By using Lami's theorem, Mass m = 1.75 kg approximately
Given that a strong weightless rope has a mass, m, hanging from the middle of it. If the tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees to the horizontal.
By using Lami's theorem, we can get how much mass is hanging from the rope.
Let the angle between the rope = α = 180 - 40
α = 140 degrees
The angle between one of the rope and mass = β = 20 + 90
β = 110 degrees
The angle between the mass and the other rope = γ = 360 - (140 + 110)
γ = 360 - 250
γ = 110 degrees
W/ sinα = T/ sinβ = T/sinγ
W/ sinα = T/ sinβ
Substitute all the necessary parameters
W/sin140 = 25/sin 110
W / 0.643 = 25 / 0.939
W = 17.1 N
Weight W = mg
17.1 = 9.8m
mass m = 17.1/9.8
Mass m = 1.7455 kg
Mass m = 1.75 kg approximately
Therefore, 1.75 kg mass is hanging from the rope.
Learn more about resolution of forces here: https://brainly.com/question/1858958
Choose the option that best suit the question .
An inclined plane of angle Ɵ, acting as a simple machine has a velocity ratio :
(a) sinƟ
(b) 1/sinƟ
(c) cosƟ
(d) 1/cosƟ
(e) sinƟcosƟ
Answer:
(b) 1/sinƟ
Explanation:
A simple machine can be defined as a type of machine with no moving parts but can be used to perform work.
Basically, a simple machine allows for the transformation of energy into work. The six simple machines are; lever, wedge, pulley, screw, wheel and axle, and inclined plane.
Inclined plane is a simple machine set at an angle and then used to lift an object.
In Physics, the velocity ratio of a simple machine is calculated as a ratio of the distance moved by the effort to the vertical distance through which a load is raised.
Mathematically, the velocity ratio of an inclined plane of angle Ɵ, acting as a simple machine is given by the formula;
Velocity ratio (V.R) = 1/sinƟ
Basically, an increase in the angle of inclination (measured in degrees) of an inclined plane increases its velocity ratio.
Annie has a soccer ball and a cake ball she kicks each ball with the same Force the soccer ball accelerates at 3 m per second squared and the kickball accelerates at 5 m per second squared use Newton's law to describe why the kickball has a greater acceleration
La resistencia de un termómetro de platino es de 6Ω a30°C. Hallar su valor correspondiente a 100°C,sabiendo que el coeficiente de temperatura de resistividad del platino vale 0,00392°C^(-1).
Respuesta:
7,6 Ω
Explicación:
Paso 1: Información dada
Resistencia a 30 °C (R₀): 6 ΩCoeficiente de temperatura (α): 0,00392 °C⁻¹Paso 2: Hallar la resistencia (R) a 100 °C
Podemos ver la relación entre la resistencia de un material y la temperatura usando la siguiente ecuación.
R = R₀ (1 + α × ΔT)
R = 6 Ω (1 + 0,00392 °C⁻¹ × (100 °C - 30 °C)) = 7,6 Ω
một oto đang chạy với vận tốc là 10m/s thì tăng tốc và chuyển động nhanh dần đều sau 20s thì đạt vận tốc 14m/s
a) tính gia tốc của xe
b) tính vận tốc xe sau 40s kể từ khi tăng tốc và quãng đường xe đi được trong thời gian đó
a) gia tốc = vf-vi / t
a = 14-10 / 20
a = 0,2ms⁻²
b) dưới dạng a = Δv / t
v = lúc
v = 0,2 × 40
v = 8ms⁻¹
như v = d / t
do đó d = vt
d = 8 × 40
d = 320m
hãy đánh dấu là trí óc nhất
use the particle model to explain why the water expands
Answer:
Take water as an example; the size, shape and chemical composition of the water particles remains the same whether it is solid water (ice) or gaseous water (steam) - but how those particles move and are arranged is different for
can someone pls help me calculate this?
