how many protons does a neutral atom of oxygen-16 have

Answers

Answer 1

Answer:

eight

Explanation:

it's atomic number is 8 which mean that an oxygen atom has eight protons in it's nucleus


Related Questions

An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth

Answers

Answer:

0 N

Explanation:

This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N

A small block, with a mass of 0.05 kg compresses a spring with spring constant 350 N/m a distance of 4 cm. It is released from rest, then slides around the loop and up the incline before momentarily comes to rest at point A. The radius of the loop is 0.1 m.

Required:
Find the elastic potential energy.

Answers

Answer:

The elastic potential energy of the spring is 0.28 J

Explanation:

Given;

mass of the block, m = 0.05 kg

spring constant, k = 350 N/m

extension of the spring, x = 4 cm = 0.04 m

The elastic potential energy of the spring is calculated as;

[tex]U_x = \frac{1}{2}kx^2\\\\U_x = \frac{1}{2} \times 350 \times (0.04)^2\\\\U_x = 0.28 \ J[/tex]

Therefore, the elastic potential energy of the spring is 0.28 J

an object is 70 um long and 47.66um wide. how long and wide is the object in km?​

Answers

Answer:

length =  7*10^(-8)km

width = 4.666*10^(-8) km

Explanation:

We know that:

1 μm = 1*10^(-6) m

and

1km = 1*10^3 m

or

1m = 1*10^(-3) km

if we replace the meter in the first equation, we get:

1 μm = 1*10^(-6)*1*10^(-3) km

1 μm = 1*10^(-6 - 3)km

1 μm = 1*10^(-9)km

Now with this relationship we can transform our measures:

Length: 70 μm is 70 times 1*10^(-9)km, or:

L = 70*1*10^(-9)km = 7*10^(-8)km

And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:

W = 46.66*1*10^(-9)km = 4.666*10^(-8) km

Equal masses of two different liquids are put into identical beakers.
Liquid 1 is heated for 100s and liquid 2 is heated for 200s by heaters of the same power.
The temperature of both liquids increases by the same amount.
Which statement is correct?
A Both liquids receive the same amount of energy.
B. Liquid 1 receives more energy than liquid 2.
C. Both liquids have equal thermal capacity.
D. The thermal capacity of liquid 1 is less than the thermal capacity of liquid 2.

Answers

Answer:

C

Explanation:

Because they both received the same temperature

Electricity is the result of moving electrons, so it's classified as

A. Kinetic Energy
B. Gravitational Energy
C. Potential Energy
D. Elastic Energy

Answers

The answer should be A
The correct answer is A

calculate the voltage that is being applied across a 10W bulb if a current of 0.2A flows through it​

Answers

Answer:

below

Explanation:

from P= I * V

v = p/I

v = 10/0.2

v = 50 volts

PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?

a) 8.03 x 10^16 nuclei

b) 4.01 x 10^16 nuclei

c) 2.02 x 10^16 nuclei

d) 1.61 x 10^17 nuclei

Answers

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

c) 2.02 x 10^16 nuclei

PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?

A) 1.67 x 10^-4 s^-1

B) 5.43 x 10^-4 s^-1

C) 1.40 x 10^-4 s^-1

D) 2.22 x 10^-4 s^-1

Answers

OPTION C is the correct answer.

Which pair of magnets has the strongest attraction between them?

Answers

I think it’s B. I apologize if I’m wrong.

The colors that make up white light are called what?​

Answers

Answer:

The ROYGBIV

Explanation:

R - red

O - orange

Y - yellow

G - green

B - blue

I - indigo

V - violet


What is Velocity in physics

Answers

Answer:

hii

Explanation:

i hope this helps you

Answer:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. ... Velocity is a physical vector quantity; both magnitude and direction are needed to define it

Explanation:

hope it helps

pls maek me as brainliest thanks❤

Two resistors, A and B, are connected in parallel across a 6.0-V battery. The current through B is found to be 2.0 A. When the two resistors are connected in series to the 6.0- V battery, a voltmeter connected across resistor A measures a voltage of 4.0 V. Find the resistances of A and B

Answers

Answer:

The resistance of A is 6 ohms and the resistance of B is 3 ohms

Explanation:

Step 1: For the first connection (parallel connection), the resistance of B will be calculated.

