Answer:
Spring Constant = 279.58 N/m
Explanation:
We are given;
Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg
Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m
Maximum height; h = 4.46 m
The formula for the energy in the spring is given by;
E = ½kx²
where:
k is the spring constant
x is the distance the spring is compressed.
Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;
½kx² = mgh
Plugging in the relevant values, we have;
½ * k * 0.0801² = 0.0205 * 9.81 * 4.46
0.003208005k = 0.8969
k = 0.8969/0.003208005
k = 279.58 N/m
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?
Answer:
Explanation:
a )
from lens makers formula
[tex]\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})[/tex]
f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face
putting the values
[tex]\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})[/tex]
1.462 = 2 - 1 / r₂
1 / r₂ = .538
r₂ = 1.86 cm .
= 18.6 mm .
b )
object distance u = 25 cm
focal length of convex lens f = 1.8 cm
image distance v = ?
lens formula
[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}[/tex]
[tex]\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}[/tex]
.5555 - .04
= .515
v = 1.94 cm
c )
magnification = v / u
= 1.94 / 25
= .0776
size of image = .0776 x size of object
= .0776 x 10 mm
= .776 mm
It will be a real image and it will be inverted.
What is the velocity of a car that travels 556km northwest in 3.2 hours
Answer:
173.75 km/hr in the NW direction.
Explanation:
Velocity is the time rate of change in displacement of a body. Mathematically:
v = d / t
where d = displacement
t = time
Therefore, the velocity of the car is:
v = 556 / 3.2 = 173.75 km/hr
The velocity of the car is 173.75 km/hr in the NW direction.
The velocity of a car will be "173.75 km/hr".
Displacement and Velocity,The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.
Displacement, d = 556 km
Time, t = 3.2 hours
We know the relation,
→ Velocity = [tex]\frac{Displacement}{Time}[/tex]
or,
→ V = [tex]\frac{d}{t}[/tex]
By substituting the values, we get
= [tex]\frac{556}{3.2}[/tex]
= [tex]173.75[/tex] km/hr
Thus the response above is right.
Find out more information about velocity here:
https://brainly.com/question/6504879
What is the answer for this question
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.
a research submarine what is the maximum depth it can go
Answer: 36, 200 feet deep according to information on google
Explanation:
A small submarine, the bathyscape Trieste, made it to 10,916 meters (35,813 feet) below sea level in the deepest point in the ocean, the Challenger Deep in the Marianas Trench, a few hundred miles east of the Philippines. This part of the ocean is 11,034 m (36,200 ft) deep, so it seems that a submarine can make it as deep as it's theoretically possible to go
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165 cmcm , but its circumference is decreasing at a constant rate of 14.0 cm/scm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 TT , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
If the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina
Answer:
The distance from the cornea vertex to the retina is 2.36 cm
Explanation:
The question is incomplete.
The complete question is as follows;
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
Solution.
We use image-object reaction to calculate the distance from the cornea vertex to the retina.
Mathematically;
n1/s + n2/s’ = n2-n1/R
From the question, we identify the following;
n1 ; Refractive index of air = 1
n2 ; Refractive index of lens = 1.4
S ; Object Distance = 36 cm
S’ = ?
R ; Radius of curvature of the cornea = 0.65
Substituting these values into the equation above;
1/36 + 1.4/S’ = (1.4-1)/0.65
{S’+ 36(1.4)}/36S’ = 0.4/0.65
{S’ + 50.4}/36S’ = 0.62
S’ + 50.4 = 22.32S’
50.4 = 22.32S’ -S’
21.32S’ = 50.4
S’ = 50.4/21.32
S’ = 2.36 cm
To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.35 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.
Answer:
[tex]T_{1}[/tex] = 14.88 N
Explanation:
Let's begin by listing out the given variables:
M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,
g = 9.8 m/s²
At equilibrium, the sum of all external torque acting on an object equals zero
τ(net) = 0
Taking moment about [tex]T_{1}[/tex] we have:
(M + m) g * 0.5L - [tex]T_{2}[/tex](L - d) = 0
⇒ [tex]T_{2}[/tex] = [(M + m) g * 0.5L] ÷ (L - d)
[tex]T_{2}[/tex] = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)
[tex]T_{2}[/tex]= 59.535 ÷ 2.4
[tex]T_{2}[/tex] = 24.80625 N ≈ 24.81 N
Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N
Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N
Using sum of equilibrium in the vertical direction, we have:
[tex]T_{1}[/tex] + [tex]T_{2}[/tex] = W + w ------- Eqn 1
Substituting T2, W & w into the Eqn 1
[tex]T_{1}[/tex] + 24.81 = 26.46 + 13.23
[tex]T_{1}[/tex] = 14.88 N
ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attached at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
50 points!! please help :((
Answer:
Loudness: decreases
Amplitude: decreases
Pitch: stays the same
Frequency: stays the same
Explanation:
1.
