Answer:
Explanation:
Let the stretch produced in first spring and second spring be x₁ and x₂ respectively. 300 N is the restoring force created in them .
k₁x₁ = k₂x₂ = 300 N .
k₁ = 11000 N/m
11000 x₁ = 300N
x₁ = 300 N / 11000
= 0.027 m
= 2.7 cm .
a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air
Answer:
Explanation:
The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:
initial velocity is 20 m/s
Here's what we know that we are NOT told:
a = -9.8 m/s/s and
final velocity is 0 at an object's max height in parabolic motion.
We will use the equation:
[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:
0 = 20 + (-9.8)t and
-20 = -9.8t so
t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.
8. A mass of 10 Kg is accelerating at 3 m/s2. What is the applied net force?
Answer:
Explanation:
F = ma
F = (10)(3)
F = 30 N
Answer:
[tex]\boxed {\boxed {\sf 30 \ Newtons}}[/tex]
Explanation:
We are asked to find the applied net force. According to Newton's Second of Law, force is the product of mass and acceleration.
[tex]F= m \times a[/tex]
The object has a mass of 10 kilograms and it is accelerating at 3 meters per second squared.
m= 10 kg a= 3 m/s²Substitute the known values into the formula.
[tex]F= 10 \ kg \times 3 \ m/s ^2[/tex]
Multiply.
[tex]F= 30 \ kg \times m/s^2[/tex]
1 kilogram meter per second squared is equal to 1 Newton, so our answer of 30 kg × m/s² is equal to 30 N.
[tex]F= 30 \ N[/tex]
The applied net force is 30 Newtons.
A 1050 kg car accelerates from 11.3 m/s to 26.2 m/s . What impulse does the engine give?
Answer:
I = 15,645. kg*m/s or 15,645 N*s
Explanation:
I = m(^v)
I = 1050kg((26.2m/s-11.3m/s)
I = 15,645. kg*m/s
A block of mass 2 kg starts from rest at the top of a friction quarter of a circle of radius R. The block then slides over frictionless curved surface in the shape of a eventually comes to rest 8 m from the beginning s a horizontal rough surface where e of the horizontal surface. The coefficient kinetic friction between the rough surface and the block is 0.4 . determine the acceleration of the block over the rough surface length 8m
The acceleration of the block over the rough surface is 1.22625 m/s²
The process through which the acceleration is obtained is presented as follows of approach to
The given parameters are;
Mass of block, m = 2 kg
Nature of the surface of the quarter circle = Frictionless
The length of the horizontal, d = 8 m
The coefficient of friction of the horizontal surface, μ = 0.4
The unknown parameter;
The acceleration of the block over the rough surface
Method;
Find the work done by friction to stop the block and divide the result by the mass of the block
The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)
[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d
[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ
Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block
[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N
Therefore;
The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d
[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J
The work done by the block, W = Force, F × d
Force, F = m × a
Where;
a = The acceleration of the block
According to the principle of conservation of energy, we have;
[tex]\mathbf{W_f}[/tex] = W
∴ 19.62 J = 2 kg × a × 8 m
a = 19.62/(2 kg × 8 m) = 1.22625 m/s²
The acceleration of the block over the rough surface, a = 1.22625 m/s²
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Imagine a spaceship traveling at a constant speed through outer space. The length of the ship, as measured by a passenger aboard the ship, is 28.2 m. An observer on Earth, however, sees the ship as contracted by 16.1 cm along the direction of motion. What is the speed of the spaceship with respect to the Earth
[tex]3.20×10^7\:\text{m/s}[/tex]
Explanation:
Let
[tex]L = 28.2\:\text{m}[/tex]
[tex]L' = 28.2\:\text{m} - 0.161\:\text{m} = 28.039\:\text{m}[/tex]
The Lorentz length contraction formula is given by
[tex]L' = L\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)}[/tex]
where L is the length measured by the moving observer and L' is the length measured by the stationary Earth-based observer. We can rewrite the above equation as
[tex]\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)} = \dfrac{L'}{L}[/tex]
Taking the square of the equation, we get
[tex]1 - \left(\dfrac{v^2}{c^2}\right) = \left(\dfrac{L'}{L}\right)^2[/tex]
or
[tex]1 - \left(\dfrac{L'}{L}\right)^2 = \left(\dfrac{v}{c}\right)^2[/tex]
Solving for v, we get
[tex]v = c\sqrt{1 - \left(\dfrac{L'}{L}\right)^2}[/tex]
[tex]\:\:\:\:=(3×10^8\:\text{m/s})\sqrt{1 - \left(\dfrac{28.039\:\text{m}}{28.2\:\text{m}}\right)^2}[/tex]
[tex]\:\:\:\:=3.20×10^7\:\text{m/s} = 0.107c[/tex]
A mass is hanging from the end of a horizontal bar that pivots around an axis through its center, but it is being held stationary. The bar is then released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar: ________
Answer:
The torque decreases because as the hanging mass goes down, the moment arm about the pivot point decreases. Since the torque is directly proportional to the length of the moment arm, torque decreases.
