Answer:
v = 12 m/min
Explanation:
By definition, the average velocity is the rate of change of the position with respect to time, as follows:[tex]v_{avg} =\frac{x_{f} -x_{o} }{t_{f}-t_{o} } (1)[/tex]
Choosing x₀ = 0 and t₀ =0, (1) reduces to :[tex]v_{avg} = \frac{x_{f} }{t_{f} } (2)[/tex]
From the givens, we have:tif = 6 min + 9 min = 15 min
In order to get xf, we know that during the first part, vavg = 15 m/min, so solving for xf:[tex]x_{f1} = v_{avg1}* t_{1} = 15 m/min * 6 min = 90 m (3)[/tex]
For the following 9 min, we know that the average speed was 10m/min, so the distance traveled during the second part of the trip was simply:[tex]x_{f2} = v_{avg2} *t_{2} = 10m/min * 9 min = 90 m (4)[/tex]
Adding (3) and (4):[tex]x_{f} = 90 m + 90 m = 180 m (5)[/tex]
Replacing xf and tif in (2), we finally get:[tex]v_{avg} =\frac{x_{f} }{t_{f}} =\frac{180m}{15 min} = 12 m/min (6)[/tex]
A student kicks a soccer ball upward at a 30º angle with an initial speed of 20 m∕s. What expression should the student use to calculate the magnitude of the ball’s initial velocity in the horizontal direction?
Answer:
[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]
[tex]y=10\ m/s[/tex]
Explanation:
Rectangular coordinates of vectors in 2D
Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:
[tex]x=v\cos\theta[/tex]
[tex]y=v\sin\theta[/tex]
The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.
The rectangular components of the vector are:
[tex]x=20\cos 30^\circ[/tex]
[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]
Operating:
[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]
[tex]y=20\sin 30^\circ[/tex]
[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]
Operating:
[tex]\mathbf{y=10\ m/s}[/tex]
If two balloons are charged and brought close to each other while hanging you observe that they move away from each other. What would the observation be if the balloons have larger charge?
A. There is no further effect. The repulsion magnitude is always the same.
B. There is no further effect. The magnitude of the force is determined by the charge signs not their magnitude
C. The balloons will lift in addition to separating as now they can start to overcome gravitational forces
D. The balloons will separate further as the repulsion magnitude increases
E. The balloons will come closer as the charges create larger polarization forces
F. The balloons will spin around each other since the electric force can produce rotational motion
Answer:
D. The balloons will separate further as the repulsion magnitude increases.
An object moves along the x-axis with its position x given as a function of time t by x(t)= Ht^2 - Ft + G
What is the object's velocity
Answer:
v(t) = 2Ht - F
Explanation:
Since, the position of the object is given in terms of time (t) as follows:
x(t) = Ht² - Ft + G
where,
H, F, G are constants.
Therefore, the velocity of the object can also be found in terms of the time (t), by simply taking the derivative of the given position equation with respect to time. So, the velocity can be found as follows:
(d/dt) x(t) = (d/dt)(Ht² - Ft + G)
v(t) = (d/dt)(Ht²) - (d/dt)(Ft) + (d/dt)(G)
v(t) = H (d/dt)(t²) - F (d/dt)(t) + (d/dt)(G)
v(t) = H(2t) - F(1) + 0
v(t) = 2Ht - F
A 0.20 kg mass (m1) hangs vertically from a spring and an elongation of the spring of 9.50 cm (r1) is recorded. With a mass (m2) of 1.00 kg hanging on the spring, a second elongation (r2) of 12.00 cm is recorded. Calculate the spring constant k in Newtons per meter (N/m). (Note: The equilibrium position is not zero.)
Answer:
k=320N/m
Explanation:
Step one:
given data
Let the initial/equilibrum position be x
mass m1= 0.2kg
F1= 0.2*10= 2N
elongation e= 9.5cm= 0.095m
mass m2=1kg
F2=1*10= 10N
elongation e= 12cm= 0.12m
Step two:
From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F
F=ke
2=k(0.095-a)
2=0.095k-ka----------1
10=k(0.12-a)
10=0.12k-ka----------2
solving equation 1 and 2 simultaneously
10=0.12k-ka----------2
- 2=0.095k-ka----------1
8=0.025k-0
divide both side by 0.025
k=8/0.025
k=320N/m
In contact forces, _____.
A.) objects do not touch each other
B.) objects must touch each other
C.) more work is done than in other forces
Answer:
B is the best answer for this
What is the relationship between resistance and current in a circuit with no change in voltage?
A. Current and resistance must be equal in a circuit.
B. A circuit that has more resistance will have smaller current.
C. Current does not depend on resistance in a circuit.
D. A circuit that has more resistance will have a greater current.
Answer:
A
Explanation:
answer plz answer plzzz I am a little confused with full time
The mass of a paper-clip is 0.50 g and the density of its material is 8.0g/cm'. The total volume of
a number of clips is 20 cm.
How many paper-clips are there?
Answer:
320 paper clips
Explanation:
mass = volume × density = 20cm³ × 8g/cm³ = 160g
mass of 1 paper clip = 0.50g
mass of x paper clips = 160g
x = 160/0.50 = 320
a piping system consists of 100 ft of 2-inch pipe, a sudden expansion to 3-inch pipe, and then 50 ft of 3-inch pipe. Water is flowing at 100 gal/min through the system. What is the pressure difference from one end of the pipe to the other
Answer:
16+15+19= ??
Am just messign with u lol
Explanation:
anwser s 19 inches
i
When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) what is the effective force constant of the springs? in N/m (b) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?
Answer:
Explanation:
F=kx
x=F/k
F=2000 kg
x=100 cm=9*10^-3
effective spring constant=k=F/x
k=2000/9*10^-3=2.2*10^-5
now frequency
f=1/2π√k/m
f=1/2*3.14√2.2*10^-5/310
f=1/6.28√7.097*10^-8
f=1/6.28*2.7*10^-4
f=0.16*2.7*10^-4
f=4.32*10^-5
The effective spring constant of the springs is 33755.55 N/m.
The frequency of the car's vibration is 2.07 Hz.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
Weight of the four people: F = 310 × 9.80 N = 3038 Newton.
The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.
Hence, the effective spring constant of the springs: k= force/compression
= 3038 N/0.90 × 10⁻² m
= 33755.55 N/m.
The frequency of the car's vibration is: f = 1/2π√(k/m)
=1/2π√(33755.55/2000)
= 2.07 Hz.
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What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?
PLEASE SHOW WORK.
Answer:
The answer is 5.4 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{135}{25} = \frac{27}{5} \\ [/tex]
We have the final answer as
5.4 kgHope this helps you
What is the value of the normal force of the coefficient of kinetic friction friction is 0.22 and kinetic friction force is 40 N
Answer: 1.8x10 sqaured newtons for all plato users
Explanation:
The value of the normal force of reaction is 181.8 N
Friction is defined as the resistance to motion caused by the contact between a surface and the object moving against it.
The normal reaction force is the weight of the body experiencing friction and normally acts vertically downwards.
The frictional force, the coefficient of kinetic friction and the normal force are related by the formula:
coefficient of friction = frictional force / normal reaction
Therefore, normal reaction = frictional force / coefficient of friction
Frictional force = 40 N; coefficient of kinetic friction friction is 0.22
Normal reaction = 40 N/ 0.22 = 181.8 N
Therefore, the value of the normal force of reaction is 181.8 N
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Which statement is true about two isotopes of the same element?
Answer:
atoms that have the same # of protons but a DIFFERENT # of NEUTRONS
Explanation:
Answer:
D-They have different number of neutrons
Explanation:
brainliest? Plz
Two blocks with different masses are dropped, hitting the ground with the same velocity. Which of the following is true?
They have same change in velocity but different changes in kinetic energy
The lighter object started at a smaller height.
The heavier object started at a smaller height
They started at the same height
They have same change in kinetic energy but different changes in velocity
Answer: • They have same change in velocity but different changes in kinetic energy
•They started at the same height.
Explanation:
First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.
Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.
Therefore, based in the explanation, the correct options are:
• They have same change in velocity but different changes in kinetic energy
•They started at the same height.
A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick together and continue moving at a new velocity. Assume that friction is negligible. What was the momentum of the shopping cart before the collision? a) 66.0 kg-m/s b) 0 kg.m/s c) 126 kg.m/s d) 378 kg.m/s What was the momentum of the box before the collision? a) 66.0 kg.m/s b) 0 kg.m/s c) 378 kg-m/s d) 126 kg-m/s What is the velocity of the combined shopping cart-box wreckage after the collision? a) 6.0 m/s b) 3.9 m/s c) 0 m/s d) 11.5 m/s
Answer:
a) 126 kgm/s
b) 0 kgm/s
c) 3.9 m/s
Explanation:
To solve this question, we will use the law of conservation of momentum.
Momentum before collision = momentum after collision
m1v1 + m2v2 = (m1 + m2)v, where
m1 = mass of the shopping cart, 21 kg
m2 = mass of the box, 11 kg
v1 = initial velocity of the shopping cart, 6 m/s
v2 = initial velocity of the box, 0 m/s
v = final velocity of the box+cart
a)
Momentum of the shopping cart before collision = P
P = mv
P = 21 * 6
P = 126 kgm/s = c
b)
Momentum of the box before collision
Like in question a above, the momentum of the box is P
P = mv
P = 11 * 0
P = 0 kgm/s = b
c)
Velocity of the combined shopping cart wreckage after collision is
m1v1 + m2v2 = (m1 + m2)v
(21 * 6) + (11 * 0) = (21 + 11)v
126 + 0 = 32v
32v = 126
v = 126/32
v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.
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A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant of the spring?
Answer:
Force constant of the spring (k) = 24.07 N/m
Concept/Theory:
The period [tex] \sf (T_s) [/tex] of a spring-mass system is proportional to the square root of the mass (m) and inversely proportional to the square root of the force constant of the spring (k).
Equation of period:
[tex] \boxed{ \bf{T_s = 2 \pi \sqrt{\dfrac{m}{k}}}}[/tex]
Explanation:
Mass = 0.250 kg
Period = 0.640 s
By substituting values in the equation, we get:
[tex] \rm \longrightarrow 0.640 = 2 \pi \sqrt{\dfrac{0.250}{k}} \\ \\ \rm \longrightarrow 2 \times 3.14 \sqrt{ \dfrac{0.25 0}{k} } = 0.640 \\ \\ \rm \longrightarrow 6.28 \sqrt{ \dfrac{0.250}{k} } = 0.640 \\ \\ \rm \longrightarrow \sqrt{ \dfrac{0.250}{k} } = \frac{0.640}{6.28} \\ \\ \rm \longrightarrow \frac{0.250}{k} = { \bigg(\frac{0.640}{6.28} \bigg) }^{2} \\ \\ \rm \longrightarrow \frac{k}{0.250} = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \\ \\ \rm \longrightarrow k = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \times 0.250 \\ \\ \rm \longrightarrow k = 24.07 \: N/m[/tex]
The force constant of the spring is approximately 24.038 newtons per meter.
As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant ([tex]k[/tex]), in newtons per meter, from the equation of the Period ([tex]T[/tex]), in seconds, which is described below:
[tex]T = 2\pi\cdot \sqrt{\frac{m}{k} }[/tex] (1)
Where [tex]m[/tex] is the mass of the moving element, in kilograms.
If we know that [tex]T = 0.640\,s[/tex] and [tex]m = 0.250\,kg[/tex], then the spring constant of the spring is:
[tex]0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }[/tex]
[tex]\sqrt{\frac{0.250}{k} } \approx 0.102[/tex]
[tex]\frac{0.250}{k} \approx 0.0104[/tex]
[tex]k \approx 24.038\,\frac{N}{m}[/tex]
The force constant of the spring is approximately 24.038 newtons per meter.
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A kid runs and slides down a slip-n-slide. Once the kid hits the slide they have 200N of friction force acting on them, and they decelerate at 2.5m/s. What is the mass of the kid?
A hockey player skating at 21 m/s comes to a complete stop in 12.0 m.
a) What is its acceleration during this displacement?
b) How long did it take the hockey player to come to a stop?
Answer:
a. 18.375m/s²
b. 1.142s
Explanation:
A. Using v² = u² + 2as
Where;
V = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)
s = distance (m)
In this question, v = 21m/s, u= Om/s, s = 12m
21² = 0² + 2 × a × 12
441 = 0 + 24a
441 = 24a
a = 18.375m/s²
B). Using V = u + at
21 = 0 + 18.375t
21 = 18.375t
t = 21/18.375
t = 1.142s
A horse trots away from its trainer in a straight line, moving away 8m away in 8.0s it then turns abruptly gallops halfway back in 2.0’s what is the magnitude of the horses average velocity for the entire trip
Answer:
v = 0.4 m/s
Explanation:
By definition, the average velocity is the rate of change of the position with respect to time, as follows:[tex]v_{avg} = \frac{x_{f}-x_{o}}{t_{f} -t_{o}} (1)[/tex]
Choosing arbitrarily x₀ = 0 and t₀ = 0, (1) reduces to:[tex]v_{avg} = \frac{x_{f}}{t_{f} }} (2)[/tex]
During the first part of the trip, the horse moves 8m away from its trainer, and during the the second part, gallops halfway back, which means that it finishes 4m away from its trainer.Since total time is 8.0 s + 2.0 s = 10s, replacing xf = 4m and tif = 10,0 s in (2) we get:[tex]v_{avg} =\frac{x_{f}}{t_{f} } = \frac{4.0m}{10s} = 0.4 m/s (3)[/tex]
the distance from Alex home to his school is 1km and 560 cm. what is this distance?
Answer:
Required Answer:-1meter=100cm1km=1000m[tex]{:}\longrightarrow[/tex][tex]\sf 1km=1000×100cm [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 1km=100000cm [/tex]
Total distance
[tex]{:}\longrightarrow[/tex][tex]\sf 100000+560=1000560cm [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 1000.56m[/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 1.560km [/tex]
what body parts were scientists wanting to image that prompted the development of the CT scanner
Answer:
The head
Explanation:
A beam contains 4.9 × 108 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 4.6 × 105 m/s. What is (a) the magnitude of the current density and (b) direction of the current density (c) what additional quantity do you need to calculate the total current i in this ion beam?
Answer:
[tex]72.12\ \text{A/m}^2[/tex]
south
cross sectional area of the beam
Explanation:
v = Velocity of ions = [tex]4.6\times 10^5\ \text{m/s}[/tex]
Number of ions per [tex]\text{cm}^3[/tex] = [tex]4.9\times 10^8[/tex]
Charge density would be the product of number of ions per [tex]cm^3[/tex] and the charge of electrons multiplied by 2 as they are doubly charged.
[tex]\rho_q=4.9\times 10^8\times 10^6\times 2\times 1.6\times 10^{-19}\\\Rightarrow \rho_q=0.0001568\ \text{C/m}^3[/tex]
Current density is given by
[tex]J=\rho_qv\\\Rightarrow J=0.0001568\times 4.6\times 10^5\\\Rightarrow J=72.12\ \text{A/m}^2[/tex]
The current density is [tex]72.12\ \text{A/m}^2[/tex]
The direction of the current density is opposite to the movement of the charged particle. The particles are moving north so the direction of current density will be to the south.
Current is given by
[tex]I=JA[/tex]
where A is the cross sectional area of the beam .
So the cross sectional area of the beam is required to determine the total current in this ion beam.
At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1500 kg car traveling at 1.0 m/s?
Answer:
15m/sExplanation:
Step one:
given data
mass of bicycle m=100kg
the velocity of bicyle v=?
mass of car M=1500kg
the velocity of car V=1.0m/s
Step two:
we know that the momentum is expressed as
P=mv
since the momentum of the bicycle must be equal to car then
mv=MV---------1
100*v=1500*1
divide both sides by 100
v=1500/100
v=15m/s
The velocity of the bicycle should be 15m/s
A child and sled with a combined mass of 54.8
kg slide down a frictionless hill that is 11.5 m
high at an angle of 36 degrees
from horizontal.
The acceleration of gravity is 9.81 m/s^2.
If the sled starts from rest, what is its speed
at the bottom of the hill?
Answer in units of m/s.
Answer:
15.02 m/s.
Explanation:
Given that the height of the hill, h= 11.5 m.
