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3 years ago

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Where can you find the neutrons of an atom? in the nucleus with the protons orbiting the nucleus In the nucleus with the electro

ns this is not part of an atom
ANSWER PLEASE ITS A EXAM 10 POINTS PLEASE FASTTTTTTTTTTT!!!!!!!1

1 answer:

cricket20 [7]3 years ago

3 0

Answer:

Nuclease is the answer I know

I hope this is the answer

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The sound from a single source can reach point O by two different paths. One path is 20.0 m long and the second path is 21.0 m l

aleksandrvk [35]

Answer:

minimum frequency = 170 Hz

Explanation:

given data

One path long = 20 m

second path long = 21 m

speed of sound = 340 m/s

solution

we get here destructive phase that is path difference of minimum $\frac{\lambda}{2}$

here λ is the wavelength of the wave

so path difference will be

21 - 20 = $\frac{\lambda}{2}$

λ = 2 m

and

velocity that is express as

velocity = frequency × wavelength .............1

frequency = $\frac{340}{2}$

minimum frequency = 170 Hz

7 0

3 years ago

How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?

lbvjy [14]

During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
$Q=mL_e$
where
m is the mass of the water
$L_e$ is the latent heat of evaporation.

For water, the latent heat of evaporation is $L=2257 kJ/kg$ , while the mass of the water is
$m=1234 g=1.234 kg$

so, the amount of heat released in the process is
$Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ$

6 0

3 years ago

Protons and neutrons grouped in a specific pattern

alexgriva [62]

Answer b protons and electrons

5 0

3 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

a

$\theta _2 = 13^o$

b

$\theta _1 =32.94^o$

c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

3 years ago

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago