What's an air column

The thylacine is an extinct marsupial that once roamed Tasmania and Australia. There are several preserved specimens in existenc

grin007 [14]

Answer:

B.

Explanation:

Its just right

4 0

4 years ago

What colors are included in our visible spectrum?

777dan777 [17]

ALL colors are in the visible spectrum.

"Color" is a word we use to describe light that our eyes can detect. If the light has a "color", that means we can see it, and somebody gave it a name. If we can't see it, then we don't have a "color" to describe it.

Every color that has ever been perceived by human eyes is present in the humanly-visible spectrum, except "black".

There are at least hundreds of them. Probably thousands. Maybe millions.

7 0

3 years ago

Proton traveling across a capacitor A proton enters with a velocity v between the plates of a capacitor as shown in the figure.

yulyashka [42]

Answer:

Please see the direction of force in the attachment

Explanation:

As the particle enters the capacitor, it experiences electric. In order to move in the same direction as the electron was moving before entering the capacitor, the force due to the electric field of capacitor must be balanced by the magnetic field.

The direction of particle is shown in the attachment.

As we know,

Force = q (v *B)

3 0

3 years ago

A block of mass m = 2.20 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.80

Bogdan [553]

Answer:

(a) The speed of the lighter block is $v_{2x} = 3.7~m/s$ .

The speed of the heavier block is $v_{3x} = 3.55~m/s$ .

(b) The smaller block goes up to 0.69 m.

Explanation:

We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.

In order to solve this question, I will assume that the smaller mass is initially at rest.

Part A:

We will use conservation of energy.

$K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv_1^2 + 0\\(2.2)(9.8)(3.6) = \frac{1}{2}(2.2)v_1^2\\v_1 = 8.4 ~m/s$

This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.

Part B:

Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.

$P_1 = P_2\\mv_{1x} = mv_{2x} + Mv_{3x}\\(2.2)(8.4\cos(30^\circ)) = (2.2)v_{2x} + (6.8)v_{3x}\\16 = 2.2v_{2x} + 6.8v_{3x}$

During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.

$K_{initial} = K_{final}\\\frac{1}{2}mv_1^2 = \frac{1}{2}mv_{2}^2 + \frac{1}{2}Mv_{3x}^2\\(2.2)(8.4)^2 = (2.2)v_{2}^2 + (6.8)v_{3x}^2\\155.23 = 2.2v_{2}^2 + 6.8v_{3x}^2$

The following relation will be used when combining the two equations:

$v_{2x} = v_2\cos{30^\circ}$

The following equation is useful for combining the two equations:

$v_{3x} = \frac{2m}{(m+M)}v_{1x} = \frac{2(2.2)}{(2.2 + 6.8)}(8.4\cos(30^\circ)) = 3.55~m/s$

Therefore from the first equation,

$16 = 2.2v_{2x} + 6.8v_{3x} = 2.2v_{2x} + 6.8(3.55) \\v_{2x} = -3.7~m/s$

Part C:

We will again use the conservation of energy to find the highest point that the mass can go:

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv_{2x}^2 + 0 = 0 + mgH\\\frac{1}{2}(2.2)(-3.7)^2 = (2.2)(9.8)H\\H = 0.69 ~m$

4 0

4 years ago

a large heavy truck and a baby carriage roll down a hill. Neglecting friction, at the bottom of the hill, the baby carriage will

Lisa [10]

Answer:

please give me brainlist and follow

Explanation:

At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.

5 0

3 years ago