If a rock is dropped from the top of a tower at the front of and it takes 3.6 seconds to hit the ground. Calculate the final velocity of the penny in m/s.
Answer:
36 m/s
Explanation:
t = 3.6s
u = 0m/s
a = +g = 10m/s²
v = ?
using,
v = u + at
v = 0 + 10(3.6)
v = 36 m/s
(A star if you answer this question) A school bus is traveling at 11.1 m/s and has a
momentum of 152,625 kgm/s. What is the mass of the bus?
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Kinematics in real world.
So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,
P = m * V
momentum = mass * velocity
here, P= 152625 kgm/s and v= 11.1 m/s
so substituting we get as,
m = 152625 ÷ 11.1 => 13,750 kg
hence,the mass of the bus is 13,750 kg.
A good baseball pitcher can throw a baseball toward home plate at 97 mi/h with a spin of 1540 rev/min. How many revolutions does the baseball make on its way to home plate
Answer:
10778292789403987593790
Explanation:
I am a Cow'
a bus with a mass of 5000kg is uniformly acceleration from rest. The net force acting on the bus is
Answer:
50,000N
Explanation:
According to Newton's second law of motion;
Net Force = Mass * acceleration
Given
Mass = 5000kg
Let the acceleration = 10m/s²
Net force = 5000 * 10
Net force = 50,000N
Hence the net force acting on the bus is 50000N
A uniform metal meter-stick is balanced with a 1.0 kg rock attached to the left end of the stick. If the support is located 0.25 m from the left end, what is the mass of the meter-stick? (Hint: meter-stick is 1 m long). a) 0.6 kg b) 2.0 kg c) 1.0 kg d) 3.0 kg
Answer:
c) 1.0 kg
Explanation:
The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support
According to law of moment
Sum of clockwise moment = sum of anti clockwise moments
Clockwise moment = M×0.25(mass of metre rule is M)
CW moment = 0.25M
Anti clockwise moment = 0.25×1
ACW moments = 0.25kgm
Equate;
0.25M = 0.25
M = 0.25/0.25
M = 1.0kg
Hence the mass of the metre rule is 1.0kg
I went for a walk one day. I walked north 6.0 km at 6.0 km/h and then west 10 km at 5.0 km/hr.
Determine the total distance of the entire trip
Answer:
The total distance of the entire trip is 16 Km
Explanation:
Distance and Displacement
A moving object constantly travels for some distance at certain periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:
dtotal = d1 + d2 + d3 + ... + dn
This sum is obtained independently of the direction the object moves.
The displacement only takes into consideration the initial and final positions of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.
The described movements are 6 Km North and 10 Km West. Regardless of the speed or time taken, the total distance is:
d = 6 Km + 10 Km = 16 Km
The total distance of the entire trip is 16 Km
At what distance on the axis of a current loop is the magneticfield half the strength of the field at the center of the loop? Give your answer as a multiple of R. z=?R
Answer:
x = 1.26 R
Explanation:
For this exercise let's find the magnetic field using the Biot-Savart law
B = μ₀ I/4π ∫ ds x r^ / r²
In the case of a loop or loop, the quantity ds is perpendicular to the distance r, therefore the vector product reduces to the algebraic product and the direction of the field is perpendicular to the current loop
suppose that the spiral eta in the yz plane, therefore the axis is in the x axis
B = μ₀ I/4π ∫ ds / (R² + x²)
The total magnetic field has two components, one parallel to the x axis and another perpendicular, this component is annual when integrating the entire loop, so the total field is
B = Bₓ i^
using trigonometry
Bₓ = B cos θ
we substitute
Bₓ = μ₀ I/4π ∫ ds cos θ / (x² + R²)
the cosine function is
cos θ = R /√(x² + R²)
