Answer:
Answer is in the picture.
Explanation:
Answer is in the picture.
5a. i) Define pressure.
Answer:
Explanation:
How much force is acting on a certain area
4
Type the correct answer in the box. Use numerals instead of words.
Anne has a sample of a substance. Its volume is 20 cm and its mass is 100 grams. What is the sample's density?
The sample's density is
g/cm?
Reset
Next
Answer:
5g/cm
Explanation:
denisty=mass/volume
100/20
5g/cm
uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.00 cm. (a) What is the work done by the field on the electron? 1.753e-18 J (b) What is the change in potential energy associated with the electron? J
Answer:
a) W = - 1.752 10⁻¹⁸ J, b) U = + 1.752 10⁻¹⁸ J
Explanation:
a) work is defined by
W = F . x
the bold letters indicate vectors, in this case the force is electric
F = q E
we substitute
F = q E x
the charge of the electron is
q = - e
F = - e E x
let's calculate
W = - 1.6 10⁻¹⁹ 365 3 10⁻²
W = - 1.752 10⁻¹⁸ J
b) the change in potential energy is
U = q ΔV
the potential difference is
ΔV = - E. Δs
we substitute
U = - q E Δs
the charge of the electron is
q = - e
U = e E Δs
we calculate
U = 1.6 10⁻¹⁹ 365 3 10⁻²
U = + 1.752 10⁻¹⁸ J
The thermal efficiency (in %) of a system that undergoes a power cycle while receiving 1000 kJ of energy by heat transfer from a hot reservoir at 1000 K and discharging 500 kJ of energy by heat transfer to a cold reservoir at 400 K is:
Answer:
η = 0.5 = 50%
Explanation:
The efficiency of the power cycle is given by the following formula:
[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]
where,
where,
η = efficiency = ?
Q₁ = heat received from hot reservoir = 1000 KJ
Q₂ = heat discharged to cold reservoir = 500 KJ
Therefore,
[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]
η = 0.5 = 50%
A spinning electron produces a(n)
a. element.
b. magnetic field.
c. proton.
d. piece of iron.
Answer:
A spinning electron produces a magnetic field.
Explains the fine structure of Hydrogen lines.
What distance do I cover if I travel 10 m E, then 6 mW, then 12 m E?
A. 16 m
B. 28 m
C. 16 m E
D. 28 m E
Answer:
C. 16 m E
Explanation:
Applying,
The law of addition of vector: Vector in the same direction are added while vector in opposite direction are substracted
From the question above,
Step 1: Total distance covered towards east = 10+12 = 22 m E
Step2: Total distance covered towards west = 6 m W
Therefore, the resultant distance traveled = 22-6 = 16 m E
Hence the right option is C. 16 m E
Using the data provided below, calculate the corrected wavelength for a spectroscope reading of 6.32.
Spectroscope Readings Known Helium Wavelengths (nm)
3.65 388.9
4.75 486.6
4.90 501.6
5.65 587.6
6.45 667.8
7.00 706.5
Answer:
646.6 nm
Explanation:
Using the data given, we fit a linear model :
Spectroscope Readings Known Helium Wavelengths (nm)
3.65 388.9
4.75 486.6
4.90 501.6
5.65 587.6
6.45 667.8
7.00 706.5
Using technology like excel to fit the model, the regression model obtained is :
Y = 97.96582X + 27.48458
Where, y is the predicted or corrected wavelength value.
This mod could be used to calculate the corrected wavelength for a given Spectroscope value :
Given a Spectroscope value of 6.32 ; the corrected wavelength is obtained by replacing x in the equation by 6.32 and calculate y ;
Y = 97.96582X + 27.48458
x = 6.32
Y = 97.96582(6.32) + 27.48458
Y = 619.1439824 + 27.48458
Y = 646.6285624
Corrected wavelength value is : 646.6
How many joules of energy are required to accelerate one kilogram of mass from rest to a velocity of 0.866c?
Answer:
the amount of energy needed is 1.8 x 10¹⁷ J.
Explanation:
Given;
mass of the object, m₀ = 1 kg
velocity of the object, v = 0.866 c
By physics convection, c is the speed of light = 3 x 10⁸ m/s
The energy needed is calculated as follows;
E = Mc²
As the object approaches the speed of light, the change in the mass of the object is given by Einstein's relativity formula;
[tex]M = \frac{M_0}{\sqrt{1- \frac{v^2}{c^2} } } \\\\ M = \frac{1}{\sqrt{1- \frac{(0.866c)^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{1- \frac{0.74996c^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{0.25} } \\\\ M = 2 \ kg[/tex]
The energy required is calculated as;
E = 2 x (3 x 10⁸)²
E = 1.8 x 10¹⁷ J
Therefore, the amount of energy needed is 1.8 x 10¹⁷ J.
