Answer:
A bridge suspended by cables
Explanation:
Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.
While a mason was working concrete into formwork, the formwork collapses. Who is BEST suites to rectify this problem? Mason Carpenter Project Manager O Construction Technician A device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop
Answer:
1. Carpenter
2. True
Explanation:
While a mason was working concrete into the formwork, the formwork collapses. The best person to rectify this problem is CARPENTER.
This is because it is the job of the Carpenter to design and build formwork, most especially wooden formwork. Formwork is like casing built to receive concrete and reinforcement during construction. Hence, when formwork collapses either due to stress, tension, or improper construction, it is the job of Carpenter to reconstruct the formwork or rectify the problem.
It is TRUE that when a device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop. However, this communication will be an instruction on what to do next, and it usually directs those responsible to take action where necessary. For example, a workshop manager communicates to a carpenter about the need to rectify a chair or table that has a defect.
Planets closer to a star will have what type of average temperature
Answer:
Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018
a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
Answer:
The surface tension is 190.2 N/m.
Explanation:
Initial radius, r = 4 cm
final radius, r' = 6 cm
Work doen, W = 15 J
Let the surface tension is T.
The work done is given by
W = Surface Tension x change in surface area
[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]
The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light
Answer:
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Explanation:
The energy of a photon is calculated using the following equation;
E = hf
where;
h is Planck's constant = 6.63 x 10⁻³⁴ Js
f is frequency of the photon
[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]
[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Question 18/55 (2 p.)
A vibrating object produces ripples on the surface of a liquid. The object completes 20 vibrations
every second. The spacing of the ripples, from one crest to the next, is 3.0 cm.
What is the speed of the ripples?
D
C 60 cm/s
120 cm/s
A 0.15cm/s
B 6.7 cm/s
Answer:
the correct answer is C v = 60 cm / s
Explanation:
The speed of a wave is related to the frequency and the wavelength
v = λ f
They indicate that the object performs 20 oscillations every second, this is the frequency
f = 20 Hz
the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength
λ = 3 cm = 0.03 m
let's calculate
v = 20 0.03
v = 0.6 m / s
v = 60 cm / s
the correct answer is C
a 50kg skater on level ice, has built up her speed to 30km/h. how far will she coast before sliding friction dissipates her energy?
Answer:
belpw
Explanation:
The distance prior to the sliding friction dispersing her energy would be:
- The distance will remain unaffected by the sliding friction i.e. 354m
As we know, When Sliding friction dissolves her energy, leading her Kinetic Energy to turn 0 on coming to the state of rest. So,
[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex] (∵ Work in -ve denotes it is done opposite to friction)
Given that,
m(mass) [tex]= 50 kg[/tex]
v(velocity) [tex]= 30 km/hr[/tex] or [tex]8.33 m/s[/tex]
The coefficient of Kinetic Friction [tex]= 0.01[/tex]
g(gravitational force) [tex]= 9.8 m/s^2[/tex]
Initial Velocity(u) [tex]= 30[/tex] × [tex]1000/3600 m/s[/tex]
[tex]= 8.33 m/s[/tex]
Now by employing the provided values,
[tex]F =[/tex] μ[tex]mg[/tex]
[tex]= (0.01) (50) (9.8)[/tex]
[tex]= 4.9[/tex]
∵ [tex]F = 4.9 N[/tex]
By using the above expression, we will find the distance;
[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]
⇒ [tex]1/2 (50) (0)^2 - 1/2 (50) (8.33)^2 = -4.9(S)[/tex]
⇒ [tex]1734.7225 = 4.9S[/tex]
⇒ [tex]S = 1734.7225/4.9[/tex]
∵ [tex]S = 354 m[/tex]
Because [tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex] [tex]= -[/tex] μmgS
⇒ [tex]S = (u^2 - v^2)[/tex]/2μ[tex]g[/tex]
Thus, the distance will remain unaffected by the sliding friction i.e. 354m
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Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of fall at the earth's surface is given by g = Gm/R2 . What is the acceleration of a satellite moving in a circular orbit round the earth of radius 2R
Explanation:
The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.
[tex]W=F_G[/tex]
[tex]mg = G \dfrac{mM}{R^2}[/tex]
which gives us an expression for the acceleration due to gravity g as
[tex]g = G\dfrac{M}{R^2}[/tex]
At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity [tex]g_h[/tex] at this height is
[tex]mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}[/tex]
Simplifying this, we get
[tex]g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g[/tex]
Calculate the forces that the supports \rm A and \rm B exert on the diving board shown in when a 58-\rm kg person stands at its tip.
4. Water stands 12.0 m deep in a storage tank whose top is open to the atmosphere at
1.00 atm. The density of water is given as 1000 kg/m² and some pressure conversion
are 1 Pa = 1 N/m² while 1 atm = 101 325 Pa.
a) What is the absolute pressure at the bottom of the tank?
b) What is the gauge pressure at the bottom of the tank?
[4]
[4]
Answer:
[tex]P=217600Pa[/tex]
Explanation:
From the question we are told that:
Density [tex]\rho=1000kg/m^3[/tex]
Depth of Water [tex]d=12.0m[/tex]
Generally the equation for Pressure is mathematically given by
[tex]P=\rho gh[/tex]
[tex]P=1000*9.8*12[/tex]
[tex]P=117600N/m^2[/tex]
Therefore
Absolute Pressure=P+P'
Where
P=Pressure under water
P'=Atmospheric Pressure
Therefore
[tex]P_A=P+P'[/tex]
[tex]P_A=117,600+10^5[/tex]
[tex]P=217600Pa[/tex]
What is the "best" explanation for why the universe is the way it is?
A) god created the universe
B) there is a multiverse and this one happens to be perfect for life.
C) this is the only universe and it happens to be perfect for life.
D) It is all in illusion and none of it exists.
E) none of the above, they are all just guesses.
I know the answer I just wanna see what you guys think.
i will give brainly if you get it right.
Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is . ms
Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution
answer:
P = 141.21 N
Explanation:
Given data:
Mass of crate = 50 kg
coefficient of static friction ( μ ) = 0.25
Calculate minimum horizontal force ( P ) that holds the crate from sliding
∑fx = 0
= P + Fcos θ - N*sinθ = 0
= P + 0.25N cos 30° - Nsin30° = 0
∴ P = 0.2835 N = 0
P - 0.2853 N = 0 ------- ( 1 )
∑fy = 0
- 50g + Ncosθ + Fsinθ
- 50*9.81 + Ncos30° + 0.25Nsin30°
∴ N = 494.942 N ----- ( 2 )
input 2 into 1
P - 0.2853 ( 494.942 ) = 0
P = 141.21 N
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
Explanation:
From the given information:
the car's momentum = momentum of the truck
∴
(a) 816 kg × v = 2650 kg × 16.0 km/h
v = (2650 kg × 16.0 km/h) / 816 kg
v = 51.96 km/hr
(b) 816 kg × v = 9080 kg × 16.0 km/h
v = (9080 kg × 16.0 km/h) / 816 kg
v = 178.04 km/hr
The human ear can respond to an extremely large range of intensities - the quietest sound the ear can hear is smaller than 10-20 times the threshold which causes damage after brief exposure. If you could measure distances over the same range with a single instrument, and the smallest distance you could measure was 1 mm, what would the largest be, in kilometers?
Answer:
the largest distance we can measure is 10¹⁴ km
Explanation:
Given the data in the question;
Threshold hearing = 10⁻²⁰
smallest distance measured = 1 mm
Largest distance measured will be;
⇒ ( threshold hearing )⁻¹ × smallest distance
= ( 1 / 10⁻²⁰ ) × 1 mm
= 10²⁰ × 1mm
= 10²⁰ mm
we know that; 1000 mm = 10⁶ km
Largest distance = ( 10²⁰ / 10⁶ ) km
= 10¹⁴ km
Therefore, the largest distance we can measure is 10¹⁴ km
Explain how the Laws of planetary motion and Newton’s laws allow the hotel to keep moving in space.
Answer:
Explanation:
i am sorry i needed points
Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:
a. Sine wave;
b. Square wave,
c. Triangle wave.
