Answer:
a)Car E = Car D > (Car F = Car B = Car C) > Car A
b)Car E = Car D > (Car F = Car B = Car C) > Car A
Explanation:
Car A: mass = 500 kg; speed = 10 m/s
Car B: mass = 2000 kg;speed = 5 m/s
Car C:mass = 500 kg; speed = 20 m/s
Car D: mass = 1000 kg; speed = 20 m/s
Car E:mass = 4000 kg; speed = 5 m/s
Car F: mass = 1000 kg; speed = 10 m/s
Part a) Now we know that momentum of each car is product of mass and velocity , so we will have
CarA:
[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]
Car B:
[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]
Car C:
[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]
Car D:
[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]
Car E:
[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]
Car F:
[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]
So the momentum is given as ,
Car E = Car D > (Car F = Car B = Car C) > Car A
Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car
so the order of impulse from largest to least is given as
Car E = Car D > (Car F = Car B = Car C) > Car A
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement. Ф -10.3 103 N-m2/C for
0 for r<2 4:
-36.8 x 10נ N-m2/c
0 for r > R2
36.8 x 10נ N-m2/c
Determine the magnitude ond sign of the point chorge ond the charge on the surface of the two shels point charge inner shell outer shel.
Answer:
the magnitude is 7 and sign of the point charge on the surface shell is -13
Explanation:
A dimension is a physical nature of a quantity.
(i) give two (2) limitations of dimensional analysis..
(ii) if velocity (v), time (T) and force (F) were chosen as basic quantities, find the dimensions of mass?
Answer:
i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:
Dimesnional analysis is unable to derive relation when a physical quantity depends on more than three factors with dimensions. It is unable to derive a formula that contain exponential function, trigonometric function, and logarithmic function.ii) Given:
Velocity = v
Time = t
Force = F
Force = mass x acceleration
= mass x velocity/time
So, mass= (force x time) / velocity
[mass] = Ftv^-1
Hence, dimesnion of mass is Ftv^-1.
A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on top. A large load of coal is suddenly dumped into the car. What happens to the speed of the freight car
Answer:
The speed of the freight car decreases.
Explanation:
According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received
In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.
Therefore the speed of the freight car decreased by applying the law of conservation of momentum i
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determine the time necessary for the projectile to reach its maximum height. (b) Determine the maximum height reached by the projectile. (c) Determine the horizontal and vertical components of the velocity vector at the maximum height. (d) Determine the horizontal and vertical components of the acceleration vector at the maximum heigh
Answer:
(a) t = 3.74 s
(b) H = 136.86 m
(c) Vₓ = 41.83 m/s, Vy = 0 m/s
(d) ax = 0 m/s², ay = 9.8 m/s²
Explanation:
(a)
Time to reach maximum height by the projectile is given as:
t = V₀ Sinθ/g
where,
V₀ = Launching Speed = 55.6 m/s
Angle with Horizontal = θ = 41.2°
g = 9.8 m/s²
Therefore,
t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)
t = 3.74 s
(b)
Maximum height reached by projectile is:
H = V₀² Sin²θ/g
H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)
H = 136.86 m
(c)
Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:
Vₓ = V₀ₓ = V₀ Cos θ
Vₓ = (55.6 m/s)(Cos 41.2°)
Vₓ = 41.83 m/s
Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.
Vy = 0 m/s
(d)
Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.
ax = 0 m/s²
The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:
ay = 9.8 m/s²
A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train then moves at a constant velocity of 42 m/s for 420 s. The train then slows down uniformly at 0.065 m/s^2, until it is brought to a halt. What is the acceleration during the first 5.6 km of travel?
Answer:
0.1575 m/s^2
Explanation:
Solution:-
- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).
- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.
- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).
- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:
[tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]
- We will plug in the given parameters in the equation of motion given above:
[tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]
Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2
A block is supported on a compressed spring, which projects the block straight up in the air at velocity VVoj The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block?
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.
Answer:
the correct answer is B
Explanation:
We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.
For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.
