglecting air resistance).

Then:

vx = 18.77 m/s · cos (31.51°)

vx = 16.00 m/s

0.100 s before the ball is caught, the horizontal component of the velocity is 16.00 m/s.

Now let´s calculate the vertical component of the velocity:

vy = 18.77 m/s · sin (31.51°) - 9.807 m/s² · 1.900 s

vy = -8.823 m/s

The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.

d) Let´s use the same equations we have used in part a).

x = x0 + v0 · t · cos α

x = 18.77 m/s · 1.900 s · cos (31.51°)

x = 30.40 m

The horizontal component of the position vector at time t = 1.900 s is 30.40 m

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 18.77 m/s · 1.900 s · sin (31.51°) - 1/2 · 9.807 m/s² · (1.900 s)²

y = 0.9375 m

The vertical component of the position vector at time t = 1.900 s is 0.9375 m

HIGH SCHOOL

With which field of science is Albert Einstein associated? biology chemistry medecine physics

Answer

Albert Einstein associated with physics field of science .

Albert Einstein is a physicist, his best known theory are

(1 )Theory of relativity and the equation i.e E = MC²

(2) Photo electric effect

For his discovery of the law of the photoelectric effect Albert einstein was awarded by Nobel Prize in Physics for his work in theoretical physics,

He was a theoretical physicist

MIDDLE SCHOOL

In line 19 inertness most nearly means

Answer:

Inertness refers to the non-reactive property of a substance. In chemistry, a substance is said to be inert if it does not react with another substance. For example, noble gases like Helium and Neon are inert gases.

In physics, inertia refers to the property of a body by virtue of its mass where in it does not change its state of motion unless acted upon by an external force.

It most nearly refers to non-activity.

If you provided the aforementioned line 19, that would kinda be great.

COLLEGE

Alcohols are organic compounds that contain a. carbon and oxygen only. c. carbon, oxygen, and hydrogen. b. carbon and hydrogen only. d. carbon, nitrogen, and hydrogen.

Answer: Option (c) is the correct answer.

Explanation:

Compounds which contain carbon and hydrogen atoms are known as organic compounds.

As general chemical formula of alcohols is R-OH, where R = any alkyl or aryl group.

For example, is known as ethanol and it is an alcohol as it contains the functional group "-OH".

So, alcohols contains elements carbon, hydrogen and oxygen.

Thus, we can conclude that alcohols are organic compounds that contain carbon, oxygen, and hydrogen.

C

I think that it is carbon, oxygen, and hydrogen

I think that it is carbon, oxygen, and hydrogen