Which element has 7 valence electrons?

A coin is dropped in a 15.0 m deep well.

labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0

3 years ago

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b. $v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

c.Work don=Final K.E-Initial K.E

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

4 0

3 years ago

Read 2 more answers

How many days are in a year on mars

Triss [41]

687 days
hope this helps
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brainliest is appreciated

7 0

3 years ago

A closed cylinder with a 0.15-m radius ends is in a uniform electric field of 300 n/c, perpendicular to the ends. the total flux

bixtya [17]

The total flux through the cylinder is zero.

In fact, the electric flux through a surface (for a uniform electric field) is given by:

$\Phi = E A \cos \theta$

where

E is the intensity of the electric field

A is the surface

$\theta$ is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.

We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because $\theta=90^{\circ}$ and $\cos \theta=0$ ).

On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

$\Phi_1 = E \pi r^2$

where r is the radius, and where we have taken $\theta=0^{\circ}$ since the perpendicular to the surface is parallel to the direction of the electric field, so $\cos \theta=1$ . On the second surface, however, the perpendicular to the surface is opposite to the electric field, so $\theta=180^{\circ}$ and $\cos \theta=-1$ , therefore the flux is