Let's write the kinematic equation for the two bodies
The block. At the highest point of the path
y = - ½ g t2
The ball, in its vertical movement
y = vo t - ½ g t2
therefore the correct answer is B
1. In 214 BC, Archimedes invented a large spherical-type mirror used to focus the sun's intense rays onto far away enemy boats, which would eventually light them on fire. If the boats were travelling in a nearby channel approximately 1,000 m from the river bank, what would the radius of curvature of his mirror need to be? Show your work.
Answer:
2000 m
Explanation:
since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m
for a spherical mirror, the focal length is given by
f = R/2
where R is the radius of curvature
1000 = R/2
R = 2000 m
R = 2000 m
this means that the radius of curvature must be 2000 m
How many significant figures does 0.09164500561 have?
Answer:
10 Sig Figs
Explanation:
Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.
Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m
Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]
Calculate the inlet and exit velocity of water jet
[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]
The conservation of mass equation of steady flow
[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]
[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]
[tex]A_e\ \texttt {is the exit area of the jet}[/tex]
since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal
The expression for thickness of the jet
[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]
(b)
[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]
[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]
[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
What is the relationship between electric force and distance between charged objects and the amount of charge?
Explanation:
The relationship between electric force and distance between charged objects is given by the formula as follows :
[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]
k is electrostatic constant and d is distance between charges
The electric force between charges is inversely proportional to the square of distance between them.
Work out the velocity v at the end of a rollercoaster ride (0). (rearrange the equation for KE to make velocity v the subject)
KE=1/2mv^2
Explanation:
If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :
[tex]K=\dfrac{1}{2}mv^2[/tex]
m is mass of the object
We can rearrange the above equation such that,
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
Hence, this is the velocity at the end of a rollercoaster ride.
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 4.23 m/s, strikes the rod at angle θ = 46° from the normal at a distance D = 2/3 L, where L = 0.95 m, from the point of rotation and sticks to the rod after the collision.
Required:
What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
Convert from scientific notation to standard form
9.512 x 10-8
Answer:
0.00000009512
Explanation:
Scientific notation is a very useful and abbreviated way of writing quantities that are very large or small. It consists of placing the number with an integer and multiplying by an exponent to arrive at the same number.
let's pass the number 9,512 10⁻⁸ to decimal notation
9,512 / 10⁸ = 9,512 / 100000000
0.00000009512
As we see writing this number, it is very easy to make mistakes
An object will sink in a liquid if the density of the object is greater than that of the liquid. The mass of a sphere is 0.723 g. If the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density
Answer:
= 0.0532 cm^3
Explanation:
The computation of volume of the sphere is shown below:-
[tex]Density = \frac{Mass}{Volume}[/tex]
Where,
Density = 13.6 g/cm^3
Mass of sphere = 0.723 g
now we will put the values into the above formula to reach volume of the sphere which is here below:-
[tex]Volume = \frac{0.723}{13.6}[/tex]
= 0.0532 cm^3
Therefore for computing the volume of the sphere we simply applied the above formula.
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy Stored = 36000 J = 36 KJ
Explanation:
The power of a battery is given by the formula:
P = IV
where,
P = Power delivered by the battery
I = Current Supplied to the battery
V = Potential Difference between terminals of battery = 12 volt
Now, we multiply both sides by the time period (t):
Pt = VIt
where,
Pt = (Power)(Time) = Energy Stored = E = ?
It = Battery Current Rating = 50 A.min
Converting this to A.sec;
It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
Which symbol in a chemical equation separates the reactants from the products?
Answer:
the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔
Explanation:
i hope this will help you
If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the
runner's kinetic energy will become:
(a) four times as great as it was.
(b) half the magnitude it was.
(c) energy is conserved.
(d) twice as great as it was.
(e) four times less than it was.
Answer:
It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4
A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , and an output piston of cross-sectional area 1.2 m2 . The pistons can slide up or down without friction while keeping the fluid sealed inside. What is the maximum weight that can be lifted when a force of 60 N is applied to the input piston
Answer:
Maximum weight that can be lifted = 18,000 N
Explanation:
Given:
Cross-sectional area of input (A1) = 0.004 m²
Cross-sectional area of the output (A2) = 1.2 m ²
Force (F) = 60 N
Computation:
Pressure on input piston (P1) = F / A1
Assume,
Maximum weight lifted by piston = W
Pressure on output piston (P2) = W / A2
We, know that
P1 = P2
[F / A1] = [W / A2]
[60 / 0.004] = [W / 1.2]
150,00 = W / 1.2
Weight = 18,000 N
Maximum weight that can be lifted = 18,000 N
One end of an insulated metal rod is maintained at 100c and the other end is maintained at 0.00 c by an ice–water mixture. The rod has a length of 75.0cm and a cross-sectional area of 1.25cm . The heat conducted by the rod melts a mass of 6.15g of ice in a time of 10.0 min .find the thermal conductivity k of the metal?k=............ W/(m.K)
Answer:
The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].
Explanation:
This is a situation of one-dimensional thermal conduction of a metal rod in a temperature gradient. The heat transfer rate through the metal rod is calculated by this expression:
[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]
Where:
[tex]\dot Q[/tex] - Heat transfer due to conduction, measured in watts.
[tex]L_{rod}[/tex] - Length of the metal rod, measured in meters.
[tex]A_{c,rod}[/tex] - Cross section area of the metal rod, measured in meters.
[tex]k_{rod}[/tex] - Thermal conductivity, measured in [tex]\frac{W}{m\cdot K}[/tex].
Let assume that heat conducted to melt some ice was transfered at constant rate, so that definition of power can be translated as:
[tex]\dot Q = \frac{Q}{\Delta t}[/tex]
Where Q is the latent heat required to melt the ice, whose formula is:
[tex]Q = m_{ice}\cdot L_{f}[/tex]
Where:
[tex]m_{ice}[/tex] - Mass of ice, measured in kilograms.
[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.
The latent heat of fussion of water is equal to [tex]330000\,\frac{J}{g}[/tex]. Hence, the total heat received by the ice is:
[tex]Q = (6.15\,g)\cdot \left(330\,\frac{J}{g} \right)[/tex]
[tex]Q = 2029.5\,J[/tex]
Now, the heat transfer rate is:
[tex]\dot Q = \frac{2029.5\,J}{(10\,min)\cdot \left(60\,\frac{s}{min} \right)}[/tex]
[tex]\dot Q = 3.382\,W[/tex]
Turning to the thermal conduction equation, thermal conductivity is cleared and computed after replacing remaining variables: ([tex]L_{rod} = 0.75\,m[/tex], [tex]A_{c,rod} = 1.25\times 10^{-4}\,m^{2}[/tex], [tex]\Delta T = 100\,K[/tex], [tex]\dot Q = 3.382\,W[/tex])
[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]
[tex]k_{rod} = \frac{\dot Q \cdot L_{rod}}{A_{c,rod}\cdot \Delta T}[/tex]
[tex]k_{rod} = \frac{(3.382\,W)\cdot (0.75\,m)}{(1.25\times 10^{-4}\,m^{2})\cdot (100\,K)}[/tex]
[tex]k_{rod} = 202.92\,\frac{W}{m\cdot K}[/tex]
The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].
If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways, you can feel a resistive force, indicating the presence of eddy currents in the surface.
A)Estimate the magnetic field strength Bof the magnet to be 5 mTand assume the magnet is rectangular with dimensions 4 cmwide by 2 cmhigh, so its area A is 8 cm2. Now estimate the magnetic flux ΦB into the refrigerator door behind the magnet.
Express your answer with the appropriate units.
B)If you move the magnet sideways at a speed of 2 cm/s, what is a corresponding estimate of the time rate at which the magnetic flux through an area A fixed on the refrigerator is changing as the magnet passes over? Use this estimate to estimate the emf induced under the rectangle on the door's surface.
Express your answer with the appropriate units.
Answer:
(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v
Explanation:
Solution
Given that:
A refrigerator magnet about = 2 mm
The estimated magnetic field strength of the magnet is = 5 m T
The Area = 8 cm²
Now,
(A) The magnetic flux ΦB = BA
Thus,
ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²
So,
ΦB = 4* 6 ^ ⁻6 T m²
(B)By applying Faraday's Law we have the following formula given below:
Ε = Bℓυ
Here,
ℓ = 2 cm the same as 2 * 10 ^⁻2 m
B = 5 m T = 5 * 10 ^ ⁻3 T
υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s
Thus,
Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v
E =2 * 10 ^ ⁻6 v
A) The magnetic flux ΦB into the refrigerator door behind the magnet :
4 * 6⁻⁶ Tm²B) The estimated emf induced under the rectangle on the door's surface ;
2 * 10⁻⁶ vGiven data :
magnetic field strength of magnet ( B ) = 5 mT
size of refrigerator magnet = 2 mm
Area of magnet ( A ) = 4 * 2 = 8 cm²
A) Determine the magnetic flux ΦBwhere ; ΦB = BA
ΦB = ( 5 * 10⁻³ ) * ( 4 * 10⁻² ) * ( 2 * 10⁻² ) Tm²
= 4 * 6⁻⁶ Tm²
B) Determine estimated emf inducedTo determine the estimated emf we will apply Faraday's law
Ε = Bℓυ ---- ( 2 )
where : B = 5 * 10⁻³ T, ℓ = 2 * 10⁻² m, υ = 2 * 10⁻² m/s
insert values into equation 2
E = ( 5 * 10⁻³ ) * ( 2 * 10⁻² ) * ( 2 * 10⁻² )
= 2 * 10⁻⁶ v
Hence we can conclude that The magnetic flux ΦB is 4 * 6⁻⁶ Tm² and The estimated emf induced is 2 * 10⁻⁶ v
Learn more about magnet flux : https://brainly.com/question/4721624
The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm1cm and the IMA of the machine is 66 , what is the radius of the handle
Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?
Answer:
Radius of the handle = 3 cm = 0.03 m
Explanation:
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)
Diameter of the rod = 1 cm
Radius of the rod = Diameter/2
Radius of the rod = 1/2
Radius of the rod = 0.5 cm
Mechanical Accuracy of the machine, MA = 6
Substitute the values into equation (1)
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
The Radius of the handle is = 3 cm = 0.03 m
Calculation of the radius of the handle:Since
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)
Here,
Diameter of the rod = 1 cm
We know that
The radius of the rod = Diameter/2
So,
Radius of the rod = 1/2
So,
Radius of the rod = 0.5 cm
Now
Mechanical Accuracy of the machine, MA = 6
Now
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
Learn more about radius here: https://brainly.com/question/18648019
The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its
A.climate.
b.weather.
C. global warming
d. seasons
Answer:
a. global warming
Explanation:
that's the definitain of global warming
Answer:
A climate
Explanation:
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track and the train reaches it 76s later. The acceleration is -0.26s^2. Find its speed at the next signal.
Answer:
3.2 m/s
Explanation:
Given:
Δx = 1000 m
v₀ = 23 m/s
a = -0.26 m/s²
t = 76 s
Find: v
This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:
v = at + v₀
v = (-0.26 m/s²) (76 s) + (23 m/s)
v = 3.2 m/s
Or:
Δx = ½ (v + v₀) t
(1000 m) = ½ (v + 23 m/s) (76 s)
v = 3.3 m/s
Or:
v² = v₀² + 2aΔx
v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)
v = 3.0 m/s
Or:
Δx = vt − ½ at²
(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²
v = 3.3 m/s
As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.
If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4% Between 4% and 48% 48% More than 48%
Answer:
More than 48%
Explanation:
If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...