An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.
2.
Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.
3 and 4.
The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.
Hope this helps!
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
1. Describe what must happen to an atom to make it
A. A cation
B. An anion
2. Describe why some acids are strong while other acids are weak
3. Compare protons, neutrons and electron, listing their similarities and differences
4. Explain why you breathe faster and deeper when exercising
Answer:
Explanation:
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...
Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...
An electron and a positron collide head on, annihilate, and create two 0.804 MeV photons traveling in opposite directions. What was the initial kinetic energy of an electron? What was the initial kinetic energy of a positron?
Answer:
Ke- = Ke+ = 0.294MeV
Explanation:
To fins the kinetic energy of both electron and positron you use the following formula, for the case of annihilation of one electron an positron:
2[tex]E_p=2E_o+K_{e^-}+K_{e^+}[/tex] (1)
Ep: photon energy = 0.804MeV
Eo: rest energy of one electron (and positron) = 0.51MeV
Ke-: kinetic energy of electron
Ke+: kinetic energy of positron
You replace the values of Ep and Eo in the equation (1):
[tex]K_{e^-}+K_{e^+}=2E_p-2E_o=2(0.804MeV-0.51MeV)=0.588MeV[/tex]
Iy you assume both positron and electron have the same speed, then, the kinetic energy of them are equal, and the kinetic energy of each one is:
[tex]K_{e^-}=K_{e^+}=\frac{0.588MeV}{2}=0.294MeV[/tex]
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
The friends now feel ready to try a problem. Suppose an Atwood machine has a mass of m1 = 2.5 kg and another mass of m2 = 8.5 kg hanging on opposite sides of the pulley. Assume the pulley is massless and frictionless, and the cord is ideal. Determine the magnitude of the acceleration of the two objects and the tension in the cord.
Answer:
a = 5.34 m/s²
T = 37.86 N
Explanation:
This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:
a = g(m₂ - m₁)/(m₂ + m₁)
where,
a = acceleration of both masses = ?
g = 9.8 m/s²
m₂ = heavier mass = 8.5 kg
m₁ = lighter mass = 2.5 kg
Therefore,
a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)
a = (9.8 m/s²)(6 kg)/(11 kg)
a = 5.34 m/s²
The formula for tension in cable is derived to be:
T = 2m₁m₂g/(m₁ + m₂)
T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)
T = 37.86 N
A camera takes a picture that is the correct brightness and the correct zoom level, but the depth-of-focus is too small. One way to increase the depth-of-focus is to increase the f-number. Assuming that we will make changes that have the overall effect to:
1. increase the f-number, and
2. keep the brightness and the zoom level the same, which changes should we make to the aperture diameter and to the shutter time? (keep in mind we're talking about the time the shutter is open; we aren't talking about the shutter speed)
a. Increase the aperture diameter, decrease the shutter time
b. Decrease the aperture diameter, increase the shutter time
c. Increase both the aperture diameter as well as the shutter time
d. Decrease both the aperture diameter as well as the shutter time
A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 21.9 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.(a) Use approximate relationships to find thenet charge on the shell.
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from theaxis, measured radiallyoutward from the midpoint of the shell.
Answer:
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.
A 148 g ball is dropped from a tree 11.0 m above the ground. With what speed would it hit the ground
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy
[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]
[tex]V^2 - U^2 = 2gh[/tex]
[tex]V^2 - 0 = 2gh[/tex]
[tex]V = \sqrt{2 g h}[/tex]
[tex]= \sqrt{2\times9.8\times11}[/tex]
= 14.68m/s
Convert from standard form to scientific notation:
0.00000013
A)1.3 x 10-7
B)13 x 108
C)1.3 x 107
D)13 x 10-8
A skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stopA skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stop
Before the bus starts moving, the bus and the skater are both standing still.
When the bus starts moving and pulls away from the bus-stop, the skater stays right where she is.
The people outside on the sidewalk see her standing still, and they see the bus moving out from under her.
The other passengers on the bus see her rolling backwards down the aisle, toward the back of the bus.
To practice Problem-Solving Strategy 6.1: Circular motion A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?
Answer:
The curve should be banked at an angle of 13 degrees.
Explanation:
We have,
Radius of a highway curve is 274 m
Speed of car on this curve is 25 m/s
Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.
[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]
g = 10 m/s²
Plugging all the values in above formula,
[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]
So, the curve should be banked at an angle of 13 degrees.
