All categories

3 years ago

Science fair project

Physics

2 answers:

victus00 [196]3 years ago

8 0

For Science fair project you can do convection current

Explanation:

Show how to do the process

and then make a model to demonstrate

Materials required:

Potassium Permanganate,beaker,bunsen burner,water and retort stand

kondaur [170]3 years ago

5 0

Answer:

Explanation:

You can make a projector using convex lens and a mobile phone back of it

For more details search in you. tubeyou will get videos based on this

You might be interested in

A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f

uysha [10]

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

$\Sigma F=ma$

Replacing the values:

$mg\sin 40=ma$

We may cancel out the mass:

$g\sin 40=a$

Using the gravity constant as 9.8 meters per square second:

$9.8\frac{m}{s^2}\sin 40=a$

Solving the operations:

$6.3\frac{m}{s^2}=a$

Therefore, the acceleration is 6.3 meters per square second.

8 0

1 year ago

When light behaves like a particle, it is called a ???

nirvana33 [79]

It is called a photon i believe

7 0

3 years ago

Read 2 more answers

If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe

RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

$P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}$

where,

P = Friction Loss per foot of pipe = $\frac{0.4}{1000\ ft}$ = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

$4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}$

5 0

3 years ago

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

Archy [21]

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4 ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a) final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5 ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a) final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀ ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a) final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

3 0

3 years ago

A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

$\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s$

5 0

3 years ago