Answer:
surface
Explanation:
You need friction created by your tires and the road surface
to control your speed and direction.
10 POINTS!! SPACE QUESTION!
Answer:
The Gas Giants have more moons.
Explanation:
Mercury-0
Venus-0
Earth-1
Mars-2
Jupiter-66
Saturn-62
Uranus-27
Neptune-13
What happens when Earth rotates on its axis and how long does it take
Answer:
You get Day and Night
It takes 24 hour
Answer:
Explanation:
The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled?
Answer:
A boat travels for three hours with a... A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in still water?
Explanation:
While visiting Planet Physics, you toss a rock straight up at 15 m/s and catch it 2.7 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the planet's radius every 250 min .
Part A What is the mass of Planet Physics?
Part B What is the radius of Planet Physics?
Answer:
R = 7.915 10⁶ m, M = 1.04 10³⁵ kg
Explanation:
Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations
v = v₀ - g t
the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls
v = -v₀
-v₀ = v₀ - g t
g = 2v₀ / t
g = 2 15 / 2.7
g = 11.11 m / s²
I now write the law of universal gravitation and Newton's second law
F = m a
G m M / R² = m a
a = g
g = G M / R²
Now let's work with the cruiser in orbit
F = ma
acceleration is centripetal
a = v² / r
G m M / r² = m v² / r (1)
the distance from the center of the planet is
r = R + h
r = R + R = 2R
we substitute in 1
G M / 4R² = v² / 2R
G M / 2R = v²
The modulus of the velocity in a circular orbit is
v = d / T
the distance is that of the circle
d = 2π r
v = 2π 2R / T
v = 4π R / T
G M / 2R = 16pi² R² / T²
T² = 32 pi² R³ / GM
let's write the equations
g = G M / R² (2)
T² = 32 pi² R³ / GM
we have two equations and two unknowns, so it can be solved
let's clear the most on the planet and equalize
g R² / G = 32 pi² R³ / GT²
g T² = 32 pi² R
R = g T² / 32 pi²
let's reduce the period to SI units
T = 250 min (60 s / 1 min) = 1.5 104 s
let's calculate
R = 11.11 (1.5 10⁴) ² / 32 π²
R = 7.915 10⁶ m
from equation 2 we can find the mass of the planet
M = g R² / G
M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹
M = 1.04 10³⁵ kg
Determine the poles of the magnet. Look at the three compass readings that are on top of the magnet. Label the
end the compass points away from as "S" (south), and the other end that the compass points toward as "N" (north).
Record these poles in Figure 1.
Continue
Intro
Answer:
the red pointer on the magnet ( grey region) : points towards north
red pointer outside the magnet ( white region) is pointing towards south
Explanation:
please see the attached image
Saved Which of the following is NOT an important function of facial display? Multinio Choic
A. emotion
B. attractiveness
c. Primacy
d. identity
Answer:
C
Explanation:
Primacy means being first or important so thats not an important facial display as the others.
An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k = 450 N/m. The object is determined to have a velocity of 0.3 m/s when passing through the equilibrium.
1. Find the amplitude of the motion
2. Find the total energy of the object at any point of its motion
Answer:
1) The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object at any point of its motion is 16.892 joules.
Explanation:
1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ([tex]U_{e}[/tex]), in joules, is equal to maximum translational kinetic energy ([tex]K[/tex]), in joules:
[tex]U_{e} = K[/tex]
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]A[/tex] - Amplitude, in meters.
[tex]m[/tex] - Object mass, in kilograms.
[tex]v[/tex] - Speed of the object at equilibrium, in meters per second.
If we know that [tex]k = 450\,\frac{N}{m}[/tex], [tex]m = 0.25\,kg[/tex] and [tex]v = 0.3\,\frac{m}{s}[/tex], then the amplitude of the motion is:
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]k\cdot A^{2} = m\cdot v^{2}[/tex]
[tex]A = v\cdot \sqrt{\frac{m}{k} }[/tex]
[tex]A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }[/tex]
[tex]A \approx 0.274\,m[/tex]
The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object ([tex]E[/tex]), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ([tex]k = 450\,\frac{N}{m}[/tex], [tex]A \approx 0.274\,m[/tex])
[tex]E = U_{e}[/tex]
[tex]E = \frac{1}{2}\cdot k\cdot A^{2}[/tex]
[tex]E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}[/tex]
[tex]E = 16.892\,J[/tex]
The total energy of the object at any point of its motion is 16.892 joules.
state newton first law of motion
Newton’s first law of motion states that there must be a cause—which is a net external force—for there to be any change in velocity, either a change in magnitude or direction. An object sliding across a table or floor slows down due to the net force of friction acting on the object.
got it off g lol..
Answer:
it state that everybody in the universe is state that" universe continues its state of rest or uniform motion in a straight path unless it is acted upon by external force."
