Answer:
a= -80.357 m/s
Explanation:
use the formula
vf^2=vi^2+2a(xf-xi)
Plug in givens
0=(7.50)^2+2a(0.350m)
solve for acceleration
a= -80.357 m/s
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigger than the object and the screen is 5 m away from the mirror as shown in fig 5.2, calculate the focal length of the mirror.
Answer:
f = 1 m
Explanation:
The magnification of the lens is given by the formula:
[tex]M = \frac{q}{p}[/tex]
where,
M = Magnification = 4
q = image distance = 5 m
p = object distance = ?
Therefore,
[tex]4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m[/tex]
Now using thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\[/tex]
f = 1 m
In what direction is the centripetal force directed?
Answer:
towards the center
Explanation:
that is the solution above
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant pressure until 60 percent of it, by mass, condenses. Determine (a) the work (W) done during the process. (b) What-if Scenario: What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v[tex]_f[/tex] + xv[tex]_{fg[/tex]
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v[tex]_f[/tex] + x₂v[tex]_{fg[/tex]
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ
write the formulae of magnesium chloride and sodium sulfate
Answer:
Magnesium Chloride: MgCl2
Sodium Sulfate: Na2SO4
The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?
Answer:
Acosθ
Explanation:
The x-component of a vector is defined as :
Magnitude * cosine of the angle
Maginitude * cosθ
The magnitude is represented as A
Hence, horizontal, x - component of the vector is :
Acosθ
Furthermore,
The y-component is taken as the sin of the of the angle multiplied by the magnitude
Vertical, y component : Asinθ
If the source moves, the wavelength of the sound in front of the direction of motion is____than the wavelength behind the direction of motion.
a. the same.
b. smaller than.
c. unrealted to.
d. larger then.
Answer:
B. Smaller than
Explanation:
This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.
You may have been surprised to learn that Olympic gold medals are not made from solid gold, but instead have a coating of • Saved gold on the outside.
To see a possible reason why, determine the value of the medal the size (not mass) of the Olympic gold medal if it were made of solid gold. Hint: As of mid-2018, the cost of gold is about $40 per gram.
Answer:
A gold medal has the (minimum) dimensions of:
diameter = 60mm
thickness = 3mm
So we will work with those dimensions.
The medal is then a cyinder of diameter
D = 60mm = 6cm
and height:
H = 3mm = 0.3cm
Remember that the volume of a cylinder is:
V = pi*(D/2)^2*H
where pi = 3.14
Then the volume of a medal is:
V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3
The density of the gold in g/cm^3 is:
d = 19.3 g/cm^3
And remember that:
density = mass/volume
So, if the volume is 3.768 cm^3
Then the mass will be:
mass = density*volume = 19.3 g/cm^3*3.768 cm^3 = 72.7 g
So, a single gold medal would weight 72.7 grams
And each gram of gold costs $40
Then the total cost of the gold medal would be:
value = $40*72.7 = $2,908
Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.
And each gold medal costs $2,908
So the total cost (only for the gold medals, ignoring the others) would be to high.
This is why the gold medals are made mostly of silver.
Where would the normal force exerted on the rover when it rests on the surface of the planet be greater
Answer:
Normal force exerted on the rover would be greater at a point on the surface of the planet where the weight of the rover is experienced to be greater.
Explanation:
Since weight is a vector quantity, it can vary with position. Weight is the amount of force the planet exerts on the rover centered towards the planet.
Such a force is the result of gravitational pull and is quantified as:
[tex]F=G\times \frac{M.m}{R^2}[/tex]
and [tex]M=\rho\times \frac{4\pi.r^3}{3}[/tex]
where:
R = distance between the center of mass of the two bodies (here planet & rover)
G = universal gravitational constant
M = mass of the planet
m = mass of the rover
This gravitational pull varies from place to place on the planet because the planet may not be perfectly spherical so the distance R varies from place to place and also the density of the planet may not be uniform hence there is variation in weight.
