Answer:
[tex] \boxed{\sf Acceleration \ (a) = 5.7 \ m/s^2} [/tex]
Given:
Initial velocity (u) = 10 m/s
Final velocity (v) = 30 m/s
Time taken (t) = 3.5 seconds
To Find:
Acceleration (a) of the car
Explanation:
From equation of motion:
[tex] \boxed{ \bold{ v = u + at}}[/tex]
Substituting value of v, u & t in the equation:
[tex] \sf \implies 30 = 10 + a(3.5) \\ \\ \sf \implies 30 - 10 = 3.5a \\ \\ \sf \implies 20 = 3.5a \\ \\ \sf \implies 3.5a = 20 \\ \\ \sf \implies a = \frac{20}{3.5} \\ \\ \sf \implies a = 5.7 \: m/s^2[/tex]
[tex] \therefore[/tex]
Acceleration (a) of the car = 5.7 m/s²
A merry-go-round is rotating at constant angular speed. Two children are riding the merry-go-round: Ana is riding at point A and Bobby is riding at point B.
Part A Which child moves with greater magnitude of velocity? Ana has the greater magnitude of velocity. Bobby has the greater magnitude of velocity. Both Ana and Bobby have the same magnitude of velocity.
Part B Who moves with greater magnitude of angular velocity? Ana has the greater magnitude of angular velocity. Bobby has the greater magnitude of angular velocity. Both Ana and Bobby have the same magnitude of angular velocity.
Part C Who moves with greater magnitude of tangential acceleration? Ana has the greater magnitude of tangential acceleration. Bobby has the greater magnitude of tangential acceleration. Both Ana and Bobby have the same magnitude of tangential acceleration.
Part D Who has the greater magnitude of centripetal acceleration? Ana has the greater magnitude of centripetal acceleration. Bobby has the greater magnitude of centripetal acceleration. Both Ana and Bobby have the same magnitude of centripetal acceleration.
Part E Who moves with greater magnitude of angular acceleration? Ana has the greater magnitude of angular acceleration. Bobby has the greater magnitude of angular acceleration. Both Ana and Bobby have the same magnitude of angular acceleration.
Answer:
bobby has a greater magnitude of velocity because because when angular speed is constant linear velocity is proportional to radius of the circular path
B. They both have same magnitude of angular velocity since the angular speed of the merrygoround is constant
C. Also they both have the same tangential acceleration because the angular speed is constant and tangential is zero for both of them
D. Centripetal acceleration of Bobby is greater
E.they both have the same angular acceleration because angular Speed I constant so angular acceleration is zero for both
Which particle is most likely to interact with your hand? Select one: a. Alpha particle b. Beta particle c. Gamma particle d. Neutrino
Answer:
The correct option is a
Explanation:
The alpha particle has the lowest penetrating power of the trio of alpha, beta and gamma particles and can be stopped by a sheet of paper and hence cannot penetrate a human skin. Beta particle has a higher penetrating power than alpha particle (some of it penetrates the human skin and some do not) while the gamma particle has the highest penetrating power (with all of it penetrating the human skin).
From the above description, it can be deduced that the alpha particle will stay and interact with the hand (because of its low penetrating power) as the remaining particles move through the skin.
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed on a screen that is 2.40 m from the grating. Part A In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.92 mm . What is the difference between these wavelengths? Express your answer in meters.
Answer:
13.51 nm
Explanation:
To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians
y/L=tan θ ≈ θ
and ∆θ ≈∆y/L
Where ∆y= wavelength distance= 2.92 mm =0.00292m
L=screen distance= 2.40 m
=0.00292m/2.40m
=0.001217 rad
The grating spacing is d = (90000 lines/m)^−1
=1.11 × 10−5 m.
the small-angle
approx. Using difraction formula with m = 1 gives:
mλ = d sin θ ≈ dθ →
∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad
=0.000000001351m
= 13.51 nm
What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A
Complete Question
A very long, straight solenoid with a cross-sectional area of 2.39cm2 is wound with 85.7 turns of wire per centimeter. Starting at t= 0, the current in the solenoid is increasing according to i(t)=( 0.162A/s2) t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?
