Answer:
a. V = 19.1m/s
b. The mass of the car does not matter
Explanation:
A.
KE = 1/2mv² = fd --------(1)
Fd = umgd ---------(2)
Therefore,
1/2mv² = umgd ---------(3)
M will cancel itself out from both sides of equation 3.
Then we will have:
1/2v² = ugd
Then we cross multiply to make v² the subject of the formula
V² = 2ugd
V = √2ugd -------(4)
U = 0.38
g = 9.81
d = 98
When we input these values into equation 4, we will have:
V = √2x0.38x9.81x98
V = √730.6488
V = 27.03m/s
B.
The mass of the car does not actually matter as the mass was cancelled out on the both sides of equation 3
We will now determine the indexes of refraction for two Mystery materials, A and B. These materials can be selected from the list of materials on the right. Be sure to set your laser pointer to a frequency of 589 nm. Questions:A. Devise an experiment for determining the indices of refraction for these. Explain your methodology. B. What are the indices of refraction for the two mystery materials, A and B?
Answer:
A) refraction experiment n = n₁ sin θ₁ / sin θ₂
B) n_A = 1.19 , n_B = 1.53
Explanation:
A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index
n₁ sin θ₁ = n₂ sinθ₂
n₂ = n₁ sin θ₁₁ /sin θ₂
If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is
n = n₁ sin θ₁ / sin θ₂
B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be
material A
n_A = sin 50 / sin 40
n_A = 1.19
material B
n_B = sin 50 / sin30
n_B = 1.53
A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy
The diagram below shows two bowling balls, A and B, each having a mass of 7.0 kg, placed 2.00 m apart between their centers.
Answer:
F = 1.63 x 10⁻⁹ N
Explanation:
Complete question is as follows:
The diagram below shows two bowling balls, A and B, each having a mass of 7.0 kg, placed 2.00 m apart between their centers. Find the magnitude of Gravitational Force?
Answer:
The gravitational force is given by Newton's Gravitational Law as follows:
F = Gm₁m₂/r²
where,
F = Gravitational Force = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = m₂ = mass of each ball = 7 kg
r = distance between balls = 2 m
Therefore,
F = (6.67 x 10⁻¹¹ N.m²/kg²)(7 kg)(7 kg)/(2 m)²
F = 1.63 x 10⁻⁹ N
A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?
Answer:
163.35
__________________________________________________________
We are given:
Mass of the object (m) = 36.3 kg
Velocity of the object (v) = 3 m/s
Kinetic Energy of the object:
We know that:
Kinetic Energy = 1/2(mv²)
KE = 1/2(36.3)(3)² [replacing the variables with the given values]
KE = 18.15 * 9
KE = 163.35 Joules
Hence, the cart has a Kinetic Energy of 163.35 Joules
A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.173. (indicate the direction with signs of your answer) (a) What is the force of kinetic friction in N acting on the block? (b) What is the block's acceleration in /s^2? (c) How Far will it slide (in m) before coming to rest? Plz answer as soon as possible
Answer:
Explanation:
a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force
R = mg where m is mass of the block
Force of friction F = μ x mg
= .173 x 12.2 x 9.8
= 20.68 N
b ) Only force of friction is acting on the body so
deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²
acceleration = - 1.7 m /s²
c )
v² = u² - 2 a s
v = 0 , u = 3.9 m /s
a = 1.7 m /s
0 = 3.9² - 2 x 1.7 x s
s = 4.47 m
If it requires 7.0 J of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be required to stretch it an additional 3.6 cm?
Answer:
56 J
Explanation:
The following data were obtained from the question:
Energy 1 (E₁) = 7 J
Extention 1 (e₁) = 1.8 cm
Extention 2 (e₂) = 1.8 + 3.6 = 5.4 cm
Energy 2 (E₂) =?
Energy stored in a spring is given by the following equation:
E = ke²
Where E is the energy.
K is the spring constant.
e is the extension.
E = ke²
Divide both side by e²
K = E/e²
Thus,
E₁/e₁² = E₂/e₂²
7/ 1.8² = E₂/ 5.4²
7 / 3.24 = E₂/ 29.16
Cross multiply
3.24 × E₂ = 7 × 29.16
3.24 × E₂ = 204.12
Divide both side by 3.24
E₂ = 204.12 / 3.24
E₂ = 63 J
Thus, the additional energy required can be obtained as follow:
Energy 1 (E₁) = 7 J
Energy 2 (E₂) = 63 J
Additional energy = 63 – 7
Additional energy = 56 J
If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (3a1)
Question 10 options:
There cannot be any forces applied to the ball.