The torque decreases because as the hanging mass goes down, the moment arm about the pivot point decreases. Since the torque is directly proportional to the length of the moment arm, torque decreases.
What is the difference between atomic number and atomic mass ?Atomic number of an element is defined as total number of protons present in the nucleus, neutrons carry no net electrical charge, so it is the charge number of the nucleus.
atomic mass of an element can be defined as the atomic weight is measured total mass of an element’s atom, the total number of neutrons and protons in the nucleus of an atom.
Both Atomic mass and an atomic number of elements are closely related if atomic number is high, then the atomic mass is also said to be high.
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What is utilization of energy
Explanation:
Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector
You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. How much force must you exert to keep the pole motionless in a horizontal position?
Answer:
[tex]F=133N[/tex]
Explanation:
From the question we are told that:
Length [tex]l=3.0m[/tex]
Mass [tex]m=24kg[/tex]
Distance from Tip [tex]d=35cm[/tex]
Generally, the equation for Torque Balance is mathematically given by
[tex]mg(l/2)=F(l-d)[/tex]
[tex]2*9.81(3/2)=F(3-35*10^-2)[/tex]
Therefore
[tex]F=133N[/tex]
trong cùng một nhiệt độ, lượng năng lượng trên mỗi mol của chất khí nào lớn nhất
a) Khí đơn nguyên tử
b) Khí có từ ba nguyên tử
c) Khí lưỡng nguyên tử
An energy efficient light bulb uses 15 W of power for an equivalent light output of a 60 W incandescent light bulb. How much energy is saved each month by using the energy efficient light bulb instead of the incandescent light bulb for 4 hours a day? Assume that there are 30 days in one month
A. 7.2 kW⋅hr
B. 21.6 kW⋅hr
C. 1.8 kW⋅hr
D. 5.4 kW⋅hr
E. 1.35 kW⋅hr
Answer: (d)
Explanation:
Given
15 W is equivalent to 60 W light that is, it save 45 W
So, for 4 hours it is, [tex]4\times 45=180\ W.hr[/tex]
For 30 days, it becomes
[tex]\Rightarrow 180\times 30=5400\ W.hr\\\Rightarrow 5.4\ kWh[/tex]
Thus, [tex]5.4\ kWh[/tex] is saved in 30 days
option (d) is correct.
what is simple machine?
Explanation:
Those tools that helps to make our work easier ,faster and more convenient in our daily life it is called simple Machine.
An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s at 60° above the horizontal. Calculate (a) the maximum height, (b) the time required to reach its highest point, (c) the total time of flight, (d) the components of its velocity just before striking the ground, and (e) the horizontal distance traveled from the base of the cliff.
a) y(max) = 337.76 m
b) t₁ = 5.30 s the time for y maximum
c)t₂ = 13.60 s time for y = 0 time when the fly finish
d) vₓ = 30 m/s vy = - 81.32 m/s
e)x = 408 m
Equations for projectile motion:
v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )
v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s
a) Maximum height:
The following equation describes the motion in y coordinates
y = y₀ + v₀y*t - (1/2)*g*t² (1)
To find h(max), we need to calculate t₁ ( time for h maximum)
we take derivative on both sides of the equation
dy/dt = v₀y - g*t
dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g
v₀y = 60*sin60° = 60*√3/2 = 30*√3
g = 9.8 m/s²
t₁ = 5.30 s the time for y maximum
And y maximum is obtained from the substitution of t₁ in equation (1)
y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²
y (max) = 200 + 275.40 - 137.64
y(max) = 337.76 m
Total time of flying (t₂) is when coordinate y = 0
y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²
0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0
The above equation is a second-degree equation, solving for t₂
t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8
t = [51.96 ±√2700 + 3920]/9.8
t = [51.96 ± 81.36]/9.8
t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)
t₂ = 13.60 s time for y = 0 time when the fly finish
The components of the velocity just before striking the ground are:
vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant
vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)
vy = 51.96 - 133.28 vy = - 81.32 m/s
The sign minus means that vy change direction
Finally the horizontal distance is:
x = vₓ * t
x = 30 * 13.60 m
x = 408 m
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If a person is 18 % shorter than average, what is the ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height? Assume that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same.
f(short)/f(avg)=?