Answer:
The value is [tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 2.39 \ cm^2 = \frac{2.39}{10000} = 0.000239 \ m^2[/tex]
The number of turns is [tex]N = 85.7 \ turns/cm = 8570 \ turns / m[/tex]
The initial time is t = 0s
The current on the solenoid is [tex]I(t) = (0.162 \ A/s^2) t^2[/tex]
The number of turns of the secondary winding is [tex]n = 5 \ turns[/tex]
Generally At I = 3.2 A
[tex]3.2 = (0.162 )t^2[/tex]
=> [tex]t^2 = 19.8[/tex]
=> [tex]t = 4.4 \ s[/tex]
Generally induced emf is mathematically represented as
[tex]\epsilon = A * \mu_o * n * N \frac{d(I)}{dt}[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2t[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2 * 4.4[/tex]
[tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
We will use the same coordinate system for a number of the questions on this exam.
Place a sheet of paper flat on a table in front of you as you sit facing it.
In the horizontal plane toward the right side of the paper is the positive x direction.
In the horizontal plane toward the top of the paper ( away from you in the direction you are facing ) is the positive y direction.
In the vertical direction toward the ceiling is the positive z direction
A spatially uniform magnetic field is steadily changing its size. When a clock reads 2.00s the field is 1.75T in the positive z direction. In the x y plane there is a coil of wire with 55 loops and an area of 0.12m%. The EMF ( voltage change) from one end of the coil to the other is a steady 12.7 Volts. One of these is the magnetic field when the clock reads 2.37s. Which is it?
Answer:
Explanation:
Let the magnetic field after .37 s is B
magnetic flux associated with coil = n BA , where B is magnetic field and A is area of coil and n is no of turns
= 55 x .12 x 1.75
= 11.55 Tm²
magnetic flux after .37 s
= 55 x B x .12
= 6.6 B
rate of change of flux = (66B - 11.55) / .37 = EMF induced
(6.6B - 11.55) / .37 = 12.7
6.6 B - 11.55 = 4.7
6.6 B = 16.25
B = 2.46 T .
Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal projection Pa of R onto axi
Answer: Hello your question is incomplete attached below is the complete question
Ra = 1132 N
Rb = 522.6 N
Pa = 679.7 N
Explanation:
To determine the scalar components Ra
[tex]\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}[/tex]
therefore : Ra = [tex]\frac{sin120^o * 750}{sin 35^o}[/tex] = 1132 N
To determine the scalar component Rb
[tex]\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}[/tex]
therefore : Rb = [tex]\frac{sin 25^o * 750}{sin 35^o}[/tex] = 522.6 N
To determine the orthogonal projection Pa of R onto
Pa = 750 cos[tex]25^o[/tex] = 679.7 N
(a) A standard sheet of paper measures 8 1/2 by 13 inches. Find the area of one such sheet of paper in m2.8.5(!meter/39.37in) — 0.2159 13 (!meter/39.37in ) — 0.3302 0.2159 X 0.3302 = 0.0713
= 0.0713 m2 (b) A second sheet of paper is three-quarters as long and three-quarters as wide as the one described in part (a). By what factor is its area less than the area found in part(a)?= _____________ times less
Answer:
a) A = 0.07129 m²
b) A / A ’= 1.77
Explanation:
In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.
width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)
a = 0.2159 m
length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)
l = 0.3302 m
The area of a rectangle is
A = l a
A = 0.3302 0.2159
A = 0.07129 m²
b) we have a second sheet with reduced dimensions
a ’= 3/4 a
l ’= ¾ l
Let's find the area of this glossy sheet
A ’= l’ a ’
A ’= ¾ l ¾ a
A ’= 9/16 l a
To find the factor we divide the two quantities
A / A ’= l a 16 / (9 l a
A / A ’= 1.77
which of the following is not a method of caculating percentage of body fat
Body mass index, known as BMI, is not a method of calculating body fat percentage, but it is a technique used to check nutritional status and to see if a person is within normal range with respect to weight and height.. This technique is measured by the formula: BMI = Weight (Kg) / (Height (m)) ².
Ergo, the answer is BMI!!!!!!!!!!!
Hope this helped you!
hor and Hulk are fighting eachother, back and forth in a straight line in the master's colosseum. Hulk punches thor into a wall 30 meters away from the start. Thor then hit's hulk with a hammer, hurling Hulk 45 meters back from the wall. This all happens in the course of 3 seconds. What is the speed of their fight across the arena?
a. 25 m/s
b. 10 m/s forward from Hulk and Thor's starting point.
c. 5 m/s backwards from hulk's starting point
d. 5 m/s
Answer:
A
Explanation:
cant be c because its asking for speed and speed doesnt have direction
Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and g, the magnitude of the acceleration due to gravity.
Answer:
(a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Explanation:
Given that,
Gauge pressure at bottom = p₁
Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.
Find the speed of the fluid in the left end of the main pipe.
(a). We need to calculate the gauge pressure at the bottom of tube 1
Using bernoulli equation
[tex]P_{1}=\rho g h_{1}[/tex]
(b). We need to calculate the speed of the fluid in the left end of the main pipe
Using bernoulli equation
Pressure for first pipe,
[tex]P_{1}=\rho gh_{1}[/tex].....(I)
Pressure for second pipe,
[tex]P_{2}=\rho gh_{2}[/tex].....(II)
From equation (I) and (II)
[tex]P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
Put the value of P₁ and P₂
[tex]\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
[tex]gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2[/tex]....(III)
We know that,
The continuity equation
[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]
[tex]v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})[/tex]
Put the value of v₂ in equation (III)
[tex]2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2[/tex]
Here, [tex]\dfrac{A_{1}}{A_{2}}=\gamma[/tex]
So, [tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)[/tex]
[tex]v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Hence, (a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
When sunlight interacts with a raindrop to produce a rainbow, how many refractive events occur for each drop
Answer:
The correct answer will be "2".
Explanation:
The mechanism of rainbow formation starts whenever sunlight is shone on a raindrop. When the sun's light waves hit and go through a droplet of water, their pace changes a little. This induces bending or "refracting" of the direction of light.Light moves in such a single direction, unless someone represents something, refracts, or diffracts everything that. When some of these things have happened, the balanced information of color is isolated, in which each could still be seen.What are two examples of healthcare fraud ?
A truck took 4 hours to complete a journey. At the first 1 h 45
min, it travelled at an average speed of 70 km/h. For the rest of
the journey, it travelled at an average speed of 80 km/h. What
was the total distance of the journey?
Answer:302.5
Explanation:
Distance is the length of a path followed by a particle. The displacement of a particle is defined as its change in position in some time interval.
a. True
b. False
Answer:
a. True
Explanation:
Distance is described with only magnitude. It is defined as the total path covered by an object, in other words it is the length of a path followed by a particle.
Displacement is described with both magnitude and direction. It is distance traveled in a specified direction or change in position in some time interval.
Therefore, the correct option is " a. True"
does anyone know the answer to these 3 questions?
Answer:
independent
dependent
dependent
The effective spring constant describing the potential energy of the HBr molecule is 410 N/m and that for the NO molecule is 1530 N/m.A) Calculate the minimum amplitude of vibration for the NO molecule.
B) Calculate the minimum amplitude of vibration for the HCl molecule.
Answer:
a. the minimum amplitude of vibration for the NO molecule A [tex]\simeq[/tex] 4.9378 pm
b. the minimum amplitude of vibration for the HCl molecule A [tex]\simeq[/tex] 10.9336 pm
Explanation:
Given that:
The effective spring constant describing the potential energy of the HBr molecule is 410 N/m
The effective spring constant describing the potential energy of the NO molecule is 1530 N/m
To calculate the minimum amplitude of vibration for the NO molecule, we use the formula:
[tex]\dfrac{1}{2}kA^2= \dfrac{1}{2}hf[/tex]
[tex]kA^2= hf[/tex]
[tex]A^2= \dfrac{hf}{k}[/tex]
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 5.63 \times 10^{13} \ s^{-1})}{1530 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{3.730438 \times 10^{-20} \ m}{1530 }[/tex]
[tex]A = \sqrt{2.43819477 \times 10^{-23}\ m[/tex]
[tex]A =4.93780799 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 4.9378 pm
The effective spring constant describing the potential energy of the HCl molecule is 480 N/m
To calculate the minimum amplitude using the same formula above, we have:
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 8.66 \times 10^{13} \ s^{-1})}{480 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{5.738116\times 10^{-20} \ m}{480 }[/tex]
[tex]A = \sqrt{1.19544083\times 10^{-22}\ m[/tex]
[tex]A = 1.09336217 \times 10^{-11} \ m[/tex]
[tex]A = 10.9336217 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 10.9336 pm
Convert 252 cL into uL
Answer:
Explanation:
252,000
Using the lensmaker's formula (equation (5) of your lab manual), calculate the index of refraction of the acrylic lens. You should use the f_are you calculated in part (1) above instead of the value for the focal length of the concave lens that you measured. Remember that the focal length of a concave lens is negative so in this case, f = -f_are.