(1 +4%)^12 -1 = 60.1% . . . . more than 48%
The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%
Calculation of Annual Interest rateThe formula used to calculate annual Interest rate =
[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]
where i= nominal interest rate = 4%
n= number of periods= 12 months
Annual Interest rate=
[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]
= (1+0.333)^12 -1
= (1.333)^12-1
= 31.56 - 1
= 30.56%
Therefore, the effective annual or yearly interest rate would be= 30.56%
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When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³
so here P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex] [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]
solve it we get
P2 = 87.9 lb/ft³
A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal to the interface of the two liquids is light totally reflected?
a. 78 degree
b. 88 degree
c. 68 degree
d. 49 degree
e. the critical angle isundefined
Answer:
a. 78 degree
Explanation:
According to Snell's Law, we have:
(ni)(Sin θi) = (nr)(Sin θr)
where,
ni = Refractive index of medium on which light is incident
ni = Refractive index of ethyl alcohol = 1.361
nr = Refractive index of medium from which light is refracted
nr = Refractive index of ethyl alcohol = 1.333
θi = Angle of Incidence
θr = Angle of refraction
So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:
θi = θc
when, θr = 90°
Therefore, Snell's Law becomes:
(1.361)(Sin θc) = (1.333)(Sin 90°)
Sin θc = 1.333/1.361
θc = Sin⁻¹ (0.9794)
θc = 78.35° = 78° (Approximately)
Therefore, correct answer will be:
a. 78 degree
The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
From the information given;
the refractive index of the ethyl alcohol [tex]\mathbf{n_1= 1.361}[/tex]the refractive index of the water [tex]\mathbf{n_2 = 1.333}[/tex] the angle of incidence is the critical angle [tex]\theta_i = \theta_c[/tex] the angle of refraction [tex]\theta _r = 90^0[/tex]According to Snell's Law of refraction;
[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]
[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]
[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]
[tex]\mathbf{ \theta _c =78.35^0}[/tex]
[tex]\mathbf{ \theta _c \simeq78^0}[/tex]
Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
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A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)
Answer:
The angular momentum is [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The mass of the woman is [tex]m = 50 \ kg[/tex]
The angular speed of the rim is [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]
The mass of the disk is [tex]m_d = 110 \ kg[/tex]
The radius of the disk is [tex]r_d = 4.0 \ m[/tex]
The moment of inertia of the disk is mathematically represented as
[tex]I_D = \frac{1}{2} m_d r^2_d[/tex]
substituting values
[tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]
[tex]I_D = 880 \ kg \cdot m^2[/tex]
The moment of inertia of the woman is
[tex]I_w = m * r_d^2[/tex]
substituting values
[tex]I_w = 50 * 4^2[/tex]
[tex]I_w =800\ kg[/tex]
The moment of inertia of the system (the woman + the large disk ) is
[tex]I_t = I_w + I_D[/tex]
substituting values
[tex]I_t = 880 +800[/tex]
[tex]I_t =1680 \ kg \cdot m^2[/tex]
The angular momentum of the system is
[tex]L = I_t w[/tex]
substituting values
[tex]L = 1680 * 5.024[/tex]
[tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
The Great Lakes are all part of what? The Mississippi River The St. Lawrence Seaway A large body of salt lakes The Missouri River
Answer:
St Lawrence Sea way
Explanation:
The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.
Assume the three blocks (m. = 1.0 kg, m = 20 kg and m = 40 ko) portrayed in the figure below move on a frictionless surface and a force F: 36w acts as shown on the 4.0 kg block.
a) Determine the acceleration given this system (in m/s2 to the right). m/s2 (to the right)
b) Determine the tension in the cord connecting the 4.0 kg and the 1.0 kg blocks in N). Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N). N (a) What If How would your answers to parts (a) and (b) of this problem change if the 2.0 kg block was now stacked on top of the 1.0 kg block? Assume that the 2.0 kg block sticks to and does not slide on the 1.0 kg block when the system is accelerated.
(Enter the acceleration in m/s2 to the right and the tension in N.) acceleration m/s (to the right) tension
Answer:
a) 5.143 m/s^2
b) T = 15.43 N
c) Fr = 10.29 N
d) 5.143 m/s^2 , T = 15.43 N
Explanation:
Given:-
- The mass of left most block, m1 = 1.0 kg
- The mass of center block, m2 = 2.0 kg
- The mass of right most block, m3 = 4.0 kg
- A force that acts on the right most block, F = 36 N
Solution:-
a)
- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.
- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:
[tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]
Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2
b)
- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).
- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.
- We will again apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]
Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )
c)
- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).
- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]
- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )
Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.
d)
- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2
- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).
- We apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]
Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.