We have that the ratio of his walking pace to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
given the assumption and the calculation given below
From the question we are told that:
Consider a person 18\% shorter than average
Let average height of a person be [tex]10m[/tex]
Therefore
The height of an [tex]18\%[/tex] shorter man is mathematically given as
H=10*0.18
H=8.2m
Generally, the equation for velocity is mathematically given by
[tex]v=\frac{1}{2\pi} \sqrt{{g}{l}}[/tex]
Where we have the Assumption that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same
Therefore
[tex]\frac{V_1}{V_2}=\frac{l_1}{l_2}[/tex]
[tex]\frac{V_1}{V_2}={82}{100}[/tex]
[tex]\frac{V_2}{V_1}=1.10[/tex]
In conclusion
The ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
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the 2kg block slids down a firctionless curved ramp starting from rest at heiht of 3m what is the speed of the block at the bottemvof the ramp
A
Explanation:
1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.
Vặt nhỏ được ném lên từ điểm A trên mặt đất với vận tốc đầu 20m/s theo phương thẳng đứng. Xác định độ cao của điểm O mà vật đạt được. Bỏ qua ma sát
Explanation:
mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.
14. What's one of the two requirements electric current?
A. There must be an electric potential between two bodies
B. There must be no valence electrons that make their element unstable
C. There must be a carbon element present in the electric current
D. There must be a magnetic force between two bodies
Marko
One of the two requirements of electric current is there must be an electric potential between two bodies
For electric current to flow, there must be an electric potential between two bodies.
This is because electric charge flows from a higher electric potential to a lower electric potential just as, water flows from a higher gravitational potential to a lower gravitational potential.
The difference between the electric potential between the two bodies causes the electric charge to flow between the two bodies.
This flow of electric charge constitutes electric current and electric current will only flow when there is an electric potential between two bodies.
So, one of the two requirements of electric current is there must be an electric potential between two bodies.
So, the answer is A
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a bullet is dropped from the same height when another bullet is fired horizontally they will hit the ground
Answer:
simultaneously
Time taken to reach the ground depends on the vertical component of velocity, not horizontal component of velocity.
A train starts from rest (at position zero) and moves with constant acceleration. On the first observation, its velocity is 20m/s and 80seconds later the velocity became 60m/s. At 80s calculate the position, average velocity, and the constant acceleration over the interval.(7-points)
The value of the acceleration is a = 0.5 m/s². The position at 80 s is x = 3200 m and finally the average velocity is v = 40 m/s.
Acceleration:
We can use the fallowing kinematic equation to get the acceleration at 80 s.
[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]
Where:
v(i) is the initial velocity (20 m/s)v(f) is the final velocity (60 m/s)t is the interval (80 s)The, we have:
[tex]\vec{a}=\frac{60-20}{80}[/tex]
[tex]\vec{a}=0.5\: m/s^{2}[/tex]
Position:
Knowing the acceleration we can find the position using the falling equation.
[tex]\vec{x}=v_{i}t+0.5at^{2}[/tex]
[tex]\vec{x}=20*80+0.5*0.5*80^{2}[/tex]
[tex]\vec{x}=3200 m[/tex]
Average velocity:
The definition of the average velocity is:
[tex]\vec{v}=\frac{\Delta x}{t}[/tex]
[tex]\vec{v}=\frac{x_{f}-x_{i}}{t}[/tex]
[tex]\vec{v}=\frac{3200-0}{80}[/tex]
[tex]\vec{v}=40\: m/s[/tex]
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Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )
1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )
1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) = [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ = [tex]\frac{V}{i+0.5}[/tex]
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ = [tex]\frac{V}{i+0.25}[/tex]
R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]
R₂ = 19.9 Ω
1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of 2 m above the elevator floor when the elevator floor when the elevator is 28 m above the ground.
a. What's the maximum height?
b. How long does it take for the ball to return to the elevator floor?
