What is the atomic number for the following Bohr model?
e
P= 8
N= 8
---
a
8
b
2
OOOO
с
6
d
16

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

Answer:

What?

Explanation:

Complete Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.

Answer:

The value is  [tex]N = 109 \ rev[/tex]      

Explanation:

From the question we are told that

    The speed of the car is  [tex]u = 28.4 \ m/s[/tex]

     The constant deceleration experienced is  [tex]a = 1.92 \ m/s^2[/tex]

      The radius of the tire is  [tex]r = 0.307 \ m[/tex]

Generally from kinematic equation we have that

      [tex]v^2 = u^2 + 2as[/tex]

Here  v is the final velocity which is  0 m/s

   So

         [tex]0^2 = 28.4^2 + 2 * 1.92 * s[/tex]

=>      [tex]s = 210.04 \ m[/tex]

Generally the circumference of the tire is mathematically represented as

         [tex]C = 2 \pi r[/tex]

=>      [tex]C = 2 * 3.142 * 0.307[/tex]    

=>      [tex]C = 1.929 \ m[/tex]

Generally the number of revolution is mathematically represented as

         [tex]N = \frac{ s}{C}[/tex]    

=>     [tex]N = \frac{210.04}{1.929}[/tex]

=>     [tex]N = 109 \ rev[/tex]      

Answer:

15N

Explanation:

F=ma so F=.3*50 therefore F=15N

The force of an egg that is thrown at a brick wall is equal to 15 N.

Force can be defined as the influence or effect that changes the state of the body of from motion to rest or vice versa. The S.I. unit of force is Newton (N) as well as force is a vector quantity. Force can change the direction or the speed of the moving object.

The force acting on an object can be calculated from the multiplication of the mass(m) and acceleration(a). The mathematical form of the second law of motion for force can be written as follows:

F = ma

Given, the mass of the egg, m = 0.3 Kg

The acceleration of the egg with which it is thrown on the wall, a = 50 m/s²

The force of an egg that is thrown at a brick wall can be calculated as:

F = ma = 50 ×0.3 = 15 N

Learn more about force, here:

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Heya!!

For calculate aceleration, let's applicate second law of Newton:

[tex]\boxed{F=ma}[/tex]

⇒ Being:

→ F = Force = 12 N

→ m = Mass = 3 kg

→ a = aceleration = ?

Lets replace according formula and leave the "a" alone:

[tex]12\ N = 3\ kg * \textbf{a}[/tex]

[tex]\textbf{a} = 12\ N / 3\ kg[/tex]

[tex]\textbf{a} = 4\ m/s^{2}[/tex]

Result:

The aceleration of the object is of 4 m/s²

Answer:

it’s c not d

Explanation:

took the test

Answer: D!!!

Explanation: jus got it wrong from the other answer.

Answer:

The bell has a potential energy of 8550 [J]

Explanation:

Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.

[tex]E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J][/tex]

Answer:

The half-life is [tex] t_{1/2} = 1.005 h[/tex]

Explanation:

Using the decay equation we have:

[tex]A=A_{0}e^{-\lambda t}[/tex]

Where:

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]

[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]

[tex]0.5=e^{-\lambda*1 h}[/tex]

Taking the natural logarithm on each side we have:

[tex]ln(0.5)=-\lambda[/tex]

[tex]\lambda=0.69 h^{-1}[/tex]

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]

[tex] t_{1/2} = 1.005 h[/tex]

I hope it helps you!

Answer:

1.2 kg

__________________________________________________________

We are given:

Mass of the block = 2 kg

Coefficient of Static Friction = 0.6

__________________________________________________________

Friction Force on the Block:

Finding the Normal Force:

We know that the normal force will be equal and opposite to the weight of the 2 kg block

So, Normal Force = mg

replacing the variables with the given values

Normal Force = (2)(9.8)                    [Taking g = 9.8]

Normal Force = 19.6 N

Friction force on the Block:

We know that:

Coefficient of Static Friction =  Static Friction Force/Normal Force

replacing the variables

0.6 = Static Friction force / 19.6

Static Friction force = 0.6*19.6 N                 [Multiplying both sides by 19.6]

Static Friction force = 11.76 N

__________________________________________________________

Largest Mass that can Hang:

We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest

So, if the weight of the second block is less than the static friction force, it will hang

Weight of the second block ≤ 11.76

We know that weight = mg

mg ≤ 11.76

m(9.8) ≤ 11.76                                                   [since g = 9.8]

m ≤ 1.2 kg                                                        [dividing both sides by 9.8]

From this, we can say that the maximum mass of the second block is 1.2 Kg

Answer:

A. 28.42 m/s

B. 41.21 m.

Explanation:

A. Determination of the initial velocity of the ball:

Time (t) to reach the maximum height = 2.9 s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) =?

Thus, we can obtain the initial velocity of the ball as follow:

v = u + gt

0 = u + (–9.8 × 2.9)

0 = u – 28.42

Collect like terms

u = 0 + 28.42

u = 28.42 m/s

Therefore, the initial velocity of the ball is 28.42 m/s.

B. Determination of the maximum height reached.

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) = 28.42 m/s.

Maximum height (h) =?

