Answer:
Fc = [ - 4.45 * 10^-8 j ] N
Explanation:
Given:-
- The masses and the position coordinates from ( 0 , 0 ) are:
Sphere A : ma = 80 kg , ( 0 , 0 )
Sphere B : ma = 60 kg , ( 0.25 , 0 )
Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15
- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2
Find:-
what is the gravitational force on C due to A and B?
Solution:-
- The gravitational force between spheres is given by:
F = G*m1*m2 / r^2
Where, r : The distance between two bodies (sphere).
- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-
Determine the angle (α) between vectors rac and rab using cosine rule:
[tex]cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}[/tex]
Determine the angle (β) between vectors rbc and rab using cosine rule:
[tex]cos ( \beta ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta ) = 0.6\\\\\beta = 53.13^{\circ \:}[/tex]
- Now determine the scalar gravitational forces due to sphere A and B on C:
Between sphere A and C:
Fac = G*ma*mc / rac^2
Fac = (6.674×10−11)*80*0.2 / 0.2^2
Fac = 2.67*10^-8 N
vector Fac = Fac* [ - cos (α) i + - sin (α) j ]
vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]
vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N
Between sphere B and C:
Fbc = G*mb*mc / rbc^2
Fbc = (6.674×10−11)*60*0.2 / 0.15^2
Fbc = 3.56*10^-8 N
vector Fbc = Fbc* [ cos (β) i - sin (β) j ]
vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]
vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N
- The Net gravitational force can now be determined from vector additon of Fac and Fbc:
Fc = vector Fac + vector Fbc
Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8
Fc = [ - 4.45 * 10^-8 j ] N
what happens when a gas is exposed to a small flame. list as many results.
Answer:
Explanation:
What happens when a flame is deprived of oxygen? Carbon dioxide molecules are heavier than air. Because of this, they push the oxygen and other molecules in the air out of the way as they sink down over the flame and candle. When oxygen is pushed away from the wick, it can't react with the wax anymore. This makes the flame go out.
So if the flame is exposed to air, the flame will rise.
Answer: When gas is exposed to a small flame it explodes into a firerer blow torch at an unbelievable temperature of almost 450 degrees ^F
Explanation: This is just like igniting a lighter then placing an aerosol can of
fumeable gas then spraying the flammeable gas from the aerosol can into the flame of the lighter which will then give you a flamethrower.
A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coasting. Describe each of the following events as an elastic, an inelastic, or a perfectly inelastic collision between the skater and the Frisbee.
Question:
(a) She catches the Frisbee and holds it.
Answer:
The correct option is;
A perfectly inelastic collision
Explanation:
A perfectly inelastic collision is one in which there is maximum amount of loss of kinetic energy in the system. In a perfectly inelastic collision, the colliding members lose their initial speed and they stick together resulting in a loss of kinetic energy.
Since she catches and holds on to the Frisbee, the kinetic energy of the Frisbee is lost as she holds on to it so as to combine her mass to that of the Frisbee.
What r photons?
........................,........,..........
Answer:
in physics, a photon is a bundle of electromagnetic energy. It is the basic unit that makes up all light
Explanation:
Answer:
The quantum of light and other electromagnetic energy, regarded as a discrete particle having zero rest mass, no electric charge, and an indefinitely long lifetime. It is a gauge boson.
Why does sound travel faster in water than in air?
A.
because water is a denser medium than air
B.
because air is a denser medium than water
C.
because water has a larger wavelength than air
D.
because air has a larger wavelength than water
Answer:
C.
Explanation:
The pressure at the bottom of a jug filled with water does not depend on
A mass m = 1.2 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 130 N/m and negligible mass. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(Ït â Ï). The positive y-axis points upward.a. Find the angular frequency of oscillations in radians per second.b. Determine the value of A in meters.c. Determine the value of Ï in radians.d. Enter an expression for velocity along y axis as function of time in terms of A, Ï and t using the value of Ï from part c.e. What is the velocity of mass at time t = 0.25 s?f. What is the magnitude of mass's maximum acceleration?
Answer:
a. 3π/2 b. 0.36 m c. 0.234 m/s d. 42.55 m/s²
Explanation:
Here is the complete question
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 130 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y ( t ) = A c o s ( ω t − ϕ ) , with the positive y-axis pointing upward. At time t = 0 the mass is observed to be passing through its equilibrium height with an upward speed of v 0 = 3.9 m/s.
A. Find the smallest positive value of ϕ , in radians.
B. Calculate the value of A in meters.
C. What is the mass's velocity along the y-axis in meters per second, at the time t = 0.15 s?
D. What is the magnitude of the mass's maximum acceleration, in meters per second squared?
Solution
a. Since y ( t ) = A c o s ( ω t − ϕ ), the smallest possible value for ϕ is gotten when c o s ( ω t − ϕ ) = 0 ⇒ ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0, ω t − ϕ = ω 0 − ϕ = 0 − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
Since this is a negative angle, we add 2π to the right side.
So, ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
b. Since v = Aω = A√(k/m) where v = maximum velocity at time t = 0 = 3.9 m/s. A = amplitude, k = spring constant = 130 N/m and m = mass = 1.1 kg
A = v/√(k/m) = 3.9 m/s/√(130 N/m/1.1 kg) = 3.9/√118.18 = 3.9/10.87 = 0.36 m
c. To find its velocity, we differentiate y(t)
So, v = dy(t)/dt = dA c o s ( ω t − ϕ )/dt = -'ωAsin( ω t − ϕ ) = v₀sin( ω t − ϕ )
v = v₀sin( ω t − ϕ ) = v₀sin( ω t − ϕ)
Substituting the value of the variables,
v = 3.9sin( 10.87 t − 3π/2)
At t = 0.15 s,
v = 3.9sin( 10.87 × 0.15 − 3π/2)
v = 3.9sin( 1.6305 − 4.7124)
v = -'3.9sin( -3.0819)
v = -'3.9 × - 0.06
v = 0.234 m/s
d. The maximum acceleration, a
a = Aω² = Ak/m = 0.36 × 130/1.1 = 42.55 m/s²
Answer:
a) F = 10.4 rad/s
b) A = 0.375 m
c) ϕ = 3π/2
d) V(t) = -ωAsin( ω t - 3π/2 )
e) V = 0.144 m/s
f) a = 40.625 m/s²
Explanation: Given that
mass m = 1.2 kg
The spring constant k = 130 N/m Time t = 0
Distance d = 0.35 m
y( t ) = A c o s ( ω t − ϕ )
At time t = 0
Speed of Vo = 3.9 m/s.
a) Find the angular frequency of oscillations in radians per second
W = √(k/m)
2πF = √(k/m)
F = 1/2π√(k/m)
F = 1/2π √(130/1.2)
F = 1.66Hz
1 Henz = 6.28 rad/s therefore,
F = 1.66 × 6.28
F = 10.4 rad/s
b) Calculate the value of A in meters.
V = Aω = A√(k/m)
where V = 3.9 m/s the maximum velocity at time t = 0
A = amplitude
A = v/√(k/m)
A = 3.9/√(130/1.2)
A = 3.9/10.4
A = 0.375 m
c. Determine the value of ϕ in radians
If y( t ) = A c o s ( ω t − ϕ ) We can obtain the smallest possible value of ϕ when c o s ( ω t − ϕ ) = 0
ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0,
ω(0) − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
This is a negative angle, let us add 2π to the right side. So,
ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
d. Enter an expression for velocity along y axis as function of time in terms of A, ϕ and t using the value of ϕ from part c.
To find expression for velocity, we differentiate y(t) with respect to time t So,
V = dy/dt = dA c o s ( ω t − ϕ )/dt
V = -ωAsin( ω t − ϕ )
Therefore
V(t) = -ωAsin( ω t + π/2 )
Or
V(t) = -ωAsin( ω t - 3π/2 ) ...... (1)
e. What is the velocity of mass at time t = 0.25 s?
From equation (1)
V(t) = V₀sin( ω t − ϕ )
Substituting the value of the variables,
V = 3.9sin( 10.4t − 3π/2)
At t = 0.25 s,
V = 3.9sin( 10.4 × 0.25 − 3π/2)
V = 3.9sin( 2.6 − 4.7124)
V= -3.9sin( -2.1124)
V= -3.9 × - 0.037
V = 0.144 m/s
f. What is the magnitude of mass's maximum acceleration?
The maximum acceleration a = Aω²
a = Aω² =
a = Ak/m
a = 0.375 × 130/1.2
a = 40.625 m/s²
If love is the answer, then what is the question?
Answer:
what is happiness
Explanation:
Answer:
Love is the answer to the question of how do you make someone do something that is not to their benefit. How do we stop bad people from doing bad things? How do we overcome greed? How can we find happiness if we always know that we have a limited time on earth?
Explanation:
Several years from now you have graduated with an engineering/physics degree from OSU and have been hired by a nanoengineering firm as an intern. You have been assigned to work under a top engineer from the company. Their current project is to design a microscopic oscillator as a time keeping device. The engineering design involves placing a negative charge at the center of a very small positively charged metal ring. Your boss claims that the negative charge will undergo simple harmonic motion of displaced away from the center of the ring. Furthermore, they claim they can change the period (timing) of oscillation by adjusting the amount of charge on the ring. The first task they give you is to check the validity of their design.Consider a charge −???? located a small distance z above the center of a positively charged ring with total charge +Q and radius R. Write an expression for the net force exerted on the charge −???? due to the ring of charge. What is the magnitude of the force on the charge −???? if it is at the location z = 0?
Complete Question in order
See the first image attached
Answer and Explanation:
The three images attached discussed the solution to this question
second Image
Third Image
and Fourth Image
If the images are not clear enough right-click on it and open in a new tab
1. The current through a light bulb connected across the terminals of a 120 V outlet is 0.50 A. At what rate does the bulb convert electric energy to light? 2. A 12.0 V battery causes a current of 2.0 A to flow through a lamp. What is the power used by the lamp? 3. What current flows through a 100. W light bulb connected to a 120. V outlet? 4. The current through a motor is 210 A. If a battery keeps a 12.0 V potential difference across the motor, what electric energy is delivered to the motor in 10.0 s?
Answer:
1. 60 W
2. 24 W
3. 0.83 A
4. 25200 J
Explanation:
1. What we are simply asked to look for is Electrical Power. It is the rate electrical energy is being transferred.
It is given as:
[tex]P = IV[/tex]
where I = current and V = voltage.
Therefore, Power is:
P = 0.50 * 120 = 60 W
2. Power, as given in the formula above is:
P = 2.0 * 12
P = 24.0 W
3. According to the formula of Power, current is given as:
[tex]I[/tex] = [tex]\frac{P}{V}[/tex]
Power is 100 W and voltage is 120 V, therefore, current is:
[tex]I = \frac{100}{120} \\\\\\I = 0.83 A[/tex]
4. Recall that power is the time rate of transfer of electrical energy. Mathematically:
[tex]P = \frac{E}{t}[/tex]
where t = time
This means that Electrical energy is:
[tex]E = Pt[/tex]
Recall that Power is:
[tex]P = IV[/tex]
Therefore, Electrical energy is:
[tex]E = IVt[/tex]
[tex]E = 210 * 12 * 10\\\\\\E = 25200 J[/tex]
1. At "60 W" the bulb convert electric energy to light
2. Power used by lamp will be "24.0 W".
3. The current flow will be "0.83 A".
4. The electric energy delivered to the motor will be "25200 J".
Potential difference and CurrentAccording to the question,
1. Current, I = 0.50 A
Voltage, V = 120 V
We know,
→ Power (P) = Current (I) × Voltage (V)
= 0.50 × 120
= 60 W
2. Current, I = 2.0 A
Voltage, V = 12.0 V
then,
→ P = 2.0 × 12
= 24.0 W
3. Power, P = 100 W
Voltage, V = 120 V
We know,
→ Current, I = [tex]\frac{P}{V}[/tex]
= [tex]\frac{100}{120}[/tex]
= 0.83 A
4. Current, I = 210 A
Potential difference, V = 12.0
We know,
→ E = Pt
P = IV
or,
The electrical energy be:
→ E = IVt
= 210 × 12 × 10
= 25200 J
Thus the above answer is correct.
Find out more information about power here:
https://brainly.com/question/14306881
Un móvil, que sale desde un punto situado 3 metros a la izquierda del origen y lleva un movimiento uniforme, se sitúa a 12 metros a la derecha del origen al cabo de 3 segundos. Tras esto invierte el sentido del movimiento, empleando 4 segundos más en llegar al origen. Realiza una gráfica que represente el movimiento descrito. Obtén el desplazamiento en cada tramo. (Sol: 15 m; −12 m) Determina la distancia total recorrida por el móvil. (Sol: 27 m) Determina la velocidad del móvil en cada tramo. (Sol: 5 m/s; −3 m/s) Representa gráficamente la velocidad del móvil frente al tiempo.
Answer:
x₁ = 15 m, x₂ = 12 m , x_total = 27m, v₁ = 5 m / s , v₂ = - 3 m / s
Explanation:
In this exercise we will use the kinematics of uniform motion
v = d / t
let's apply this equation for the first move
v₁ = Δx / t = (x₂-x₀) / t
v₁ = (12- (-3)) / 3
v₁ = 5 m / s
the distance traveled is x₁ = 15 m
Now let's analyze the return movement
v₂ = Δx / dt
v₂ = (0 - 12) / 4
v₂ = - 3 m / s
The negative sign indicates that the vehicle is moving to the left
the distance traveled is x₂ = 12 m
The total dystonia is
x_total = x₁ + x₂
x_total = 15 +12
x_total = 27m
In the attached we have the graphics of the movement
A charged particle is surrounded by an electric field and a magnetic field
Answer:
yeah
Explanation:
electric fields help the charged particles interact
and isn't magnetic field the same as electric field.
A roadway for stunt drivers is designed for racecars moving at a speed of 97 m/s. A curved section of the roadway is a circular arc of 420 m radius. The roadway is banked so that a vehicle can go around the curve with the friction force from the road equal to zero. At what angle is the roadway banked?
Answer:
Explanation:
Given that,
The speed of the car is
Vc = 97m/s
The radius of circular path of the car is
Rc = 420m
We want to find the angle of roadway banked β?
To determine the angle of roadway banked, we will use the formula
tanβ = Vc² / Rc•g
Where Vc = 97m/s, Rc = 420m and
g = 9.8m/s²
Then
tanβ = 97² / (420 × 9.8)
tanβ = 2.28596
β = ArcTan ( 2.28596)
β = 66.37°
The railway banked at an angle of 66.37°
Answer:
Banking angle is 66.35°
Explanation:
Given radius r=420m
Speed=97m/s
banking angle is A
Note before
(V)=√(r*gtanA)
√97=√420*9.81*tanA)taking square of both sides
97^2=420*9.81*tan A.
tanA=66.35°
A=66.35°
All particles made of quarks are called _____.
electrons
gluons
hadrons
nucleons
Answer:
All particles made of quarks are called hadrons.
All particles madeup of two or more quarks held together by a force are called ; ( C ) Hadrons
What are Hadrons ?Hadrons are subatomic particles which are made of two or more quarks which are held firmly by a force similar to electric force. An example of Hadron is Baryon which contains three quarks which are held closely together.
Hence we can conclude that All particles made of quarks are called Hadrons
Learn more about Hadrons : https://brainly.com/question/3734714
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in the bearings causes the wheel to stop in just 13 s . If the moment of inertia of the wheel about its axle is 0.33 kg⋅m2 , what is the magnitude of the frictional torque?
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.44 m long. The block is initially at rest. A bullet with mass 0.0134 kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m , the tension in the cord is 4.76 N .
What was the initial speed v0 of the bullet?
Answer:
v₀ = 240 m / s
Explanation:
This problem must be solved in two parts, first we must use the conservation of the moment, then the conservation of energy.
Let's start by applying moment conservation, to the system formed by the block and bullet, in this case the forces during the crash are internal and the moment is conserved
Instant starts. Before the crash
p₀ = m v₀
Final moment. Right after the crash
[tex]p_{f}[/tex] = (m + M) v
The moment is preserved
Po =p_{f}
M v₀ = (m + M) v
v = m / (m + M) v₀ (1)
This is the speed with which the bullet block system comes out, now we can use energy conservation
Starting point. Right after the crash
Em₀ = K = ½ (m + M) v²
Final point. Highest point of the path
[tex]Em_{f}[/tex] = U = (m + M) g y
Em₀ = Em_{f}
½ (m + M) v² = (m + M) g y
v = √2 g y (2)
We substitute 1 in 2
m / (m + M) v₀ = √ 2gy
v₀ = (m + M) / m √ 2gy
Let's calculate
v₀ = (0.0134 +0.800) /0.0134 √ (2 9.8 0.8)
v₀ = 240 m / s
Answer:
298.04 m/s
Explanation:
Let m = mass of bullet = 0.0134 kg and M = mass of block = 0.800 kg.
Since the bullet becomes embedded in the block and rises a vertical height,h = 0.800 m
The kinetic energy change of mass + block = potential energy of mass + block at height, h
ΔK = -ΔU
So, 1/2(m + M)(v² - V²) = -[(m + M)gh - 0] where v is the velocity of the bullet + block at height, h. Since the tension, T is the centripetal force at height, h, it follows that
T = (m + M)v²/r r = length of cord = 1.44 m
v = √(Tr/(m + M)) = √4.76 N × 1.44 m/(0.800 + 0.0134)kg = √(6.8544/0.8134) = √8.427 = 2.9 m/s
So. 1/2(v² - V²) = -gh
v² - V² = -2gh
V = √(v² + 2gh) = √((2.9 m/s)² + 2 × 9.8 m/s² × 0.8 m) = √(8.41 + 15.68) = √24.09 = 4.91 m/s
This is the velocity of the bullet plus block at collision.
From the law of conservation of momentum,
momentum of bullet = momentum of bullet plus block
mv₀ = (m + M)V where v₀ = initial speed of bullet
v₀ = (m +M)V/m = (0.0134 kg + 0.800 kg)4.91 m/s ÷ 0.0134 kg = 3.994 ÷ 0.0134 kg = 298.04 m/s
From the edge of a cliff, a 0.55 kg projectile is launched with an initial kinetic energy of 1550 J. The projectile's maximum upward displacement from the launch point is +140 m. What are the (a) horizontal and (b) vertical components of its launch velocity?
Answer:
(a) 38.5m/s
(b) 64.4m/s
Explanation:
First, we can obtain the launch speed from the definition of kinetic energy:
[tex]K=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2K}{m}}\\\\[/tex]
Plugging in the given values, we obtain:
[tex]v=\sqrt{\frac{2(1550J)}{0.55kg}}\\\\v=75.0m/s[/tex]
Now, from the conservation of mechanical energy, considering the instant of launch and the instant of maximum height, we get:
[tex]E_0=E_f\\\\K_0=U_g_f+K_f\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0_x^2\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0^2\cos^2\theta\\\\\implies \cos\theta=\sqrt{1-\frac{2gh_f}{v_0^2}}[/tex]
And with the known values, we compute:
[tex]\cos\theta=\sqrt{1-\frac{2(9.8m/s^2)(140m)}{(75.0m/s)^2}}\\\\\cos\theta=0.513\\\\\theta=59.12\°[/tex]
Finally, to know the components of the launch velocity, we use trigonometry:
[tex]v_0_x=v_0\cos\theta=(75.0m/s)\cos(59.12\°)=38.5m/s\\\\v_0_y=v_0\sin\theta=(75.0m/s)\sin(59.12\°)=64.4m/s[/tex]
It means that the horizontal component of the launch velocity is 38.5m/s (a) and the vertical component is 64.4m/s (b).
Work is done on a ball when a soccer player kicks it. Is the player still doing work on the ball as it rolls across the ground? Explain
Answer:
Energy is conserved.
Explanation:
According to law of conservation of energy ,energy can neither be created nor be destroyed .
When the player kicks the soccer ball, the kinetic energy is transferred to the ball and it will roll for a while and then stops. The combination of force and distance is called as work.Greater the force faster the ball will go.
The force with which the soccer player player kicked the ball will be having it effects on the ball. As the ball rolls in the ground kinetic energy which the ball got from the player will be lost in the form of heat due to friction.
According to the Newton first law of motion,the object continues to remain in rest or move with constant speed unless acted upon by the external force.
A 5.00-kg box is suspended from the end of a light vertical rope. A time dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to v(t)=(2.00m/s2)t+(0.600m/s3)t2. What is the tension in the rope when the velocity of the box is 9.00 m/s?
Answer:
T = 74.3N
Explanation:
We are given;
v(t) = (2.00m/s²)t+(0.600m/s³)t²
So, when v = 9 m/s;
9 = 2t + 0.6t²
0.6t² + 2t - 9 = 0
Solving this quadratic equation,
t = -5.88 or 2.55
We'll pick t = 2.55 s
Now, kinematic acceleration will be the derivative of the acceleration.
Thus, a = dv/dt = 2 + 1.2t
So, acceleration at that time t = 2.55s is; a = 2 + 1.2(2.55) = 5.06 m/s²
Since the rope is subject to both acceleration and gravity, Tension is;
T = mg + ma
T = m(g + a)
T = 5(9.8 + 5.06)
T = 74.3N
Think about multiplying the mass of each student by a factor to calculate each student's kinetic energy. Is there a
common factor that works for every student? If so, what's this factor?
Answer:
Multiplying the mass of any student by a factor of 4.5 gives the kinetic energy of the student.
Explanation:
because
Answer: the question was asked to think about how to multiply a mass speech student by a factor to include the student. Kinetic energy, are there any common factors that appeal to them? The table shows mass energy and constant velocity. Let's calculate a constant factor, multiply it by mass, get the energy and see if we can do it. It's very easy. You can do it just in case. Only kinetic energy divided by mass can be produced. That is, the first is 100 and 21.5 divided by 27. Now you have the next 4.5. This is one. Then it is 162 divided by 36. This is 4.5. January 3, 202
Divide 5 by 45 to get 4.5. And hopefully you can see the patterns that are appearing here. In the fourth one, 243 divided by 54 gives 4.5. The fifth 288 divided by 64. Since it is 4.5, you can guess what it will be like. And finally, in the sixth case, it's 3 to 8.5 divided by 73. This is also equivalent to 4. Well, this common factor 4.5 was found.
What is kinetic energy?Kinetic energy is a form of energy that an object or particle has due to its movement. When work is done on an object by applying a net force that transfers energy, the object accelerates, thereby gaining kinetic energy. In physics, the kinetic energy of an object is the energy that the object has due to its motion.
This is defined as the work required to accelerate an object of a given mass from a stationary state to a specified velocity. After the body gains this energy during acceleration, the body retains this kinetic energy unless the velocity changes.
All moving objects use kinetic energy. Examples of kinetic energy are walkers, thrown baseball, bread crumbs falling from a table, and charged particles in an electric field.
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A migrating salmon heads in the direction N 45° E, swimming at 5 mi/h relative to the water. The prevailing ocean currents flow due east at 2 mi/h. Find the true velocity of the fish as a vector. (Assume that the i vector points east, and the j vector points north.)
Answer:
Explanation:
velocity of salmon with respect to water, v(s,w) = 5 mi/h at N 45° E
velocity of water with respect to ground, v(w,g) = 2 mi/h due east
Let the true velocity of salmon is velocity of salmon with respect to water is v(s,g)
First write the velocities in vector from
[tex]\overrightarrow{v}_{s,w}=5(Cos 45\widehat{i}+Sin 45\widehat{j})=3.54\widehat{i}+3.54\widehat{j}[/tex]
[tex]\overrightarrow{v}_{w,g}=2\widehat{i}[/tex]
Using the formula of relative speed,
[tex]\overrightarrow{v}_{s,w}=\overrightarrow{v}_{s,g}-\overrightarrow{v}_{w,g}[/tex]
[tex]3.54\widehat{i}+3.54\widehat{j}=\overrightarrow{v}_{s,g}-2\widehat{i}[/tex]
[tex]\overrightarrow{v}_{s,g}=5.54\widehat{i}+3.54\widehat{j}[/tex]
This i the true velocity of salmon.
The true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].
The given parameters;
velocity of the salmon relative to the water, = 5 mi/hdirection of the velocity, θ = 45⁰ocean currents, = 2 mi/hThe true velocity of the fish as a vector is calculated as follows;
[tex]v_f = v\ cos(\theta)i \ + \ v\ sin(\theta)j \ + \ 2i\\\\v_f = 5cos(45) i \ + 5sin(45)j \ + \ 2i\\\\v_f = 3.54i \ + \ 3.54j \ + 2i\\\\v_f = 5.54i \ + \ 3.54 j[/tex]
Thus, the true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].
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How does the eye and brain work together to give you perception of color
Answer:
The human eye and brain together translate light into color. Light receptors within the eye transmit messages to the brain, which produces the familiar sensations of color. ... Rather, the surface of an object reflects some colors and absorbs all the others. We perceive only the reflected colors.
Explanation:
Final answer:
The eye and brain work together to perceive color. Light enters the eye and is detected by cones in the retina, which send signals to the brain. The brain processes these signals and combines information from different cones to create the perception of color.
Explanation:
The eye and brain work together to give you perception of color through a complex process. When light enters the eye, it is detected by special cells called cones in the retina. These cones are sensitive to different wavelengths of light and send signals to the brain. The brain then processes these signals and combines the information from different cones to create the perception of color.
For example, if you see a red object, the red cones in your eye will be activated and send signals to the brain. The brain then interprets these signals as the color red. Similarly, the green cones are activated by green light and the blue cones by blue light. By combining the information from all three types of cones, the brain is able to perceive a wide range of colors.
