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2 years ago

How can isotopes of a given element be identified?

1 answer:

6 0

Answer: All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom. ... Each atomic number identifies a specific element, but not the isotope; an atom of a given element may have a wide range in its number of neutrons.

Explanation: Hopefully this helps. Have a great day.

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What happens inside someone's body when they danve

LUCKY_DIMON [66]

The body releases endorphins

6 0

2 years ago

Which of these hazmat products are allowed in your FC?

aliina [53]

Answer: Hazmat products are allowed in your FC are:

A GPS unit (lithium batteries)
A subwoofer (magnetized materials)

Explanation:

Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.

Hazardous material allowed in FC are as follows.

Magnetized material products like as speakers.
Non-spillable battery products like toy cars.
Lithium-ion battery containing products like laptops, mobile phones etc.
Non-flammable aerosol.

So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).

Thus, we can conclude that hazmat products are allowed in your FC are:

A GPS unit (lithium batteries)
A subwoofer (magnetized materials)

7 0

2 years ago

Metric conversions

sergij07 [2.7K]

1:-

1L=1000mL

$\\ \sf\longmapsto 45.23L$

$\\ \sf\longmapsto 45.23\times 1000$

$\\ \sf\longmapsto 45230mL$

2:-

1hL=10000cL

$\\ \sf\longmapsto .035hL$

$\\ \sf\longmapsto 0.035\times 10000$

$\\ \sf\longmapsto 350cL$

3:-

1mm=0.0001m

$\\ \sf\longmapsto 27.32$

$\\ \sf\longmapsto 27.32\times 0.0001$

$\\ \sf\longmapsto 0.0027m$

7 0

3 years ago

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

3 years ago

I NEED HELP!!!!!!!!!!!

nadezda [96]

1. 168.1 Hz

To find the apparent frequency heard by the driver in the car, we can use the formula for the Doppler effect:

$f'=(\frac{v\pm v_o}{v\pm v_s})f$

where

f is the original sound of the horn

v is the speed of sound

$v_o$ is the velocity of the observer (the driver and the car), which is positive if the observer is moving towards the source and negative if it is moving away