There must be exactly one force applied to the ball.
The net force applied to the ball is zero.
The net force applied to the ball is directed to the right.
Answer:
C. The net force applied to the ball is zero.
Explanation:
From Newton's second law of motion;
F = ma
Where F is the force on an object, m is its mass and a is its acceleration.
Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.
Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.
So that;
F = m x 0
= 0
No force is applied on the object.
Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.
A student throws a 110 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks.
What is the magnitude of the average force on the wall if the duration of the collision is 0.19 s ?
Answer:
3.6NExplanation:
Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time.
the expression is Ft=mv
where F= force
m= mass
t= time
v= velocity
Step one:
given data
mass m= 110g= 0.11kg
velocity v= 6.5m/s
time t= 0.19seconds
Step two:
we also know that the force on impulse is given as
Ft=mv
F=0.11*6.5/0.19
F=0.715/0.19
F=3.76N
The magnitude of the average force on the wall if the duration of the collision is 0.19 is 14N
I NEED THIS ASAP!!
Which formula defines the unit for electrical power?
Answer:
1 W = 1 V x 1 A
Explanation:
Other dude is wrong this is right
A boxer is punching the heavy bag. The impact of the glove with the bag is 0.10s. The mass of the glove and his hand is 3kg. The velocity of the glove just before impact is 25m/s. What is the average impact force exerted on the glove?
Answer:
750NExplanation:
Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time.
the expression is Ft=mv
where F= force
m= mass
t= time
v= velocity
Step one:
given data
mass m=3kg
velocity v= 25m/s
time t= 0.10seconds
Step two:
we also know that the force on impulse is given as
Ft=mv
F=3*25/0.10
F=75/0.10
F=750N
The magnitude of the average force on the heavy bag if the duration of the collision is 0.1 is 750N
The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in seconds. What is the angle in radians through which the propeller rotates from t = 1.00 s to t = 6.10 s?
Answer:
The value is [tex]\theta =407.3 \ radian[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]
The first time is [tex]t_1 = 1.00 \ s[/tex]
The second time [tex]t_2 = 6.10 \ s[/tex]
Generally the angular velocity is mathematically represented as
[tex]w = \int\limits {\alpha } \, dt[/tex]
=> [tex]w = \int\limits {7t + 8 } \, dt[/tex]
=> [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]
=> [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]
=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]
=> [tex]\theta =407.3 \ radian[/tex]
If a penny is dropped from rest from a building takes 2 seconds to hit the ground, calculate the velocity of the penny right before it touches the ground?
a. 19.6 m/s
b 9.8 m/s
c. 0 m/s
d 29.4 m/s
Answer:
20m/s
Explanation:
u = 0m/s
t = 2s
a = +g = 10m/s²
t = 2d
v = ?
v = ut + 1/2at²
v = 0(2) + 1/2(10)(2)²
v = 0 + 5(4)
v = 20m/s
Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
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gravities limit is under which sphere as the perimeter?
Section 4.1- Newton's First Law
Answer:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. ... If that velocity is zero, then the object remains at rest.
Explanation:
Answer:
Newton's First Law is about inertia; objects at rest stay at rest unless acted upon and objects in motion continue that motion in a straight line unless acted upon. The amount of inertia an object has is simply related to the mass of the object.
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel?A) The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.B) Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.C) Pressure fluctuations travel along the direction of propagation of the sound wave.D) Propagation of energy that passes through empty spaces between the particles that comprise the mediumDoes air play a role in the propagation of the human voice from one end of a lecture hall to the other?a) yesb) no
Answer:
None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave
Explanation:
An extraterrestrial creature is standing in front of plane mirror. The height of this creature is H and we know that this creature has eyes positioned h below the top of its head. This creature sees its reflection which fit exactly the mirror, it means, this creature can just see the top of head and the bottom of its feet (or whatever it uses for motion). We can conclude that the top of a mirror is exactly:________
a. H/2 above the ground
b. H above the ground
c. (H-h/2) above the ground
d. (H-h) above the ground
e. We can not guess anything without information about the nature of this creature.
Answer:
c. (H-h/2) above the ground
Explanation:
The mirror must be at least half as tall as the alien, and its base must be located at half of the distance between the alien's eyes and the ground (assuming that the alien doesn't float or levitate).