The differential is
ds = R dθ
we substitute
Bₓ = μ₀ I/4π ∫ (R dθ) R /√( (x² + R²)³ )
we integrate from 0 to 2π
Bₓ =μ₀ I/4π R² / √(x² + R²)³ 2pi
therefore the final expression is
B = μ₀ I R²/ 2√(x² + R²)³ i^
In our case the distance is requested where B is half of B in the center of the bone loop x = 0
Spire center field x=0
B₀ = μ₀ I/2R
Field at the desired point (x)
B = B₀ / 2
we substitute
R² /√(x² + R²)³ = ½ 1 /R
2R³ =√(x² + R²)³
(x² + R²)³ = 4 (R²)³
(x²/R² + 1)³ = 4
The exact result is the solution of this equation, but it is quite laborious, we can find an approximate result assuming that the distance x is much greater than R (x »R)
B = μ₀ I/2x³
we substitute
R² / x³ = 1/2 1 / R
2R³ = x³
x = ∛2 R
x = 1.2599 R
The distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of R is 0.766 R
Suppose we consider a magnetic field located at point z on the axis of the current loop with radius R carrying a current (I), then the magnetic field can be represented as:
[tex]\mathbf{B = \dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{^{\dfrac{3}{2}}}}}[/tex]
And the field situated at the center of the loop is:
[tex]\mathbf{B_{z=0} =\dfrac{\mu_o}{2} \dfrac{I}{R} }[/tex]
Let consider a distance (z) on the axis of the loop, in which the magnetic field as a result of the loop is equal to half the strength of the magnetic field at the center of the loop;
Then;
[tex]\mathbf{B(z) = \dfrac{1}{2}B_{z=0}}[/tex]
[tex]\mathbf{\dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{\frac{3}{2}}} =\dfrac{1}{2} \Big (\dfrac{\mu_o}{2} \dfrac{I}{R} \Big )}[/tex]
Multiply both sides by (2);
[tex]\mathbf{\dfrac{R^2}{(z^2+R^2)^{\frac{3}{2}}} = \dfrac{1}{2R}}[/tex]
Cross multiply;
[tex]\mathbf{2R^3 = (z^2 +R^2)^{\dfrac{3}{2}}}[/tex]
[tex]\mathbf{4R^6 = (z^2 +R^2)^3}[/tex]
[tex]\mathbf{z = \sqrt{4^{1/3} R^2 -R^2 }}[/tex]
[tex]\mathbf{z = R\sqrt{4^{1/3} -1 }}[/tex]
[tex]\mathbf{z = R\sqrt{0.587401052 }}[/tex]z = 0.766 R
Therefore, we can conclude that the distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of R is 0.766 R
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explain approach in volleyball
Answer:
This refers when a spiker quickly strides towards the net before they jump in the air for an attack.
How long will it take an object to accelerate from a velocity of 4.8 m/s to a velocity of 15.6 m/s if it is accelerating at a rate of 0.9 m/s2
Initial velocity of the object (u) = 4.8 m/s
Final velocity of the object (v) = 15.6 m/s
Acceleration of the object (a) = 0.9 m/s²
By using equation of motion, we get:
[tex] \bf \longrightarrow v = u + at \\ \\ \rm \longrightarrow 15.6 = 4.8 + 0.9t \\ \\ \rm \longrightarrow 15.6 - 4.8 = 4.8 - 4.8 + 0.9t \\ \\ \rm \longrightarrow 10.8 = 0.9t \\ \\ \rm \longrightarrow 0.9t = 10.8 \\ \\ \rm \longrightarrow \dfrac{0.9}{0.9} t = \dfrac{10.8}{0.9} \\ \\ \rm \longrightarrow t = 12 \: s[/tex]
[tex] \therefore [/tex] Time taken by the object (t) = 12 s
1. Black Panther ran 567.5 ks to the east in 2.3 hours then tumed around and ran 2218 km to
the west in 1.2 hours. What was his average velocity during the trip in km/h?
Answer:Average Velocity of the trip=98.77km/h
Explanation:
Given that:
Black Panther started his journey with
Distance ran towards east = 567.5km
Time ran towards east = 2.3 hours
and then turned back,
Distance ran towards west, = 2218 km
Time taken towards west, = 1.2 hours
Let us treat the westward movement as negative movement of the east
So total displacement =567.5 km east + (-2218km east )= 1,650.5km
But it seems the value of the distance ran west is wrong , because i do not think he can run 2218km under 1.2hours considering he ran 567.5m due east at 2.3hours .
So Let we use 221.8 km as distance ran due west.
So that Total Displacement becomes = 567.5 km east + (-221.8km east )= 345.7km
Total time = 2.3 + 1.2 = 3.5hrs.
Average velocity = Total displacement / total time
.345.7km / 3.5hr = 98.77km/h
a scuba diver in a pool looks at his instructor. the angle between the ray in the water and the perpendicular to the water is 25. find the height of the instructors head
Answer:
y= y' 0.67
Explanation:
This is an exercise in refraction of light,
n₁ sin θ₁ = n₂ sin θ₂
where subscript 1 is used for the incident medium and subscript 2 for the refracted medium
sin θ₁ = n2 /n1 sin θ₂
the incident medium is air with refractive index n1 = 1 and the medium where the ray is refracted is water with n = 1.33
let's calculate
sin θ₁ = 1.33 / 1 sin 25
θ₁ = sin⁻¹ (0.56208)
θ₁ = 34.2º
when the ray is refracted we can assume that the adjacent leg (water surface) is the same for the two media
let's use the trigonometry relationship
tan θ₁ = x / y
tan θ₂ = x / y '
tan θ₁ = y’ tan θ₂
y = y ’ tanθ₂ / tan θ₁
to finish exercise you must know the depth of the object
y =y' tan 25/ tan 34.2
y= y' 0.67
The angle of incidence is the angle between the the normal of the surface. and A. point of incidence B. reflected ray C. incident ray D. refracted ray
Answer: c. incident ray
Explanation:
A cable that weighs 4 lb/ft is used to lift 1000 lb of coal up a mine shaft 700 ft deep. Find the work done.
Answer:
980000ft-lbsExplanation:
Step one:
given data
mass of cable= 4lb/ft
mass of coal= 1000lb
dept of mine= 700ft
Step two:
Required
the work-done to lift the coal and the rope combined
Work-done to lift coal
Wc=1000*700= 700,000 lb-ft
Work-done to lift rope
[tex]Wr=\int\limits^{700} _0 {4(700-y)} \, dx \\\\Wr=4(700y-\frac{1}{2}y^2 )\limits^{700}_0[/tex]
substitute y=700 we have, since y=0 will result to 0
[tex]Wr=4(700*700-\frac{1}{2}*700^2 )\\\\Wr=4(490000-245000)\\\\Wr=4(245000)\\\\Wr=980000ft-lbs[/tex]
An echo is sound that returns to you after being reflected from a distant surface (e.g., the side of a cliff). Assuming that the distances involved are the same, an echo under water and an echo in air return to you _____________________
a. at different times, the echo under water returning more slowly.
b. at different times, the echo under water returning more quickly.
c. at the same time
Answer:
The correct answer is B
An echo underwater and an echo in the air will return at different times. The echo underwater will return more quickly than the echo in the air.
Explanation:
The physics of this is simple.
Water and air are both made up of particles. The particles for water are more closely or densely arranged that those of the air molecules. Hence sound travels faster in water than in air. When measured, the speed actually differs by as much as 5 times with water being the fastest medium.
Think of it like this. Assume you have two stacks of dominoes, one closely packed than the other but exactly the same amount of dominos, you'd notice that the stack that is more tightly arranged will be the first to topple over because it takes less time for the kinetic energy from the first domino to reach the next and on and on like that until the last domino.
Cheers
Answer:
b. at different times, the echo under water returning more quickly.
An electron and a proton have charges of an equal magnitude but opposite sign of 1.60 x 10^-19 C. If the electron and proton in a hydrogen atom are separated by a distance of 4.20 x10^-11 m, what are the magnitude and direction of the electrostatic force exerted on the electron by the proton?
Answer:
i. F = 1.3 x [tex]10^{-7}[/tex] N
ii. The direction of the force of attraction exerted by the proton on the electron is towards the itself (i.e a pull).
Explanation:
Since the given charges are opposite, then the force of attraction is experienced. The force of attraction between the two charges can be determined by:
F = [tex]\frac{kq_{1} q_{2} }{d^{2} }[/tex]
where F is the force, k is the constant, [tex]q_{1}[/tex] is the charge of the electron, [tex]q_{2}[/tex] is the charge on the proton, and d is the distance between them.
So that; k = 9.0 x [tex]10^{9}[/tex] N[tex]m^{2}[/tex][tex]C^{-2}[/tex] , [tex]q_{1}[/tex] = 1.6 x [tex]10^{-19}[/tex] C, [tex]q_{2}[/tex] = 1.6 x
Thus,
F = [tex]\frac{9.0*10^{9}*1.6*10^{-19}*1.6*10^{-19} }{(4.2*10^{-11}) ^{2} }[/tex]
= [tex]\frac{2.304*10^{-28} }{1.764*10^{-21} }[/tex]
= 1.3061 x [tex]10^{-7}[/tex]
F = 1.3 x [tex]10^{-7}[/tex] N
The force between the charges is 1.3 x [tex]10^{-7}[/tex] N.
ii. The direction of the force of attraction exerted by the proton on the electron is towards the itself.
Calculating the Velocity of a Wave Quick Check What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest whole number. m/s I need answers
Answer:
448m/s
Explanation:
The speed of the wave with a given wavelength and frequency can be calculated using the formula:
λ = v/f
Where;
λ = wavelength (m)
v = speed of wave (m/s)
f = frequency of wave (Hz)
In this question, λ = 2.59 metres, f = 173 Hz, v = ?
λ = v/f
v = λ × f
v = 2.59 × 173
v = 448.07m/s
To the nearest whole number, 448.07 can be written as 448
Hence, the speed of the wave (v) is 448m/s.
Answer:
What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest whole number.
448
m/s
Explanation:
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m^2. What is the maximum value of the magnetic field in the wave?
a. 77.9 nT
b. 55.1 nT
c. 38.9 nT
d. 108 nT
Answer:
a. 77.9 nT
Explanation:
Given;
average value of the electromagnetic wave, E = 0.724 W/m^2
The average value of a Poynting vector is given by;
[tex]S = \frac{1}{2\mu_o} *E_oB_o\\\\But, E_o = cB_o\\\\S = \frac{1}{2\mu_o} *cB_o^2\\\\cB_o ^2= 2\mu_o S\\\\B_o^2 = \frac{2\mu_o S}{c}\\\\B_o = \sqrt{ \frac{2\mu_o S}{c}} \\\\B_o = \sqrt{ \frac{2(4\pi*10^{-7}) (0.724)}{3*10^8}}\\\\B_o = 7.79*10^{-8} \ T\\\\B_o = 77.9*10^{-9} \ T\\\\B_o = 77.9 \ nT[/tex]
Therefore, the maximum value of the magnetic field in the wave is 77.9 nT.
Answer:
A I think : )
Explanation:
A Ferris wheel with 60 spokes has a diameter of 100 m. It makes one rotation every 60 seconds. Find the speed of the passengers when the Ferris wheel is rotating at this rate.
A. 5.24 m/s
B. 6.23 m/s
C. 7.45 m/s
D. 8.01 m/s
Answer:
v = 5.24 m/s
Explanation:
Given that,
No of spokes of a Ferris wheel = 60
The diameter of a wheel, d = 100 m
Radius, r = 50 m
It makes one rotation every 60 seconds.
We need to find the speed of the passengers when the Ferris wheel is rotating at this rate. Let it is v. It can be calculated as follows :
[tex]v=r\omega\\\\=\dfrac{2\pi r }{T}\\\\=\dfrac{2\pi \times 50 }{60}\\\\v=5.235\ m/s[/tex]
or
v = 5.24 m/s
So, the speed of the passengers is 5.24 m/s. Hence, the correct option is (A).
. A crane does 62,500 joules of work to lift a boulder a distance of 25.0 meters. How much did the boulder
weigh? (Hint: The weight of an object is considered to be a force in units of newtons.)
13
Answer:
2500 NExplanation:
The weight of the object can be found by using the formula
[tex]f = \frac{w}{d} \\ [/tex]
w is the workdone
d is the distance
From the question we have
[tex]f = \frac{62500}{25} \\ [/tex]
We have the final answer as
2500 NHope this helps you
What is the main reason why the age of the oldest rocks can vary from one part of the continent to another?
Answer:
The options are
A. Older rocks are commonly remitted over huge regions
B. Older rocks have been uplifted and eroded away
C. Large parts of the continent are subducted deep within the mantle
D. Parts of the continent have been added by the accretion of tectonic terraces.
The answer is D. Parts of the continent have been added by the accretion of tectonic terraces.
The major reason why the age of the oldest rocks can vary from one part of the continent to another is that parts of continent have been added by the accretion of tectonic terraces.
What is Solar Energy used for? Use in your own words.
Answer:
TO POWER ELECTRIC GADGETS AND SAVE EARTH FROM POLLUTION
Explanation:
What kind of acceleration occurs when an object speeds up? positive negative neutral zero
Answer:
POSITIVE
Explanation:
The acceleration of an object is given by the rate of change of velocity.
u and v are initial and final velocities
If the final velocity is more than that of the initial velocity, the acceleration of the object is positive. It means positive acceleration occurs when an object speeds up. Hence, the correct option is (a) "positive".
Answer:
Positive
Explanation:
Calculate the force of Earth's gravity on a spacecraft 2.00 Earth radii above the Earth's surface if its mass is 1650 kg .
Answer:
F = 1.8 KN
Explanation:
Given that the spacecraft is 2.00 Earth radii above the Earth's surface, the force of Earth's gravity on it can be determined by;
F = [tex]\frac{GMm}{(R + h)^{2} }[/tex]
Where: F is the force, G is the universal gravitation constant, M is the mass of the earth, m is the mass of the spacecraft, and R is the radius of the earth and h is the distance of the spacecraft to the surface of the earth.
G = 6.67 x [tex]10^{-11}[/tex] N[tex]m^{2}[/tex][tex]kg^{-2}[/tex], M = 5.972 x [tex]10^{24}[/tex] kg, m = 1650 kg, R = 6371 km, h = (2.0 x 6371 km) = 12742 km.
Thus,
F = [tex]\frac{6.67*10^{-11}*5.972*10^{24} *1650 }{(6371*10^{3}+12742*10^{3}) ^{2} }[/tex]
= [tex]\frac{6.5725*10^{17} }{(19113000)^{2} }[/tex]
= 1799.17
F = 1.8 KN
The force of the Earth's gravity on the spacecraft is 1.8 KN.
two object, one of mass m and the other of mass 2m, are dropped from the top of a building. if there is no air resistance, when they hit the ground
Answer:
They hit the ground at the same time
Explanation:
Mass doesn't matter in free fall. As long as they were released with the same velocity and the same height, they'll hit the ground at the same time
A graph of the carbon dioxide levels in the Earth's atmosphere is shown below. Which of the following is true? A. The level of carbon dioxide in the atmosphere did not change before 1900. B. Carbon dioxide levels in the atmosphere have been dropping steadily since 1750. C. The steepest increase in carbon dioxide levels took place in the 18th century. D. The carbon dioxide levels in the atmosphere began increasing rapidly around 1950.
Answer:
D. The carbon dioxide levels in the atmosphere began increasing rapidly around 1950.
Explanation:
Study Island
The carbon dioxide levels in the atmosphere began increasing rapidly around 1950. Therefore, the correct option is option D.
What is carbon dioxide?One carbon atom is covalently doubly connected to the two oxygen atoms in each of the molecules that make up carbon dioxide. At room temperature, it exists as a gas.
In the atmosphere, carbon dioxide serves as a greenhouse gas because it absorbs infrared radiation despite being transparent to visible light. It has increased from which was before values of 280 ppm to be a gas in the atmosphere in the Stratosphere at 431 parts per thousand, or roughly 0.04% by volume. The carbon dioxide levels in the atmosphere began increasing rapidly around 1950. This statement is true.
Therefore, the correct option is option D.
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Two kids are roller skating. Amy, with a mass of 55 kg, is traveling forward at 3 m/s. Jenny, who has a mass of 40 kg, is traveling in the opposite direction at 5 m/s. They crash into each other and hold onto each other so that they move as one mass. How fast are they traveling?
Answer:
-7/19
Explanation:
A block slides down an inclined plane. Which force does zero work?
a. normal force
b. friction
c. the weight
d. all of the above
e. none of the above
Answer:
e
Explanation:
cause you have any attachments are not sure if you can get it
When a block slides down an inclined plane, normal force does zero work. Hence, option (a) is correct.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity.
An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.
As normal force always acts perpendicular to the motion, this force does zero work.
But friction force the weight acts in opposite direction of the motion, this force does work against the motion.
In an inclined plane, the parallel component of weight does work along the motion.
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A 25 gram bullet travels at 240 m/s, penetrates into the wooden block and move together with the wooden block with a velocity of 27m/s. What is the mass of the wooden block?
Answer:
m2=0.2kg approx.Explanation:
Step one:
given data
mass of bullet m1=25g= 0.025kg
initial velocity of bullet u1=240m/s
mass of block m2=?
initial velocity of bullet u2 of block=0m/s
final velocity of the system after impact v= 27m/s
Required; the mass of the block m2
Step two:
applying the conservation of linear moment/elastic collision we have
m1u1+m2u2=v(m1+m2)
substituting we have
0.025*240+m2*0=27(0.025+m2)
6=0.675+27m2
6-0.675=27m2
5.325=27m2
divide both sides by 27
m2=5.325/27
m2=0.197
m2=0.2kg approx.
A bus is moving with
a velocity of lom/s. After 2 sec its velocity
becames 20/s find the distance from travelled
by a bus at that this time of interval
Answer:
30 m
__________________________________________________________
Explanation:
We are given:
Initial velocity (u) = 10 m/s
Final velocity (v) = 20 m/s
Time interval (t) = 2 seconds
Distance travelled by the bus (s) = s meters
Solving for the distance travelled:
Solving for the acceleration:
v = u + at [first equation of motion]
20 = 10 + a(2) [replacing the given values]
2a = 10 [subtracting 10 from both sides]
a = 5 m/s² [dividing both sides by 2]
Solving for the distance:
s = ut + 1/2 (at²) [second equation of motion]
s = 10(2) + 1/2(5)(2)² [replacing the given values]
s = 20 + 10
s = 30 m
Therefore, the bus travelled 30 m in the given time interval
Given F1: a force of magnitude 6 N at an angle of 30°
F2: a force of magnitude 8 N at an angle of 50°C
a. Find F1+ F2 analytically (using equations instead of graphing) and write it in the form Fr1i + Fr2 j
b. Find the magnitude FR and θ_resultant
Answer:
13.8 N
[tex]41.44^{\circ}[/tex]
Explanation:
[tex]F_1=6\ \text{N}[/tex]
[tex]F_2=8\ \text{N}[/tex]
[tex]F_1\cos\theta_1\hat{i}+F_1\sin\theta_1\hat{j}\\ =6\cos30^{\circ}+6\sin30^{\circ}\hat{j}\\ =5.2\hat{i}+3\hat{j}[/tex]
[tex]F_2\cos\theta_2\hat{i}+F_2\sin\theta_2\hat{j}\\ =8\cos50^{\circ}+8\sin50^{\circ}\hat{j}\\ =5.14\hat{i}+6.13\hat{j}[/tex]
[tex]F_R=F_1+F_2=10.34\hat{i}+9.13\hat{j}[/tex]
[tex]|F_R|=\sqrt{10.34^2+9.13^2}=13.8\ \text{N}[/tex]
The magnitude of the resultant is 13.8 N
Direction is given by
[tex]\tan^{-1}=\dfrac{y}{x}=\tan^{-1}\dfrac{9.13}{10.34}=41.44^{\circ}[/tex]
The angle of the resultant is [tex]41.44^{\circ}[/tex]