A covalent bond is formet by of electrons..?
Answer:
The covalent bond is formed by pairs of electrons that are shared between two atom
Explanation:
The covalent bond is formed by pairs of electrons that are shared between two atoms, in general the electrons must have opposite spins to have a lower energy state.
In this bond, the electrons are between the two atoms and are shared between them in such a way that there is a configuration of eight electrons in the orbit.
How does radiation from the sun spread throughout Earth's atmosphere?
- through convection currents
- through conduction currents
- through solar panels
- through hot water systems
Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statement correctly describes the change in momentum of the two balls?
a. |ΔpBl<|ΔPA|
b. |ΔpBl=|ΔPA|
c. |ΔpB|>|ΔPA|
d. ΔpB > ΔPA
Answer:
Option A
Explanation:
From the question we are told that:
Mass [tex]m=0.20kg[/tex]
Velocity [tex]v=4m/s[/tex]
Generally the equation for momentum for Ball A is mathematically given by
Initial Momentum
[tex]M_{a1}=mV[/tex]
[tex]M_{a1}=0.2*4[/tex]
[tex]M_{a1}=0.8[/tex]
Final Momentum
[tex]M_{a2}=-0.8kgm/s[/tex]
Therefore
[tex]\triangle M_a=-1.6kgm/s[/tex]
Generally the equation for momentum for Ball B is mathematically given by
Initial Momentum
[tex]M_{b1}=mV[/tex]
[tex]M_{b1}=0.2*4[/tex]
[tex]M_{b1}=0.8[/tex]
Final Momentum
[tex]M_{b2}=-0 kgm/s[/tex]
Therefore
[tex]|\triangle M_a|>|\triangle Mb|[/tex]
Option A
The ratio of atoms in a compound is shown by the _________.
A. subscript
B. chemical formula
C. chemical equation
D. word equation
The ratio of atoms in a compound is shown by the subscripts: option A.
Formulae of compoundsA molecule of a compound is represented by the symbols of its component elements and such a representation is known as a chemical formula.When writing the chemical formula of a compound, the ratio of atoms in the compound is shown by the subscripts attached to the element in the compound.
For example, the chemical formula of the compound water is H₂O
From the formula, the ratio of hydrogen to oxygen is shown by the subscript:
The ratio of hydrogen to oxygen is H : O = 2 : 1
Therefore, the ratio of atoms in a compound is shown by the subscripts: option A.
Learn more about chemical formula at: https://brainly.com/question/1307605
The ratio of atoms in a compound is shown by the chemical formula.
Chemical formulaChemical formula refers to a representation of a compound using the symbols of the elements that make up the compound. The ratio of atoms in the compound can easily be observed by looking at the subscripts in the chemical formula.
Hence, the ratio of atoms in a compound is shown by the chemical formula. The chemical formula of a compound serves as the identity of the compound.
Learn more about chemical formula: https://brainly.com/question/4102170
5. A big wheel has a diameter of 5 m and a mass of 1500 kg when fully laden with people. a) Work out the moment of inertia of the big wheel. (Hint: which shape from the ones given on p114 would be most suitable? b) When the wheel is rotating at full speed, a person has a linear velocity of 3 m/s. What is the angular velocity of this person? c) What is the rotational kinetic energy at this speed? d) A motor takes 10 seconds to accelerate the wheel from rest to a linear velocity on the circumference of 3 m/s. What is the power of the motor?
Answer:
a) I = 3.75 10⁴ kg m², b) w = 0.6 rad / s, c) K = 6.75 10³ J, d) P = 6.75 10² W
Explanation:
This is a rotations exercise
a) the proper shape for a wheel is that of a rim where most of the weight is in the circumference plus the point weights of the people sitting on its periphery.