Answer:
n = 1 + R / f
Explanation:
The equation of the constructor is optical is
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and image, respectively
The exercise tells us that it is a concave lens with focal length fo, in these lenses the focal length is negative. The relationship to calculate the focal length is
1 / f = (n -n₀) (1 /R₁ - 1 /R₂)
where is n₀ the refractive index of the medium that surrounds the lens in this case it is air with n₀ = 1, you do not indicate the type of lens, but the most used lens is the concave plane, in this case R₂ = ∞, so which 1 / R₂ = 0, let's substitute
1 / f = (n-1) / R₁
n - 1 = R₁ / f
let's calculate
n = 1- R₁ / f
remember that the radius of curvature is negative, so the equation is
n = 1 + R / f
en un experimento se retiro a un pardela de su nido se le llevo a 5150 km de distancia y luego fue liberada su regreso a su nido fue 13.5 dias despues de haberse soltado si el origen es el nido y extendemos el eje +x al punto de liberacion ¿cual fue la velocidad media del ave?a)en el vuelo de regreso b) desde que se retirodel nido hasta que regreso
Answer:
a) v = - 4.4168 m / s , b) V = 0
Explanation:
The average speed is the variation of the displacement in time used.
v = Δx /Δt = (x₂-x₁) / (t₂-t₁)
let's apply this equation to our case
Let's reduce the magnitudes of the system Yes
Δx = 5150 km (1000 m / 1km) = 5.15 106 m
Δt = 13.5 day (86,400 s / 1 day) = 1,166 10 6 s
a) return trip
the vector is negative because it points long towards the center of the system
v = - 5.15 106 / 1.166 106
v = - 4.4168 m / s
the negative sign indicates that he is coming back, to the lair
b) In a complete trip the distance is zero, because it is a vector, consequently the mean fickleness is also zero
V(j_ = 0
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
Answer:
6787.5 V
Explanation:
From the question,
P = IV..................... Equation 1
Where P = Power, I = rms current, V = rms voltage.
make V the subject of the equation
V = P/I................. Equation 2
Given: P = 1500 W, I = 6.4/√2 = 4.525 A
Substitute these values into equation 2
V = 1500(4.525)
V = 6787.5 V
Hence the rms voltage = 6787.5 V
1) A plane lands on a runway with a speed of 125 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2) and direction of the plane's average acceleration during this time interval?2) Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +20.8 km/s to −17.1 km/s over a period of 2.01 years.(a) What is the total change in the asteroid's velocity? (Indicate the direction with the sign of your answer.)(b) What is the asteroid's average acceleration during this interval? (Indicate the direction with the sign of your answer.)
Answer: 1) a = 9.61m/s² pointing to west.
2) (a) Δv = - 37.9km/s
(b) a = - 6.10⁷km/years
Explanation: Aceleration is the change in velocity over change in time.
1) For the plane:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]
[tex]a=\frac{125}{13}[/tex]
a = 9.61m/s²
The plane is moving east, so velocity points in that direction. However, it is stopping at the time of 13s, so acceleration's direction is in the opposite direction. Therefore, acceleration points towards west.
2) Total change of velocity:
[tex]\Delta v = v_{f}-v_{i}[/tex]
[tex]\Delta v = -17.1-(+20.8)[/tex]
[tex]\Delta v= -37.9[/tex]km/s
The interval is in years, so transforming seconds in years:
v = [tex]\frac{-37.9}{3.15.10^{-7}}[/tex]
[tex]v=-12.03.10^{7}[/tex]km/years
Calculating acceleration:
[tex]a=\frac{-12.03.10^{-7}}{2.01}[/tex]
[tex]a=-6.10^{7}[/tex]
Acceleration of an asteroid is a = -6.10⁷km/years .
A NASA spacecraft measures the rate of at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is:
Complete Question
A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is: [tex]R = 0.0498 \ kPAkm^{-1}[/tex] Convert R to [tex]kJ*m^{-4}[/tex]
Answer:
The value is [tex]R = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]
Explanation:
From the question we are told that
The altitude is [tex]R = 0.0498 \ kPAkm^{-1}[/tex]
Generally
[tex]1 k PA = 1000 PA[/tex]
So
[tex]R = 0.0498 \frac{1000PA}{ km}[/tex]
Also
1 km = 1000 m
So
[tex]R = 0.0498 \frac{1000PA}{ 1000m}[/tex]
=> [tex]R = 0.0498 \frac{1 PA}{ 1 m}[/tex]
Now PA is Pascal which is mathematically represented as
[tex]PA = \frac{N}{m^2 }[/tex]
So
[tex]R = 0.0498 \frac{\frac{N}{m^2} }{m}[/tex]
[tex]R = 0.0498 \frac{N}{m^3}[/tex]
Looking the unit we are arrive at we see that it contains J which is mathematically represented as
[tex]J = N * m[/tex]
So
[tex]R = 0.0498 \frac{ N \frac{m}{m} }{m^3}[/tex]
=> [tex]R = 0.0498 \frac{\frac{J}{m} }{m^3}[/tex]
=> [tex]R = 0.0498 \frac{J}{m^4}[/tex]
Generally
[tex]1 J \to 1.0*10^{-3} kJ[/tex]
[tex]0.0498 J \to x kJ[/tex]
=> [tex]x = \frac{0.0498 * 1.0*10^{-3}}{1}[/tex]
=> [tex]0.0498 *10^{-3} kJ[/tex]
So
[tex]R = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]
Question 8 of 10
It takes a person 22 seconds to swim in a straight line from the south end of
a pool to the north end of the pool, a distance of 28 meters. What is the
swimmer's velocity?
A. 1.3 m/s south
B. 1.3 m/s north
C. 0.8 m/s south
D. 0.8 m/s north
The correct answer is B. 1.3 m/s north
Explanation:
Velocity is a factor that describes how fast or slow the motion of a body occurs and its direction. Moreover, this can be calculated by dividing the total displacement into the time of movement, and the final result is expressed in units such as meters per second followed by the direction, for example, 152 m/s south. The process to calculate the velocity of the swimmer is shown below.
[tex]v = \frac{d}{t}[/tex]
[tex]v = \frac{28 meters}{22 seconds}[/tex]
[tex]v = 1.27 m/s[/tex]
This means the velocity of the swimmer is 1.27 m/s, which can be rounded as 1.3 m/s. Additionally, if the direction is considered it would be 1.3 m/s north because the swimmer went from the south of the pool to its north.
Answer:
the answer is B
Explanation:
confirmed
In lifting a heavy weight from the floor, one should use the power of the __________ in order to avoid straining the lower back.
Answer:
Hip and knee extensors
Explanation:
These are gluteus muscles and hamstring muscles the there are the major movers for your body and are very important in pelvic alignment and lower back support during weight lifting
i need help Mr or ms tutor
Explanation:
Height is the x-axis, and gravitational potential energy is the y-axis. As the height increases, the gravitational potential energy increases linearly.
A clock battery wears out after moving 10,900 C of charge through the clock at a rate of 0.450 mA. (a) How long (in s) did the clock run
Answer:
2.42×10⁷ s
Explanation:
From the question above,
Applying,
Q = It.................... Equation 1
Where Q = quantity of charge in coulombs, I = electric current in Ampere, t = time in seconds
make t the subject of the equation
t = Q/I................ Equation 2
Given: Q = 10900 C, I = 0.450 mA = 4.5×10⁻⁴ A
Substitute these values into equation 2
t = 10900/(4.5×10⁻⁴)
t = 2.42×10⁷ s
Hence the clock runs for 2.42×10⁷ s
When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon.The last 10m it freely falls to the surface of the Moon. i) How long does it take for the Eagle to touch down
Answer:
3.5s
Explanation:
Using equation of motion number 2
S= ut + 1/2 gt²
Substituting
10= 0.5+1.62t²
t= 3.5seconds
scott and jafar are ready to work a problem. in this simulation, assume the coordinates of the points are as follows. a (0 s, 30 m); b (10 s, 50 m); c (20 s, 38 m); d (30 s, 0 m); e (40 s, −38 m); and f (50 s, −50 m) recall the definitions of average speed, vavg ≡ d δt , and average velocity in the x direction, vx ,avg ≡ δx δt . find the average velocity from circled a to circled b.
Answer:
Explanation:
point a represents time 0 and position coordinate 30
point b represents time 10 s , and position coordinate 50 m .
time elapsed = 10 - 0 = 10 s .
displacement = 50 m - 30 m
= 20 m
average velocity = displacement / time elapsed
= 20 / 10
= 2 m /s .
List two scientific questions that would be best explained using a model.
Answer:
Is a flame hottest when it is blue? Is it cold today?
Explanation:
Verification questions. These are basic data collecting questions. They are useful in building knowledge.
A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest. What is the common velocty of the two-car train after the collisions?
Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
V is the common velocity after the collisions
[tex]V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s[/tex]
So, the two car train will move with a common velocity of 6.79 m/s.