If you buy an amateur-sized reflecting telescope, say around 10 inches (25cm) aperture, it'll have something in it that sends the gathered starlight out the side of the telescope tube. What do we call this thing
Answer: objective lens
Explanation:
Light enters a refra
Light enters a telescope through a lens at the upper end, which focuses the light near the bottom of the telescope. An eyepiece then magnifies the image so that it can be viewed by the eye, or a detector like a photographic plate can be placed at the focus. The upper end of a reflecting telescope is open, and the light passes through to the mirror located at the bottom of the telescope. The mirror then focuses the light at the top end, where it can be detected. Alternatively, a second mirror may reflect the light to a position outside the telescope structure, where an observer can have easier access to it.
A bullet with mass 5.35 g is fired horizontally into a 2.174-kg block attached to a horizontal spring. The spring has a constant 6.17 102 N/m and reaches a maximum compression of 6.34 cm.
(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.
Answer:
a)[tex]V=1.067\: m/s[/tex]
b)[tex]v=434.65\: m/s [/tex]
Explanation:
a)
Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.
[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]
Where:
M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)V is the speed of the systemk is the spring constant (6.17*10² N/m)Δx is the compression of the spring (0.0634 m)Then, let's find the initial speed of the bullet-block system.
[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]
[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]
[tex]V=1.067\: m/s[/tex]
b)
Using the conservation of momentum we can find the velocity of the bullet.
[tex]mv=MV[/tex]
[tex]v=\frac{MV}{m}[/tex]
[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]
[tex]v=434.65\: m/s [/tex]
I hope it helps you!
A block of mass M is connected by a string and pulley to a hanging mass m.
The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg.
b. Find the acceleration of the system and tensions on the string.
c. How far will block m drop in the first seconds after the system is released?
d. How long will block M move during the above time?
e. At the time, calculate the velocity of block M
f. Find out the deceleration of block M if the connection string is removal by cutting after the first second. Then, calculate the time taken to contact block M and pulley
How far will block m drop in the first seconds after the system is released?
(b) Use Newton's second law. The net forces on block M are
• ∑ F (horizontal) = T - f = Ma … … … [1]
• ∑ F (vertical) = n - Mg = 0 … … … [2]
where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.
Right away, we see n = Mg, and so f = µn = 0.2Mg.
The net force on block m is
• ∑ F = mg - T = ma … … … [3]
You can eliminate T and solve for a by adding [1] to [3] :
(T - 0.2Mg) + (mg - T ) = Ma + ma
(m - 0.2M) g = (M + m) a
a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)
a = 1.96 m/s²
We can get the tension from [3] :
T = m (g - a)
T = (10 kg) (9.8 m/s² - 1.96 m/s²)
T = 78.4 N
(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of
1/2 (1.96 m/s²) t
(e) Assuming block M starts from rest, its velocity at time t is
(1.96 m/s²) t
(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have
∑ F = -f = Ma
The effect of friction is constant, so that f = 0.2Mg as before, and
-0.2Mg = Ma
a = -0.2g
a = -1.96 m/s²
Then block M slides a distance x such that
0² - (1.96 m/s²) = 2 (-1.96 m/s²) x
x = (1.96 m/s²) / (2 (1.96 m/s²))
x = 0.5 m
(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)
Meanwhile, block m would be in free fall, so after 1 s it would fall a distance
x = 1/2 (-9.8 m/s²) (1 s)
x = 4.9 m
A source emits sound at a fixed constant frequency f. If you run towards the source, the frequency you hear is
Answer:
increased because as you run into each sound wave the time between each sound decreases meaning the period of each wave decreases to your years and since f=1/T and T is decreasing by greater than 0, f must increase.
Explanation:
Consider a tall building of height 200.0 m. A stone A is dropped from the top (from the cornice of the building). One second later another stone B is thrown vertically up from the point on the ground just below the point from where stone A is dropped.Birthstones meet at half the height of the tower. (a) Find the initial velocity of vertical throw of stone B.(b) Find the velocities of A and B, just before they meet.
Answer:
a) v₀ = 44.27 m / s, b) stone A v = 44.276 m / s, stone B v = 0.006 m / s
Explanation:
a) This is a kinematics exercise, let's start by finding the time it takes for stone A to reach half the height of the building y = 100 m
y = y₀ + v₀ t - ½ gt²
as the stone is released its initial velocity is zero
y- y₀ = 0 - ½ g t²
t = [tex]\sqrt{ -2(y-y_o)/g}[/tex]
t = [tex]\sqrt{ -2(100-200)/9.8}[/tex]
t = 4.518 s
now we can find the initial velocity of stone B to reach this height at the same time
y = y₀ + v₀ t - ½ g t²
stone B leaves the floor so its initial height is zero
100 = 0 + v₀ 4.518 - ½ 9.8 4.518²
100 = 4.518 v₀ - 100.02
v₀ = [tex]\frac{100-100.02}{4.518}[/tex]
v₀ = 44.27 m / s
b) the speed of the two stones at the meeting point
stone A
v = v₀ - gt
v = 0 - 9.8 4.518
v = 44.276 m / s
stone B
v = v₀ -g t
v = 44.27 - 9.8 4.518
v = 0.006 m / s
For waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths.
a) true
b) false
Answer:
false.
Explanation:
We know that for a wave that moves with velocity V, with a wavelength λ, and a frequency f, we have the relation:
V = λ*f
So, if the velocity is constant and we increase the frequency to:
f' > f
we will have a new wavelength λ'
Such that:
V = f'*λ'
And V = f*λ
Then we have:
f'*λ' = f*λ
Solvinf for λ', we get:
λ' =(f/f')*λ
And because:
f' > f
then:
(f/f') < 1
Then:
λ' =(f/f')*λ < λ
So, if we increase the frequency, we need to decrease the wavelength.
So, for higher frequency waves, we must have proportionally shorter wavelengths.
Then we can conclude that the given statement:
"or waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths"
is false.
Light of intensity I0 and polarized horizontally passes through three polarizes. The first and third polarizing axes are horizontal, but the second one is oriented 20.0� to the horizontal. In terms of I0, what is the intensity of the light that passes through the set of polarizers?
A) 0.442 I0
B) 0.180 I0
C) 0.780 I0
D) 0.883 I0
Answer:
Option C.
Explanation:
Suppose that we have light polarized in some given direction with an intensity I0, and it passes through a polarizer that has an angle θ with respect to the polarization of the light, the intensity that comes out of the polarizer will be:
I(θ) = I0*cos^2(θ)
Ok, we know that the light is polarized horizontally and comes with an intensity I0
The first polarizer axis is horizontal, then the intensity after this polarizer is:
then θ = 0°
I(0°) = I0*cos^2(0°) = I0
The intensity does not change. The axis of polarization does not change.
The second polarizer is oriented at 20° from the horizontal, then the intensity that comes out of this polarizer is:
I(20°) = I0*cos^2(20°) = I0*0.88
And the axis of polarization of the light that comes out is now 20° from the horizontal
Now the light passes through the last polarizer, which has an axis oriented horizontally, so the final intensity of the light will be:
note that here the initial polarization is I0*0.88
and the angle between the axis is 20° again.
Then the final intensity is:
I(20°) = I0*0.88*cos^2(20°) = I0*0.78
Then the correct option is C.
A television tube can accelerate electrons to 2.00 · 104 ev. Calculate the wavelength of emitted X-rays with the highest energy.
λ = _____ m
9.9 x 10 -30
6.2 x 10 -11
1.6 x 10 10
7.1 x 10 -57
Answer:
6.2 × 10^-11 m
Explanation:
1 eV = 1.602 × 10-19 joule
2.00 × 104 ev. = 2.00 × 10^4 eV × 1.602 × 10^-19 joule/1eV
= 3.2 × 10^-15 J
From;
E= hc/λ
λ = hc/E
λ = 6.6 × 10^-34 × 3 × 10^8/3.2 × 10^-15
λ = 6.2 × 10^-11 m
A 100-m long transmission cable is suspended between two towers. If the mass density is 18.2 g/cm and the tension in the cable is 6543 N, what is the speed (m/s2) of transverse waves on the cable
After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits
Answer:
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Explanation:
Given :
The first dark fringe is for m = 0
[tex]$\theta_1 = \pm 19^\circ$[/tex]
Now we know for a double slit experiments , the position of the dark fringes is give by :
[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]
The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :
[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex] (since, m = 0)
[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex] or 1 : 1.54
A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should thisresult be recorded in a table of results? a. 0.2000m b. 0.200m c. 0.20m d. 0.2m
Answer:
C
Explanation:
20 cm = 0.2m
since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty
therefore it's 0.200m
A 2.90 m segment of wire supplying current to the motor of a submerged submarine carries 1400 A and feels a 2.00 N repulsive force from a parallel wire 4.50 cm away. What is the direction and magnitude (in A) of the current in the other wire? magnitude A direction
Answer:
[tex]I_2=30.9A[/tex]
Explanation:
From the question we are told that:
Wire segment [tex]l_s=2.9m[/tex]
Initial Current [tex]I_1=1400A[/tex]
Force [tex]F=2.00N[/tex]
Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]
[tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]
[tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]
[tex]I_2=30.9A[/tex]
A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster
Answer:
on the moon, they will fall at the timeon earth, the coin will fall faster to the groundExplanation:
A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.
If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).
Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k
Answer:
parametric representation: x = u, y = v - u , z = - v
Explanation:
Given vectors :
i - j , j - k
represent the vector equation of the plane as:
r ( u, v ) = r₀ + ua + vb
where: r₀ = position vector
u and v = real numbers
a and b = nonparallel vectors
expressing the nonparallel vectors as :
a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )
hence we can express vector equation of the plane as
r(u,v) = ( x₀ + u, y₀ - u + v, z₀ - v )
Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0
( x, y , z ) = ( x₀ + u, y₀ - u + v, z₀ - v )
x = 0 + u ,
y = 0 - u + v
z = 0 - v
∴ parametric representation: x = u, y = v - u , z = - v
How do the magnitude and direction of the electric field on the left side of the dipole compare to the right side for the same distance
Answer:
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
Explanation:
The direction of the electric field due to the dipole on the axial line is same as the direction of dipole moment.
The magnitude of the electric field due to an electric dipole on its axial line is
[tex]E=\frac{2kp}{r^3}[/tex]
where, k is the constant, p is the electric dipole moment and r is the distance from the center of dipole.
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
a tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?
Answer: 2.4 km/hr
Explanation:
Distance = 600m
Time= 15 minutes = 15 x 60 second/minute = 900 seconds
Speed = [tex]\frac{Distance}{Time}[/tex] = [tex]\frac{600}{900}[/tex] = [tex]\frac{2}{3}[/tex] m/sec
⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)
When an object is in free fall, ____________________.
Answer:
Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.
Explanation:
Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.
If an object travels on a circular path is an acceleration? What is changing to cause an acceleration?
93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
A chimpanzee sitting against his favorite tree gets up and walks 51 m due east and 39 m due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound
Answer:
64.20m
Explanation:
As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.
[tex]a^{2} +b^{2} = c^{2}[/tex]
where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x
[tex]51^{2} +39^{2} =x^{2}[/tex]
2,601 + 1,521 = [tex]x^{2}[/tex]
4,122 = [tex]x^{2}[/tex] ... square root both sides
64.20 = x
Finally, we see that the shortest distance is 64.20m
Find the coefficient of kinetic friction between a 3.80-kg block and the horizontal surface on which it rests if an 87.0-N/m spring must be stretched by 6.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.
Answer:
μ = 0.15
Explanation:
Let's start by using Hooke's law to find the force applied to the block
F = k x
F = 87.0 0.065
F = 5.655 N
Now we use the translational equilibrium relation since the block has no acceleration
∑ F = 0
F -fr = 0
F = fr
the expression for the friction force is
fr = μ N
if we write Newton's second law for the y-axis
N -W = 0
N = W = mg
we substitute
F = μ mg
μ = F / mg
μ = [tex]\frac{5.655}{3.8 \ 9.8}[/tex]
μ = 0.15
An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 47.2 H inductor, determine the inductive reactance and the rms current through the inductor.
Answer:
The inductance is 17784.96 ohm and rms current is 4.77 mA.
Explanation:
Voltage, V = 120 V
frequency, f = 60 Hz
Inductance, L = 47.2 H
The rms voltage is
[tex]V_{rms}=\frac{V_o}{\sqrt 2}\\\\V_{rms}=\frac{120}{\sqrt 2}\\\\V_{rms} = 84.87 V[/tex]
The reactance is given by
[tex]X_L = 2\pi f L\\\\X_L = 2\times 3.14\times 60\times 47.2 \\\\X_L = 17784.96 ohm[/tex]
The rms current is
[tex]I_{rms} =\frac{V_{rms}}{X_L}\\\\I_{rms}=\frac{84.87}{17784.96}\\\\I_{rms} = 4.77\times 10^{-3} A = 4.77 mA[/tex]
Why is the melting of ice a physical change?
A. It changes the chemical composition of water.
B. It does not change the chemical composition of water.
C. It creates new chemical bonds.
D. It forms new products.
E. It is an irreversible change that forms new products.
It does not change the chemical composition of water.
You simultaneously release two balls: one you throw horizontally, and the other you drop straight down. Which one will reach the ground first? Why?
(a) The ball dropped straight down lands first, since it travels a shorter distance.
(b) Neither. Their vertical motion is the same, so they will reach the ground at the same time.
(c) It depends on the mass of the balls—the heavier ball falls faster so lands first
Answer:
Option B.
Explanation:
Remember that we can think on any movement as a sum of a movement in the y-axis, the movement in the x-axis, and the movement in the z-axis. And these are not related, this means that, for example, the movement in x does not affect the movement in y.
So, when we analyze the problem of "how long takes an object to hit the ground"
We do not care for the horizontal motion of the object, we only care for the vertical motion of the object.
So, if an object is dropped, and another has a given initial velocity in the x-axis, in both cases the initial velocity in the y-axis will zero.
And in both cases, the only vertical force acting on the balls will be the gravitational force (so both objects will have the same vertical acceleration and the same vertical initial velocity) with this, we already know that the vertical motion of both objects will be exactly the same.
So, both objects will hit the ground at the same time.
(notice that here we are ignoring things like air resistance and other complex forces)
So here the correct option is b: Neither. Their vertical motion is the same, so they will reach the ground at the same time.
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 mls2, and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars A, B, and C change if 39 kg of luggage were removed from car 2 and placed in (a) car I and (b) car 3
Answer:
a) ΔT₁ = -4.68 N, ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N
Explanation:
In this exercise we will use Newton's second law.
∑F = m a
Let's start with the set of three cars
F_total = M a
F_total = M 0.12
where the total mass is the sum of the mass of each charge
M = m₁ + m₂ + m₃
This is the force with which the three cars are pulled.
Now let's write this law for each vehicle
car 1
F_total - T₁ = m₁ a
T₁ = F_total - m₁ a
car 2
T₁ - T₂ = m₂ a
T₂ = T₁ - m₂ a
car 3
T₂ = m₃ a
note that tensions are forces of action and reaction
a) They tell us that 39 kg is removed from car 2 and placed on car 1
m₂’= m₂ - 39
m₁'= m₁ + 39
m₃ ’= m₃
they ask how much each tension varies, let's rewrite Newton's equations
The total force does not change since the mass of the set is the same F_total ’= F_total
car 1
F_total ’- T₁ ’= m₁’ a
T₁ ’= F_total - m₁’ a
T₁ ’= (F_total - m₁ a) - 39 a
T₁ '= T₁ - 39 0.12
ΔT₁ = -4.68 N
car 2
T₁’- T₂ ’= m₂’ a
T₂ ’= T₁’- m₂’ a
T₂ '= (T₁'- m₂ a) + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
b) in this case the masses remain
m₁ '= m₁
m₂ ’= m₂ - 39
m₃ ’= m₃ + 39
we write Newton's equations
car 3
T₂ '= m₃' a
T₂ ’= (m₃ + 39) a
T₂ '= m₃ a + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
car 1
F_total - T₁ ’= m₁’ a
T₁ ’= F_total - m₁ a
car 2
T₁' -T₂ '= m₂' a
T₁ ’= T₂’- m₂’ a
T₁ '= (T₂'- m₂ a) + 39 a
T₁ '= T₁ + 39 0.12
ΔT₁ = 4.68 N
The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.What is tension force?Tension is the pulling force carried by the flexible mediums like ropes, cables and string.
Tension in a body due to the weight of the hanging body is the net force acting on the body.
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.
The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,
[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]
On solving the above 3 equation, we get the values of tension in each bar as,
[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]
Case 1- When 39 kg of luggage were removed from car 2 and placed in car IThe tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]
Case 2- When 39 kg of luggage were removed from car 2 and placed in car IIIThe tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]
Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.
Learn more about the tension here;
https://brainly.com/question/25743940
~~~NEED HELP ASAP~~~
Please solve each section and show all work for each section.
Explanation:
Forces on Block A:
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as
[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]
Substituting (2) into (1), we get
[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]
where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]
Forces on Block B:
Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as
[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]
From (5), we can solve for N as
[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]
Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by
[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]
Substituting (7) into (4) we get
[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]
Collecting similar terms together, we get
[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]
or
[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]
Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]
Receptor elétrico 5 pontos Dispositivo que converte energia elétrica em outra forma de energia, não exclusivamente térmica. Exemplos: motores elétricos, ventiladores, liquidificadores, geladeiras, aparelhos de sons, vídeos, celulares, computadores?
Answer:
Electromechanical transducer and Electrical receiver.
Explanation:
Electromechanical transducer is the part of a communication system which converted electrical waves or electrical energy into sound waves. The most common example loudspeaker while on the other hand, Electrical receiver is a device that converts electrical energy into another form of energy, except thermal. Examples are cell phones, computers and television.
Why is the temperature constant during the melting of water?
[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}[/tex]
Why is the temperature constant during the melting of water?
[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]
[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{ REASON}}}}}}}[/tex]
THE HEAT WE R SUPPLYING TO THE WATER TO RAISE THE TEMP OF THE WATER IS USED BY THE MOLECULES TO BREAK INTERMOLECULAR BONDS WHICH HELP IN THE CHANGING OF THE LATTICE (STRUCTURE) OF THE WATER .
ICE HAS A HEXAGONAL RING LIKE STRUCTURE WHICH IS CONVERTED INTO REGULAR CRYSTALLINE STRUCTURE WHICH CAN ONLY BE FORMED WITH THE HELP OF FORMATION OF NEW BONDS AND BREAKDOWN OF OLDER ONES
THE AMOUNT OF ENERGY WHICH IS USED IN CONVERSATION OF THE STATE OF FROM SOLID TO LIQUID IS KNOWN AS LATENT HEAT OF FUSION.
SO TEMP REMAIN CONSTANT DURING CHANGE IN STATE .
[tex] \red \star{Thanks \: And \: Brainlist} \blue\star \\ \green\star If \: U \: Liked \: My \: Answer \purple \star[/tex]
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 54000 m/s. The experiment is repeated with a He+ ion (charge e, mass 4 u).What is the ion's speed at the negative plate?
•. What is called the error due to the procedure and used apparatuses?
a. Random error
b. Index error
c. Systematic error
d. Parallax error.
Answer:
[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]
