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3 years ago

Which part of the atom is not found in the nucleus

1 answer:

7 0

The electrons are not found in the nucleus. Electrons are in constant movement in the electron cloud. The only sub-atomic particles in the nucleus are the protons and neutrons.

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A liter is the same as a cubic meter. <br> a. True<br> b. False

sammy [17]

It is false ...........

8 0

3 years ago

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During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient

ElenaW [278]

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

$g=9.8 m/s^2$

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

$W=F_f d$

where

$F_f$ is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

$K_f = U_i -E_t$

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

7 0

3 years ago

I need help in this pls i really need a answer

IrinaK [193]

Answer:

In terms of distance, average speed is 20 km/h

In terms of displacement, average speed is 0 km/h.

Explanation:

Total distance:

$= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\ = 30.0 + 10.0 + 20.0 \\ = 60.0 \: km$

Total time is 3.0 hours

but:

$average \: speed = \frac{total \: distance}{total \: time} \\$

In terms of distance.

substitute:

$average \: speed = \frac{60.0}{3.0} \\ \\ = 20 \: {kmh}^{ - 1}$

$displacement = ( - 30.0) + 10 .0+ 20.0 \\ = 0$

In terms of displacement:

$speed = \frac{0}{3} \\ \\ = 0 \: {kmh}^{ - 1}$

3 0

3 years ago

In which scenario will the two objects have the greatest gravitational force

Brilliant_brown [7]

Answer: I think the answer C

Explanation:

7 0

3 years ago

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Studying this brochure from nasa, which explains more detail the instruments carried by the Juno spacecraft which scientific act

adoni [48]

Answer:

Juno scientific payload includes:

A gravity/radio science system (Gravity Science)
A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
A vector magnetometer (MAG)
Plasma and energetic particle detectors (JADE and JEDI)
A radio/plasma wave experiment (Waves)
An ultraviolet imager/spectrometer (UVS)
An infrared imager/spectrometer (JIRAM)

Explanation:

Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.

3 0

2 years ago