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pychu [463]
3 years ago
8

When a front wheel drops off the roadway you should?

Physics
1 answer:
nikdorinn [45]3 years ago
6 0
<span>braking and returning suddenly to the roadway</span>
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a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²​
tigry1 [53]

At the player's maximum height, their velocity is 0. Recall that

{v_f}^2-{v_i}^2=2a\Delta y

which tells us the player's initial velocity v_i is

0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}

The player's height at time t is given by

y=v_it-\dfrac g2t^2

so we find their airtime to be

0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s

6 0
3 years ago
In some cases fixture wires may be used for
zalisa [80]

You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.

<h3>What are some uses of fixture wires?</h3>

Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG

In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

To know more about fixture wires, refer

brainly.com/question/26098282

#SPJ4

3 0
1 year ago
One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the t
natka813 [3]

This question involves the concepts of density, volume, and mass.

The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".

<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>

It is given that:

Mass of one mole = 24 grams

Mass of 6 x 10²³ atoms = 24 grams

Mass of 1 atom = \frac{24\ grams}{6\ x\ 10^{23}\ atoms} = 4 x 10⁻²³ grams

<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

\rho = \frac{m}{V}\\\\V=\frac{m}{\rho}

where,

  • \rho = density = 1.7 grams/cm³
  • m = mass of single atom = 4 x 10⁻²³ grams
  • V = volume of single atom = ?

Therefore,

V=\frac{4\ x\ 10^{-23}\ grams}{1.7\ grams/cm^3}

V = 2.35 x 10⁻²³ cm³

<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>

The atom is in a spherical shape. Hence, its Volume can be given as follows:

V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}

d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m

Learn more about density here:

brainly.com/question/952755

7 0
2 years ago
1) if you could eliminate one item from your diet to improve your heath, what would it be?
Makovka662 [10]

Answer:

Rice

Explanation:

Because I can't control eating lots of rice

3 0
3 years ago
Read 2 more answers
Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul
Yuki888 [10]

Answer:

63.4^{\circ}

Explanation:

When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so

I_1 = \frac{1}{2}I_0 (1)

where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.

When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle \theta with respect to the first one, the intensity of the light coming out is

I_2 = I_1 cos^2 \theta (2)

If we combine (1) and (2) together,

I_2 = \frac{1}{2}I_0 cos^2 \theta (3)

We want the final intensity to be 1/10 the initial intensity, so

I_2 = \frac{1}{10}I_0

So we can rewrite (3) as

\frac{1}{10}I_0 =  \frac{1}{2}I_0 cos^2 \theta

From which we find

cos^2 \theta = \frac{1}{5}

cos \theta = \frac{1}{\sqrt{5}}

\theta=cos^{-1}(\frac{1}{\sqrt{5}})=63.4^{\circ}

6 0
3 years ago
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