We are going to approximate the reda with a thin ring
I = M r²
I = 1500 5²
I = 3.75 10⁴ kg m²
b) angular and linear velocity are related
v = w r
w = v / r
w = 3/5
w = 0.6 rad / s
c) the expression for kinetic energy is
K = ½ I w²
K = ½ 3.75 10⁴ 0.6²
K = 6.75 10³ J
d) the power is
P = W / t
to find the work we use the relationship between work and the variation of kinetic energy
W = ΔK = K_f - K₀
the system part of rest wo = 0
W = K_f
W = 6.75 10³ J
we calculate
P = 6.75 10³/10
P = 6.75 10² W
El punto de ebullición de azufre es de 444.5°C.
c cual es la temperatura Correspondiente en la escala
Fahrenheit?
°F = 1.8°C + 32°
444.5°C = 832.1°F
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
The answer is continuity ( D )
Explanation:
PLZ MARK AS BRAINLIEST
The human eye can readily detect wavelengths from about 400 nm to 700 nm. Part A If white light illuminates a diffraction grating having 910 lines/mm , over what range of angles does the visible m
Answer:
The correct answer is "[tex]21.344^{\circ}[/tex]" and "[tex]39.56^{\circ}[/tex]".
Explanation:
According to the question,
Slit width,
[tex]d=\frac{1}{910 \ lines/mm}[/tex]
[tex]=\frac{1}{910\times 10^3}[/tex]
[tex]=1.099\times 10^{-6} \ m[/tex]
The condition far first order maxima will be:
⇒ [tex]d Sin \theta = 1 \lambda[/tex]
Now,
⇒ [tex]\Theta_{min} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{400\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=21.344^{\circ}[/tex]
⇒ [tex]\Theta_{max} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{700\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=39.56^{\circ}[/tex]
The mass is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time
Answer:
0.28 J
Explanation:
Let the mass of the object is 0.5 kg
The maximum velocity of the object is 1.06 m/s.
We need to find the kinetic energy at that time. It is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]
So, the required kinetic energy is equal to 0.28 J.
If I tell you that you have a right triangle with one leg having length a and another leg having length b , if we
call the hypotenuse c, express the length of c in terms of legs a and b.
C is found using the Pythagorean theorem:
C = sqrt(a^2 + b^2)
Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If
they started from the same place at the same time, what would be the distance between them after 2 min? Find
the distance covered by each car at the same time. (2 min).
Answer:
Distance between them is 4,200 meters.
Explanation:
Consinder car A:
[tex]{ \bf{distance = speed \times time }}[/tex]
substitute:
[tex]distance = 20 \times (2 \times 60) \\ = 2400 \: m[/tex]
Consider car B:
[tex]distance = 15 \times (2 \times 60) \\ = 1800 \: m[/tex]
since these cars move in opposite directions, distance between them is their summation:
[tex]distance \: between = { \sum(distance \: of \: each \: car)} \\ = 2400 + 1800 \\ = 4200 \: m[/tex]
A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a 60.0 N force for the first 16.4 m at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 6.88 m. How much total work did the person do on the box
Over the first 16.4 m, the person performs
W = (60.0 N) (16.4 m) = 984 J
of work.
Over the remaining 6.88 m, they perform a varying amount of work according to
F(x) ≈ 60.0 N + (-8.72 N/m) x
where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :
W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J
(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)
So the total work performed by the person on the box is
984 J + 206.4 J = 1190.4 J ≈ 1190 J
light of wavelength 687 nm is incident on a single slit 0.75 mm wide at what distance from the slit should a screen
Explanation:
The wavelength of light λ=687 nm.
width of the slit, w=0.75 mm.
The diffraction formula for single slit diffraction is;
[tex]Sin\theta=\frac{m\lambda}{w}[/tex]
m here is an integer.
For small θ, sinθ≅ θ.
The formula changes to;
[tex]\theta=\frac{m\lambda}{w}[/tex].
The data in the question is missing. However, one put data in the above formula to find the unknown.
What is the net force on the side of the container The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.
Answer:
26467.21 N
Explanation:
Initial height of water ( h1 ) = 5 m
diameter of container ( d1 )= 100 cm
pressure inside the container ( p1 )= 0.850 atm
Diameter of faucet ( d2 )= 1 inch
Calculate the value of the net force on the side of container
lets assume ; pressure outside the container ( p2 ) = 1 atm
Fnet = ( P1A1 + mg ) - ( P2A2 )
= [ ( 0.85 * 101325 ) ( π(1/2)^2 ) + mg ) - [ ( 101325 )( π )(0.0127)^2 ]
= [ 16902.2766 + pvg ] - [ 51.3161 ]
where ; pvg = pAhg = 1000 * π ( 1/2 )^2 * 5 * 9.8 = 9616.25
= [ 16902.2766 + 9616.25 ] - [ 51.3161 ] = 26467.21 N ( downwards )
HEELLPPPPPpppppppppppppppp
Explanation:
Given:
[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]
[tex]v_1[/tex] = 40 cm/s
[tex]v_2[/tex] = 90 cm/s
[tex]A_2[/tex] = ?