Note: in a parallel connection, the voltage through each resistor is the same.

[tex]V = I_AR_A = I_BR_B\\\\R_B = \frac{V}{I_B} = \frac{6}{2} = 3 \ ohms[/tex]

Step 2: The resistance of A will be calculated from the second connection (series connection)

Note: in series connection, the current flowing in each resistor is the same

[tex]V = V_A + V_B\\\\V = IR_A + IR_B\\\\The \ voltage \ drop \ in \ B; \ V_B = V- V_A\\\\V_B = 6 - 4 = 2 \ V\\\\IR_B = 2\ V\\\\I = \frac{2 \ V}{R_B}= \frac{2}{3} \ A\\\\The \ resistance \ of \ A \ is \ calculated \ as ;\\\\IR_A = 4 \ V\\\\R_A = \frac{4}{I} = \frac{4 \times 3}{2} = 6 \ ohms[/tex]

List and briefly explain the incidents leading to the occurrence of any five nuclear accidents that have taken place in different parts of the world.​

Answers

Answer:

Chernobyl Nuclear Disaster  Nuclear Disaster. Japan 2011  Kyshtym Nuclear Disaster. Russia 1957  Windscale Fire Nuclear Disaster. Sellafield, UK 1957 Three Mile Island Nuclear Accident. Pennsylvania, USA 1979

Explanation:

Hope this helps... pls vote as brainliest

Effects of global warming is

A-decrease in temperature
B-melting of polar ice caps
C-breathing problems

Answers

Answer:

B- the melting of polar ice caps

Explanation:

As the world's temperature increases, polar ice caps will no longer be able to remain solid.

The power in an electrical circuit is given by the equation P= RR, where /is the current flowing through the circuit and Ris the resistance of the circuit. What is the current in a circuit that has a resistance of 100 ohms and a power of 15 watts?

[pleas ee helpppp)​

Answers

I= 0.39 A

OPTION B is the correct answer.

PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)

A) 8.18 x 10^-14 J

B) 2.73 x 10^-22 J

C) 1.5053 x 10^-10 J

D) 1.5032 x 10^-10 J

Answers

Answer:

djfjci3jsjdjdjdjdjddndn

ds

identify the word being referred to choose your answer from the words below​

Answers

Answer:

1:Rotation

2:Axis

3:Aphelion

4:orbit

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?

a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min

Answers

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]

let's calculate

       v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]

       v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the bucket is gradually increased by the addition of sand. At some point, the bucket will accumulate enough sand to set the block in motion. The coefficients of static and kinetic friction are 0.60 and 0.50 respectively.

Required:
a. Determine the mass of sand in [kg], including the bucket, needed to start the block moving.
b. Find the blocks acceleration, in [m/s^2] up the plane?

Answers

Answer:

a). M = 20.392 kg

b). am = 0.56 [tex]m/s^2[/tex] (block),  aM = 0.28 [tex]m/s^2[/tex] (bucket)

Explanation:

a). We got  N = mg cos θ,

                  f = [tex]$\mu_s N$[/tex]

                    = [tex]$\mu_s mg \cos \theta$[/tex]

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + [tex]$\mu_s mg \cos \theta$[/tex]   .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

[tex]$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$[/tex]

[tex]$M=2(m \sin \theta + \mu_s mg \cos \theta)$[/tex]

[tex]$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$[/tex]

M = 20.392 kg

b). [tex]$(h-x_m)+(h-x_M)+(h'+x_M)=l$[/tex]  .............(iii)

   Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

[tex]$-\ddot{x}-2\ddot x_M=0$[/tex]

[tex]$\ddot x_M=\frac{\ddot x_m}{2}$[/tex]

[tex]$a_M=\frac{a_m}{2}$[/tex]   .....................(iv)

We got,   N = mg cos  θ

                [tex]$f_K=\mu_K mg \cos \theta$[/tex]

∴ [tex]$T-(mg \sin \theta + f_K) = ma_m$[/tex]

  [tex]$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$[/tex]  ................(v)

Mg - 2T = M[tex]a_M[/tex]