Thus, we can obtain the maximum height reached by the ball as follow:

v² = u² + 2gh

0² = 28.42² + (2 × –9.8 × h)

0 = 807.6964 + (–19.6h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = –807.6964 / –19.6

h = 41.21 m

Therefore, the maximum height reached by the ball is 41.21 m

Answer:

The initial horizontal and vertical components are 12.5 m/s and 21.7 m/s, respectively?

Explanation:

Answer:10000000

Explanation:

Answer:

18.9s

Explanation:

Using the formula;

ω = v/r

Where;

ω = angular velocity (rad/s)

v = linear velocity (m/s)

r = radius of the circular track (m)

According to the given information, v = 30m/s, r = 90m

ω = v/r

ω = 30/90

ω = 3/9

ω = 0.3333 radians/seconds.

Since ω = 2π/T

Where;

π = 3.142

T = period (s)

ω = angular velocity

0.333 = 2 × 3.142/T

T = 2 × 3.142/0.333

T = 6.284/0.333

T = 18.87s

T = 18.9s

Answer:

D.-1500Joules

Explanation:

The change in kinetic energy of the object s equivalent to the workdone by the body in the west direction (negative x direction)

Workdone = Force * Distance

Given

Force = 30N

Distance moved by the object = 30m

Required

Kinetic energy

Kinetic energy = 30 * 50

Kinetic energy = 1500Joules

Since the body moves in the negative  direction, hence the kinetic energy will be -1500Joules

Answer:

t = 66.7 s

Explanation:

Given that,

Speed of a hare, v = 15 m/s

Speed of a turbocharged tortoise, v' = 3 m/s

The hare in a 250 meter race

Let the Hare takes time t. It can be calculated as follows :

[tex]t=\dfrac{250}{15}=16.67\ s[/tex]

Let a turbocharged tortoise takes t'. It can be calulated as follows :

[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]

To tie the race, required time is given by :

[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]

Hence, the correct option is (b) i.e. 66.7 seconds.

Answer: 0.3m/s/s

(i'm really sorry if i'm wrong)

:(

Answer:

  about 2398 seconds

Explanation:

The relation between time, distance, and acceleration is ...

  d = (1/2)at²

  t = √(2d/a) = √(2·10·1.852 km/(83.5 km/h²)) ≈ √0.4436 h ≈ 0.6660 h

That is about ...

  (0.6660 h)(3600 s/h) ≈ 2397.7 s

The cruise ship takes about 2397.7 seconds to cruise 10 nautical miles, accelerating all the way.

Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then

(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v

where v is the velocity of the 2.22 kg block after collision. Solve for v :

6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v

(2.22 kg) v = 2.70 kg•m/s

v = (2.70 kg•m/s) / (2.22 kg)

v ≈ 1.22 m/s

i.e. a velocity of about 1.22 m/s to the right.

[tex]d = s \times t \\ d = 90 \times \frac{30}{60} \\ d = 90 \times \frac{1}{2 } \\ d = 45km[/tex]

Answer:

25

Explanation:

Answer:

The fundamental resonance frequency is 172 Hz.

Explanation:

Given;

velocity of sound, v = 344 m/s

total length of tube, Lt = 1 m = 100 cm

height of water, hw = 50 cm

length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm

For a tube open at the top (closed pipe), the fundamental wavelength is given as;

Node to anti-node (N ---- A) : L = λ / 4

λ = 4L

λ = 4 (50 cm)

λ = 200 cm = 2 m

The fundamental resonance frequency is given by;

[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]

Therefore, the fundamental resonance frequency is 172 Hz.

Answer:

The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.

Explanation:

Hope this helps Plz mark brainliest

Answer:

724.3J/Kg.K

Explanation:

CHECK THE COMPLETE QUESTION BELOW

Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.

The specific heat capacity can be computed by using expression below

c= CV/M

Where c= specific heat capacity

M= molar mass

CV= molar hear capacity

Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R

Where R= 8.314J/mol.k

Molar mass= 28 ×10^-3Kg/mol

If we substitute to the expression, we have

c= (5R/2)/(M)

=5R/2 × 1/M

=(5×8.314) /(2×28 ×10^-3)

=724.3J/Kg.K

Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is

724.3J/Kg.K

The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.

Answer:

Transparent or Translucent

Explanation:

Answer:

1.92 m/s2

Explanation:

Answer:

transition metals im sorry if this was too late

Answer: the car that is moving downhill

Explanation:

Answer:

There are 25,000 marbles in the container

Explanation:

There is a certain number of marbles in a container.

We know each marble has a mass of 80 mg and we also know the total mass of the marbles in the container is 2 Kg.

Since the data is given in different units of mass, we convert them into one unit in common.

Let's convert 2 Kg to milligrams. There are 1,000 grams in a kilogram, and there are 1,000 milligrams in a gram, thus there are one million milligrams in one kilogram, that is:

1 Kg = 1,000,000 mg

And 2 Kg = 2,000,000 mg

The number of marbles can be found by dividing the total mass by the individual mass:

No. of marbles = 2,000,000 / 80 = 25,000

There are 25,000 marbles in the container

Answer:

P = 33.33 [Pa]

Explanation:

The pressure can be calculated by the relationship of the force over the area.

[tex]P =F/A[/tex]

where:

F = force = 20 [N]

A = area = 1 x 0.6 = 0.6 [m²]

Now replacing:

[tex]P=20/0.6\\P=33.33 [Pa][/tex]