Average Amplitude of Waves

Why does the sample on a microscope slide need to be very thin?

ziro4ka [17]

B. so light can shine through it from below.

4 0

2 years ago

Read 2 more answers

What 2 environments once existed where the Grand Canyon is located today?

patriot [66]

The entire park area is considered to be a semi-arid desert, but distinct habitats are located at different elevations along the 8,000-foot elevation gradient. Near the Colorado River, riparian vegetation and sandy beaches prevail.

5 0

2 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

2 years ago

What is E(r)E(r)E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the dist

andriy [413]

Answer:

E(r) = λ/2πrε0

Explanation:

If we consider an infinitely long line of charge with the charge per unit length being λ, we can take advantage of the cylindrical symmetry of this situation.

By symmetry, i mean that the electric fields all point radially away from the line of charge and thus there is no component parallel to the line of charge.

Niw, let's use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface.

Doing that will mean that the electric field would be perpendicular to the curved surface of the cylinder. Therefore, the angle between the electric field and area vector is equal to zero and cos θ = cos 0 = 1

Now, the top and bottom surfaces of the cylinder will lie parallel to the electric field. Therefore, the angle between the area vector and the electric field would be 90° and cos θ = cos 90 = 0

Now, we know that according to Gauss Law,

Electric Flux, Φ = E•dA

Thus,

Total Φ = Φ_curved + Φ_top + Φ_bottom

Thus,

Φ = ∫E•dA cos 0 + ∫E•dA cos 90° + ∫E•dA cos 90°

We now have ;

Φ = ∫E . dA × 1

Since we are dealing with the radial component, the curved surface would be equidistant from the line of charge and the electric field in the surface will be the same magnitude throughout.

Thus,

Φ = ∫E•dA = E∫dA = E•2πrl

The net charge enclosed by the surface is given by:

q_net = λl

So using gauss theorem, we have;

Φ = E•2πrl = q_net/εo = λl/εo

E•2πrl = λl/ε0

Making E the subject, we obtain ;

E = λ/2πrε0

4 0

2 years ago

Give two examples of vernier calliper

omeli [17]

Answer:

Example 1, if a vernier caliper output a measurement reading of 2.13 cm, this means that: The main scale contributes the main number(s) and one decimal place to the reading

E.g. 2. 1 cm, whereby 2 is the main number and 0.1 is the one decimal place number

Explanation:

plz mark as brainliest and hope it helps you

4 0

2 years ago