Answer:
g ±Δg = (9.8 ± 0.2) m / s²
Explanation:
For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use
T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]
T² = [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g
g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]
They indicate the average time of 20 measurements 1,823 s, each with an oscillation
let's calculate the magnitude
g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2
g = 9.7766 m / s²
now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation
for the period
T = t / n
ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn
In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently
ΔT = Δt / n
ΔT = Δt
now let's look for the uncertainty of g
Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex] ΔT
Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex] ΔL + 4π²L (-2 T⁻³) ΔT
a more manageable way is with the relative error
[tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]
we substitute
Δg = g ( \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}DL / L + ½ Dt / T)
the error in time give us the stanndard deviation
let's calculate
Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])
Δg = 9.7766 (0.001215 + 0.0184)
Δg = 0.19 m / s²
the absolute uncertainty must be true to a significant figure
Δg = 0.2 m / s2
therefore the correct result is
g ±Δg = (9.8 ± 0.2) m / s²
Wave 1 has a wavelength 0.5 m and wave 2 has a wavelength= 0.05m. which wave carries higher energy?
Answer & Explanation:
The energy of a wave depends upon its amplitude, frequency and wave speed.
Assuming that both these waves are light waves, they have the same speed and amplitude can be assumed to be constant.
Wave 2 has a smaller wavelength and hence a higher frequency as c = λf, where
c is the speed of light,
λ is the wavelength and
f is the frequency of the wave.
Since wave 2 has a higher frequency, it has more energy.
unpolarized light of intensity Io is incident on an ideal linear polariser (no absorption) . what is the transmitted intensity?
Answer:
A Polarizing sheet transmits only the component of light polarized along a particular direction and absorbs the component perpendicular to that direction.
Consider a light beam in the z direction incident on a Polaroid which has its transmission axis in the y direction. On the average, half of the incident light has its polarization axis in the y direction and half in the x direction. Thus half the intensity is transmitted,and the transmitted light is linearly polarized in the y direction.
If a truck has a mass of 7,692 kg and travels at a speed 51 m/s what is the momentum of the truck
Answer:
Explanation:
Given the following data;
Velocity = 51 m/s
Mass = 7,692 kg
To find the momentum;
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
Substituting into the equation, we have;
Momentum = 7692 × 51
Momentum = 392292 Kgm/s
A skater with an initial speed of 5.90 m/s stops propelling himself and begins to coast across the ice, eventually coming to rest. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is 0.0500. Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest
Answer:a) - 0.4905 m/s² b) distance = 35.48 m
Explanation:
Given that
The initial velocity of the skater = 5.90 m/s
kinetic friction coefficient = 0.0500
final velocity = 0 m/s(since it comes to rest)
deceleration cause by the kinetic friction = ?
we know that
F = μN
and N= mg
Therefore;
F = μ m g....................(1)
also that
F = m a........................(2)
with our common Force, F, equating (1) and (2), we have that
m a = - μ m g
a = - μ g
a = - 0.05 × 9.81
a = - 0.4905 m/s²
The deceleration cause by the kinetic friction is a = - 0.4905 m/s²
b)
The distance the skater travels before stopping
is given as
Vf² = v₀² - 2 a x
final velocity = 0 m/s(since it comes to rest)
Therefore We have that
0 = v₀² - 2 a x
x = - v₀² / 2 a
x = 5.90² / (2 x 0.4905 )
34.81/0.981
x = 35.48 m
Or
using
v²-u² = 2aS final velocity = 0 m/s(since it comes to rest)
0²-5.90² = -2×0.4905×S
34.81=0.981S
S= 34.81/0.981
S=35.48m
A 20 cm radius ball is uniformly charged to 70 nC.
(a) What is the ball's charge density?
(b) How much charge is enclosed by spheres of radii 5, 10 and 20 cm?
(c) What is the electric field strength at points 5, 10 and 20 cm from the center?
Answer:
Explanation:
A)
Density= charge/total volume .......eqn(1)
But volume= 4/3πr^3
r= radius= 20 cm= 0.20m
If we substitute into the volume equation, we have
volume= 4/3 * 3.142 *( 0.20)^3
= 0.0335 m^3
The volume= 0.0335 m^3
Charge=71 nC= 71×10^-9
If we substitute into eqn(1) we have
Density= (71 *10^-9C )/0.0335
= 2.11µc/m^3
B) charge enclose= Density × volume
spheres of radii are
5cm
10 cm
20 cm
Volume for 5cm
V= 4/3 * 3.142 *( 0.05)^3 = 0.0005237 m^3
charge enclose=2.11µc/m^3×0.0005237
charge enclose= 2.110 nC
Volume for 10cm
V= 4/3 * 3.142 *( 0.10)^3 = 0.004189 m^3
charge enclose= 2.11µc/m^3 ×0.004189
=8.9 nC
Volume for 20cm
V= 4/3 * 3.142 *( 0.20)^3 = 0.0335 m^3
charge enclose= 71nC
Name the nutrients required for the
body.