Combined mass, m= 54.8 kg
The initial velocity of the combined mass, u=0
Acceleration due to gravity, [tex]g = 9.81 m/s^2[/tex].
Angle of the path the horizontal, [tex]\theta = 36[/tex] degree.
Let A be the initial position and B be the final position of the sled as shown in the figure.
The path is frictionless so the drag force =0
The gravitational force acting on the combined mass in the downward direction, [tex]F= mg\cdots(i)[/tex]
The component of force acting in the direction of motion = [tex]F\sin \theta.[/tex]
Let [tex]a[/tex] be the acceleration of the combined mass, m, So,
[tex]F\sin \theta= ma[/tex]
[tex]\Rightarrow mg \sin \theta= ma[/tex] [ from equation (i)]
[tex]\Rightarrow a = g \sin \theta \cdots(ii).[/tex]
Let v be the final velocity of the combined mass.
Now, by using the equation of motion,
[tex]v^2=u^2+2as\\\\\Rightarrow v^2=0^2+2as\\\\ \Rightarrow v^2=2as\cdots(iii)[/tex]
Here, s is the displacement in the direction of motion,
So, s= AB
Now, in the right-angled triangle ABO,
[tex]\sin\theta = OA/AB= h/AB\\\\\Rightarrow AB = h/ \sin\theta\\\\\Rightarrow s = h/ \sin\theta\cdots(iv)[/tex]
Now, from equations (ii), (iii) and (iv), we have
[tex]v^2= 2\times g \sin \theta \times \frac {h}{\sin\theta}\\\\\Rightarrow v^2= 2gh\\\\\Rightarrow v= \sqrt{2gh}[/tex]
By using the given values, we have
[tex]v= \sqrt{2\times 9.81\times 11.5}=\sqrt {225.63}\\\\\Rightarrow v = 15.02 m/s[/tex]
Hence, the speed of the combined mass at the bottom = 15.02 m/s.
The speed of the sled at the bottom of the hill is [tex]15.02m/s[/tex]
The speed of sled is calculated by using Newton's law of motion,
[tex]v^{2} =u^{2} +2gh[/tex]
where u is initial velocity, v is final velocity , g is acceleration due to gravity and h is height.
Given that, [tex]u = 0, g = 9.81m/s^{2}[/tex] and [tex]h = 11.5 m[/tex]
Substitute values in above equation.
[tex]v^{2}=0^{2}+2*9.81*11.5\\\\v^{2}=225.63\\\\v=\sqrt{225.63}=15.02m/s[/tex]
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What is the difference between renewable energy non-renewable energy? Use in your own words.
A ball is thrown 24 m/s into the air. How high does it go?
556.4 m
0 m
29.4 m
-556.4 m
Answer:
option c is correct
Explanation:
we know that
2as=vf^2-vi^2
vf=24 m/s
vi= 0 m/s
a=g= 9.8 m/s^2
s=vf^2-vi^2/2a
s=(24)²-(0)²/2*9.8
s=576/19.6
s=29.4 m
therefore option c is correct
During takeoff, an airplane goes from 0 to 42 m/s is 5 s.
How far is it going after 4 s. Calculate answer to one decimal
Answer:
33.6 m
Explanation:
In order to solve the distance of the airplane traveled, we have to write each statement in the question:
1. The distance covered by the plane after 5s: 42 m ( This was calculated by getting the difference of the first and second speed recorded)(42 - 0 = 42)
2. Time taken for the plane to reach this speed: 5s
The question asks for:
Distance of the plane after 4s.
We then solve this by saying:
If 42 m are covered in 5s, how many are covered in 4s? or:
42 m = 5s
x m = 4s
We then cross multiply:
42 × 4 = 168 and 5 × (x) = 5x
After that we divide the "x" side on both sides:
168÷5 and 5x ÷ 5
= 33.6 m = x
Since its already rounded to one decimal place already, it remains unchanged.
According to the Law of Reflection, the angle of incidence the angle of reflection. O A. is greater than B. is less than C. equals D. is opposite from
Answer:
C. Equals
Explanation:
Law of reflection Equals the angle of incidence
A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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The atomic of nitrogen is 7. The number of electrons a neutral atom has is_, and its atomic mass is approximately_amu.