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
The given parameters include:
constant velocity of the elevator, u₁ = 10 m/sinitial velocity of the ball, u₂ = 20 m/sheight of the boy above the elevator floor, h₁ = 2 mheight of the elevator above the ground, h₂ = 28 mTo calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.
[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]
The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.
[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]
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Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
As a skydiver accelerates downward, what force increases? A. Gravity B. Thrust C. Air resistance D. Centripetal
Answer:
(A) Gravity is you're answer.
Explanation:
When an object or human is falling at an increased rate, The force of gravity is taking place.
Serena wants to play a trick on her friend Marion. She adds salt, sugar, and vinegar into her glass of water when Marion is out of the room. Why does she know that Marion will drink the water?
Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules
Answer:
the change in the kinetic energy of the system is -42.47 J
Explanation:
Given;
mass A, Ma = 2 kg
initial velocity of mass A, Ua = 15 m/s
Mass M, Mm = 4 kg
initial velocity of mass M, Um = 7 m/s
Let the common velocity of the two masses after collision = V
Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;
[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]
The initial kinetic of the two masses;
[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]
The final kinetic energy of the two masses;
[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]
Therefore, the change in the kinetic energy of the system is -42.47 J
Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in J) is stored in this inductor when 21.0 A of current flows through it? J (c) How fast (in s) can it be turned off if the induced emf cannot exceed 3.00 V? s
Answer:
(a) The self inductance, L = 21.95 mH
(b) The energy stored, E = 4.84 J
(c) the time, t = 0.154 s
Explanation:
(a) Self inductance is calculated as;
[tex]L = \frac{N^2 \mu_0 A}{l}[/tex]
where;
N is the number of turns = 1000 loops
μ is the permeability of free space = 4π x 10⁻⁷ H/m
l is the length of the inductor, = 45 cm = 0.45 m
A is the area of the inductor (given diameter = 10 cm = 0.1 m)
[tex]A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2[/tex]
[tex]L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH[/tex]
(b) The energy stored in the inductor when 21 A current ;
[tex]E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J[/tex]
(c) time it can be turned off if the induced emf cannot exceed 3.0 V;
[tex]emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s[/tex]
When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?
Answer:
The angle is 153.9 degree.
Explanation:
Let the magnetic field is B and the charge is q. Angle = 26.1 degree
The force is F.
Let the angle is A'.
Now equate the magnetic forces
[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]
What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
Air is compressed polytropically from 150 kPa, 5 meter cube to 800 kPa. The polytropic exponent for the process is 1.28. Determine the work per unit mass of air required for the process in kilojoules
a) 1184
b) -1184
c) 678
d) -678
Answer:
wegkwe fhkrbhefdb
Explanation:B
please answer all of them
I'll give brainly if answer for points will be reported
Answer:
Level 1-
1. It depends of the sense and the magnitude of the force
2. Electric force
3. A contact force need to touch to act in the object, like push a box for example. A non-contact force don't need to touch to act in the object, like an magnet attracting other magnet
4. The pressure is the force divided by the area. The unit for pressure in the international system is Pascal
5. Because the pressure is applied in all the surface of our bodies, so the force is divided by the surface area of our bodies.
Level 2-
1. The balloons stick to the walls because when she rubbed they in her clothes they earned eletric charge, and when they touched the wall, the electric charges of the wall got polarized and it creats a attraction force.
The same happened with the water stream. The balloons were charged with electric charges and the water was attracted by it.
2. Mass is the amount of matter, it's an scalar quantity. Weight is the force created by the attraction of a massive body as the Earth, and another body as a human, and a force is a vector
3. It's for increase the surface area, so the pressure will be decreased
4. When a person pulls up the syringe plunger the pressure inside the syringe is smaller than the pressure outside, so the pressure push the liquid into the syringe
5. a) The stream of the top is falling closer than the stream from the bottom, causa in the top the pressure is lower than in the bottom. In the bottom, beyond the air pressure, it has also the whole column of water making more pressure, so it goes far.
b) The streams are all near because the holes are in the same height, so the pressure is divided for all the holes.
Level 3-
1. The girl.
pressure of the girl: 50/1 = 50
pressure of the man: 100/25 = 4
pressure of the elephant: 4500/250 = 18
So, the girl exerts more pressure.
2. When the can is heated the air inside expands and get out of it. If you seal the mouth of the can, the air cannot return to inside it, and when it get colder the air inside will shrink back to the normal volume, so it will occupy less space and the outside pressure will exerts a force and deform the can.