a) The continuity equation is given by
[tex]A_1v_1 = A_2v_2[/tex]
Solving for [tex]A_2[/tex],
[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]
[tex]= 2\:\text{cm}^2[/tex]
b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of
[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]
A ball rolled along a horizontal surface comes to rest in a distance of 72m in 6s. Its initial velocity
and deceleration are
and
Answer:
1. Initial velocity = 24 m/s
2. Deceleration = –4 m/s²
Explanation:
From the question given above, the following data were obtained:
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
1. Determination of the initial velocity.
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) =?
s = (u + v)t / 2
72 = (u + 0) × 6 / 2
72 = u × 3
Divide both side by 3
u = 72 / 3
u = 24 m/s
2. Determination of the deceleration.
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) = 24 m/s
Deceleration (a) =?
v = u + at
0 = 24 + (a × 6)
0 = 24 + 6a
Collect like terms
0 – 24 = 6a
–24 = 6a
Divide both side by 6
a = –24 / 6
a = –4 m/s²
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by a 12.0 - V battery, what total energy in Joules is delivered to the lightbulb filament during 2.00 s
Answer:
E = 20.03 J
Explanation:
Given that,
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,
Voltage, V = 12 V
We need to find the energy delivered to the lightbulb filament during 2.00 s.
The energy delivered is given by :
[tex]E=I^2Rt[/tex]. ....(1)
As,
[tex]I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A[/tex]
As per Ohm's law, V = IR
[tex]R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega[/tex]
Using formula (1).
[tex]E=0.835^2\times 14.37\times 2\\\\=20.03\ J[/tex]
So, the energy delivered to the lightbulb filament is 20.03 J.
5. For the speaker in this circuit, the voltage across it is always proportional to the current through it. Find the maximum amount of power that the circuit can deliver to the speaker.
Answer:
speaker64
--------
34x
Explanation:
64-34
x
speaker
4
2
4
788
- circuit
voltage
100000
x.34
Sorry but you have no picture shown
Place each description under the correct theory
Gravity is an attractive force.
Universal Law of Gravitation
General Theory of Relativity
Mass and distance affect force.
Time and space are absolute,
Time and space are relative.
Gravity is due to space-time curving.
Mass affects space-time curving.
Answer:
1) Law of Universal Gravitation Gravity is an attractive force
5) General relativity Gravity is due to the curvature of spacetime
Explanation:
In this exercise you are asked to relate the correct theory and its explanation
Theory Explanation
1) Law of Universal Gravitation Gravity is an attractive force
2) Law of universal gravitation Mass and distance affect force
3) Classical mechanics time and space are absolute
4) Special relativity Time and space are relative
5) General relativity Gravity is due to the curvature of
spacetime
6) General relativity Mass affects the curvature of space - time
Answer:
Explanation:
edge2022
An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
[tex]p.d' = 37.6 V[/tex]
Explanation:
From the question we are told that:
Potential difference [tex]p.d=18.8V[/tex]
New Capacitor [tex]C_1=C_2/2[/tex]
Generally the equation for Capacitor capacitance is mathematically given by
[tex]C=\frac{eA}{d}[/tex]
Generally the equation for New p.d' is mathematically given by
[tex]C_2V=C_1*p.d'[/tex]
[tex]p.d' = 2V[/tex]
[tex]p.d'= 2 * 18.8[/tex]
[tex]p.d' = 37.6 V[/tex]
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 3.22 x 104 V
Answer:
[tex]E=3.22*10^6 N/C[/tex]
Explanation:
From the question we are told that:
Separation Distance [tex]d=1.0cm =0.01m[/tex]
Potential difference [tex]V=3.22 * 10^4 V[/tex]
Generally the equation for Electric Field strength is mathematically given by
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{3.22*10^4}{0.01}[/tex]
[tex]E=3.22*10^6 N/C[/tex]
write physical quantities and its unit
length= metre
mass= kg
time= second
temperature = kelvin
current= ampere
luminous intensity= candela
Amount of substance = mole
etc
I hope this will help you
stay safe