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3 years ago

What is the basic atomic difference between isotopes of the same element?

1 answer:

4 0

The atomic mass. By taking two different types of the same element and calculating the average based on the amount of each type in the world, you get the basic atomic difference.

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23. How does the microwave appliance work?

kaheart [24]

Answer: The microwaves are reflected within the metal interior of the oven where they are absorbed by food. Microwaves cause water molecules in food to vibrate, producing heat that cooks the food.

Explanation:

7 0

3 years ago

A race car was is moving at a constant speed of 35 m/s. A security car was

Andrej [43]

Answer:

Final speed of security car v = 65 m/s

Explanation:

Given:

Speed of race car u1 = 35m/s

Speed of security car u2 = 5 m/s

Acceleration = 5 m/s²

Find:

Final speed of security car v

Computation:

Assume, they chase S meter

S = u1t + [1/2]at²

S = 35t

S = u2t + [1/2]at²

so,

35t = 5t + [1/2](5)t²

t = 12 s

v = u + at

v = 5 + 5(12)

Final speed of security car v = 65 m/s

7 0

3 years ago

Ender and Shen are flying at each other during a battle in space. Ender moves with a velocity v1 of 12 m/s and Shen, with a mass

Shalnov [3]

Answer:

Ender's mass = 2.25 kg

Explanation:

using law of conservation of momentum .

since there is inelastic collision

given.

$let \ \ m_{1} = m \\m_{2} = 45 kg \\v_{1} = 12 m/s \\v_{2} = 9 m/s \\\\V_{3} =6.4 m/s\\using \ \ \\m_{1}v_{1} + m_{2}v_{2} = (m_{1} +m_{2})V\\m\times12 + 45\times9 = (m+45)\times6.4\\64m -12m = 405 -288\\52m =117\\m =\frac{117}{52}\\m =2.25kg$

6 0

3 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

At what phase or phases could water exist at 4.579mm pressure and 0.0098°C?

zimovet [89]

The correct answer is A. The water at this conditions is in the solid phase. This can be verified by using a phase diagram. A phase diagram is a figure which shows the phases of a certain substance at a specific temperature and pressure.

7 0

4 years ago