Answer:
10N is the answerrrerreer
Answer:
f= 100N
Explanation: F=m×(v₀-vf/t)
=0.05ₓ(200-0/0.1)
=0.05ₓ2000
=100N
please mark as brainliest
What's an area of the earth that suggest that " the Earth is like one giant jigsaw" idea could be true.
Two children sitting on a see saw such that they can't swing.what is the net torque in this situation ??? Need an appropriate answer
Answer:
two childrens are sitting on a see saw but can't swing.it means both weight is same (a=b)
Explanation:
hope it will help u please mark me brillientAnswer:
which subject is that question
A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the starting temperature of the cold water is 16'c? The specific heat capacity of water is 4200 J/Kg'c
Answer:
[tex]8.4 \times 10^{5}\; \rm J[/tex], assuming that there's no heat exchange between the washing machine and the environment.
Explanation:
Let [tex]m[/tex] denote the mass of water and [tex]c[/tex] the specific heat capacity of water. The energy required to raise the temperature of that much water by [tex]\Delta T[/tex] would be:
[tex]Q = c \cdot m \cdot \Delta T[/tex].
Washing at [tex]30\; \rm ^{\circ} C[/tex] would require a temperature change of [tex]\Delta T = 30\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 14\; \rm K[/tex].
Washing at [tex]50\; \rm ^{\circ} C[/tex] would require a temperature change of [tex]\Delta T = 50\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 34\; \rm K[/tex].
In both situations, [tex]c = 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1}[/tex] while [tex]m = 10\; \rm kg[/tex].
Calculate the energy required in either situation:
Washing at [tex]30\; \rm ^{\circ} C[/tex]:
[tex]\begin{aligned}& Q({30\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 14\; \rm K \\ &= 588000 \times 10^{5}\; \rm J\end{aligned}[/tex].
Washing at [tex]50\; ^{\circ} {\rm C}[/tex]:
[tex]\begin{aligned}& Q({50\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 34\; \rm K \\ &= 1428000 \; \rm J\end{aligned}[/tex].
[tex]1428000\; \rm J - 588000\; \rm J = 8.4 \times 10^{5}\; \rm J[/tex].
personal and unbiased errors belong to
personal to systematic errors
unbiased to random errors
a. A man of 100 kg mass is standing on the surface of the earth of mass 6 × 1024 kg and
radius 6400km. Calculate the gravitational force between them. [Ans: 977.05 N]
Answer:
m1=100kg
m2=6×10^24kg
r=6400km
G=6.6×10^-11
F=?
as we know that
F=G m1.m2/r²
F=6.6×10^-11×100×6×10^24/(6400)²
F=6.6×10^-11×6×10^26/4.096×10^7
F=6.6×10^-11×1.465×10^19
F=9.67×10^8N
A body having uniform velocity has zero acceleration? give reason
Answer:
A body having uniform velocity has zero acceleration because
there is not change in velocity.
a car moves at a speed of 40km/h. it is stopped by applying brake which produce a uniform acceleration of-0.5m/s^2. how much distance will it move before coming to stop ?
Answer:
Explanation:
We first need to convert the 40 km/h to m/s. Going by the fact that 40 has only 1 significant figure in it, 40 km/h = 10 m/s. The rest of the values are in their proper labels. We will use the equation:
[tex]v^2=v_0^2+2a[/tex]Δx where the final velocity is 0 because the car is coming to a stop at the end; the initial velocity is 10 m/s, the acceleration (or, rather, deceleration) is -.5 m/s/s, and our unknown which is displacement. Filling in:
[tex]0=(10)^2+2(-.5)[/tex]Δx and solving for Δx:
Δx = [tex]\frac{-100}{2(-.5)}[/tex] which ends up being simply that
Δx = 100 m
steel wire 8m long and 4mm in diameter is fixed to two rigid b. [1] supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steel, a 12x10 %/K, Young's modulus for steel, Y = 2x10¹¹ N/m².
Answer:
301.6 N
Explanation:
The length of the wire L₀ = 8 m and its diameter, d = 4 mm = 4 × 10⁻³ m. Since its temperature drops by 10°C, it will have a change in length ΔL = L₀αΔθ where α = linear expansivity of steel, a 12 × 10⁻⁶ /K, and Δθ = temperature change = -10°C = -10 K(negative since it is a drop)
So, the strain, ε = ΔL/L₀ = αΔθ = 12 × 10⁻⁶ /K × 10 K = 12 × 10⁻⁵
Now the Young's modulus of steel, Y = σ/ε where σ = stress = T/A where T = increase in tension in steel wire and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 4 × 10⁻³ m and ε = strain = 12 × 10⁻⁵
So, σ = Yε
Since Y = 2 × 10¹¹ N/m².
Substituting the values of the variables into the equation, we have
σ = Yε
σ = 2 × 10¹¹ N/m² × 12 × 10⁻⁵
σ = 24 × 10⁶ N/m²
Since σ = T/A
T = σA
T = σπd²/4
Substituting the values of the variables into the equation, we have
T = σπd²/4
T = 24 × 10⁶ N/m² × π × (4 × 10⁻³ m)²/4
T = 24 × 10⁶ N/m² × π × 16 × 10⁻⁶ m²/4
T = 24 × 10⁶ N/m² × π × 4 × 10⁻⁶ m²
T = 96 N × π
T = 301.59 N
T ≅ 301.6 N
So, the increase in tension in the steel wire is 301.6 N
When the system pressures are lower than the pressure in the refrigerant tank, such as when the system has just been evacuated or if the system is out of refrigerant, refrigerant can be added to both the high and low pressure sides of the system. A. True B. False
Answer: True
Explanation:
It should be noted that when refrigerant is added in vapor form to the operating refrigeration system, then the addition of the refrigerant should be to the low-pressure side of the refrigeration system.
Furthermore, when system pressures are lower than the pressure in the refrigerant tank as stated in the question, then the refrigerant can be added to both the high and low pressure sides of the system.
Therefore, the correct option is true.
1. The hydrogen balloons which are used to collect weather information from the atmosphere is made of plastic and never completely filled. Thus the pressure inside and outside are same. The balloon is filled with 150 litres of hydrogen, the air temperature is 27°C and the atmospheric pressure is 98 kPa. The balloon rises to a height where it radios back that the pressure is 30kPa and the temperature is - 33°C. i. What is the Kelvin temperature equivalent to 27°C and -33°C? (1 mark) ii. What is the volume of hydrogen at this height? (2 marks)
Answer:
Answer:- Volume of the balloon is 5.78 L.
Solution:- There are 0.24 moles of hydrogen gas in a balloon at 35 degree C and 1.05 atm pressure. It asks to calculate the volume of the balloon.
This problem is based on ideal gas law equation:
P = 1.05 atm, n = 0.24 mole, T = 35 + 273 = 308 K
R =
V = ?
The equation could be rearranged for the volume as:
Let's plug in the values and do the calculations to get the volume of the balloon:
V = 5.78 L
So, the volume of the gas balloon is 5.78 L.
A jet airplane is in level flight. The mass of the airplane is m=8950 kg. The airplane travels at a constant speed around a circular path of radius R=9.33 mi and makes one revolution every T=0.123 h. Given that the lift force acts perpendicularly upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane?
Answer:
The net force is 91780.8 N.
Explanation:
mass, m = 8950 kg
Radius, R = 9.33 miles = 15015.2 m
Time, T = 0.123 h = 442.8 s
There are two forces acting on the plane.
Horizontal force is the centripetal force and the vertical force is the weight.
[tex]Fx =m R w^2\\\\Fx = m R \frac{4\pi^2}{T^2}\\\\Fx = 8950\times 15015.2\times \frac{4\times 3.14\times 3.14}{442.8\times 442.8}\\\\Fx = 27030.8 N \\\\Fy = m g \\\\ Fy = 8950\times 9.8 \\\\Fy = 87710 N[/tex]
The net force is
[tex]F = \sqrt{Fx^2 + Fy^2}\\\\F = \sqrt {27030.8^2 + 87710^2}\\\\F = 91780.8 N[/tex]
A cyclist competes half round of a circular track of a diameter 636m in 1.5 minutes . find the velocity
Answer:
Given information:-
Radius of the circular track is 636 mThe time elapsed in the journey is 1.5 min or 90 sThe distance traveled by the cyclist is semi circumference of the circle = πr[tex]\clubsuit[/tex]The displacement of the cyclist will be shortest distance from his initial to final positions which is nothing but diameter of the circle = 2r
We know that,
[tex] \underline{ \boxed{ \sf{Speed = \frac{Distance \: covered }{Time \: taken}}}}[/tex]
[tex]\longrightarrow \sf{\frac{636 \times 3.14 m}{90s}} \\ \\ \longrightarrow \sf22.18 m/s[/tex]
Speed of the cyclist is 22.18 m/s[tex]\sf{Velocity = \frac{Displacement }{ Time} }[/tex]
[tex]\longrightarrow \sf{ \frac{2 \times 636m}{90s} } \\ \\ \longrightarrow \sf{14.13 \: m/s}[/tex]
Velocity of the cyclist is 14.13 m/sThank You :)
diffusion in physics
Answer:
sorry but which class your talking 'bout
Una rueda tiene 4 cm de diámetro y realiza
200 vueltas cada 5 minutos. Calcular:
a. la frecuencia
b. el período
c. la velocidad lineal o tangencial
d. La velocidad angular
e. La aceleración centrípeta.
Answer:
Una rueda tiene 4 cm de diametro y realiza 200 vueltas cada 4 minutos calcular a frecuencia b el periodo c la velocidad lineal l tangencial
20. A semiconductor is a
A) crystalline solid that conducts current under any condition
B) metallic solid that conducts current under any condition
C) metallic solid that conducts current under certain conditions
D) crystalline solid that conducts current under certain conditions
Answer:
D) crystalline solid that conducts current under certain conditions
Explanation:
A semiconductor is any crystalline solid material whose electrical conductivity is intermediary between that of insulators and conductors. That is, a semiconductor does not conduct like a conductor but does more than an insulator.
Based on this, a semiconductor is said to only conduct electricity under certain conditions as opposed to conductors that do so under any condition. Example of semiconductors are silicon, carbon, antimony, arsenic etc.
A 4000 N force acts on an object that initially has a momentum of 400 kg-m/s for 0.9 seconds. What is the final momentum of the system?
Answer:
4360 Kgm/s
Explanation:
Applying,
Ft = M-M'................. Equation 1
Where F = force, t = time, M = Final momentum, M' = Initial momentum.
make M the subject of the equation
M = Ft+M'............ Equation 2
From the question,
Given: F = 4000 N, t = 0.9 seconds, M' = 400 kg-m/s
Substitute these values into equation 2
M = 4000+(0.9×400)
M = 4000+360
M = 4360 kgm/s
Hence the final momentum is 4360 kgm/s
The energy required to change the state of a substance was determined to be E(H).
If the mass of the substance was DOUBLED, the value of E(H) will
(A) be halved
(B) be doubled
(C) be quadrupled
(D) remain constant
Answer:
ccccccccc or bbbbbbb or aaaaaaaa or ddddddd
a car of mass 500 kg travelling with a speed of 54km/hr is brought to rest over a distance of 40 m. find the acceleration and force applied on brakes
Answer:
Explanation:
[tex]a=\frac{v_f-v_0}{t}[/tex] We don't yet need the mass of the car, as we can see by this equation. What we do need is the velocity of the car, the initial velocity, in meters per second, and right now it's in km/hr. Not good. We need to convert. The conversion is as follows:
[tex]54\frac{km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex] Ok, that's good. Now we have everything we need but the time element. If the car traveled a distance of 40 meters at 15 m/s, then we can use the d = rt equation to solve for t, and when we find t we plug it into the acceleration equation:
40 = 15t and
t = 2.7 seconds. The car traveled for 2.7 seconds to go that 40 meters. That's the only reason we were given the displacement. We need it for nothing else but that.