To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is 5.0×103kg/m35.0×10
3
kg/m
3
. What fraction of the specimen's apparent volume is solid?
Answer:
Fraction of the specimen's is 0.4.
Explanation:
We know,
Mass = volume × density
Weigh= mass × g
= volume × density× g
= density× g × volume
[tex]=\rho.g.V[/tex]
An object weighs less submerged due to buoyant force acting on it.
[tex]\therefore W_{wet}= W_{dry}-B[/tex]
[tex]B= W_{dry}-W_{wet}[/tex]
[tex]=W_{\textrm{fluid displaced}}[/tex]
[tex]=\rho_{fluid}. g.V_{submerged}[/tex]
Given that, the weighs of the specimen in dry is twice of the weighs in air.
[tex]W_{wet}=\frac 12W_{dry}[/tex]
Then ,
[tex]B= W_{dry}-W_{wet}[/tex]
[tex]= W_{dry}-\frac12W_{dry}[/tex]
[tex]=\frac12W_{dry}[/tex]
[tex]=\rho_{Rock}. g.V_{Rock}[/tex]
Therefore,
[tex]\rho_{fluid}. g.V_{submerged}=\frac12\rho_{Rock}. g.V_{Rock}[/tex]
[tex]\Rightarrow \rho_{Rock}. g.V_{Rock}=2\rho_{fluid}. g.V_{submerged}[/tex]
[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\rho_{fluid}. g}{\rho_{Rock}.g}[/tex]
[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\rho_{fluid}}{\rho_{Rock}}[/tex]
[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\times 1.0 \times 10^3\ kg /m^3}{5.0\times 10^3 \ kg/m^3}[/tex]
=0.4
Fraction of the specimen's is 0.4.
105. A cable with a linear density of μ=0.2kg/m is hung from telephone poles. The tension in the cable is 500.00 N. The distance between poles is 20 meters. The wind blows across the line, causing the cable resonate. A standing waves pattern is produced that has 4.5 wavelengths between the two poles. The speed of sound at the current temperature T=20°C is 343.00m/s . What are the frequency and wavelength of the hum?
Answer: 11.5 Hz, 29.83 m
Explanation:
Given
Linear density of the cable, μ = 0.2 kg/m
Tensión in the cables, T = 500 N
Wavelength of the wave, = 4.5 Waves
Distance between the poles, L = 20 m
Temperature of, t = 20° C
Speed of sound, v = 343 m/s
λ = length / number of waves =
λ = 20 / 4.5
λ = 4.44 m
Frequency of a standing wave is the same as frequency of a hum. Calculated using the formula
F = n/2L * √(T/μ)
F = 1/λ * √(T/μ)
F = 1/4.44 * √(500/0.2)
F = 0.23 * √2500
F = 0.23 * 50
F = 11.5 Hz
Wavelength of the hum,
λ = v/f
λ = 343 / 11.5
λ = 29.83 m
The wavelength of the hum produced by the cable is approximately 4.44 meters, and the frequency is approximately 11.26 Hz, calculated using the known linear density, tension, and distance between poles.
Explanation:The student is asking about the properties of a standing wave created when the wind blows across a cable hung between two poles. Given the linear density (µ) of 0.2 kg/m, the tension in the cable (500 N), and the distance between poles (20 meters), we are to find the frequency and wavelength of the hum produced by the cable.
Firstly, we can calculate the wavelength (λ) of the wave using the information that 4.5 wavelengths fit between the two poles that are 20 meters apart:
λ = 20 meters / 4.5 wavelengths = 4.44 meters (approx)
Next, we can find the speed (v) of the wave on the cable using the formula:
v = sqrt(T / µ)
Plugging in the given values, we get:
v = sqrt(500 N / 0.2 kg/m) = sqrt(2500 m2/s2)
v = 50 m/s
Now, using the wave speed (v) and the wavelength (λ), we can calculate the frequency (f) of the wave using the formula:
f = v / λ
f = 50 m/s / 4.44 m = 11.26 Hz (approx)
Therefore, the wavelength is approximately 4.44 meters, and the frequency of the hum is approximately 11.26 Hz.
find the final speed and the time taken for a skier who skies 75 m along a slope that is 15°, measured from horizontal, in the following situations (you may neglect friction). How long (in s) does it take for the skier to reach the bottom of this hill starting from rest?