A car is moving with speed 30 m/s and acceleration 4 m/s2 at a given instant. (a) Using a second-degree Taylor polynomial, estimate how far the car moves in the next second.
Answer:
68 meters moved in the next seconds
Explanation:
Given
[tex]u= 30m/s[/tex]
[tex]a = 4m/s^2[/tex]
Required
Distance covered by the car in the next second
At a point in time t, the current distance is calculated as:
[tex]s(t) = ut + \frac{1}{2}at^2[/tex]
Substitute values for a and u in the above equation.
[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]
[tex]s(t) =30t + 2t^2[/tex]
Next, we generate the second degree Taylor polynomial as follows;
Calculate velocity (s'(t))
Differentiate s(t) to get velocity
[tex]s(t) =30t + 2t^2[/tex]
[tex]s'(t) =30 + 4t[/tex]
Calculate acceleration (s"(t))
Differentiate s'(t) to get acceleration
[tex]s'(t) =30 + 4t[/tex]
[tex]s"(t) =4[/tex]
When t = 0
We have:
[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]
[tex]s'(0) =30 + 4*0 = 30[/tex]
[tex]s"(0) = 4[/tex]
So, the second degree tailor series is:
[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]
To see the distance moved in the next second, we set t to 1
So, we have:
[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]
Solving s(1), s'(1) and s"(1)
We have:
[tex]s(1) =30*1 + 2*1^2 = 32[/tex]
[tex]s'(1) =30 + 4*1 = 34[/tex]
[tex]s"(1) =4[/tex]
Hence:
[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]
[tex]T_2(1) = 32 + 34 + 2[/tex]
[tex]T_2(1) = 68[/tex]
which statement can best describe the energy transformations that occur when a pendulum swings back and forth?
a. gravitational energy is converted to spring energy and back.
b. gravitational energy is converted to kinetic energy and back
c. kinetic energy is converted to spring energy and back
d. none of the above.
Answer:
I think it is C, I'm not for sure.
The density of 1 kilogram of gold is
Answer:
0.02 kg/cm³
Explanation:
1. Objects become electrically charged as a result of the transfer of
Answer:
Electron
Explanation:
An object can become electrically charged when it gains or loses an electron. Because an electron is negatively charged, when an object gains an electron it becomes negatively charged. Also, when it gives up an electron, it becomes positively charged. This positive charge is because the atom has one proton more than electron. In a neutral atom, the number of the proton is equal to the number of the electron. An electron is negatively charged, and a proton is positively charged.
two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg
and the incline is µ1 = 0.20 and that between the block of mass 4.0 kg and the
incline is µ2 = 0.30. Find the acceleration of 2.0 kg block. ( g = 10m/s^2).
Answer:
The acceleration of 2.0 kg block is 2.7 m/s²
Explanation:
Since, µ₁ < µ₂ acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:
Force down the plane on the system
= (4 + 2) g sin30°
= (6)(10)(½)
= 30N
Force up the plane on the system
= µ₁ (2)(g)cos30° + µ₂ (4)(g)cos30°
= (2µ₁ + 4µ₂) g cos30°
= (2 × 0.2 + 4 × 0.3)(10)(√3/2)
≈ 13.76 N
∴ Net force down the plane is F
F = 30 - 13.76
F = 16.24 N
∴Acceleration of both the blocks down the
plane will b a
a = F ÷ (4 + 2)
a = 16.24 ÷ 6
a = 2.7 m/s²
Thus, The acceleration of 2.0 kg block is
2.7 m/s²
-TheUnknownScientist
Explanation:
2.7m/s2
I hope its helpful
Which of the following best describes what occurs in a fission reaction?
A.
Two low mass nuclei are joined to form one nucleus.
B.
Electrons are shared between the nuclei.
C.
A single nucleus divides into two or more nuclei and gives off energy.
D.
A chemical reaction occurs between the nuclei.
Answer:
C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.
Answer:
C.
A single nucleus divides into two or more nuclei and gives off energy.
hope it is helpful to you
According to the Traditional Square of Opposition: If "All S are P" is true, then is "Some S are P" true or false?
[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the concept of Logic.
Since in the above statement it is given that,
All S are P ==> True,
then obviously Some S are also P always, hence it is true.
Answer is True.
How much work will a 500 watt motor do in 10 seconds?
Answer:
50j
Explanation:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Answer:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Explanation:
A radioactive material produces 1160 decays per minute at one time, and 4.0 h later produces 170 decays per minute. whats the half life
Answer:
Half life is 3.23 hours
Explanation:
Given
Decay rate at starting = 1160 decays per minute
Decay rate after 4 hours = 170 decays per minute
As we know know
[tex]N = N_0 *e ^{\Lambda *T}[/tex]
Substituting the given values, we get -
[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]
Also
[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]
Substituting the given values we get -
[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours
The unit of work done is called derived unit why
Drag each statement to the correct location on the table.
