Answers
1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4
Related Questions
5.000 km =
3.125
mi
8.000 fl oz =
mL
Answers
Answer:
236.588 mL
Explanation:
The formula for an approximate result is to multiply the volume value by 29.574
[tex]8.000 \times 29.574 = 236.588[/tex]
Is that what you were asking for?
What causes the difference in the angle of the sun on the Earth's surface throughout the year?
Answers
Answer:
The axis is tilted and points to the North Star no matter where Earth is in its orbit. Because of this, the distribution of the Sun's rays changes. ... It also means that the angle at which sunlight strikes different parts of Earth's surface changes through the year.
Explanation:
Pls sub to bdoggaming if this helped
If a 35 kg box collides with a stationary 120 kg box with a force of 90 N, what must be true of the magnitude of the reaction force?
Answers
Newton's third law allows to find the result for the value of the reaction force during the collision is:
The reaction force is F = 90 N and is applied to the lighter body.Newton's third law stable that the forces appear in pairs or ea that when two bodies interact, the interaction forces appear in the two bodies simultaneously, in general they are called action and reaction forces.
These furas are of the same magnitude, but in the opposite direction, each one applied to one of the bodies.
They indicate that the most lighter body collides with the one with the greatest mass with a force of F = 90 N. If we call this the action, the larger body must react with a force of equal magnitude on the lighter body.
Consequently, the reaction force is F = 90 N directed towards the lighter body.
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What are sound detectors?
Answers
Answer:
A sound detector comes in the shape of a rectangular board which comprises a microphone as well as a processing circuitry.
Answer:
Sound detection sensor works similarly to our Ears, having diaphragm which converts vibration into signals.
A 2-kg object moving at 10 m/s has a 4-N force applied to it. Can you predict how the force will affect the object? A) yes, it will slow it down at a rate of 2 m/s2. B) yes, it will speed it up at a rate of 2 m/s2. yes, it will speed it up at a rate of 4 m/s2. D) It cannot be determined without more information.
Answers
Answer:
D) It cannot be determined without more information.
Explanation:
Velocity is a vector meaning it has both magnitude and direction.
Force and acceleration are also vectors.
Without knowing the directions of each, we cannot know if the mass has a speed (scalar value) change.
We can know that the mass will have a velocity change at the rate of 2 m/s² in the direction of the applied force.
If that force is applied in the direction of initial velocity, the velocity (and speed) will increase in magnitude in the same direction.
If that force is applied opposite of the initial velocity, the the velocity (and speed) will decrease in magnitude in the same direction. If the acceleration lasts long enough, velocity will eventually become zero and then become negative. At the same time, speed will become zero, and then increase again as speed is the absolute value of the magnitude of velocity.
If the force is applied at any other angle, both the velocity and the speed will change in both magnitude and direction.
Explain how the linear rate spring operates?
Answers
I need a short answer ?
Answers
Answer:
Explanation:
7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s
7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m
or
h = (45sin14.5)² / (2(9.81)) = 6.47 m
which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.
12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]
Answers
Hi there!
We can use the following:
W = ΔKE = F · d
Find the work done on the cart:
W = 200 · 10 = 2000 J
Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:
2000J = KEf - KEi
KE is given as:
[tex]KE = \frac{1}{2}mv^2[/tex]
2000J = 1/2(55)v²
4000 = 55v²
√(4000/55) = 8.53 m/s
The diagram below shows a 5.00-kilogram block
at rest on a horizontal, frictionless table.
5.00-kg
block
Table
Which of the following is the correct name and strength of the force holding the block up?
Answers
The name and strength of the force holding the block up is 50 N upward - Normal force.
The given parameters:
Mass of the block, m = 5 kgThe weight of the block acting downwards due to gravity is calculated as follows;
W = mg
where;
g is acceleration due to gravity = 10 m/s²W = 5 x 10
W = 50 N (downwards)
Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.
Fₙ = 50 N (upwards)
Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.
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The block will remain on the table because the normal force balances with the weight of the block. The correct answer is 50 N upward normal force
From the diagram shown a 5.00-kilogram block at rest on a horizontal, frictionless table. The weight of the block will act downward which will be
Weight W = mg
let g = 10 m/[tex]s^{2}[/tex]
W = 5 x 10
W = 50 N
The block will also produce an equal but in opposite direction of a normal force which is equal to the weight of the block. That is,
Normal force N = 50 N
The block will remain on the table because the normal force balances with the weight of the block.
Therefore, the correct name and strength of the force holding the block up is 50 N upward normal force.
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An object, initially traveling at a velocity of 73 m/s, experiences an acceleration of -9.8 m/s^2. How much time will it take it to come to rest?