John pushes Hector on a plastic toboggan.The free-body diagram is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = negative 490 N. The second vector is pointing right, labeled F Subscript t Baseline = 735 N. The third vector is pointing upward, labeled F Subscript N Baseline = 490 N. The fourth vector is pointing left, labeled F Subscript f Baseline = negative 245 N. The up and down vectors are the same length. The right vector is longer than the left vector. What is the net force acting on Hector and the toboggan?
Answer:
490 N
Explanation:
is the correct answer
If the up and down vectors are the same length. The right vector is longer than the left vector, then the net force acting on Hector and the toboggan would be 490 Newtons.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
As given in the problem John pushes Hector on a plastic toboggan .The free-body diagram is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = negative 490 N. The second vector is pointing right, labeled F Subscript t Baseline = 735 N. The third vector is pointing upward, labeled F Subscript N Baseline = 490 N. The fourth vector is pointing left, labeled F Subscript f Baseline = negative 245 N.
The net force acting on the vertical direction = 490-490
=0
The net force acting on the horizontal direction = 735 -245
=490 Newtons
Thus, the net force acting on Hector and the toboggan would be 490 Newtons.
Learn more about Newton's second law from here, refer to the link ;
brainly.com/question/13447525
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You have just landed on Planet X. You take out a ball of mass 100 gg , release it from rest from a height of 16.0 mm and measure that it takes a time of 2.90 ss to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X?
Answer:
0.173 N.
Explanation:
We will calculate the mass and then use the following calculations on the surface of planet X that is :
[tex]W=mg[/tex]
We would use the following equation to get the value of g for planet X that is :
[tex]y_f-y_i=v_{yi}t+\frac{1}{2}gt^2[/tex]
Then, put the values in the above equation.
[tex]16=0+\frac{1}{2}\times g\times(2.90)^2[/tex]
[tex]\bf\mathit{g=3.80\;m/s^2}[/tex]
Now, we will measure the ball weight on planet X's surface:
[tex]m=\frac{100}{1000} \;\;\;\;\;\;\;\;\;\;[1kg=1000g][/tex]
Then, we have to put the value in the above equation.
[tex]W=0.1\times 1.73=0.173\:N[/tex]
250cm3 of fres
er of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *
Answer:
1008.57kg/m3
Explanation:
Now the mass of fresh water is 250×1000 /1000000 = 0.25kg
Now the mass of salt water is
100×1030 /1000000 = 0.103kg
Note Density = mass / volume
Mass = volume × density
Note that converting from cm3 to m3 we divide by 1000000
Total mass = 0.25kg +0.103kg= 0.353kg.
Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3
Hence the density of the mixture= total mass / total volume
0.353kg/35 × 10^{-5}m3=1008.57kg/m3
A sulfur dioxide molecule has one sulfur
atom and two oxygen atoms. Which is its
correct chemical formula?
A. SO2
C. S2O2
B. (SO)
D. S20
Answer:
a. SO2
Explanation:
Two vectors having magnitudes of 5.00 and 9.00 respectively. If the value of their dot product is 12.0, find the angle between the two vectors.
Answer:
C = 74.53°
Explanation:
Let the magnitudes of 5.00 and 9.00 be vectors A and B respectively, hence the dot product of this vector is defined as
A.B = |A||B|cosC; let C be the angle between the vectors
12 = 5×9 cos C
Hence cos C = 12/45
C = cos^-1(12/45)
C = 74.53°
A 2 kg object is subjected to three forces that give it an acceleration −→a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = −(12.0N)ˆi + (8.00N)ˆj, find the third force.
Answer:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
Explanation:
You have three forces F1, F2 an F3 that produce the following acceleration:
a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj
you know that force F1 and F2 are:
F1 = (30.0N)ˆi + (16.0N)ˆj
F2 = −(12.0N)ˆi + (8.00N)ˆj
and the force F3 is unknown:
F3 = F3x ˆi + F3y ˆj
The second Newton law is given by the following equation:
[tex]\vec{F}=m\vec{a}[/tex]
F: the total force = F1 +F2 + F3
m: mass of the object = 2 kg
By the properties of vectors you have:
[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]
Both x and y component must be equal in the previous equality, then you have:
[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]
Hence, the vector F3 is:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?
Speed = (distance) / (time)
Speed = (
Velocity = speed, and its direction
The velocity of the plane is 10.2 miles per second East.
(about 48 times the speed of sound)
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal
Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is [tex]H = 318.5 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the plane takes off is [tex]u = 49 \ m/s[/tex]
The angle at which it takes off is [tex]\theta = 30 ^o[/tex]
The time taken is [tex]t = 13 s[/tex]
The vertical distance traveled is mathematically represented as
[tex]H = u sin \theta t[/tex]
Substituting values
[tex]H = (49) * sin (30) *13[/tex]
[tex]H = 318.5 \ m[/tex]