Answer:
1- water
2- fat
3- carbohydrates
4- vitamins
5- minerals
It is necessary to to secure an inflated balloon tightly give reason
Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.50 cm. Assume that the lens is limited only by diffraction and λ = 600 nm
Answer:
R = 1.2295 10⁵ m
Explanation:
After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body
θ = 1.22 λ / D
where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture
how angles are measured in radians
θ = y / R
where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens
[tex]\frac{y}{R} = 1.22 \frac{ \lambda}{D}[/tex]
R = [tex]\frac{ y \ D}{1.22 \lambda}[/tex]
let's calculate
R = [tex]\frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}[/tex]
R = 1.2295 10⁵ m
A bicyclist rides 5.0 km due east, while the resistive forcefrom the air has a magnitude of 3.0 N and points due west. Therider then turns around and rides 5.0 km due west, back to herstarting point. The resistive force from the air on the return triphas a magnitude of 3.0 N and points due east.
a) Find the work done by the resistive force during the roundtrip.
Based on answer in part A.
b) Is the resistive force a conservative force? explain.
Answer:
a) Find the work done by the resistive force during the roundtrip.
W=-30kJ
b) Is the resistive force a conservative force? explain.
The resistive force is not a conservative force since the work done during the round trip is not zero
Explanation:
The worf done on object y a constant force F is given by:
W= (F cos ∅)S
Where S is the displacement and ∅ is the angle between the force and the displacement.
The displacement of the bicycle during each part of the trip is s=5000m and teh magnitude of teh resistance force is F=3.0N
∅1=180° he angle between the displacement and the force
W1=W2
W1 = (3.0 cos180) 5000m
W1=-15.O kJ
W=W1+W2
W=-30kJ
The resistive force is not a conservative force since the work done during the round trip is not zero
(a) The work done by the resistive force is 15,000 J
(b) The work done the resistive force is non-conservative since the resultant resistive force in not zero.
Work doneWork is said to be when an applied force displaces an object from its initial position.
Work done by resistive forceThe work done by the resistive force is calculated as follows;
W = FΔr
W = 3 x (5,000 - 0)
W = 15,000 J
Thus, the work done the resistive force is non-conservative since the resultant resistive force in not zero.
Learn more about conservative force here: https://brainly.com/question/15357875
How much work must be done to raise a 1100kg car 2m above the ground?
Answer:
21560 J
Explanation:
Work = mg*h = 1100*9.8*2 = 21560 J
21560 J work must be done to raise an 1100 kg car 2m above the ground.
Work = mass * gravity * height
= 1100 * 9.8 * 2
= 21560 J
What is work done?In precis, work is carried out whilst pressure acts upon an item to purpose a displacement. 3 portions must be regarded in the way to calculate the quantity of work. The ones 3 portions are force, displacement, and the perspective between the pressure and the displacement.
Paints carried out are elaborated in this type of manner that includes both forces exerted on the body and the whole displacement of the body. This block is preceded by a steady force F. The purpose of this pressure is to move the body a certain distance d in an immediate route in the route of the pressure.
Learn more about work here: https://brainly.com/question/1382377
#SPJ2
what was Thomas Edison first major invented?
Answer:
Thomas Edisons most famous invention was the phonograph
Thomas Edison announces his invention of the phonograph, a way to record and play back sound. Edison stumbled on one of his great inventions—the phonograph—while working on a way to record telephone communication at his laboratory in Menlo Park, New Jersey.
Explanation:
Hope I helped
A cork dropped into a water filled beaker floats with volume V1 representing the portion of cork above water. When it is dropped in a beaker containing corn syrup, it floats with its volume V2 (again, the portion of cork above syrup). How do these volumes compare?A. V1 = V2.B. V1 > V2.C. V1 < V2.D. V1 ≥ V2.