Weight is basically a force that a mass on the surface of the planet experiences.
According to Newton's third law the there is an equal and opposite reaction force on the body (here rover) which is the normal force.
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s, exactly how fast (in m2/s) is the area of the spill increasing when the radius is 39 m?
Explanation:
The area of a circle of radius r is given by
[tex]A = \pi r^2[/tex]
Taking the derivative of A with respect to time t, we get
[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]
We also know that
[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]
[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]
15.1.2 Exam: Semester Exam
ŽA
This graph shows the energy of a reaction over time. Which statement is
true?
Potential energy
N
Reaction progress
O A. G represents the activation energy
OB. H represents the energy of the products,
O C. G represents the energy of the products.
< PREVIOUS
Answer: D. F represents the activation energy
Explanation:
The activation energy is the energy required to get the reactants to begin reacting with one another such that products are created. This energy ranges from the minimum to the maximum energy required.
F is therefore the activation energy because it shows the range between the minimum energy it took for the reaction to start and the maximum energy that was required to continue the reaction.
Question: A NEO distance from the Sun is 1.17 AU. What is the speed of the NEO (round your answer to 2 decimal places)
Answer:
v = 2.75 10⁴ m / s
Explanation:
For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system
F = ma
where force is the force of gravity
F = [tex]G \frac{m M}{r^2}[/tex]
acceleration is centripetal
a = [tex]\frac{v^2}{r}[/tex]
we substitute
G m M / r² = m v² / r
[tex]\frac{GM}{r}[/tex] = v²
v = [tex]\sqrt{GM/r}[/tex]
indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system
r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m
let's calculate
v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11
v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8
v = 2.75 10⁴ m / s
Images formed by a convex mirror are always
Answer:
Images formed by a convex mirror are always virtual
Explanation:
A virtual image is always created by a convex mirror, and it is always situated behind the mirror. The picture is vertical and situated at the focus point when the item is far away from the mirror. As the thing approaches the mirror, the image follows suit and increases until it reaches the same height as the object.
OAmalOHopeO
А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?
Answer:
I am not able to answer this question please don't mind...Explanation:
please marks me as brainliests...Monique walks 360 meters to get to lanier on days when she is early and doesn't get caught by traffic it takes her 60 seconds to get to school how fast was she running
Answer:
6m/s
Explanation:
We are to calculate the speed of Monique
Speed = Distance/Time
Given
Distance = 360m
Time = 60secs
Substitute
Speed = 360m/60s
Soeed = 6m/s
Hence she was running at 6m/s
A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips
Answer:
I could not find the answer or do it myself if I did find it I would defenetly share
1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-34).S)(3.0 x 108m/s)
Answer:
E = 2,964 10⁻¹⁹ J
Explanation:
The energy of the photons is given by the Planck relation
E = h f
the speed of light is related to wavelength and frequency
c = λ f
we substitute
E = h c /λ
let's reduce the magnitude to the SI system
λ = 671 nm = 671 10⁻⁹ m
let's calculate
E = 6.63 10⁻³⁴ 3 10⁸ /671 10⁻⁹
E = 2,964 10⁻¹⁹ J
Express the unit of force in terms of fundamental unit
Answer:
The fundamental unit of force is kg.m/s²
Explanation:
According to Newton's second law of motion, force is given as the product of mass and acceleration.
Mathematically, force can be expressed as; F = ma
where;
F is the force
M is mass of the object, unit of mass = kg
a is acceleration of the object, unit of acceleration = m/s²
Force = kg x m/s²
Force = kg.m/s² = Newton [N]
Therefore, the fundamental unit of force is kg.m/s²
If 2cm³ of wood has a mass 0.6g what would be its density
we know density = mass/ volume
as mass = 0.6 g
and volume = 2cm³
so density = (6/20)(g/cm³)
0.3g/cm³ (ans)
Hope it helps
1. Draw four illustrations of a globe and paper that are positioned to yield equatorial, transverse, oblique, and polar aspect projections. Label the equator in each. Use your textbook or lecture material if you need a reference.2. On any map, why is there distortion at areas that do not fall on lines of tangency or secancy?