This question is about the Law of Reflection which states that the angle of reflection = angle of incidence.
I attached an image that can help you understand the concept, although the alien is not included.
g When the movable mirror of the Michelson interferometer is moved a small distance X while making a measurement, 246 fringes are counted moving into the field of the viewing mirror. What is X if the wavelength of the light entering the interferometer is 562 nm
Answer:
X = 69.1 x 10⁻⁶ m = 69.1 μm
Explanation:
The relationship between the motion of the moveable mirror and the fringe count of the Michelson's Interferometer is given by the following formula:
d = mλ/2
where,
d = distance moved by the mirror = X = ?
m = No. of Fringes counted = 246
λ = wavelength of light entering interferometer = 562 nm = 5.62 x 10⁻⁷ m
Therefore,
X = (246)(5.62 x 10⁻⁷ m)/2
Therefore,
X = 69.1 x 10⁻⁶ m = 69.1 μm
In 1993, Wayne Brian threw a spear at a record distance of 201.24 m. (This is not an official sports record because a special device was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with respect to the horizontal. What was the initial speed of the spear? 2. Find the maximum height and time of flight of the spear in problem #1.
I really don't know how to do any of this please help me :(
Answer:
V₀ = 45.81 m/s
H = 70.45 m
T = 5.36 s
Explanation:
The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 201.24 m
V₀ = Initial Speed = ?
θ = Launch Angle = 35°
g = 9.8 m/s²
Therefore,
201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²
V₀ = √[(201.24 m)/(0.095 m/s²)
V₀ = 45.81 m/s
Now, for maximum height:
H = V₀² Sin² θ/g
H = (45.81 m/s)² Sin² 35°/9.8 m/s²
H = 70.45 m
For the total time of flight:
T = 2 V₀ Sin θ/g
T = 2(45.81 m/s) Sin 35°/9.8 m/s²
T = 5.36 s
A mass of 100 g is tied to the end of an 80.0-cm string and swings in a vertical circle about a fixed center under the influence of gravity. The speed of the mass at the top of the swing is 3.50 m/s. What is the speed of the mass at the bottom of its swing?
Answer:
the speed of the mass at the bottom of its swing is 6.61m/s
Explanation:
Applying energy conservation
[tex]\frac{1}{2}m(Vlowest)^2 = mg(2R) + \frac{1}{2}m(Vtop)^2[/tex]
There is no potential energy at the bottom as the body will have a kinetic energy there.
h= 2R = 1.6m as the diameter of the circle will represent the height in the circle.
g = 9.8m/s^2
m will cancel out, so the net equation becomes.
[tex]\frac{(Vbottom)^2}{2} = 2gR + \frac{(Vtop)^2}{2}[/tex]
= [tex]2*9.8*0.8 + \frac{(3.5)^2}{2}[/tex]
= 15.68+ 6.125
[tex]\frac{(Vbottom)^2 }{2}[/tex] = 21.805
(Vb)^2 = 2*21.805
= 43.64
Vb = 6.61m/s
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?
A 0.31 s
B 0.56 s
C 4.3s
D 70s
Answer:
C. 4.3 seconds
Explanation:
B 0.56 s is the time period of a twirlers baton.
What is Centripetal Acceleration?Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.
Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.
The centripetal acceleration is given by:
a = 4π²R/T²
Given values are:
a = 47.8 m/s²
D = 0.76 m so , R = 0.76/2 = 0.38m
Using this formula,
47.8*T² = 4π² x0.38
T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]
T = 0.56 s
Therefore,
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have time period of 0.56 s.
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#SPJ5
.
Select the correct answer.
The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's motion?
A. It's not moving.
B.It's moving at a constant speed.
C.It's moving at a constant velocity
D.It's speeding up.
Answer:
It isn't moving
Explanation:
3. Which object has more inertia?
A. A tractor trailer rig moving at 2 m/s
B. A pingpong ball rolling a 2 m/s
C. A bowling ball rolling at 1m/s
D. A car rolling at 5 m/s
Answer:
A. A tractor trailer rig moving at 2 m/s
Explanation:
Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.
In physics, Sir Isaac Newton's first law of motion is known as law of inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.
The inertia of an object such as a tractor trailer rig is greatly dependent or influenced by its mass; the higher quantity of matter in a tractor trailer rig, the greater will be its tendency to continuously remain at rest.
Hence, the object that has more inertia is a tractor trailer rig moving at 2 m/s because it has more mass than all the other objects in the category. Also, the mass of an object is directly proportional to its inertia.