Answer:
Final speed = 148.21m.s
Time of flight = 7.82seconds
Explanation:
The motion of the body is a projectile motion.
Projectile is a motion created by an object launched in air and allowed to fall to freely under the influence of gravity.
Taking the maximums height reached H = 75m
Angle of launch = 15°
Using the maximum height formula to get the velocity U of the object
H = U²sin²theta/2g
Where g is the acceleration due to gravity = 9.81m/s²
75 = U²(sin15°)²/2(9.81)
1471.5 = U²(sin15°)²
1471.5 = 0.06699U²
U² = 1471.5/0.06699
U² = 21,965.9
U = √21,965.9
U = 148.21m/s
The time taken for the skier to reach the bottom of this hill starting from rest is the time of flight T.
T = 2Usintheta/g
T = 2(148.21)sin15°/9.81
T = 296.42sin15°/9.81
T = 76.72/9.81
T = 7.82seconds
A mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 30. cm3. If the combustion of this mixture releases 984 J of energy, to what volume (in L) will the gases expand against a constant pressure of 648 torr if all the energy of combustion is converted into work to push back the piston?
Answer:
Explanation:
Given
Original volume V1=30cm^3 converting to L
=30/1000=0.03L
Constant pressure P= 648 tors
Converting to atm; 648 tors*1atm/760 torr=0.853 atm
Work=984J= 984**1L/101.33=9.7L.atm
Note before
W= -P(Vfinal-Vinitial)
-9.7/0.853+0.03L=11.68L
Answer:
Final expanded volume 11.43 L
Explanation:
The picture attached shows the solution
The cytoplasm is the watery fluid found within cells. The cytoplasm holds all of the organelles, except _______, in place within the cell.
A.
chloroplasts
B.
mitochondria
C.
vacuoles
D.
the nucleus
Answer:
The cytoplasm holds all the organelles, except the nucleus.
Explanation:
The cytoplasm holds all organelles, with the primary exception being the nucleus, which is separated by a nuclear envelope. Other organelles like chloroplasts, mitochondria, and vacuoles, are held within the cytoplasm.
Explanation:The cytoplasm, the watery fluid inside cells, houses almost all the organelles. The primary organelle that it does not encompass is the nucleus. The nucleus stays separated from the cytoplasm by a nuclear envelope.
Other organelles such as chloroplasts, mitochondria, and vacuoles are embedded within the cytoplasm. These organelles perform various roles within cells, while the nucleus controls the cell's activities and contains the cell's genetic information.
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If the speed of a wave is 150 m/s and it’s frequency is 2 Hz , what is its wavelength ?
Answer:
75 m
Explanation:
wavelength = frequency/ speed = 75 m
If a truck driver is driving at 30 mph and she makes a U-turn then starts driving 30 mph in the opposite direction did the driver speed or velocity change after changing direction
The final velocity of the truck driver does not change after making a U-turn and driving in the opposite direction.
Explanation:The final velocity of the truck driver does not change after making a U-turn and driving in the opposite direction. While the truck's speed changes, the velocity remains the same because velocity takes into account both the speed and direction of motion. Since the speed remains constant at 30 mph and the direction changes, the velocity is still 30 mph, just in the opposite direction.
_________ is a layer of the earth that is classified not by composition
Answer:
Asthenosphere
Explanation:
The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.
You find it takes 200 J of heat to take 4 kg of an unknown substance from 200 K to 240 K. It does not change phases during this interval. If you added 300 J of heat to the substance instead of 200 J, what would be its final temperature?
Answer:
300 K
Explanation:
First, we have find the specific heat capacity of the unknown substance.
The heat gained by the substance is given by the formula:
H = m*c*(T2 - T1)
Where m = mass of the substance
c = specific heat capacity
T2 = final temperature
T1 = initial temperature
From the question:
H = 200J
m = 4 kg
T1 = 200K
T2 = 240 K
Therefore:
200 = 4 * c * (240 - 200)
200 = 4 * c * 40
200 = 160 * c
c = 200/160
c = 1.25 J/kgK
The heat capacity of the substance is 1.25 J/kgK.
If 300 J of heat is added, the new heat becomes 500 J.
Hence, we need to find the final temperature, T2, when heat is 500 J.
Using the same formula:
500 = 4 * 1.25 * (T2 - 200)
500 = 5 * (T2 - 200)
100 = T2 - 200
=> T2 = 100 + 200 = 300 K
The new final temperature of the unknown substance is 300K.