Answer:
1) attached below
2) assumption that the earth is spherical
Explanation:
1) Four illustrations of a globe
attached below
2) Reason for distortions at areas that do not fall on lines of tangency or secancy
The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true
An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 19 m/s and measures a time of 24.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet
Answer:
1.56 m/s²
Explanation:
Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.
The total time (time of flight) of an object is given by:
T = 2usinθ / g
where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity
Since the astronaut throws a rock straight up, hence θ = 90°, u = 19 m/s, T = 24.4 s.
T = 2usinθ / g
Substituting:
24.4 = 2(19)(sin90)/g
g = 2(19)(sin90) / 24.4
g = 1.56 m/s²
The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: y1(x, t) = (8.20 mm) sin(4.00πx - 430πt) y2(x, t) = (8.20 mm) sin(4.00πx + 430πt), with x in meters and t in seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?
Answer:
Explanation:
From the information given:
The angular frequency ω = 430 π rad/s
The wavenumber k = 4.00π which can be expressed by the equation:
k = ω/v
∴
4.00 = 430 /v
v = 430/4.00
v = 107.5 m/s
Similarly: k = ω/v = 2πf/fλ
We can say that:
k = 2π/λ
4.00 π = 2π/λ
wavelength λ = 2π/4.00 π
wavelength λ = 0.5 m
frequency of the wave can now be calculated by using the formula:
f = v/λ
f = 107.5/0.5
f = 215 Hz
Also, the Period(T) = 1/215 secs
The time at which particle proceeds from point A to its maximum upward displacement and to its maximum downward displacement can be computed as t = T/2;
Thus, the distance(x) covered by each wave during this time interval(T/2) will be:
x = v * t
x = v * T/2
x = λ/2
x = 0.5/2
x = 0.25 m
A 55 g soapstone cube--a whisky stone--is used to chill a glass of whisky. Soapstone has a density of 3000 kg/m3, whisky a density of 940 kg/m3. What is the approximate normal force of the bottom of the glass on a single stone?
Answer:
[tex]N=0.37N[/tex]
Explanation:
Mass [tex]m=55g=>0.055kg[/tex]
Soapstone Density [tex]\rho_s=3000kg/m^2[/tex]
Whisky Density [tex]\rho_w=940kg/m^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=U+N[/tex]
Therefore
[tex]N=m*g-(\frac{m}{\rho_s})*\rho_w*g[/tex]
[tex]N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}[/tex]
[tex]N=0.37N[/tex]
A black T-shirt is warmer in the summertime than a white T-shirt because the black T-shirt
A. Is reflecting all wavelengths of light.
B. Absorbs violet light, the highest energy wavelength.
C. Is absorbing all wavelengths of light. D. Doesn’t absorb red, the longest wavelength.
Answer:
c
Explanation:
darker colors absorb app light
Answer:
C. Is absorbing all wavelengths of light.
Explanation:
Black isn't a color, but rather the absence of color. We see a T-shirt as black because it isn't reflecting any light toward our eyes. A black T-shirt absorbs all of the wavelengths of light, causing it to absorb more energy and become warmer than white, which reflects light.
The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?
Explanation:
The period T of a simple pendulum is given by
[tex]T = 2 \pi \sqrt{\dfrac{l}{g}}[/tex]
Doubling the length of the pendulum gives us a new period T'
[tex]T' = 2 \pi \sqrt{\dfrac{l'}{g}} = 2 \pi \sqrt{\dfrac{2l}{g}}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2} \left(2 \pi \sqrt{\dfrac{l}{g}} \right)[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2}\:T = \sqrt{2}(3.5\:\text{s})= 4.95\:\text{s}[/tex]
A car is moving north at 5.2 m/s2. Which type of motion do the SI units in this
value express?
Answer:
the SI unit (meter per second square) indicates a linear type of motion.