Answers
7.4 s
Explanation:
Given:
[tex]v_0 = 73\:\text{m/s}[/tex]
[tex]v = 0[/tex]
[tex]a = -9.8\:\text{m/s}^2[/tex]
[tex]t = ?[/tex]
To solve the time it takes for the object to come to a stop, we are going to use the equation below:
[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]
Using the given values above, we get
[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]
[tex]\;\;\;\;= 7.4\:\text{s}[/tex]
A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 25 J. The maximum speed of the block is:
Answers
Answer:
Explanation:
easy way
when system is all kinetic energy, velocity is at a maximum
E = ½mv²
v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s
harder way
ω = √(k/m) = √(80/0.5) = √160 rad/s
When the system is entirely spring potential, the amplitude A is
E = ½kA²
A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m
maximum velocity is ωΑ = 0.79√160 = 10 m/s
have a difinite shape and do not easily take the shape of their containers
Answers
Ok well that’s something to think about
A 200-kg, 2.0-m-radius, merry-go-round in the shape of a flat, uniform, circular disk parallel to level ground is rotating at 1.2 cycles/second about an axis through its center of mass and perpendicular to the ground. A 50-kg boy jumps onto the edge of the merry-go round and lands at a fixed point. What is the angular velocity of the merry-go-round after the boy lands on it
Answers
Answer:
Explanation:
Conservation of angular momentum.
Disk I = ½MR²
Point mass I = mR² (boy)
Initial angular momentum
L₀ = Iω = ½MR²ω₀
Final angular momentum
L₁ = Iω = (½MR² + mR²)ω₁
as momentum is conserved, these are equal
(½MR² + mR²)ω₁ = ½MR²ω₀
ω₁ = ω₀(½MR²/ (½MR² + mR²))
ω₁ = ω₀(½M/ (½M + m))
ω₁ = 1.2(½(200)/ (½(200) + 50))
ω₁ = 1.2(⅔)
ω₁ = 0.8 cycles/second or 0.8(2π) = 1.6π rad/s
A mars surface exploration vehicle drops a rock off a 1.00 I'm high vertical Cliff. The sound of the rock landing at the base of the cliff is recorded by instruments on the vehicle 27.1 seconds later. Calculate the acceleration due to gravity on Mars given that the speed of sound on Mars is 320 m/s
Answers
The acceleration due to gravity on Mars is 11.81 m/s².
The given parameters:
Height of the cliff, h = 1 mTime of motion of the sound wave, t = 27.1 sSpeed of sound in mass, v = 320 sThe equation of motion to determine the acceleration due to gravity on the moon is calculated as follows;
[tex]s = vt + \frac{1}{2} gt^2[/tex]
where;
s is the distance traveledt is the time of motionSince the time measured is two way time, the new equation for the total distance traveled is calculated as;
[tex]v = \frac{2d}{t} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{320 \times 27.1}{2} \\\\d = 4,336 \ m[/tex]
The acceleration due to gravity is calculated as follows;
[tex]s = vt + \frac{1}{2} gt^2\\\\4,336 = 0 \ + \ \frac{1}{2} \times g \times (27.1)^2\\\\4,336 = 367.21g\\\\g = \frac{4,336}{367.21} \\\\g = 11.8 1 \ m/s^2[/tex]
Thus, the acceleration due to gravity on Mars is 11.81 m/s².
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two factors affecting the magnitude of force of gravity betwwn 2 objects are A. mass and matter B. mass and distance C. weight and mass D. distance and weight
Answers
Answer:
B. MASS & MATTER
Explanation:
Newton's law also states that the strength of gravity between any two objects depends on two factors: the masses of the objects and the distance between them. Objects with greater mass have a stronger force of gravity between them.
Define the term dimension
Answers
Answer:
Q1. A measurable extent of a particular kind, such as length, breadth, depth, or height.
Q2. A dimensional constant is a physical quantity that has dimensions and has a fixed value. Some of the examples of the dimensional constant are Planck's constant, gravitational constant, and so on.
Q3. Physical quantities which posses dimensions and have variable are called dimensional variables. Examples are length, velocity, and acceleration etc.
Q4. Dimensionless variables are the quantities which doesn't have any dimensions the the value is a variable. Eg: angle = arc/ radius. Dimensions = L/L. = 1. So angle does not have any dimensions and the value can vary.
Q5. Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.
Q6. Dimensional analysis has been around a long time, Newton called it the "Great principle of Similitude", but the modern form can be traced back to James Clerk Maxwell. It was Maxwell who distinguished mass [A/], length [£], and time [7"] as the independent dimensions from which others could be derived.
Q7. Mass, length, time, temperature, electric current, amount of light, and amount of matter.
Q8. Dimensional analysis is used to convert the value of a physical quantity from one system of units to another system of units. Dimensional analysis is used to represent the nature of physical quantity. The expressions of dimensions can be manipulated as algebraic quantities.
Hope that helps. x
A 21 newton force keeps a 3 kg object in uniform circular motion. The speed of the object is 9 m/s. The magnitude of the centripetal acceleration is
3 m/s^2
7 m/s^2
27 m/s^2
30 m/s^2
Answers
Answer:
Explanation:
F = ma
a = F/m = 21/3 = 7 m/s²
yayy here you are f, r, e, e, p, o, i, n, t, s
Answers
Answer:
Albert Einstein Albert Einstein was a German-born theoretical physicist, widely acknowledged to be one of the greatest physicists of all time. Einstein is best known for developing the theory of relativity, but he also made important contributions to the development of the theory of quantum mechanics.Explanation:
Thank you so much buddy !!The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.
