Explanation:
the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).
An electric motor converts electrical energy into which kind of energy?
potential
kinetic
stored
chemical
Answer:
Kinetic Energy
Answer:
Kinetic
Explanation:
got it on edge
Neon gas in a container was heated from 20 degrees celsius to 120 degrees celsius.Its new volume is 150ml.what was the original volume?
Answer:
110 mL
Explanation:
Ideal gas law:
PV = nRT
Assuming the container isn't rigid, and the pressure is constant, then:
V/T = V/T
Plug in values (remember to use absolute temperature).
V / 293 K = 150 mL / 393 K
V = 110 ml
Hope u understood!
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Help Please
Energy unit conversions between electron volts (eV) and joules, (J); Einstein's photon energy equation;
Compton's momentum equation;
de Broglie's wavelength equation
Using your knowledge of these equations, quantitatively compare a 3.1 eV photon and a 3.1 eV electron by completing the following data table:
Energy (J)
Rest mass (kg)
Speed (m/s)
Wavelength (m)
Momentum (kg•m/s)
Answer:
Energy=3.1times 10^-17 J
Rest mass: 6.2 kg
Speed: 47.5 m/s
Wavelength: 2.659 times 10^-6
Momentum: 67.3 kg(m/s)
Explanation:
El dormitorio de Pablo es rectangular, y sus lados miden 3 y 4 metros. Ha decidido dividirlo en dos partes triangulares con una cortina que une dos vértices opuestos. ¿Cuántos metros deberá medir la cortina?
Answer:
c = 5 m
Explanation:
this exercise you want to divide the rectangular room into two triangular rooms
the area of triangles is
A = ½ base height
A = ½ 4 3
A = 6 m²
the length of the curtain can be found using the Pythagorean theorem
c² = b² + a²
c = √ (4² + 3²)
c = 5 m
this is the length of the curtain
3) A stationary box is acted upon by 5 forces represented by letters of the alphabet as shown figure 1. Under which of these conditions will the box move in the direction shown in figure 2? Assume a frictionless surface. a) S=P+Q and T P or =P and T >Q+R
Answer:
Of course, with a triangle with a 90° angle, a right triangle, you can simply use pythagoras theorem (a2+b2=c2) then SOH-CAH-TOA to solve for the angle (θ=Cos−1(adjacenthypotenuse)) .
But what if it isn't a right triangle? The cosine rule works for all triangles, even if they don't have a 90° angle.
a2=b2+c2−2(b)(c)Cos(A)
Where a is the missing side, and A is the angle opposite to side a.
With this rule, we can find a missing angle or a missing side to any type of triangle, assuming you have the needed variables.
Explanation:
The box acted upon by the force of 5 N will move on a frictionless surface, with the condition S = P + Q.
What is Force?The effect of a push or a pull on the body is known as force. The main types of forces include friction forces, nuclear forces, and gravitational forces. For instance, when a hand strikes a wall, the wall exerts a force on the hand as well as the hand exerting a force on the wall. Newton was given several laws to comprehend force.
Given: with a triangle with a 90° angle, a right triangle, you can simply use Pythagoras theorem (a2+b2=c2) then SOH - CAH - TOA to solve for the angle (θ=Cos−1(adjacent hypotenuse)).
The cosine rule works for all triangles, even if they don't have a 90° angle.
a2=b2+c2−2(b)(c)Cos(A)
A is the angle opposite to side a,
With this rule, we can find a missing angle or a missing side to any type of triangle, assuming you have the needed variables.
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Two objects ① and ② are moving along two different circular paths of radii 10 m and 20 m respectively. If the ratio of their speeds ( V₁ : V₂) = 2:1, then the ratio of their time periods (T₁ : T₂) --------- [ Use numerals to write your answer] *
Answer:
The ratio of the time period of object 1 to the time period of object 2 is [tex]\frac{1}{4}[/tex]. (T₁ : T₂) = 1 : 4
Explanation:
Let suppose that both objects are moving along the circular paths at constant speed, such that period of rotation of each object is represented by the following formula:
[tex]\omega = \frac{2\pi}{T}[/tex]
Where:
[tex]\omega[/tex] - Angular speed, measured in radians per second.