Calculate the period of a waveform with the frequency of:
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Answer:
a) [tex]T=0.01s[/tex]
b) [tex]T=0.001s[/tex]
c) [tex]T=0.00001s[/tex]
Explanation:
From the question we are told that:
Given Frequencies
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Generally the equation for Waveform Period is mathematically given by
[tex]T=\frac{1}{f}[/tex]
Therefore
a)
For
[tex]T=100 Hz[/tex]
[tex]T=\frac{1}{100}[/tex]
[tex]T=0.01s[/tex]
b)
For
[tex]F=1kHz[/tex]
[tex]T=\frac{1}{1000}[/tex]
[tex]T=0.001s[/tex]
c)
For
[tex]F=100kHz[/tex]
[tex]T=\frac{1}{100*100}[/tex]
[tex]T=0.00001s[/tex]
ADvantage of friction
Answer:
1. Friction enables us to walk freely.
2. It helps to support ladder against wall.
3. It becomes possible to transfer one form of energy to another.
4. Objects can be piled up without slipping.
The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cross-sectional area. According to theory, what should be the ratio of the resistance of the second coil to the first coil and the fourth coil to the third
Answer:
The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Explanation:
The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.
So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
When using the process of evaporation to separate a mixture what is left behind to an evaporating dish
A. The mixture does not separate in the entire mixture remains in the dish
B. The liquid evaporates in the solid is left in the dish
C. The mixture does not separate in the entire mixture evaporates
D. None of these
Answer:
B
Explanation:
The liquid evaporates in the solid is left in the dish..
A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car,
increasing the speed to 2 m/s. How much work did the man do?
A. 640 J
B. 360 J
C. 1360 J
D. 1000 J
Work done by man will be A. 640 J
What is work energy theorem?
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
according to work energy theorem
Work done = final Kinetic energy - initial kinetic energy
= KE (final) - KE (initial )
= 1/2 m ([tex]v^{2}[/tex]) - 1/2 m ([tex]u^{2}[/tex])
= 1/2 m ([tex]v^{2}[/tex] - [tex]u^{2}[/tex])
= 1/2 * 500 * ( [tex]2^{2}[/tex] - [tex]1.2^{2}[/tex])
= 250 * 2.56 = 640 J
correct answer is A. 640 J
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Phát biểu nào sau đây là SAI?
A. Cường độ điện trường là đại lượng
đặc trưng cho điện trường về phương
diện tác dụng lực.
B. Điện trường tĩnh là điện trường có
cường độ E không đổi tại mọi điểm.
C. Đơn vị đo cường độ điện trường là
vôn trên mét (V/m).
D. Trong môi trường đẳng hướng,
cường độ điện trường giảm lần so với
trong chân không
Answer:
B.
Explanation:
sana makatulong sayo
A train is moving at a constant
speed of 55.0 m/s. After 5.00
seconds, how far has the train
gone?
cara
(Units = m)
Answer:
Distance = speed * time
55*5
275 meters.
The train would have covered a distance of 275 m
What is distance ?
We can define distance as to how much ground an object has covered despite its starting or ending point.
Distance = speed * time
given
speed= 55 m/s
time = 5 sec
Distance = 55 * 5 = 275 m
The train would have covered a distance of 275 m
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____________is obtained from the fleece of animals.
Answer:
wool and fibers
Explanation:
What is needed to Run A Brushless DC motor
Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.
CORRECT ME IF IM WRONG!!
#CARRYONLEARNING>>> JAZMINE
Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projectile motion? The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity. A force other than gravity is acting on the paper airplane. The paper airplane’s motion can be described using only one dimension. A push and a pull are the primary forces acting on the paper airplane.
highschool physics, not college physics
Answer:
Answer:
A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.
Explanation:
Edge.
Answer:
The motion of the paper airplane is best explained by horizontal inertia and vertical pull of gravity.
Explanation:
What is horizontal inertia and vertical pull of gravity?Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .
While vertical pull is due to the earth .
In a paper airplane , four forces act .these forces provide it flight.These forces are horizontal inertia , vertical pull downwards , lift by air and drag.Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.