Try to find (1) the vibration function of the particle at the origin, (2) the wave function of the planar simple harmonic wave according to the origin.
Answers
The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.
Try to find (1) the vibration function of the particle at the origin, (2) the wave function of the planar simple harmonic wave according to the origin.
Answer:
Figure 16.8 The pulse at time
t
=
0
is centered on
x
=
0
with amplitude A. The pulse moves as a pattern with a constant shape, with a constant maximum value A. The velocity is constant and the pulse moves a distance
Δ
x
=
v
Δ
t
in a time
Δ
t
.
The distance traveled is measured with any convenient point on the pulse. In this figure, the crest is used.
A wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with just enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon
Answers
Answer:
Explanation:
An impulse results in a change of momentum
If the wagon and dog both stop, they must have had equal and opposite momentums
FΔt = mΔv
F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)
v₁ = t₀ = 0
F = m(v₀)/t₁
F = 55(2.1)/0.1 = 1155 N
We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.
A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m
Answers
Answer:
Explanation:
f = [tex]\sqrt{T/(m/L)} / 2L[/tex]
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = [tex]\sqrt{120/12} /(2(3))[/tex])
f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N
A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m when it begins to roll down the incline. If rolling and sliding friction are neglected, what is the linear velocity, in m/s, of the center-of-mass of the wheel when it reached the bottom of the incline?
Answers
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
Which kind of chemical reaction is best
represented by the following equation?
BaCl2 + Na2S04 --> 2NaCl + BaSO4
A. Synthesis
B. Single displacement
C. Decomposition
D. Double displacement
Answers
Answer:
Synthesis
Explanation:
[tex] \color{lime}{ \sf better \: luck \: on \: your \: exam \: lol}[/tex]
i just want an answer please
Answers
Answer: An answer on what? I’ll never ignore you!
Explanation:
Answer:
an answer on what?
Explanation: Im here to help!!
13) A mass attached to the free end of a spring executes simple harmonic motion according to the equation y = (0.50 m) sin (18π t) where y is in meters and t is seconds. What is the period of vibration?
Answers
Hi there!
The period is given by:
[tex]T = \frac{2\pi}{w}[/tex]
T = Period (sec)
w = angular frequency (rad/sec)
According to the equation for SHM in terms of position:
y(t) = Asin(ωt + φ)
A = Amplitude (m)
ω = angular frequency (rad/sec)
t = time (sec)
φ = phase angle
In this instance, the angular frequency is given as 18π.
Plug this value into the equation for T:
[tex]T= \frac{2\pi}{18\pi} = \frac{1}{9} = \boxed{0.111 s}[/tex]
How much potential energy does an 8 kg flower pot hanging 5 m above the ground have
Answers
Answer:
Explanation:
Relative to ground level it has
PE = mgh = 8(10)(5) = 400 J
HELPPP!! Thanks!
If you only wanted to increase the particle motion of a gas without increasing any of its other properties, which would the most correct situation?
a. Keep the gas at a constant pressure and keep the temperature constant, but increase the volume of the gas
b. Keep the gas in a fixed container at constant pressure and increase the temperature
c. Keep the gas in a fixed container at constant pressure and decrease the temperature
d. Keep the gas at a constant volume and keep the temperature constant, but decrease the pressure of the gas
Answers
Answer:c i think
Explanation: not sure
explain the process of convergence and divergence ! HELPPP
Answers
Answer:
Divergence generally means two things are moving apart while convergence implies that two forces are moving together. ... Divergence indicates that two trends move further away from each other while convergence indicates how they move closer together.
Explanation:
An object accelerates from rest to 93 m/s over a distance of 49 m. What acceleration did it experience?
Answers
Answer:
Explanation:
acceleration= change in velocity/time taken
acceleration= 93/49
=2.02
During take-off a 8kg model rocket is burning fuel causing its speed to increase
at a rate of 4m/s2 despite experiencing a 90N drag.
What’s is the strength of the thrust?
(Answer unit is in N)( and the answer isn’t 212)
Answers
The strength of the thrust is 122 newtons.
The motion of the rocket is described by the second Newton's law, whose model is shown below:
[tex]\Sigma F = F - D = m\cdot a[/tex] (1)
Where:
[tex]F[/tex] - Thrust, in newtons[tex]D[/tex] - Drag, in newtons[tex]m[/tex] - Mass of the rocket, in kilograms[tex]a[/tex] - Net acceleration of the rocket, in meters per square secondIf we know that [tex]D = 90\,N[/tex], [tex]m = 8\,kg[/tex] and [tex]a = 4\,\frac{m}{s^{2}}[/tex], then the strength of the thrust is:
[tex]F = D + m\cdot a[/tex]
[tex]F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 122\,N[/tex]
The strength of the thrust is 122 newtons.
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