[tex]T[/tex] - Period, measured in seconds.
The period is now cleared:
[tex]T = \frac{2\pi}{\omega}[/tex]
Angular speed ([tex]\omega[/tex]) and linear speed ([tex]v[/tex]) are related to each other by this formula:
[tex]\omega = \frac{v}{R}[/tex]
Where [tex]R[/tex] is the radius of rotation, measured in meters.
The angular speed can be replaced and the resultant expression is obtained:
[tex]T = \frac{2\pi\cdot R}{v}[/tex]
Which means that time period is directly proportional to linear speed and directly proportional to radius of rotation. Then, the following relationship is constructed and described below:
[tex]\frac{T_{2}}{T_{1}} = \left(\frac{v_{1}}{v_{2}}\right)\cdot \left(\frac{R_{2}}{R_{1}} \right)[/tex]
Where:
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Time periods of objects 1 and 2, measured in seconds.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Linear speed of objects 1 and 2, measured in meters per second.
[tex]R_{1}[/tex], [tex]R_{2}[/tex] - Radius of rotation of objects 1 and 2, measured in meters.
If [tex]\frac{v_{1}}{v_{2}} = 2[/tex], [tex]R_{1} = 10\,m[/tex] and [tex]R_{2} = 20\,m[/tex], the ratio of time periods is:
[tex]\frac{T_{2}}{T_{1}} = 2\cdot \left(\frac{20\,m}{10\,m} \right)[/tex]
[tex]\frac{T_{2}}{T_{1}} = 4[/tex]
[tex]\frac{T_{1}}{T_{2}} = \frac{1}{4}[/tex]
The ratio of the time period of object 1 to the time period of object 2 is [tex]\frac{1}{4}[/tex].
What happens when you decrease the thrust on your scooter? A. You stop B. Nothing happens C. You fall over D. You speed up Reset Selection
Answer:
you fall over
Explanation:
its not D
i already tried it
When you decrease the thrust on your scooter by fall over. The correct option is D.
What is thrust?The thrust is the pressure exerted the on the surface area.
When you decrease the thrust on your scooter by speeding up.
When the speed increases, the thrust decreases.
Thus, the correct option is D.
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A 85 kg student stands on a bridge with a 12-m-long bungee cord tied to her feet. You can assume that the bungee cord is massless and has a spring constant of 250 N/m. The student jumps off the bridge and falls until the bungee cord is fully stretched, where she comes to a stop. How far below the bridge does the student fall before coming to a stop
Answer:
15.33 mExplanation:
According to Hooke's law "provided the elastic limit of a material is not exceeded the extension e is directly proportional to the applied force"
Given data
mass m= 85 kg
spring constant k= 250 N/m
assuming g= 9.81 m/s^2
extension e= ?
We know that
[tex]F= ke \\mg= ke[/tex]
solving for the extension e
[tex]85*9.81= 250*e\\\e= \frac{833.85}{250} \\\e= 3.33m[/tex]
to get how far bellow the bridge the student falls before coming to stop, we have to add the extension e to the length of the bungee cord.
=12+3.33= 15.33 m
The student fall below the bridge(12+3.33 = 15.33 m ) before coming to a stop. Where 12 is the length of bungee cord.
According to Hooke's law in the elastic limit of a material the the strain in the spring is directly proportional to the applied force or stress.
Formula
[tex]\bold {F_s = -kx}[/tex]
Where,
= spring force
k = spring constant = 250 N/m
x = spring stretch or compression = ?
The force,
F = mg
[tex]\bold {F = 85 \times 9.18}\\\\\bold {F = 833.85 }[/tex]
So,
[tex]\bold {x = \dfrac {F_s }{k} }\\\\\bold {x = \dfrac {833.85} {250} }\\\\\bold {x = 3.33}[/tex]
Therefore, the student fall below the bridge(12+3.33 = 15.33 m ) before coming to a stop. Where 12 is the length of bungee cord.