Explanation:
Given;
acceleration of the car, a = 5.2 m/s² North
the SI unit of the car, = m/s²
The SI unit of the given value (acceleration), indicates a linear type of motion.
Linear acceleration is the change in linear velocity with time. Also, the northwards direction indicates linear displacement of the car.
Therefore, the SI unit (meter per second square) indicates a linear type of motion.
Answer:
displacement
Explanation:
Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant force is directed vertically upward.
Answer:
how to solve this problem ???????
The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.
Resultant of the two forces
The resultant of the two forces is determined by resolving the force into x and y component as shown below;
[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]
where;
F1 = 500 NF2 = 600 NValue of Angle θThe value of Angle θ is determined from equation (1)
-500sinθ + 600sin(30) = 0
500sinθ = 600sin(30)
500sinθ = 300
sinθ = 3/5
θ = 36.87⁰
Resultant of the two forcesThe resultant of the forces is determined using the second equation;
500cosθ + 600cos(30) = R
500 x cos(36.87) + 600 x cos(30) = R
919.6 N = R
Learn more about resultant forces here: https://brainly.com/question/25239010
A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 and 1510 N, respectively. The acceleration of the cable is 0.620 m s 2 , upward. What is the tension in the cable (a) below the worker and ( b) above the worker
Answer:
Explanation:
a)
Below the worker , the tension in cable is pulling the crate . Let the tension be T₁ .
weight of crate is acting downwards .
Total weight 1510 N.
Net force acting on both = T₁ - 1510
Applying second law of Newton ,
T₁ - 1510 = 1510 / 9.8 x 0.62 [ 1510 / 9.8 = mass of crate ]
T₁ - 1510 = 95.5
T₁ = 1605.5 N.
b )
Above the worker , the tension in cable is pulling both the worker and the crate . Let the tension be T₂ .
weight of both worker and crate is acting downwards .
Total weight = 965 + 1510 = 2475 N.
Net force acting on both = T₂ - 2475
Applying second law of Newton ,
T₂ - 2475 = 2475 / 9.8 x 0.62 [ 2475 / 9.8 = mass of both worker and crate ]
T₂ - 2475 = 156.6
T₂ = 2631.6 N.
Desde el punto A sale un vehículo a 80 km/h al mismo tiempo sale un ciclista a 20km/h ¿a qué distancia se encuentra uno del otro al cabo de 5h. R:300k
Answer:
Distance between them after 5 hours is 300 km.
Explanation:
From point A a vehicle leaves at 80 km / h at the same time a cyclist leaves at 20 km / h at what distance is they from each other after 5 hours.
Distance traveled by A in 5 hours = speed x time = 80 x 5 = 400 km
Distance traveled by B in 5 hours = speed x time = 20 x 5 = 100 km
The distance between them after 5 hours = 400 - 100 = 300 km
A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
Explanation:
Given
Acceleration of the pebble is
At t=0, velocity is
considering horizontal motion
[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]
Velocity acquired during this time
[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]
Consider vertical motion
[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]
Net velocity is
[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]
Angle made is
[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]
The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6. 7 m, measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 3.0 m from the center of the circle.
Answer:
[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]
Explanation:
The centripetal acceleration is given as follows:
[tex]a_c = \frac{v^2}{r}\\[/tex]
where,
ac = centripetal acceleration
v = linear speed = rω
r = radius
ω = angular speed
Therefore,
[tex]a_c = \frac{(r\omega)^2}{r}\\\\a_c = r\omega^2[/tex]
Therefore, the ratio will be:
[tex]\frac{a_{c1}}{a_{c2}} = \frac{r_1\omega^2}{r_2\omega^2}\\\\\frac{a_{c1}}{a_{c2}} = \frac{r_1}{r_2}\\\\[/tex]
where,
r₁ = 6.7 m
r₂ = 3 m
Therefore,
[tex]\frac{a_{c1}}{a_{c2}} = \frac{6.7\ m}{3\ m}\\\\[/tex]
[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]