The image above shows the nucleus of a nitrogen atom. How can the atomic number of nitrogen be determined?A. count the humber of neutrons
B. Count the number of protons
C. Add the number of protons and neutrons
D. count the number of electrons

Answers

Answer 1
Answer: The atomic number of an atom is the
number of protons in its nucleus.

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A model rocket accelerates at 15.3 m/s2 with a force of 44 N.Calculate the mass of the rocket. Round your answer to the nearest tenth of a kg.

The mass of the rocket is
kg.

Answers

The mass of the rocket is 2.9 kg.

To calculate the mass of the rocket, you can use Newton's second law of motion, which relates force (F), mass (m), and acceleration (a) as F = ma. In this case, you have the force (44 N) and the acceleration (15.3 m/s²).

Rearrange the formula to solve for mass:

m = F / a

m = 44 N / 15.3 m/s² ≈ 2.9 kg

So, the mass of the rocket is approximately 2.9 kilograms. The mass is rounded to the nearest tenth of a kilogram as specified in the question. This mass represents the amount of matter in the rocket and is a critical factor in determining how the rocket accelerates when subjected to the given force.

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Answer:

2.9

Explanation:

In the equation you get 2.88 but you round that to the nearest tenth so you get 2.9

Calculate the maximum acceleration of a car that is heading up a 4º slope (one that makes an angle of 4º with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On dry concrete.
(b) On wet concrete.
(c) On ice, assuming that μs = 0.100 , the same as for shoes on ice.

Answers

Final answer:

The maximum acceleration of a car moving uphill can be calculated using the formula μs*g*cosθ - g*sinθ where θ is the slope angle, μs is the coefficient of static friction, and g is the acceleration due to gravity. The figures for μs differ depending on the road condition - dry concrete, wet concrete, or ice, substantially affecting the car's acceleration.

Explanation:

The maximum acceleration of a car moving uphill is determined by the force of static friction, which opposes the combined force of the car's weight component down the plane and the force utilized by the driving wheels. The maximum static friction force (F_max) is determined by the coefficient of static friction (μs) multiplied by the normal force (N), which is equivalent to the weight of the car (mg) multiplied by the cosine of the angle (cosθ).

(a) On dry concrete: Since the μs is usually 1.0 on dry concrete and half the weight of the car is supported by the drive wheels, the maximum acceleration can be calculated as μs*g*cosθ - g*sinθ

(b) On wet concrete: The μs is around 0.7 on wet concrete. Substituting this value into the formula would give us the maximum acceleration on a wet surface.

(c) On ice: With a μs value of 0.1 as given, the maximum acceleration on ice can also be calculated using the same formula.

As one can see, the road conditions significantly impact the car's maximum acceleration due to the change in the amount of friction between the tires and road surface.

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Final answer:

The maximum accelerations for the car going up a 4º slope are 9.3 m/s² on dry concrete, 6.4 m/s² on wet concrete, and -0.1 m/s² on ice.

Explanation:

The maximum acceleration of the car up the slope can be calculated using the equation: a = μs * g * cosθ - g * sinθ, where a is the acceleration, μs is the coefficient of static friction, g is the acceleration due to gravity, and θ is the angle with the horizontal.

To solve this problem, we must teach the student to take several factors into account, including the various coefficients of static friction corresponding to different road conditions, namely dry concrete, wet concrete, and ice.

Considering that each scenario has different values of μs, we fill in the equation with the angles and coefficients of static friction. As a rule of thumb, μs for dry concrete is generally taken as 1.0, for wet concrete as 0.7 and for ice (mentioned in the question) as 0.100.

  1. For dry concrete, a = 1.0 * 9.8 * cos(4) - 9.8 * sin(4) = 9.3 m/s².
  2. For wet concrete, a = 0.7 * 9.8 * cos(4) - 9.8 * sin(4) = 6.4 m/s².
  3. For ice, a = 0.100 * 9.8 * cos(4) - 9.8 * sin(4) = -0.1 m/s².

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A ball with a mass of 1.5 kilograms is tied to the end of a rope. The ball is pulled to a height of 0.5 meters above the ground and released.The ball has joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velocity of the ball at position A is meters/second. Assume there’s no air resistance. Use g = 9.8 m/s2 , PE = m × g × h, and

Answers

Solve the potential energy at point B

PE = mgh

Where m is the mass

G is the acceleration due to gravity 9.8 m/s2

H is the height

PE = ( 1.5 kg) ( 9.8 m/s2) (0.5 m)

PE = 7.35 J

 

Solve the velocity using the

KE = 0.5mv^2

Where ke is the kinetic energy

M is the mass

V is the velocity

Since all energy is converted to KE

So KE = PE

 

7.35 = 0.53(1.5) v^2

V = 3.13 m/s

The velocity of the ball at position A is equal to 3.13 m/s.

Given the following data:

  • Mass of ball = 1.5 kilograms.
  • Height = 0.5 meters.
  • Acceleration due to gravity = 9.8 m/s^2

How to calculate potential energy.

Mathematically, potential energy is given by this formula:

P.E = mgh\n\n

Where:

  • m is the mass.
  • h is the height.
  • g is acceleration due to gravity.

Substituting the given parameters into the formula, we have;

P.E = 1.5 * 9.8 * 0.5

P.E = 7.35 Joules.

Assuming there’s no air resistance, the potential energy would be equal to kinetic energy.

P.E = K.E\n\n7.35 = (1)/(2) mv^2\n\nmv^2 = 7.35 * 2\n\n1.5v^2 = 14.7\n\nv=\sqrt{(14.7)/(1.5) } \n\nv=√(9.8)

V = 3.13 m/s.

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A wave creates a strong undertow

Answers

Hey there! 

Answer: A swell

A wave which creates a strong undertow is called a swell. These types of waves are also known as surface gravity waves. 

Thank you!

Emily’s vacuum cleaner has a power rating of 200 watts. If the vacuum cleaner does 360,000 joules of work,how long did Emily spend vacuuming?

Answers

Divide 360000 by 200 to get 1800 seconds, or half of hour.

During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acceleration? Show your work.

Answers

v o = 6 m/s,
t = 2 s
v = 10 m/s
v = v o + a t
a t = v - v o
a = ( v - v o ) / t 
a = ( 10 m/s - 6 m/s ) / 2 s = 4 m/s / 2 s = 2 m/s²
Answer:
The runner`s acceleration is 2 m/s².