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PLZ HELP What is necessary condition for production of waves?
Answer:
The length of the rope must be an integral multiple of the wavelength of the wave.
Explanation:
Answer:
The condition necessary for formation or a standing wave is that the length of the rope (or the length over which the wave is distributed) must be an integral multiple of the wavelength of the wave. Therefore, l=nλ where n is a positive integer.
A ball thrown straight up into the air with a speed of 10m/s. If the ball has a mass of 0.3 Kg, how high does the ball go? Acceleration due to gravity is g=9.8 m/s2
Answer:
5.10 m
Explanation:
Given that:
Initial velocity given to the ball, v = 10 m/s
Mass of the ball, m = 0.3 kg
Acceleration due to gravity, g = 9.8m/[tex]s^2[/tex]
To find: The height upto which the ball will go = ?
Solution:
Initially, the ball will have kinetic energy and at top point the velocity will be zero, it will have only potential energy due to height.
Formula for kinetic energy and potential energy:
[tex]KE = \dfrac{1}{2}mv^2\\PE = mgh[/tex]
Applying conservation of energy principle:
[tex]\dfrac{1}{2}mv^2=mgh\\\Rightarrow \dfrac{1}{2}v^2=gh[/tex]
Putting the values of v and g to find h:
[tex]\Rightarrow \dfrac{1}{2}10^2=9.8\times h\\\Rightarrow h = \dfrac{50}{9.8}\\\Rightarrow h = 5.10\ m[/tex]
The ball will go 5.10 m high.
(01.05 LC)
Which of the following bones is classified as part of the axial skeleton?
Answer:c
Explanation:
because it is the same
Answer: skull, vertebrae, ribs and sternum.
Explanation:
You notice the flagpole at school vibrating in the breeze. You count the vibrations and find that
the flagpole makes 20 complete vibrations in 10 seconds. You read somewhere that this
flagpole vibrates at a speed of 25 m/s when the wind is blowing. Find the wavelength of these
vibrations
Answer:
Wavelength =50 m
Explanation:
Given:
n=20
t=10sec
V=25m/s
Wavelength =?
Solution:
Wavelength = frequency ×velocity ---> eq 1
Time period=time/no of vibrations
T=10/20=0.5sec
Frequency =1/Time period(T)
f=1/0.5=2
Put the value of frequency in eq 1
Wavelength =2×25
Wavelength =50m...
Hope you understand!
Have a nice day!
two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at 4 metre apart
Answer: 0.5N
Explanation:
Gravitational force is calculated using the formula :
F = Gm1m2/r^2
Where G is the gravitational constant (6.67 × 10^-11)
At a distance 'r' of 2metres apart:
Mass of objects are m1 and m2
Gravitational force 'F1' = 2N
Inputting values into the formula :
2 = Gm1m2 / 2^2 - - - - - (1)
At a distance 'r' of 4meters apart:
Mass of objects are m1 and m2
Gravitational force 'F2' = y
Inputting values
F2 = Gm1m2 / 4^2 - - - - - (2)
Dividing equations 1 and 2
2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2
2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)
2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)
2/F2 = 16 / 4
Cross multiply
2 × 4 = 16 × F2
8 = 16F2
F2 = 8/16
F2 = 0.5N
Algunas fabricas de balones de fútbol ubicadas en la costa inflan los balones que van a ser vendiéndose las ciudades como pasto,tunja o bogota por debajo de su capacidad normal y cuando los balones llegan a los estudiantes, se encuentran firmes. ¿por que sucede esto? Por favor necesito su ayuda doy hartos puntos es para hoy esta tarea por fa
Answer:
balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure
Explanation:
Este ejercicio es referente a la mecánica de fluidos, usemos la expresión para la presión
P = ρ g h
En es el caso del balón usemos la presión en la pared extrema, llamemos P la presión por el gas en el interior y P_ext la presión atmosférica del lugar
cuando se llena el valor en una ciudad de baja altura la presión atmosférica es mas alta
P_int1 < P_ext1
por lo cual la pared del balón no se mantiene rígida.
Cuando el balón es trasladado a una ciudad con mayor altura sobre el nivel del mar la presión exterior disminuye
P_ext2 = ρ g h₂ < P_ext1
en promedio la presión disminuye con la altura en 0,029 atm cada 250 m
por lo tanto como la cantidad de gas es constante en el interior la presión interior del globo es constante y en esta ciudades se hace igual o un poco mayor que la presión atmosférica, en consecuencia la pared del globo esta rígida
P_int2 >P_ext2
Traslate
This exercise is related to fluid mechanics, let's use the expression for pressure
P = ρ g h
In the case of the balloon, let's use the pressure on the extreme wall, let's call P the pressure for the gas inside and P_ext the atmospheric pressure of the place
when the value is filled in a low-lying city the atmospheric pressure is higher
P_int1 <P_ext1
therefore the wall of the ball does not remain rigid.
When the ball is transferred to a city with higher altitude above sea level, the external pressure decreases
P_ext2 = ρ g h <P_ext1
on average the pressure decreases with height by 0.029 atm every 250 m
therefore as balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure, therefore the wall of the
Pint 2> Pe
can we investigate that which fuel
is cheapest
Answer:
yeah..
as we all know that the cheapest fuel is petrol because it is the last product of petroleum.
hope it helps...
What distance does a mobile travel that with a constant acceleration of 3m / s increases its speed from 6m / s to 12m / s
Answer:
18 m
Explanation:
Given:
v₀ = 6 m/s
v = 12 m/s
a = 3 m/s²
Find: Δx
v² = v₀² + 2aΔx
(12 m/s)² = (6 m/s)² + 2 (3 m/s²) Δx
Δx = 18 m
A tub has a mass of 0.650 kg and 0.350 kg of rocks are put in it. It floats in fresh water with the top of the tub level with the water line. What is the volume of the tub? PLEASE HELP!!!
Answer: 0.001m^3
Explanation:
Given the following :
Mass of tub = 0.650kg
Mass of rock = 0.350kg
In this scenario:
Upthrust is the weight of fluid displaced
Upthrust = Total weight of tub (since the tub floats, and top of tub is level with the water line )
Recall Weight = mass * acceleration due to gravity(g)
g = 9.8m/s
Total mass = (0.650 + 0.350)kg = 1kg
Weight = 1kg × 9.8m/s = 9.8N
Upthrust = 9.8N
Recall: Upthrust = Density × acceleration due to gravity(g)× displaced body volume of fluid
Density of water = 1000kg/m^3
g = 9.8m/s^2
9.8N = 1000 × 9.8 × Volume
9.8 = 9800V
V = 9.8 / 9800
V = 0.001m^3
Answer:
0.001
Explanation:
acellus
The student concluded that the current in the LDR is inversely proportional to the potential difference across the LDR. Explain why the student’s conclusion is incorrect.
Answer:
The graph shows direct proportion because it is a straight line
through the origin.
Explanation:
Generate an explanation for the following formula: P = I²R . Explain.
Answer:
Ohm's Law:
[tex]V = I R[/tex] ------------------(1)
Where V is voltage, I is current and R is resistance.
Whereas,
P = I V --------------------(2)
Where P is power, I is current and V is Voltage.
Putting (1) into (2)
=> P = I (I R) [∴ V = I R]
=> P = I²R (Derived!)
Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He maintained this speed until he is 16.0 m from the finish line but then fades and decelerates uniformly, crossing the line with a speed of only 8.00 m/s. What is Willie’s total time for the race? Please show your steps :)
Answer:
The total time for the race is 11.6 seconds
Explanation:
The parameters given are;
Total distance ran by Willie = 100.0 m
Initial acceleration = 2.00m/s²
Top speed reached with initial acceleration = 12.0 m/s
Point where Willie start to fade and decelerate = 16.0 m from the finish line
Speed with which Willie crosses the finish line = 8.00 m/s
The time and distance covered with the initial acceleration are found using the following equations of motion;
v = u₀ + a·t
v² = u₀² + 2·a·s
Where:
v = Final velocity reached with the initial acceleration = 12.0 m/s
u₀ = Initial velocity at the start of the race = 0 m/s
t = Time during acceleration
a = Initial acceleration = 2.00 m/s²
s = Distance covered during the period of initial acceleration
From, v = u₀ + a·t, we have;
12 = 0 + 2×t
t = 12/2 = 6 seconds
From, v² = u₀² + 2·a·s, we have;
12² = 0² + 2×2×s
144 = 4×s
s = 144/4 =36 meters
Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;
Distance covered at top speed = 100 - 36 - 16 = 48 meters
Time, [tex]t_t[/tex] of running at top speed = Distance/velocity = 48/12 = 4 seconds
The deceleration from top speed to crossing the line is found as follows;
v₁² = u₁² + 2·a₁·s₁
Where:
u₁ = v = 12 m/s
v₁ = The speed with which Willie crosses the line = 8.00 m/s
s₁ = Distance covered during decelerating = 16.0 m
a₁ = Deceleration
From which we have;
8² = 12² + 2 × a × 16
64 = 144 + 32·a
64 - 144 = 32·a
32·a = -80
a = -80/32 = -2.5 m/s²
From, v₁ = u₁ + a₁·t₁
Where:
t₁ = Time of deceleration
We have;
8 = 12 + (-2.5)·t₁
t₁ = (8 - 12)/(-2.5) = 1.6 seconds
The total time = t + [tex]t_t[/tex] + t₁ =6 + 4 + 1.6 = 11.6 seconds.
Willie's total time to cover the race is 11.6 seconds.
What is a velocity-time graph?A velocity-time graph is a graph that is used to represent a way in which the motion of an object can be represented with the velocity change on the vertical (y)-axis and change in time on the horizontal (x)-axis.
From the given information:
Willie's total distance covered = 100.0 mInitial acceleration (u) when the race commence = 0 m/s²Final velocity reached = 12.0 m/sUsing the following first equation of motion:
v = u + at (where u = 0; at rest)The time during the first acceleration can be computed as:
v = u + at
12 = 0 + (2 × t)
t = 12/2
t = 6 seconds
Using the fourth equation of motion, the distance covered in the first acceleration is determined as follows:
[tex]\mathbf{v^2 = u^2 + 2as}[/tex]
[tex]\mathbf{12^2 = (0)^2 + 2(2) \times s}[/tex]
144 = 4s
s = 144/4
s = 36 meters
However, he maintained a speed of 12.0 m/s till he is 16.0 m away from the finish line when he starts to decelerate.
Hence, at his highest speed;
The distance covered = 100 - 36 - 16 = 48 meters.Recall that:
[tex]Time = \dfrac{distance}{velocity}[/tex]
Thus, the time for running at his highest speed is:
[tex]\mathbf{time = \dfrac{48 m}{12 m/s}}[/tex]
time = 4 seconds
His deceleration from his highest speed can be computed as:
[tex]\mathbf{v^2 = u^ 2 + 2as}[/tex]
here:
v = 8.00 m/su = 12.0 m/ss = h = 16.0 m[tex]\mathbf{v^2 - u^2=2as}[/tex]
[tex]\mathbf{a = \dfrac{v^2 - u^2}{2s}}[/tex]
[tex]\mathbf{a = \dfrac{8^2 - 12^2}{2(16)}}[/tex]
[tex]\mathbf{a = \dfrac{64 - 144}{32}}[/tex]
a = -2.5 m/s²
Finally, using the first equation of motion, the total time can be computed as:
[tex]\mathbf{v_1 = u_1 + a_1t_1}[/tex]
[tex]\mathbf{8= 12 + (-2.5)t_1}[/tex]
[tex]\mathbf{t_1 = \dfrac{(8- 12)}{ (-2.5)}}[/tex]
[tex]\mathbf{t_1 = \dfrac{(-4)}{ (-2.5)}}[/tex]
[tex]\mathbf{t_1 = 1.6\ seconds}[/tex]
Thus, the total time = (6+4+1.6) seconds
The total time = 11.6 seconds
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In an obstacle course, runners must jump hurdles that are .84 m high. If the speed of one of the athletes when he is 1.63 meters from the fence is 4m / s on the horizontal axis and 4m / s on the vertical axis. Calculate the maximum height reached and indicate whether or not the jump was successful for that runner. Help :c
Answer:
0.82 m
Explanation:
Given:
v₀ᵧ = 4 m/s
aᵧ = -9.8 m/s²
Find: Δy when vᵧ = 0 m/s
vᵧ² = v₀ᵧ² + 2aᵧΔy
(0 m/s)² = (4 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 0.82 m
The maximum height reached by the athlete is 0.82 meters. The jump is not successful.
In a parallel circuit, the blank
is the same for every leg in the circuit
Answer:
In a parallel circuit, the blank
is the same for every leg in the circuit why?
Explanation:
Because the voltage is common across the elements of a parallel circuit, the voltage drops are all equal to each other, and the ap- plied voltage is equal to any one of thJ individual voltage drops. are also 60v. In a series circuit, the same current flows. through every component.
Answer:
voltage
Explanation:
a p e x
Which of these results in kinetic energy of an object? a position b motion c mass d volume
Answer:
Option B
Explanation:
Kinetic Energy is the energy possessed by the body due to "motion".
The answer is:
B. MotionThree screens are placed exactly one behind the other. The light of the torch falls on all the three screens. One of the screen is a bathroom mirror which one?
Answer:
Explanation:
Three screens are placed one behind the other and light falls on all of them from a torch . In this case a mirror can be only the last screen . In case of first or second screen being a mirror , light would not have reached the last screen because the nature of mirror is to reflect the light back.
Hence , the only possibility is that , the last screen is mirror.
Identify the base quantities un the following
Speed
Area
Force
Distance
Plz help it's urgent
Answer:
Distance
Explanation:
Distance is the base quantity (measured in lengths) while speed, Area and Force are all derived quantities derived from base quantities.
A particle starts to move in a straight line from
a point with velocity 10 m/s and acceleration - 20 m/s²? Find the position and velocity of the
particle at (i) t = 5s, (ii) t' = 10 s.
Answer: s(5) = -200, v(5) = -90
s(10) = -900, v(10) = -190
Explanation:
Position: s(t)
Velocity: s'(t) = v(t) ⇒ [tex]s(t)=\int {v(t)} \, dt[/tex]
Acceleration v'(t) = a(t) ⇒ [tex]v(t)=\int {a(t)} \, dt[/tex]
We are given that acceleration a(t) = -20 and velocity v(t) = 10
[tex]v(t)=\int {a(t)} \, dt\\\\v(t)=\int{-20}\, dt\\\\v(t)=-20t + C \\\\v(t)=10\quad \longrightarrow \quad C=10\\\\v(t)=-20t+10[/tex]
[tex]s(t)=\int {v(t)} \, dt\\\\s(t)=\int {(-20t+10)} \, dt\\\\s(t)=-10t^2+10t\\\\[/tex]
(a) Input t = 5 into the s(t) and v(t) equations
s(5) = -10(5)² + 10(5) v(5) = -20(5) + 10
= -250 + 50 = -100 + 10
= -200 = -90
(b) Input t = 10 into the s(t) and v(t) equations
s(10) = -10(10)² + 10(10) v(10) = -20(10) + 10
= -1000 + 100 = -200 + 10
= -900 = -190
Calculate the pressure exerted on a surface by a brick of weight 10 N. The area in contact with the surface is 20 cm2
Answer: 5000
Explanation: pressure: force / area
SI unit of force: newton
SI unit of area: metre²
Force exerted: 10 N
Area on which it is exerted : 20 cm²
Since area has to be in m² , we convert cm² to m² by dividing it by 10000( m:100cm, 1m²: 10000 cm²)
Area=20/10000: 0.002 m²
So, pressure= 10/0.002=5000 pa( SI unit of pressure is pascal)
The pressure is the force exerted by unit area. The force on the brick by its weight is 10 N and the surface area is 20cm². Then, the pressure in the given area will be 5000 Pa.
What is pressure ?Pressure on an object is the force exerted per unit area of that object. Pressure is dependent on the volume and temperature as well. It is commonly expressed in the unit of atm. Other units of pressure are bar, pascal, torr, mmHg etc.
The normal force acting on the brick is the force by its own weight.
normal force = 10 N
surface area = 20 cm²
20 cm² /(100)² = 0.002 m²
Pressure = force/ area
p = 10 N/0.002 m²
= 5000 N/m² = 5000 pa
Therefore, the pressure exerted on the brick is 5000 Pa
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The steering wheel of a car has a radius of 36 cm, and part of rest with an acceleration of 1.8m / s. Determine
a) The angular speed after 10s
b) The number of turns the wheel gave in 10s
Answer:
a) 50 rad/s
b) 39.8 rev
Explanation:
Given:
r = 0.36 m
v₀ = 0 m/s
a = 1.8 m/s
t = 10 s
a) Find: ω
v = at + v₀
v = (1.8 m/s) (10 s) + (0 m/s)
v = 18 m/s
ω = (18 m/s) / (0.36 m)
ω = 50 rad/s
b) Find: Δθ
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1.8 m/s) (10 s)²
Δx = 90 m
Δθ = (90 m) / (2π × 0.36 m)
Δθ = 39.8 rev
a 2500kg car is travelling at a constant speed of 14 m/s along an icy, but straight and level road. the driver of the car, seeing an approaching traffic light turn red., slams on the brakes. Wheels locked and ttired skidding, the car slide to a halt in a distance of 25m. What is the coefficient of friction
Answer:
0.40
Explanation:
Given that :
the mass of the caer = 250 kg
initial speed = 14 m/s
final speed = 0 m/s
distance s = 25 m
Using the equation of motion
[tex]v^2 = u^2 + 2as[/tex]
making a the subject of the formula ; we have:
[tex]v^2-u^2 = 2as[/tex]
[tex]a= \dfrac{v^2-u^2 }{2 \ s}[/tex]
[tex]a= \dfrac{(0)^2-(14)^2 }{2 \ (25)}[/tex]
[tex]a= \dfrac{0-196 }{50}[/tex]
[tex]a= \dfrac{-196 }{50}[/tex]
a = -3.92 m/s²
However; the relation for the coefficient of the kinetic static friction can be expressed as:
[tex]f= \mu_k *mg= ma[/tex]
[tex]f= \mu_k *g= a[/tex]
[tex]f= \mu_k = \dfrac{a}{g}[/tex]
[tex]f= \mu_k = \dfrac{3.92}{9.8}[/tex]
[tex]f= \mu_k = 0.40[/tex]
HELP! I NEED HELP IN THIS PHYSICS QUESTION!!!
Answer:
79%.
Explanation:
Step 1:
Data obtained from the question. This include:
Input temperature = 100 °C.
Output temperature = 22 °C.
Step 2:
Conversion of celsius temperature to Kelvin temperature.
Temperature (Kelvin) = temperature (celsius) + 273
T (K) = T (°C) + 273
Input temperature = 100 °C.
Input temperature = 100 °C + 273 = 373 K.
Output temperature = 22 °C.
Output temperature = 22 °C + 273 = 295 K.
Step 3:
Determination of the efficiency of the locomotive.
Input temperature = 373 K.
Outpu temperature = 295 K.
Efficiency =...?
Efficiency = output /input x 100
Efficiency = 295/373 x 100
Efficiency = 79.1 ≈ 79%
Therefore, the efficiency of the locomotive is approximately 79%.