Filling in the acceleration equation now:
[tex]a=\frac{0-15}{2.7}=-5.6\frac{m}{s^2}[/tex] and the negative indicates we are in fact slowing down. That's the answer for the acceleration portion of the problem; now we need the force, F, applied to the brakes.
F = ma where m is mass (we get to use that value now!) and a is -5.6 m/s/s.
F = 500(-5.6) and
F = -2800 N and the negative here means that the force of the brakes is acting against the motion of the car: the brakes are pulling the car "backwards" to stop while the car's motion is forward. The negative indicates the direction the force is being applied.
Angie walked a distance of 90 meters east in 70 seconds. What was her
velocity?
A. 0.78 m/s east
B. 1.3 m/s east
O C. 7 m/s east
D. 9 m/s east
describe how air resistance would affect a falling object
Answer:
With air resistance, acceleration throughout a fall gets less than gravity (g) because air resistance affects the movement of the falling object by slowing it down. How much it slows the object down depends on the surface area of the object and its speed
Explanation:
PLZ MARK AS THE BRAINLIEST
HAVE A GOOD DAY
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The direction of applied force has to be .............. to the distance in order to say work is done
Answer:
Parallel
Explanation:
The work done is defined as the force applied on an object and the displacement in the position of the object in the direction of force.
W = F s cos A
where, F is the force, s is the displacement and A is the angle between force and displacement.
When the angle between the force and the displacement is 90 degree, the work done is zero.
To get the maximum work the angle between the force and the displacement is 0 degree.
So, to get the work done by the force the angle between the force and displacement is 0 degree that means the force and displacement is parallel to each other.
Si dejamos caer un objeto desde una gran altura, ¿será que tiene siempre la misma velocidad
Answer:
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A CROW BAR WITH LENGTH 200 CM IS USED TO LIFT A LOAD OF 600N . IF THE DISTANCE BETWEEN FULCRUM AND LOAD IS 0.75. CALCULATE ; a, effort b, MA c, VR
Answer:
a. Effort = 960 Newton
b. Mechanical advantage (M.A) = 0.625
c. Velocity ratio (V.R) = 1.67
Explanation:
Given the following data;
Load = 600 NLength of crowbar = 200 cmLength of load arm = 0.75 mConversion:
100 cm = 1 m
X cm = 0.75 m
Cross-multiplying, we have;
X = 0.75 * 100 = 75 cm
First of all, we would find the effort arm;
Effort arm = length of crow bar - length of load arm
Effort arm = 200 - 75
Effort arm = 125 cm
Next, we would determine the mechanical advantage (M.A) of the crow bar;
[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]
Substituting the values into the formula, we have;
[tex] M.A = \frac {125}{200} [/tex]
M.A = 0.625
To find the effort of the crow bar;
[tex] M.A = \frac {Load}{Effort} [/tex]
Making "effort" the subject of formula, we have;
[tex] Effort = \frac {Load}{M.A} [/tex]
[tex] Effort = \frac {600}{0.625} [/tex]
Effort = 960 Newton
Lastly, we would determine the velocity ratio (V.R);
[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]
[tex] V.R = \frac {125}{75} [/tex]
V.R = 1.67
Which choice shows the correct sequence of features formed by continued wave erosion?
Answer:
Wave-cut cliff, sea arch, sea stacks
Explanation:
The effect of a wave erosion is made obvious by the structures formed by the wave action.
The high areas of land adjacent to the incoming wave develop the early features or the formation of wave action, which includes the wave-cut cliff
The continuous undercutting of the cliff by the wave results in the formation of the wave cut platform
The effect of the wave further on a cliff, results in the formation of a sea arch and finally a sea stack
Therefore, the correct sequence is the wave-cut cliff, sea arch, sea stacks