[tex]$Mg-Ma_M=2T$[/tex]

[tex]$Mg-\frac{Ma_M}{2} = 2T$[/tex]    (from equation (iv))

[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}=T$[/tex]   .....................(vi)

Putting (vi) in equation (v),

[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$[/tex]

[tex]$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$[/tex]

[tex]$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$[/tex]

[tex]$a_m= 0.56 \ m/s^2$[/tex]

Using equation (iv), we get,

[tex]a_M= 0.28 \ m/s^2[/tex]

A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they stick together. What is the total momentum of the system after the collision? What is the total momentum of the system before the collision? What is the velocity of the cars after the collision?

Answers

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is the maximum vertical height the person in the cart can reach?

Answers

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

[tex]mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m[/tex]

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed in m/s of a satellite in an orbit 980 km above the Earth's surface.

Answers

Answer:

564

Explanation:

A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.

Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?

Answers

Answer:

a)   μ = 0.0136, b)   F = 22.8 N

Explanation:

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

        p₀ = m v₀

final instant. After inelastic shock

        p_f = (m + M) v

the moment is preserved

        p₀ = p_f

        m v₀ = (m + M) v

        v = [tex]\frac{m}{m + M} \ v_o[/tex]

We look for the speed of the block with the bullet inside

        v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]

        v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

         W = ΔK

          W = ½ (m + M) [tex]v_f^2[/tex]  - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

         W = - fr x

we substitute

           - fr x = 0 - ½ (m + M) vo²

           fr = ½ (m + M) v₀² / x

         

the friction force is

          fr = μ N

          μ = fr / N

equilibrium condition

          N - W = 0

          N = W

          N = (m + M) g

we substitute

         μ = ½ v₀² / x g

we calculate

          μ = ½ 0.448 ^ 2 / 0.75 9.8

          μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

          W = ΔK

initial block velocity is zero vo = 0

         F x₁ = ½ M v² - 0

         F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]

         F = ½ 2.50 0.448² / 0.0110

         F = 22.8 N

The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?

Answers

Answer: The energy released as thermal energy is 6.5 J

Explanation:

Energy stored by the spider when it relaxes is given by:

[tex]E_o=\text{Resilience}\times \text{Work}[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\text{Work done}-E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5J[/tex]

Hence, the energy released as thermal energy is 6.5 J

The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J

What is thermal energy?

Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.

Energy stored by the spider when it relaxes is given by:

[tex]\rm E_o=Resilience \ \times Work[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]\rm E_o=0.35\times 10[/tex]

[tex]E_o=3.5\ J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\rm Work done -E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5\ J[/tex]

Hence, the energy released as thermal energy is 6.5 J

To know more about thermal energy follow

https://brainly.com/question/19666326

Kilometer is a unit of length where as kilogram is a unit of mass

Answers

By George, you've nailed it, Stacy !

That's a fact, uh huh.

Truer words were never written.

Your statement is one of unquestionable veracity.

The pure truthiness of it cannot be denied.

Was there a question you wanted to ask ?

Cual es l diferencia entre ruido y sonido

Answers

Answer:

E.l soni.do es un.a sensac.ión, en el órg.ano del oído, prod.ucida por el movimie/nto ondu>latorio de un m/edio elástico (normal/mente el aire), debi.do a ra.pidísimos ca/mbios de pre.sión, generado/s por el movimiento vibrat.orio d.e un cuerpo sonoro. ... /El ruido se consid/era a to/do sonid.o / o no de.seado.

Explanation:

A ratio that compares the width and length of a garden is what type of model?

Answers

Answer:

physical

PLEASE MARK ME AS A BRAINLIEST

Answer: Mathematical

Explanation: I took the quiz

make ansentance rkdloebebjekeoejbe​

Answers

Answer:

the man has returned from his trip

Answer:

just did by typing this lol

according to the law of conservation of vhange , what must always be true in a nuclear reaction?​

Answers

Answer:

The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same

Explanation:

Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.

Calculate the Combined resistance of the Circuit voltage across each resistor Current Passing through each resistor of 6,8,12ohms​

Answers

Answer:

Sorry I don't know the answer

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