Can someone please help me with science.

NaOH + FeCl3* Na Cl + Fe 10H)3 balanced

zvonat [6]

3NaOH + FeCl3 → 3NaCl + Fe(OH)3

8 0

3 years ago

Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y

makkiz [27]

Answer:

<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:

1. Light waves are electromagnetic waves, while sound and water waves are mechanical waves, this is the first and principal difference.

2. Electromagnetic waves can propagate in vacuum (they do not need a medium or material), but mechanical waves obligatory need a material to propagate

3. Light waves are always transversal waves, this means the oscillatory movement is in a direction that is perpendicular to the propagation; but mechanical waves may be both: longitudinal waves (the oscillation occurs in the same direction as the propagation) or transversal waves.

4. Electromagnetic waves propagates at a constant velocity (Light velocity) while the velocity of mechanical waves will depend on the type of wave and the density of the medium or material.

5. Mechanical waves are characterized by the regular variation of a single magnitude, while electromagnetic waves are characterized by the variation of two magnitudes: the electric field and the magnetic field

6. Water waves are 2-dimensional waves, while the light and the sound are tridimensional spherical waves

7. Light waves transports energy in the form of radiation, while mechanical waves transport energy with material

3 0

3 years ago

A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul

zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

$\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}$

$\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}$

$\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}$

$\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }$

$\mathbf{\omega^2=\dfrac{39.24 }{2}}$

$\mathbf{\omega=\sqrt{19.62 } \ rad/sec}$

$\mathbf{\omega=4.429 \ rad/sec}$

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0

2 years ago

Read 2 more answers

Pls answer the 15a answer i cant understand it

Maksim231197 [3]

Materials required for the experiment of limiting force borne by string:-

String balance
weights
light strings
weight hanger
pan for spring balance
Sand

Steps of procedure for for the experiment of limiting force borne by string:-

First we have to tie a light string to the fixed support and then tie the other end with the weight hanger consists of weight.
Add additional weight to the hanger again and again. And continue the same until the string is broken.
Note down the weight (x) where the string is broken.
Suspend spring balance to a support.
Tie the light string at the end of the balance and at the other end suspend the pan for spring balance.
Now place the weights (x-100 grams) in pan.
Observe the reading in the spring balance.
Add a small amount of sand in the pan by observing the readings.
same is to be done till the string is broken.

Learn more about limiting force here:- brainly.com/question/11371672

#SPJ1

6 0

1 year ago

You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving

BabaBlast [244]

Y₀ = initial position of the balloon at the top of the building = 44 m

Y = final position of the balloon at halfway down the building = 44/2 = 22 m

a = acceleration of the balloon = - 9.8 m/s²

v₀ = initial velocity of the balloon = 0 m/s

v = final velocity of the balloon = ?

using the kinematics equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 0² + 2 (- 9.8) (22 - 44)

v = 20.78 m/s

4 0

4 years ago