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Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric potential difference depends on current and resistance according to this function V(I,R) = IR. Your measured current and resistance have the following values and uncertainties I = 5.9 Amps, delta I space equals space 0.4 Amps and R = 42.7 Ohms and delta R space equals space 0.6 Ohms. What is the uncertainty in the , delta V ? Units are not needed in your answer.
Answer:
ΔV = 2 10¹ V
Explanation:
The calculation of the uncertainty or error in an expression is given by
ΔV = [tex]\frac{dV}{di}[/tex] |Δi| + [tex]\frac{dV}{dR}[/tex] |ΔR |
V = i R
let's make the derivatives
[tex]\frac{dV}{di}[/tex] = R
[tex]\frac{dV}{dR}[/tex] = i
we substitute
ΔV = R | Δi | + i | ΔR |
in the exercise give the values
i = (5.9 ± 0.4) A
R = (42.7 ± 0.6) Ω
we calculate
ΔV = 42.7 0.4 + 5.9 0.6
ΔV = 20.6 V
ΔV = 2 10¹ V
the voltage is
V = i R
V = 5.9 42.7
V = 251.9 V
the result is
V = (25 ± 2) 10¹ V
HELP ME PLS!!!!
Find the location of beryllium (Be) on the periodic table. What type of ion will
beryllium form?
A. An ion with a -2 charge
B. An ion with a +6 charge
C. An ion with a +2 charge
D. An ion with a -6 charge
Answer:
the answer is c which is a+2 charge
Explanation:
Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.
The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.
What are cations and anions?In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.
This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.
Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.
The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.
Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.
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A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sled attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Answer:
[tex]\mu=0.185[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=17kg[/tex]
Force [tex]F=33N[/tex]
Velocity [tex]v=1.6m/s[/tex]
Distance [tex]d= 9.8m[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=\triangle K.E+\triangle P.E[/tex]
Where
[tex]\triangle K.E=(F-F_f)*2[/tex]
[tex]F_f=F+\frac{\triangle K.E}{d}[/tex]
[tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]
[tex]F_f=30.8N[/tex]
Since
[tex]f = \mu*m*g[/tex]
[tex]\mu= 30.8/(m*g)[/tex]
[tex]\mu= 30.8/(17*9.81)[/tex]
[tex]\mu=0.185[/tex]
please helpp!
convert 1N into dyne
In the given relation F=ma a stands for write there SI unit
Answer:
a. 1 Newton = 100000 Dyne
b. a represents acceleration.
Explanation:
Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;
1 Newton = 10⁵ Dyne
1 Newton = 100000 Dyne
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Force = mass * acceleration
[tex] F = ma[/tex]
Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.
A scooter is accelerated from rest at the rate of 8m/s
. How long will it take to cover
a distance of 32m?
Explanation:
time=Distance/speed
t=32/8
t=4 seconds
Consider the heaviest box of 150 lb that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.45, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 60.0° above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.45.
If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.)
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
What is meant by kinetic friction ?Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.
Here,
Mass of the box, m = 150 lb = 68.1 kg
Coefficient of kinetic friction, μ = 0.45
Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,
F(k) = μmg
F(k) = 0.45 x 68.1 x 9.8
F(k) = 300.32 N
Now, the box sits on a ramp inclined at 60°
Coefficient of kinetic friction, μ = 0.45
The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.
So,
Frictional force, F(k)' = μmgcosθ
F(k)' = 0.45 x M x 9.8 x cos 60
F(k)' = 2.2M
Weight of the box acting horizontally,
W = Mgsinθ
W = M x 9.8 x sin60
W = 8.5M
Therefore, net force,
Fn = W - F(k)'
Fn = 8.5M - 2.2M
Fn = 6.3M
The total force acting on the box is
F = F(k) - Fn
ma = 300.32 - 6.3M
Since, the box is moving with constant speed, the acceleration, a = 0
Therefore,
300.32 - 6.3M = 0
6.3M = 300.32
M = 300.32/6.3
M = 47.7 kg = 105.16 pound
Hence,
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
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