The image blow shows a certain type of global wind:
What best describes these winds? Polar easterlies caused by air above poles being relatively warmer.
Polar easterlies caused by air above poles being relatively cooler.
Trade winds caused by air above equator being relatively warmer.
Trade winds caused by air above equator being relatively cooler.
Answer:
i got u its a
Explanation:
A parallel plate capacitor is made up of two metal squares with sides of length 8.8 cm, separated by a distance 5.0 mm. When a voltage 187 V is set up across the terminals of the capacitor, the charge stored on the positive plate is equal to __________ nC. g
Answer:
2.56 nC
Explanation:
By definition, the capacitance is expressed by the following relationship between the charge stored on one of the plates of the capacitor and the potential difference between them, as follows:[tex]C =\frac{Q}{V} (1)[/tex]
For a parallel-plate capacitor, assuming a uniform surface charge density σ, if the area of the plates is A, the charge on one of the plates can be written as follows:[tex]Q = \sigma * A (2)[/tex]
Assuming an uniform electric field E, the potential difference V can be expressed as follows:[tex]V = E*d (3)[/tex]
where d is the distance between plates.
Applying Gauss 'Law to a closed surface half within one plate, half outside it, we find that E can be written as follows:[tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]
Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:[tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]
Taking (1) and (5), as both left sides are equal each other, the right sides are also equal, so we can write the following equality:[tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]
Solving for Q, we get:[tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]
A 0.22 caliber handgun fires a 1.9g bullet at a velocity of 765m/s. Calculate the de Broglie wavelength of the bullet. Is the wave nature of matter significant for the bullets?
Answer:
de Broglie wavelength of the bullet is 4.56 x 10⁻³⁴ mThe value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.Explanation:
Given;
mass of the bullet, m = 1.9 g = 0.0019 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
[tex]\lambda = \frac{h}{mv}[/tex]
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ is de Broglie wavelength of the bullet
[tex]\lambda = \frac{h}{mv}\\\\ \lambda =\frac{(6.626*10^{-34})}{(0.0019)(765)}\\\\ \lambda =4.56 *10^{-34} \ m[/tex]
Thus, this value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.
Given:
Mass, m = 1.9 g or, 0.0019 kgVelocity, v = 765 mPlank's constant, h = 6.626 × 10⁻³⁴ J/sThe De-Broglie wavelength,
→ [tex]\lambda = \frac{h}{mv}[/tex]
By putting the values,
[tex]= \frac{6.626\times 10^{-34}}{0.0019\times 765}[/tex]
[tex]= 4.56\times 10^{-34} \ m[/tex]
Thus the response above is right.
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A material that provides resistance to the flow of electric current is called a(n):
circuit
conductor
insulator
resistor
Answer:
it's an insulator
Explanation:
Insulators provides resistance
Answer:
C. insulator
Explanation:
if you increase the frequency of a wave by 5x whats it’s period?
We know that Period of a wave is the inverse of its Frequency
So, Period = 1 / Frequency
From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period
Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times
A typical elevator car with people has a mass of 1500.0 kg. Elevators are currently approaching speeds of 20.0 m/s - faster than the speed.
Required:
What is the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s?
Answer:
1500NExplanation:
Force = mass * acceleration
Given
Mass = 1500kg
Get the acceleration using the equation of motion;
v² = u²+2aS
20² = 0+2s(200)
400 = 400a
a = 400/400
a = 1m/s²
Get the upward force required
F = 1500 * 1
F = 1500N
Hence the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s is 1500N
When two RF signals on the same frequency arrive at a receiver at the exact same time and their peaks and valleys are in alignment, what is true about these signals?
Answer:
The two RF signals are in phase.
Explanation:
A wave is a disturbance that travels through a medium which transfers energy from one point to another in the medium without causing any permanent displacement of the particles of the medium.
Characteristics of waves include frequency, wavelength, velocity, etc.
Two types of waves are longitudinal and transverse wave. Radio Frequency (RF) signals travel in the form of transverse waves which have regions of maximum and minimum displacements called crests and troughs.
Travelling waves with the same frequency may be said to be in phase or out of phase depending on whether their crests/peaks or troughs/valleys are reached at the same instant of time.
When two RF signals on the same frequency arrive at a receiver at the exact same time and their peaks and valleys are in alignment or in step, they are said to be in phase.
The phase of a wave involves the relationship between the position of the amplitude peaks and valleys of two waveforms.