Match the characteristics with the states of matter.
does not have
a definite shape
or volume
has definite volume
but does not have a
definite shape
has a definite shape
and volume
changes to liquid
on heating
changes to liquid
on cooling
changes to solid
on cooling
Solid
Liquid
Gas
mentum. All rights reserved.
Answer:
Solid:
- has definite shape and volume.
- change to liquid on heating.
Liquid:
- has definite volume but does not have definite shape .
- changes to solid on cooling.
Gas :
- does not have definite shape or volume.
- change to liquid on cooling
An object is placed 12.0 cm from a thin diverging lens with a focal length of 4 cm. Which one of the
following statements is true concerning the image?
A. The image is virtual and 3.0 cm from the lens.
B. The image is real and 6.0 cm from the lens.
C. The image is virtual and 12 cm from the lens.
D. The image is real and 12 cm from the lens.
Answer:
soluble soluble soluble soluble
Explanation:
solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]
v₂ = 30 m/s
b)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]
m₁ = 12600 kg
c)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]
m₂ = 12600 kg
A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it
Answer:
w = 0.319 rad / s
Explanation:
This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.
initial instant. Before the squirrel jumps
L₀ = m v r
final instant. After the trough and the squirrel are together
L_f = (I_fetter + I_ardilla) w
angular momentum is conserved
L₀ = L_f
m v r = (I_fetter + I_ardilla) w
w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]
the moment inercial ofbody is
I_thed = 2.00 kg m²
We approach the squirrel to a specific mass
I_ardilla = m r²
we substitute
w = m v r / ( I_[feefer + m r²)
let's calculate
w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )
w = 0.6426 / 2.0119
w = 0.319 rad / s
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
The nucleus of a certain type of uranium atom contains
92 protons and 143 neutrons. What is the total charge of
the nucleus?
Answer:
charge = electrons + protons
=92+92
=184
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Answer:
[tex]523269.9\ \text{N/m}[/tex]
Explanation:
q = Charge
r = Distance
[tex]q_1=25\ \text{C}[/tex]
[tex]r_1=3000\ \text{m}[/tex]
[tex]q_2=40\ \text{C}[/tex]
[tex]r_2=850\ \text{m}[/tex]
The electric field is given by
[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]
The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]
While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.
Answer:
the smallest separation between two objects is 0.8067 m
Explanation:
Given the data in the question;
Altitude h = 5.75 km = 5750 m
Diameter D = 4.0 mm = 0.004 m
λ = 460 nm = 4.6 × 10⁻⁷ m
Now, Using Rayleigh criterion for Airy disks resolution.
we know that, Minimum angular separation for resolving two points is;
θ = 1.22λ / D
so we substitute
θ = (1.22 × 4.6 × 10⁻⁷) / 0.004
θ = 5.612 × 10⁻⁷ / 0.004
θ = 1.403 × 10⁻⁴ rad
so minimum separation [tex]d_{min[/tex] = θh
so we substitute
[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m
[tex]d_{min[/tex] = 0.8067 m
Therefore, the smallest separation between two objects is 0.8067 m
2(A + B)
15. The resultant of A and B is perpendicular to A
What is the angle between A and B?
(a) cos
(b) cos
La
(c) sin
(d) sin
Answer:
θ = cos^(-1) (-A/B)
Explanation:
The image of the reauktant forces A & B are missing, so i have attached it.
Now, from the attached image, we will see that;
Angle between A and B is θ
Also;
A = Bcos(180° − θ)
Now, in trigonometry, we know that;
cos(180° − θ) = -cosθ
Thus;
A = -Bcosθ
cosθ = -A/B
Thus;
θ = cos^(-1) (-A/B)
Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________
a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.
Answer:
d. the star is a member and also a part of an eclipsing binary star system.
Explanation:
If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).
The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.
Thus, option D is correct.
An object’s
✔ mass
will remain constant throughout the universe, but its
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
Answer:
mass, weight, strength of gravity increases, weight decreases
Explanation:
got it on edge
Answer:
An object’s
✔ mass
will remain constant throughout the universe, but its
✔ weight
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
✔ strength of gravity increases
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
✔ weight decreases
Explanation:
right on edge 22
In Part 5.2.3 of the experiment, you will measure the index of refraction of yellow light using Lab Manual Equation 5.2. Suppose the minimum angle of deviation is 18 degrees. What is the index of refraction
Answer:
The answer is "1.26".
Explanation:
[tex]D=18^{\circ}[/tex]
The refractive index is:
[tex]\to \mu=2\sin(30^{\circ}+\frac{D}{2})\\\\[/tex]
[tex]=2\sin(30^{\circ}+\frac{18^{\circ}}{2})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(39^{\circ})\\\\=2 \times 0.63\\\\=1.26[/tex]