Answer:
C. V1 < V2
Explanation:
The computation is shown below:
As we know that
Byoyancy force represent the displaced water weight
[tex]B = V_{in},_{w}P_{w}g[/tex]
[tex]V_{in},_{w}[/tex] denotes cork volume that inside the water
[tex]P_{w}[/tex] denotes water density
And, byoyancy force represent the displaced weight of corn syrup
[tex]B = V_{in},_{syr}P_{syr}g[/tex]
[tex]V_{in},_{syr}[/tex] denotes the cork volume that inside the water
[tex]P_{syr}[/tex] denotes syrup density
Now
[tex]P_{syr}>P_{w}\\\\V_{in},_{syr}<V_{in},_{wat}\\\\V_2>V_1 or V_1 <V_2[/tex]
Hence, the option c is correct
DOES ANYONE KNOW HOW TO DO THIS?
Question:
Convert 68 MHz to Hz.
Answer:
68000
Explanation:
1 MHz = 1000 Hz
just multiply by 1000 :)
A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.
Answer:
The speed of this particle is constantly [tex]c[/tex].
Explanation:
Position vector of this particle at time [tex]t[/tex]:
[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].
Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:
[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].
Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].
Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:
[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].
The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :
[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].
Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)
A dropped ball gains speed because
its nature is to become closer to Earth,
its velocity changes.
a gravitational force acts on it
Of inertia
Answer:
3 and 3 and 3
Explanation:
I am sure Hope for brain list
Light rays travel from one medium into another and refract away from the boundary. What changes about the light to cause this refraction?
A. Its speed increases.
B. Its frequency increases.
C. Its frequency decreases.
D. Its speed decreases.
Answer:
a
Explanation:
I think its a its speed increases.....
The magnitude obtained when adding vector A (80 N at 20 deg) with vector B (40 N at
70 deg) is:
110.06 N
89.85 N
0 130.32 N
0 141.98 N
Answer:
110.06NExplanation:
The magnitude of the force is known as the resultant.
R = √Fx²+Fy²
Fx = 80cos 20 + 40cos70
Fx = 80(0.9397)+40(0.3420)
Fx = 75.176 + 13.68
Fx = 88.856N
Fy = 80sin 20 + 40sin70
Fy = 80(0.3420)+40(0.9397)
Fy = 27.36 + 37.588
Fy = 64.948N
R = √88.586²+64.948²
R = √7,847.48+4,218.24
R = √12,065.72
R = 109.5
R = 110N
Hence the magnitude of the forces is 110N
HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!
A car with a mass of 1,200 kg accelerates at a rate of 3.0 m/s^2 forward. What is the force acting on the car?
Answer:
the force acting on the car is 3600 N
Explanation:
The computation of the force acting on the car is shown below:
As we know that
Force = mass × acceleration
= 1200 kg × 3.0 ms/^2
= 3600 N
hence, the force acting on the car is 3600 N
I don’t get my physical science hw it’s about ocean currents
A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.
a. W =68.54 Btu
b. W=145.19 Btu
c. W = 115.12 Btu
d. W=235.7 Blu
Answer:
C. [tex]W = 115.12\,Btu[/tex]
Explanation:
Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process ([tex]\eta[/tex]), no unit, is:
[tex]\eta = \left(1-\frac{T_{L}}{T_{H}} \right)[/tex] (1)
Where:
[tex]T_{L}[/tex] - Temperature of the cold reservoir, measured in Rankine.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, measured in Rankine.
If we know that [tex]T_{H} = 1809.67\,R[/tex] and [tex]T_{L} = 584.67\,R[/tex], then the energy efficiency of the ideal thermal process is:
[tex]\eta = 0.678[/tex]
By First Law of Thermodynamics, we calculate the work output:
[tex]W = Q_{H}-Q_{L}[/tex]
[tex]W = \frac{W}{\eta} -Q_{L}[/tex] (By definition of efficiency)
[tex]Q_{L} = \frac{W}{\eta}-W[/tex]
[tex]Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W[/tex](2)
Where:
[tex]Q_{H}[/tex] - Heat received by the engine, measured in Btu.
[tex]Q_{L}[/tex] - Heat rejected by the engine, measured in Btu.
[tex]W[/tex] - Work output, measured in Btu.
If we know that [tex]\eta = 0.678[/tex] and [tex]Q_{L} = 55\,Btu[/tex], then the work output of the Carnot engine is:
[tex]W = \frac{Q_{L}}{\frac{1}{\eta}-1 }[/tex]
[tex]W = 115.807\,Btu[/tex]
The work output of the Carnot engine is 115.807 Btu. (Answer: C)
How does the angle of launch affect the kinetic energy of a rubber band?
Answer:
The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.
A car turns a certain curve of radius 24.98 m with constant linear speed of
15.67 m/s. If the centripetal force experienced by that car is 34.652 kN, what is the
mass of the car?
Answer:
3525.19 kg
Explanation:
The computation of the mass of the car is shown below:
As we know that
Fc = m × V^2 ÷ R
m = Fc × R ÷ V^2
Provided that:
Fc = 34.652 kN = 34652 N
R = Radius = 24.98 m
V = speed = 15.67 m/s
So,
m = 34652 × 24.98 ÷ 15.67^2
= 3525.19 kg
23. Christina wanted to know whether the eating behavior of goldfish is affected by music. She set up a fish
tank. Every day, she played music for the fish and measured how much the fish ate. What is Christina
missing from this experiment?
A. a control group
B. a measurable result
C. a dependent variable
D. different types of fish
Answer:
a control group
Explanation:
To have a succsessful experiment you need something to compare your variable to.
A sled's mass is 9 kg. It is held in place on a frictionless 16-degree slope by a rope attached to a stake at the top of the slope. What is the tension in the rope if it is parallel to the slope?
Answer:
The tension in the rope if it is parallel to the slope is 24.31 N.
Explanation:
Given;
mass of the sled, m = 9 kg
angle of inclination of the slope, θ = 16⁰
The tension in the rope if it is parallel to the slope is calculated from the parallel component of the tension;
[tex]T_|_| = mgSin \theta\\\\T_|_| = 9 \times 9.8 \times sin(16^0)\\\\T_|_| = 24.31 \ N[/tex]
Therefore, the tension in the rope if it is parallel to the slope is 24.31 N.
Please answer my head is about to explode.
Answer:
290
Explanation:
I tried not so certain abt the answer
Answer:
Explanation:
Givens
Mass: 1450 kg
vi = 0
a = 5 m/s^2
vf=?
Formula
KE = 1/2 m v^2
d = vi * t + 1/2 a t^2
a = (vf - vi)/t
Solution
The final KE = 0 because the final velocity = 0 (The concrete wall saw to that).
Second formula
d = vi*t + 1/2 a t^2
20 = 0 + 1/2 * 5 * t^2
40 = 5*t^2 Divide by 5
40/5 = t^2
8 = t^2
2√2 = t
Third formula
5 = (vf - 0)/2√2 Multiply by 2√2
5* 2√2 = vf
10√2= vf
First formula
KE = 1/2 m * (10√2)^2
KE = 1/2 1450 * 100*2
KE = 1450 * 100 2 and 1/2 cancel out.
KE = 145000
This means that the change in KE is 145000 Joules or 145 kJ
A 107 kg boat that is 7 m in length is initially 7.3 m from the pier. A 53 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle. 7.3 m 7 m How far is the child from the pier when she reaches the far end of the boat
Answer:
11.98 m
Explanation:
Given that:
mass of the child [tex]m_c[/tex] = 53 kg
mass of the boat [tex]m_b[/tex] = 107 kg
[tex]\text{length of the boat L = 7 m}[/tex]
the distance of the boat from pies l = 7.3 m
initial momentum [tex]P_i = 0[/tex]
Final momentum [tex]P_f = mc \dfrac{L}{f}- (m_c +m_b) \dfrac{x}{l}[/tex]
where;
x = distance moved by boat towards left
t = time taken for the child to travel to the far end of the boat
[tex]P_i =P_f[/tex]
∴
[tex]m_c \dfrac{L}{t}=(m_c +m_b) \dfrac{x}{t}[/tex]
Then;
[tex]x = \dfrac{m_cL}{m_c+m_b}[/tex]
[tex]x = \dfrac{53 \times 7}{53+107}[/tex]
x = 2.32 m
The distance of the child from the pier is:
d = L +(l - x)
d = 7 m + ( 7.3 m - 2.32 m)
d = 7 m + 4.98 m
d = 11.98 m
An object is placed in material a at point P, as shown in the diagram. The light is refracted when it strikes the interface with material b. When viewed from material b, at which point will the image appear?
If the magnitude of vector A⃗ is less than the magnitude of vectorB⃗ , then the x component of A⃗ is less than the x component ofB⃗ . If the magnitude of vector is less than the magnitude of vector, then the component of is less than the component of.
a. True
b. False
pers
2. (a) Calculate the virtual depth of a black dot at the
bottom of a cubic block made of transparent glass with each
side 4 cm, while the refractive index of glass is 1.6.
Answer:
2.5 cm
Explanation:
Using the relation :
Refractive index = Real Depth / Apparent depth
Refractive index = 1.6
Real depth = 4cm
Virtual depth = apparent depth = x
1.6 = 4cm / x
1.6x = 4
x = 4 / 1.6
x = 2.5
Hence, virtual depth = 2.5cm
The spectral classification of Antares is
Answer:
M1.5Iab-Ib
Explanation: