Answer:
567.126 x 10⁻⁶ N
[tex]5.6 x 10^{-6} N[/tex]
Explanation:
Thinking process:
[tex]13pC = 13 x 10^{-12} C[/tex]
The electric field is given by E = 15000 N/C
Electric force on the charge
= charge x electric field
= 13 x 10⁻¹² x 15000
= 195 x 10⁻⁹ N.
The force acts in upward direction as force on positive charge acts in the direction in which electric field exists.
Volume of droplet = 4/3 π R³
R = 2.4 X 10⁻³ m
Volume V = 4/3 x 3.14 x ( 2.4 x 10⁻³)³
= 57.87 x 10⁻⁹ m³
density of water = 1000 kg / m³
mass of water droplet = density x volume
= 1000 x 57. 87 x 10⁻⁹ kg
= 57.87 x 10⁻⁶ kg .
Weight = mass x g
= 57.87 x 10⁻⁶ x 9.8
= 567.126 x 10⁻⁶ N.
Therefore, the weight is more than the electric force.
A 20-lb force acts to the west while an 80-lb force acts 45° east of north. The magnitu 80 lb 70 lb 067 lb 100 lb
Answer:
option C
Explanation:
given,
force act on west = 20 lb
force act at 45° east of north = 80 lb
magnitude of force = ?
∑ F y = 80 cos 45⁰
F y = 56.57 lb
magnitude of forces in x- direction
∑ F x = -20 + 80 sin 45⁰
= 36.57 lb
net force
F = [tex]\sqrt{F_x^2+F_y^2}[/tex]
F = [tex]\sqrt{56.57^2+36.57^2}[/tex]
F = 67.36 lb≅ 67 lb
hence, the correct answer is option C
Which one of the following situations is NOT Possible? A body has zero velocity and non-zero acceleration. A body travels with a northward velocity and a southward acceleration. A body travels with a northward velocity and a northward acceleration. A body travels with a constant velocity and a constant non-zero acceleration. A body travels with a constant acceleration and a time-varying velocity.
Answer:
Explanation:
A body has zero velocity and non- zero acceleration. It is possible as at the time when a body thrown upwards is at the top of the height. In that case velocity is zero but acceleration is equal to g.
A body travels with northward velocity but southward acceleration . it is also possible as in case when a body is going with northward velocity but when break is applied acceleration becomes southward.
A body travels with northward velocity and northward acceleration . it is also possible as in case when a body is going with northward velocity and when accelerator is applied . Acceleration becomes northward.
A body travels with constant velocity and a constant non zero acceleration . It is not possible . When there is acceleration , there must be a a change in velocity either in terms of magnitude or direction or both.
A body travels with a constant acceleration and a time varying velocity. It is possible. Time varying velocity represents acceleration.
In physics, a constant non-zero acceleration would necessitate a change in velocity over time. Therefore, it's impossible for a body to have a constant velocity and a constant non-zero acceleration at the same time.
Explanation:The situation that is NOT possible is a body traveling with a constant velocity and a constant non-zero acceleration.
Acceleration is the rate at which an object's velocity changes. If the velocity of the body remains constant, it means there is no change in its speed or direction. Therefore, it is not possible for the body to have a non-zero acceleration while traveling with a constant velocity.
On the other hand, scenarios with zero velocity and non-zero acceleration or a body traveling with a northward velocity and a northward acceleration are possible.
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A woman is reported to have fallen 144 ft from the 17th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18.0 in. She suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventilator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box.
Answer:
(a) 96 ft/s
(b) - 3072 ft/s^2
(c) 0.03125 s
Explanation:
h = 144 ft
u = 0 ft/s
g = 32 ft/s^2
(a) let she strikes the box with velocity v.
Use third equation of motion
[tex]v^{2}=u^{2}+2gh[/tex]
[tex]v^{2}=0^{2}+2\times 32 \times 144[/tex]
v = 96 ft/s
(b) Let the average acceleration is a.
initial velocity, u = 96 ft/s
final velocity, v = 0
h = 18 in = 1.5 ft
Use third equation of motion
[tex]v^{2}=u^{2}+2ah[/tex]
[tex]0^{2}=96^{2}+2\times a \times 1.5[/tex]
a = - 3072 ft/s^2
(c) Let the time taken is t.
Use first equation of motion
v = u + at
0 = 96 - 3072 x t
t = 0.03125 second
Using basic equations of motion, we calculated the speed of the woman just before impact with the ventilator as 29.3 m/s, her average acceleration during the impact as 930 m/s², and the time it took to crush the box as approximately 0.0315 seconds.
Explanation:The subject of this question is physics, specifically dealing with concepts of kinematics and mechanics. The falling woman problem requires application of the laws of free fall in physics along with some basic algebraic manipulation, and is quite suitable for a high school student.
(a) The speed of the woman just before she collided with the ventilator can be calculated using the formula v=√(2g h) where g is the gravity (9.8 m/s²) and h is the height (144 ft, which is approximately 43.9 m). Plugging in these values gives us v=√(2*9.8*43.9) ≈ 29.3 m/s.
(b) Knowing the depth the box was crushed (converted to meters) we can use the third equation of motion v² = u² + 2a s (s is displacement) we get a = (v² - u²) / 2s = (29.3 m/s)² / 2*0.46m = 930 m/s² as the average acceleration.
(c) Finally, using v = u + at, as initial speed is 0, we get time t = v/a = 29.3m/s / 930m/s² ≈ 0.0315 seconds.
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Two forces act on a 6.00- kg object. One of the forces is
10.0N. If the object accelerates at 2.00 m/s 2, what is the
greatestpossible magnitue of the other force?
Answer:
Fx = 22N
Explanation:
There are 2 possible scenarios for this problem:
1.- The 10N force is in the same direction of the acceleration. In this case the other force would be:
[tex]F1 - Fx = m*a[/tex] where F1 = 10N, m=6kg, a = 2m/s2
[tex]Fx = F1 - m*a = -2N[/tex] The negative result tells us that this is not possible.
2.- The 10N force is in the opposite direction of the acceleration. In this case the other force would be:
[tex]Fx - F1 = m*a[/tex] [tex]Fx = m*a + F1 = 22N[/tex]
The greatest magnitude of the other force, by which the object accelerates under the application of two forces is 22 N.
What is Newton’s second law of motion?Newton’s second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.
It can be given as,
[tex]\sum F=ma[/tex]
Here, (m) is the mass of the body and (a) is the acceleration.
One of the forces is 10.0N. . Let suppose the magnitude of the other force is [tex]F_x[/tex],
The maximum force we get when both the forces acting on the opposite direction. For this system the summation of force will be,
[tex]\sum F=F_x+(-10)\\\sum F=F_x-10[/tex]
The mass of the object is 6 kg. As the body is accelerating in at 2 m/s. Thus, plug in the values in the above formula as,
[tex]F_x-10=2\times6\\F_x=22\rm N[/tex]
Thus the greatest magnitude of the other force, by which the object accelerates under the application of two forces is 22 N.
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If a proton were traveling the same speed as electrons in a TV picture tube (about 7.1 x 10^7 m/s), what would its de Broglie wavelength be? The mass of a proton is 1.67 x 10^-27 kg. Answer must be in m.
Answer:
The wavelength is [tex]5.591\times10^{-15}\ m[/tex]
Explanation:
Given that,
Speed [tex]v= 7.1\times10^{7}\ m/s[/tex]
Mass of proton [tex]m= 1.67\times10^{-27}\ kg[/tex]
We need to calculate the wavelength
Using formula of de Broglie wavelength
[tex]p=\dfrac{h}{\lambda}[/tex]
[tex]\lambda=\dfrac{h}{p}[/tex]
[tex]\lambda=\dfrac{h}{mv}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{6.63\times10^{-34}}{1.67\times10^{-27}\times7.1\times10^{7}}[/tex]
[tex]\lambda=5.591\times10^{-15}\ m[/tex]
Hence, The wavelength is [tex]5.591\times10^{-15}\ m[/tex]
Final answer:
The de Broglie wavelength of a proton traveling at the same speed as electrons in a TV picture tube can be calculated by using the equation λ = h / p, where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the proton. The momentum of the proton can be calculated by multiplying its mass by its velocity. Once the momentum is found, the de Broglie wavelength can be calculated. In this case, the de Broglie wavelength of the proton would be 5.61 * 10^-15 m.
Explanation:
The de Broglie wavelength (λ) of a particle is given by the equation:
λ = h / p
where h is the Planck's constant
(6.62607015*10^-34 Js) and p is the momentum of the particle. To find the de Broglie wavelength of a proton traveling at the same speed as electrons in a TV picture tube, we need to first calculate the momentum of the proton. The momentum (p) is given by
p = mass * velocity
Plugging in the values, we get:
p = (1.67 * 10^-27 kg) * (7.1 * 10^7 m/s) = 1.18 * 10^-19 kg m/s
Now we can calculate the de Broglie wavelength:
λ = (6.62607015 * 10^-34 Js) / (1.18 * 10^-19 kg m/s) = 5.61 * 10^-15 m
Therefore, the de Broglie wavelength of the proton would be 5.61 * 10^-15 m.
An uncharged metal sphere hangs from a nylon thread. When a positively charged glass rod is brought close to the metalsphere, the sphere is drawntoward the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain why the sphere is first attracted and then repelled.
When a charged rod is brought near a neutral metal sphere, the opposite charges are attracted towards the rod, causing an initial attraction. However, when the sphere touches the rod, the charges redistribute, leading to like charges repelling each other with a stronger force than the attraction between opposite charges, resulting in the sphere being repelled.
Explanation:When a charged rod is brought near a neutral substance, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction.
Thus, when a positively charged glass rod is brought close to the neutral metal sphere, the opposite charges in the metal sphere are attracted towards the rod. However, when the sphere touches the rod, it suddenly flies away from the rod. This is because the charges in the metal sphere redistribute, and now the like charges repel each other with a stronger force compared to the attraction between the opposite charges.
This repulsion causes the metal sphere to be repelled away from the rod, creating a net repulsive force.
The metal sphere is first attracted to the positively charged glass rod due to induction, and then repelled after touching the rod due to the repulsion between like charges.
When the positively charged glass rod is brought close to the uncharged metal sphere, the sphere experiences electrostatic induction. The electrons in the metal sphere are repelled by the positive charges on the rod and move to the far side of the sphere, leaving the side closest to the rod with a net positive charge. This positive side of the sphere is attracted to the negative charges in the rod (or the electrons in the rod are attracted to the positive side of the sphere), causing the sphere to be drawn toward the rod.
As the sphere gets closer to the rod, the attraction increases until they make contact. When the sphere touches the rod, electrons from the rod flow onto the sphere, as the rod has a surplus of electrons due to its negative induction on the side opposite the sphere. This transfer of electrons results in the sphere acquiring a net negative charge, as it gains more electrons.
Once the sphere has the same negative charge as the side of the rod it touched, the electrostatic force of repulsion between the like charges takes over. The negatively charged sphere is now repelled by the negatively charged side of the glass rod. This repulsion is strong enough to overcome the gravitational force and the nylon thread's tension, causing the sphere to suddenly fly away from the rod.
A bird, accelerating from rest at a constant
rate,experiences a displacement of 28 m in 11s. what is the
averagevelocity?
Answer:
Average velocity of the bird is 2.54 m/s.
Explanation:
Given that,
Initial speed of the bird, u = 0
It experiences a displacement of, d = 28 m
Time taken, t = 11 s
We need to find the average velocity of the bird. Let v is the average velocity. Mathematically, it is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{28\ m}{11\ s}[/tex]
v = 2.54 m/s
So, the average velocity of the bird is 2.54 m/s. Hence, this is the required solution.
Final answer:
The average velocity of the bird is calculated by dividing the total displacement of 28 m by the total time of 11 s, resulting in an average velocity of 2.545 m/s.
Explanation:
To find the average velocity of a bird that accelerates from rest and experiences a displacement of 28 m in 11 s, we use the formula for average velocity, which is total displacement divided by total time. Since the bird starts from rest and moves in one direction, its average velocity will be the same as its average speed.
Average velocity = Total displacement / Total time = 28 m / 11 s = 2.545 m/s.
Therefore, the average velocity of the bird is 2.545 m/s.
You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring these distances is ±0.05 cm. What is the angle that the rod makes with the table?
_______ degrees
What is the uncertainty in that angle?
________ degrees
Answer:
[tex]\partial \theta = 0.003[/tex]
Explanation:
we know that
[tex]sin\theta = \frac{3.8}{146.4}[/tex]
[tex]\theta = sin^{-1} \frac{3.8}{146.4}[/tex]
[tex]\theta = 1.484°[/tex]
[tex]\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians[/tex]
as we see that [tex]sin\theta = \theta[/tex]
relative error[tex] \frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}[/tex]
Where X_1 IS HEIGHT OF ROCK
[tex]X_2[/tex] IS THE HEIGHT OF ROAD
[tex]\partial X[/tex] = uncertainity in measuring distance
[tex]\partial X = 0.05[/tex]
Putting all value to get uncertainity in angle
[tex]\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}[/tex]
solving for [tex]\partial \theta[/tex] we get
[tex]\partial \theta = 0.003[/tex]
Consider an electron that is 100 m from an alpha particle ( = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electric field (in N/C) due to the alpha particle at the location of the electron? N/C (b) What is the electric field (in N/C) due to the electron at the location of the alpha particle? N/C (c) What is the electric force (in N) on the alpha particle? On the electron? electric force on alpha particle electric force on electron
Answer:
a)[tex]E=2.88*10^{-13}N/C[/tex]
b)[tex]E=1.44*10^{-13}N/C[/tex]
c)[tex]F=4.61*10^{-32}N[/tex]
Explanation:
The definition of a electric field produced by a point charge is:
[tex]E=k*q/r^2[/tex]
a)Electric Field due to the alpha particle:
[tex]E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(100)^2=2.88*10^{-13}N/C[/tex]
b)Electric Field due to the electron:
[tex]E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(100})^2=1.44*10^{-13}N/C[/tex]
c)Electric Force on the alpha particle, on the electron:
The alpha particle and electron feel the same force magnitude but with opposite direction:
[tex]F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(100)^2=4.61*10^{-32}N[/tex]
A 50kg boy runs at a speed of 10.0m/s and jumpsonto a cart
originally at rest. the cart, with the boy on it, thentakes off in
the same direction in which the boy was running. ifthe cart with
the boy has a velocity of 2.5m/s, what is the mass ofthe
cart?
Answer:
the mass of the cart is 150 kg
Explanation:
given,
mass of boy(m) = 50 kg
speed of boy (v)= 10 m/s
initial velocity of cart (u) = 0
final velocity of cart(V) = 2.5 m/s
mass of the cart(M) = ?
m v + M u = (m + M ) V......................(1)
50× 10 + 0 = (50 + M ) 2.5
M =[tex]\dfrac{500}{2.5} - 50[/tex]
M = 150 Kg
hence, the mass of the cart is 150 kg
Answer:
Mass of the cart is 750 kg
Given:
Mass of the boy, m = 50 kg
Speed of the boy, v = 10.0 m/s
Final speed of the boy with the cart, v' = 2.5 m/s
Solution:
Initially the cart is at rest and since its on the ground, height, h = 0
Now, by the conservation of energy, mechanical energy before and after will remain conserved:
KE + PE = KE' + PE' (1)
where
KE = Initial Kinetic energy
KE' = Final Kinetic Energy
PE = Initial Potential Energy
PE' = Final Potential Energy
We know that:
Kinetic enrgy = [tex]\frac{1}{2}mv^{2}[/tex]
Potential energy = mgh
Since, potential energy will remain zero, thus we apply the conservation of Kinetic Energy only.
Let the mass of cart be M, thus the mass of the system, m' = 50 + M
Using eqn (1):
[tex]\frac{1}{2}mv^{2} = \frac{1}{2}m'v^{2}[/tex]
[tex]\frac{1}{2}\times 50\times 10^{2} = \frac{1}{2}(50 + M)\times 2.5^{2}[/tex]
[tex]5000 = 6.25(50 + M)[/tex]
M = 750 kg
A system of 1610 particles, each of which is either an electron or a proton, has a net charge of −5.376×10^−17 C. How many electrons are in this system? What is the mass of this system?
Answer:
Number of electrons in the system = 973.
Total mass of the system = [tex]\rm 1.064\times 10^{-24}\ kg.[/tex]
Explanation:
Assumptions:
[tex]\rm n_e[/tex] = number of electrons in the system.[tex]\rm n_p[/tex] = number of protons in the system.[tex]\rm q_e[/tex] = charge on an electron = [tex]\rm -1.6\times 10^{-19}\ C.[/tex][tex]\rm q_p[/tex] = charge on a proton = [tex]\rm +1.6\times 10^{-19}\ C.[/tex][tex]\rm m_e[/tex] = mass of an electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex][tex]\rm m_p[/tex] = mass of a proton = [tex]\rm 1.67\times 10^{-27}\ kg.[/tex]Given:
Total number of particles in the system, N = 1610.Net charge on the system, q = [tex]\rm -5.376\times 10^{-17}\ C.[/tex]Since, the system is comprised of electrons and protons only, therefore,
[tex]\rm N = n_e+n_p\\n_p=N-n_e\ \ \ \ \ \ ................\ (1).[/tex]
The net charge on the system can be written in terms of charges on electrons and protons as
[tex]\rm q=n_eq_e+n_pq_p\ \ \ \ \ ...................\ (2).[/tex]
Putting the value of (2) in (1), we get,
[tex]\rm q=n_eq_e+(N-n_e)q_p\\q=n_eq_e+Nq_p-n_eq_p\\q=n_e(q_e-q_p)+Nq_p\\n_e(q_e-q_p)=q-Nq_p\\n_e=\dfrac{q-Nq_p}{q_e-q_p}\\=\dfrac{-5.3756\times 10^{-17}-1610\times 1.6\times 10^{-19}}{-1.6\times 10^{-19}-1.6\times 10^{-19}}=972.98\\\Rightarrow n_e\approx 973\ electrons.[/tex]
It is the number of electrons in the system.
Therefore, the number of protons is given by
[tex]\rm n_p = N-n_e=1610-973=637.[/tex]
The total mass of the system is given by
[tex]\rm M=n_em_e+n_pm_p\\=(973\times 9.11\times 10^{-31})+(637\times 1.67\times 10^{-27})\\=1.064\times 10^{-24}\ kg.[/tex]
Final answer:
The system consists of 336 electrons and 1274 protons, given the net charge. To calculate the total mass, multiply the number of each particle by its respective mass and then sum the results., which gives 5192.362×10−31 kg as answer.
Explanation:
To find the number of electrons in a system with a net charge, we can use the formula:
Total number of electrons = (Total net charge) / (Charge per electron).Given that the net charge of the system is -5.376×10−17 C and the charge of an electron is approximately -1.6022×10−19 C, we can calculate the total number of electrons using the formula:
Number of electrons = (-5.376×10−17 C) / (-1.6022×10−19 C/electron)
Number of electrons = 3.3555×102
Since we can’t have a fraction of an electron, we round to the nearest whole number, which is 336 electrons.
To determine the mass of the system, first, we need to find the number of protons, which would be 1610 - 336 electrons = 1274 protons. Now we multiply the number of protons by the mass of a proton and the number of electrons by the mass of an electron to get the total mass:
Mass of electrons = 336 × 9.11×10−31 kg
Mass of protons = 1274 × 1.673×10−27 kg
Adding these two gives the total mass of the system, which is 5192.362×10−31 kg.
A professional diver steps off of a cliff that is 18 m high. Draw a sketch of the cliff, defining your origin and final position. (Careful with negative and positive signs.) Unlike the WB assignment, assume the diver jumps up first and has an initial vertical velocity is 4 m/s. (Ignore air resistance.) (a) How long does it take the diver to hit the water? (b) What's the diver's velocity on impact with the water? (Careful with negative and positive signs.)
Answer:
19.2 m/s
Explanation:
We set a frame of reference with the origin at the cliff top and the positive X axis pointing down.
Then the initial position is:
X0 = 0
The initial speed is:
V0 = -4 m/s
It is negative because it is speed upwards and the frame of reference is positive downwards.
Since the diver is in free fall, he is affected only by the acceleration of gravity, we can consider him moving under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
He will hit the water at X = 18 m, so:
18 = 0 - 4 * t + 1/2 * 9.81 * t^2
4.9 * t^2 - 4 * t - 18 = 0
Solving this equation electronically:
t = 2.37 s
The diver will hit the water 2.37 s after jumping.
The equation for speed under constant acceleration is:
V(t) = V0 + a * t
V(2.37) = -4 + 9.81 * 2.37 = 19.2 m/s
Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.6 cm . Two of the particles have a negative charge: q 1 = -7.7 nC and q 2 = -15.4 nC . The remaining particle has a positive charge, q 3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?
Answer:
216.97 X 10⁻⁵ N
Explanation:
Charge q₁ and q₂ will attract q₃ with force F₁ and F₂ .F₁ and F₂ will be calculated as follows
F₁ = [tex]\frac{9\times10^9\times8\times7.7\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]
F₁ = 82.01 X 10⁻⁵ N
F₂= [tex]\frac{9\times10^9\times8\times15.4\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]
F₂ = 164.02 X 10⁻⁵ N
F₁ and F₂ will act at 60 degree so their resultant will be calculated as follows
R² = (82.01 X 10⁻⁵)² +( 164.02 X 10⁻⁵ )² + 2 X 82.01 X 164.02 X 10⁻¹⁰ Cos 60
R² = 47079.48 X 10⁻¹⁰
R = 216.97 X 10⁻⁵ N
Calculate the frequency of each of the following wave lengths of electromagnetic radiation. Part A 488.0 nm (wavelength of argon laser) Express your answer using four significant figures. ν1 ν 1 = nothing s−1 Request Answer Part B 503.0 nm (wavelength of maximum solar radiation) Express your answer using four significant figures.
The frequency of electromagnetic radiation are:
Part A: 6.1432 × 10¹⁴ Hz
Part B: 5.96× 10¹⁴Hz.
The frequency of electromagnetic radiation can be calculated using the speed of light formula:
[tex]v=\frac{c}{\lambda}[/tex]
Where: v is the frequency in hertz (Hz)
c is the speed of light in a vacuum (299,792,458m/s)
λ is the wavelength in meters (m)
Given the wavelengths in nanometers (nm), we need to convert them to meters by dividing by 10⁹ (since 1 nm = 10⁻⁹m).
Part A: Wavelength = 488.0 nm
λ₁ = 488.0/10⁹
=4.88 × 10⁻⁷
v₁ = c/λ₁
=299,792,458/4.88 × 10⁻⁷
= 61432880.7377× 10⁷
=6.1432 × 10¹⁴
Part B:
Wavelength = 503.0 nm
λ₂ = 503.0/10⁹
=5.03 × 10⁻⁷m
v₂=c/λ₂
=299,792,458/5.03 × 10⁻⁷
=59600886.2× 10⁷
=5.96× 10¹⁴
Hence, the frequency of electromagnetic radiation is 5.96× 10¹⁴Hz.
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The frequency of a wave with a wavelength of 488.0 nm (argon laser) is approximately 6.1475 x 10¹⁴ Hz. The frequency of a wave with a wavelength of 503.0 nm (maximum solar radiation) is approximately 5.9682 x 10¹⁴ Hz.
Explanation:The wavelength and frequency of electromagnetic waves are inversely related by the speed of light equation, c = λ ν, where c = 3×10⁸ m/s (speed of light), λ = wavelength, and ν = frequency. To find the frequency of a wave when the wavelength is given, rearrange the equation to ν = c/λ. Applying the equation we get:
Part A: ν₁ = 3×10⁸ m/s ÷ 488.0 x 10⁻⁹ m = 6.1475 x 10¹⁴ Hz. Part B: ν₂ = 3×10⁸ m/s ÷ 503.0 x 10⁻⁹ m = 5.9682 x 10¹⁴ Hz. Learn more about Electromagnetic Waves here:
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You drive 3.4 km in a straight line in a direction 6.9º east of north. If an alternate route to this same destination takes you straight east and then turns directly north to arrive at the same point, find the distance you would have to drive north.
Answer:
You would have to drive in north direction 0.41km
Explanation:
We can solve this with trigonometry, We know that:
[tex]sin(\alpha )=\frac{opposite}{hypotenuse}\\where: \\opposite=north\\hypotenuse=distance[/tex]
[tex]North=sin(6.9^o)*3.4km=0.41km[/tex]
So the distance I will have to drive in north direction is 0.41km
A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is the force exerted on this particle by a magnetic field (a) = B(1.40 T)i and (b) È = (1.40 T)k?
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
The force on a charged particle in a magnetic field is calculated using the Lorentz force equation. For the given particle with charge -1.24 x 10⁻⁹ C and velocity vector, the force is determined by the cross product of the velocity and the magnetic field, taking the charge into account.
Explanation:A particle with a charge of -1.24 x 10⁻⁹ C is moving with an instantaneous velocity of (4.19 X 10⁴ m/s)î + (-3.85 X 10⁴ m/s)ᴇ. To find the force exerted on this particle by a magnetic field we use the Lorentz force equation, which is F = q(v x B), where F is the force on the particle, q is the charge, v is the velocity of the particle, and B is the magnetic field.
For part (a) where the magnetic field B is (1.40 T)î, the velocity vector v is perpendicular to B since v has no i-component, thus the force can be found simply by calculating the magnitude as q*v*B since sin(θ) is 1 for θ = 90°. The direction of the force is given by the right hand rule, considering that both v and B are vectors and the charged particle is negatively charged.
For part (b) where the magnetic field B is (1.40 T)ᴅ, the velocity vector v has no k-component, thus v is perpendicular to B and the same principle applies.
The magnitude of the force in both scenarios is computed using the charge, the magnitude of velocity (which is the combination of both the i and j components), and the magnitude of the magnetic field. The direction will differ based on the cross product between v and B.
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A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed of the horse is 12.8 m/s, and the woman is initially 2.23 m above the level of the saddle. How long is she in the air? The acceleration of gravity is 9.8 m/s2 Answer in units of s What must be the horizontal distance between the saddle and limb when the woman makes her move? Answer in units of m.
Answer:
0.67 seconds
8.576 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s[/tex]
Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.
Speed of the horse = 12.8 m/s
Distance = Speed × Time
⇒Distance = 12.8×0.67
⇒Distance = 8.576 m
Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.
The stuntwoman will be in the air for approximately 0.677 seconds, and the horizontal distance or freefall between the saddle and the limb when the woman makes her move must be approximately 8.668 meters.
Explanation:To find out how long the stuntwoman is in the air, we need to apply the physics concept of free fall. The equation for the time of fall under gravity is sqrt(2h/g), where h is the height and g is the acceleration due to gravity. If we substitute the given values, we get √((2 × 2.23m)/9.8m/s²) = 0.677s.
Next, we need to find horizontal distance between the saddle and the limb when the woman makes her move. Since horizontal distance is simply speed × time, and the speed of the horse is constant, we find that 0.677s × 12.8m/s = 8.668m. So, the branch from which the stunt woman drops must be 8.668m in front of the horse when she drops.
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A space vehicle is traveling at 4150 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward with a speed of 79 km/h relative to the command module (mass m). What is the speed of the command module relative to Earth just after the separation? km/h
Answer:
4213.2 Km/h
Explanation:
Given:
Initial Speed of space vehicle relative to Earth, u = 4150 km/h
Mass of rocket motor = 4m
speed of the rocket motor relative to command module , v' = 79 Km/h
Mass of the command module = m
Now,
let the speed of command module relative to earth be 'v'
From the conservation of momentum, we have
( 4m + m ) × u = m × v + (4m × (v - v'))
or
5m × 4150 = mv + 4mv - 4mv'
or
20750 = 5v - ( 4 × 79 )
or
20750 = 5v - 316
or
v = 4213.2 Km/h
An airplane pilot wishes to fly due west. A wind of 70.0 km/h is blowing toward the south. And I need to find out the speed of the plain over the ground. It is given that the speed is in still air and the airspeed is 435.0 km/h.
Answer:
Speed of the plane over the ground is 429.33 km/hr due West and 70 km/hr towards North
Solution:
According to the question:
Wind flowing in South direction, [tex]v_{s} = 70.0 km/h[/tex]
Air speed, [tex]v_{a} = 435.0 km/h[/tex]
Now,
The velocity vector is required 70 km/h towards north in order to cancel the wind speed towards south.
Therefore,
The ground speed of the plane is given w.r.t fig 1:
[tex]v_{pg} = \sqrt{v_{a^{2}} - v_{s}^{2}}[/tex]
[tex]v_{pg} = \sqrt{435.0^{2} - 70.0^{2}} = 429.33 km/h[/tex]
Final answer:
The speed of the plane over the ground, taking into consideration a wind blowing south at 70.0 km/h and a plane airspeed of 435.0 km/h, is calculated to be approximately 438.3 km/h due west.
Explanation:
An airplane pilot wishes to fly due west, with a wind of 70.0 km/h blowing toward the south, and the plane's airspeed is 435.0 km/h. To determine the speed of the plane over the ground, we must account for both the plane's airspeed and the wind's effect.
The plane's airspeed vector (435.0 km/h) is combined with the wind's vector (70.0 km/h south) to calculate the resultant vector, which represents the plane's actual velocity relative to the ground. This calculation is a vector addition problem that requires the use of the Pythagoras theorem or vector components.
The speed of the plane over the ground (groundspeed) can be found by calculating the magnitude of the resultant vector: Groundspeed = √(air speed² + wind speed²) = √(435² + 70²) km/h, which simplifies to approximately 438.3 km/h to the west. This calculation shows the combined effect of the plane's airspeed and the wind, resulting in the plane's groundspeed.
A force in the +x-direction has magnitude F = b/xn, where b and n are constants. For n > 1, calculate the work done on a particle by this force when the particle moves along the x-axis from x = x0 to infinity.
The work done on a particle by the force in the +x-direction from x = x0 to infinity can be calculated using the formula Work = ∫(b/x^n) dx. The limits of integration for the integral are from x0 to infinity.
Explanation:To calculate the work done by the force in the +x-direction, we can use the formula:
Work = ∫F dx = ∫(b/x^n) dx
Since the force is in the +x-direction and the particle moves along the x-axis from x = x0 to infinity, the limits of integration for the integral are from x0 to infinity.
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What properties does a loud, shrill whistle have? a.) high amplitude, high frequency
b.) high amplitude, low frequency
c.) low amplitude, high frequency
d.) low amplitude, low frequency
Answer:
a.) high amplitude, high frequency
Explanation:
Frequency and amplitude are properties of sound. Varying these properties changes how people perceive sound.
While hearing sound of a particular frequency we call it pitch i.e., the perception of a frequency of sound.
High pitch means high frequency and high frequency is perceived to have a shrill sound.
The loudness of a sound is measured by the intensity of sound i.e., the energy the sound possesses per unit area. As the amplitude increases the intensity increases. So, a loud sound will have higher density.
Hence, the loud shrill whistle will have high frequency and high amplitude.
Answer:
B
Explanation:
I think this is right.
A small metal bead, labeled A has a charge of 25 nC. It is touched to metal bead B, initially neutral, so that the two beads share the 25 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cm apart, the force between them is 5.4 x 10e-4 N. What are the charges qa and qb on the beads?
Answer:
15nC & 10 nC
Explanation:
We will use the formula:
[tex]q_{A}q_{B} = \frac{Fr^{2}}{k}[/tex]
Plugging values for F,r, and k, we get:
[tex]\frac{(5.4*10^{-4} N)(0.050m)^{2}}{9.0 * 10x^{9}N*m^{2}/C^{2}}=1.5 * 10 ^{-16} C^{2}[/tex]
Now, use the equation qB = 25nC - qA: (we know this from the problem)
[tex]q_{A}(25 nC - q_{A})=1.5 *10^{-16} C^{2}[/tex]
This is a quadratic equation that is solved to yield
qA = 10nC or qA = 15nC.
qB is of course the one that qA is not, but we do not know which is which, however that is irrelevant for the problem.
The charges have the same magnitude after sharing but are not necessarily equal, the charges on bead A and bead B are 0.0012 C each.
Given:
Charge on bead A, [tex]qa = 25\ nC = 25 \times 10^{-9}\ C[/tex]
Electric force between the beads, [tex]F = 5.4 \times 10^{-4}\ N[/tex]
Distance between the beads, [tex]r = 5.0\ cm = 0.05\ m[/tex]
Coulomb's constant, [tex]k = 8.99 \times 10^9\ N m^2/C^2[/tex]
Using Coulomb's law, we have on substituting the value:
[tex]F = k \times |q_a \times q_b| / r^2[/tex]
Substitute the values:
[tex]5.4 \times 10^{-4} = (8.99 \times 10^9) \times |(25 \times 10^{-9}) \times q_b| / (0.05)^2[/tex]
Now solve for qb:
[tex]|q_b| = (5.4 \times 10^{-4} \times (0.05)^2) / (8.99 \times 10^9 \times 25 \times 10^{-9})\\|q_b| = 0.0012\ C[/tex]
Since the charges have the same magnitude after sharing, but are not necessarily equal, the charges on bead A and bead B are 0.0012 C each.
To summarize:
Charge on bead A (qa) = 0.0012 C
Charge on bead B (qb) = 0.0012 C
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In a vacuum, two particles have charges of q1 and q2, where q1 = +3.2 µC. They are separated by a distance of 0.28 m, and particle 1 experiences an attractive force of 2.9 N. What is q2 (magnitude and sign)?
Answer:
q2 = - 8 × [tex]10^{-6}[/tex] C
negative sign because attract together
Explanation:
given data
q1 = +3.2 µC = 3.2 × [tex]10^{-6}[/tex] C
distance r = 0.28 m
force F = 2.9 N
to find out
q2 (magnitude and sign)
solution
we know that here if 2 charge is unlike charge
than there will be electrostatic force of attraction , between them
now we apply coulomb law that is
F = [tex]\frac{q1q2}{4\pi \epsilon *r^{2}}[/tex] ............1
here we know [tex]\frac{1}{4\pi \epsilon}}[/tex] = 9 × [tex]10^{9}[/tex] Nm²/C²
so from equation 1
2.9 = 9 × [tex]10^{9}[/tex] × [tex]\frac{3.2*10^{-6}*q2}{0.28^{2}}[/tex]
q2 = [tex]\frac{2.9*0.28^2}{9*10^9*3.2*10^{-6}}[/tex]
q2 = - 8 × [tex]10^{-6}[/tex] C
Final answer:
Using Coulomb's Law, it's found that the magnitude of charge q2 is 3.2 µC. Since the force experienced by particle 1 is attractive, q2 must have an opposite sign to q1. Therefore, q2 is -3.2 µC.
Explanation:
To determine the magnitude and sign of charge q2, we can use Coulomb's Law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for Coulomb's Law is F = k * |q1 * q2| / r², where F is the force between the charges, k is Coulomb's constant (8.988 × 10⁹ N·m²/C²), q1 and q2 are the charges, and r is the distance between the charges.
Given that q1 = +3.2 µC (+3.2 × 10⁻⁶ C), r = 0.28 m, and F = 2.9 N, we can rearrange the equation to solve for q2:
|q2| = (F * r²) / (k * |q1|).
Substituting the given values into the equation, we get:
|q2| = (2.9 N * (0.28 m)²) / (8.988 × 10⁹ N·m²/C² * 3.2 × 10⁻⁶ C),
|q2| = 3.2 µC or -3.2 µC. Because particle 1 experiences an attractive force, it implies that q2 must have an opposite sign to q1. Thus, q2 = -3.2 µC.
Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree that is 4.0 m above the gun. The base of the tree is 15 m away. The speed of the paintball as it leaves the gun is 50 m/s. How far below the knothole does the paintball strike the tree? Express your answer with the appropriate units.
Answer:
The distance between knothole and the paint ball is 0.483 m.
Explanation:
Given that,
Height = 4.0 m
Distance = 15 m
Speed = 50 m/s
The angle at which the forester aims his gun are,
[tex]\tan\theta=\dfrac{4}{15}[/tex]
[tex]\tan\theta=0.266[/tex]
[tex]\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}[/tex]
[tex]\cos\theta=0.966[/tex]
Using the equation of motion of the trajectory
The horizontal displacement of the paint ball is
[tex]x=(u\cos\theta)t[/tex]
[tex]t=\dfrac{x}{u\cos\theta}[/tex]
Using the equation of motion of the trajectory
The vertical displacement of the paint ball is
[tex]y=u\sin\theta(t)-\dfrac{1}{2}gt^2[/tex]
[tex]y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2[/tex]
[tex]y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}[/tex]
Put the value into the formula
[tex]y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})[/tex]
[tex]y=3.517\ m[/tex]
We need to calculate the distance between knothole and the paint ball
[tex]d=h-y[/tex]
[tex]d=4-3.517[/tex]
[tex]d=0.483\ m[/tex]
Hence, The distance between knothole and the paint ball is 0.483 m.
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is located on the x axis at x = 19.3 cm, moving with a speed of 49.2 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Answer:
Q = 12.466μC
Explanation:
For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:
[tex]Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}[/tex]
Solving for Q:
[tex]Q = \frac{m*V^{2}*r}{K*q}[/tex]
Taking special care of all units, we can calculate the value of the charge:
Q = 12.466μC
A rectangular container measuring 20cm x 30cm x 50cm is filled with water. What is the mass of this volume of water in kilograms? A. 30 kg B. 30,000 kg C. 30 milligrams D. 30,000 lbs
Answer:
A. 30 kg
Explanation:
As we know that,
[tex]1 litre=1000cm^{3}[/tex]
And 1 litre is equivalent to 1kg.
Given that, The volume of the rectangular container is,
[tex]V=20cm\times 30cm\times 50cm\\V=30000cm^{3}[/tex]
And this volume will be equal to [tex]V=30000cm^{3}=30litres[/tex]
And this litres in kg will be equal to,
[tex]V=30litres=30 kg[/tex]
Therefore the mass of this volume of water is 30 kg.
A solid conducting sphere of radius 2.00 cm has a charge of 8.84 μC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of −2.02 μC. Find the electric field at the following radii from the center of this charge configuration.
Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5
Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.
Region 1; 0<r<2
∫E.ds=Qinside the gaussian surface/ε0
inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.
Region 2; 2<r<4;
E.4*π*r^2=8,84/ε0
E=8,84/(4*π*ε0*r^2)
Region 3; 4<r<5
E=0 because is inside the conductor.
Finally
Region 4; r>5
E.4*π*r^2=(8,84-2.02)/ε0
The gravitational acceleration on the surface of the moon is 1/6 what it is on earth. If a man could jump straight up 0.7 m (about 2 feet) on the earth, how high could he jump on the moon?
Answer:
on moon he can jump 4.2 m high
Explanation:
given data
gravitational acceleration on moon a(m) = 1/6
jump = 0.7 m
to find out
how high could he jump on the moon
solution
we know gravitational acceleration on earth a = g = 9.8 m/s²
so on moon am = [tex]9.8 * \frac{1}{6}[/tex] = 1.633 m/s²
so if he jump on earth his speed will be for height 0.7 m i s
speed v = [tex]\sqrt{2gh}[/tex]
v = [tex]\sqrt{2(9.8)0.7}[/tex]
v = 3.7 m/s
so if he hump on moon
height will be
height = [tex]\frac{v^2}{2*a(m)}[/tex]
put here value
height = [tex]\frac{3.7^2}{2*1.633)}[/tex]
height = 4.2 m
so on moon he can jump 4.2 m high
In a historical movie, two knight on horseback start from
rest88.0 m apart and ride directly toward each other to dobattle.
Sir George's acceleration has a magnitude of0.300 m/s2
while Sir Alfred's has a magnitude of0.200 m/s2
Relative to Sir George's starting point, where do the
knightscollide?
Answer:
The knights collide 53.0 m from the starting point of sir George.
Explanation:
The equation for the position in a straight accelerated movement is as follows:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial speed
a = acceleration
t = time
The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:
Sir George´s position:
xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²
Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:
xAlfred = 88.0 m + 0 m - 1/2 * 0.200 m/s² * t²
When the knights collide:
xGeorge = x Alfred
1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200 m/s² * t²
0.150 m/s² * t² = 88.0 m - 0.100 m/s² * t²
0.150 m/s² * t² + 0.100 m/s² * t² = 88.0 m
0.250 m/s² * t² = 88.0 m
t² = 88.0 m / 0.250 m/s²
t = 18.8 s
At t = 18.8 s the position of sir George will be
x = 1/2 * 0.300 m/s² * (18.8 s)² = 53.0 m
The question asks for the calculation of the collision point of two knights charging toward each other with different accelerations from a certain distance apart, using principles of kinematics.
Explanation:The question involves kinematics in one dimension, specifically the calculation of the point of collision of two knights with different accelerations.
Sir George has an acceleration of 0.300 m/s2 and Sir Alfred an acceleration of 0.200 m/s2. They start 88.0 m apart.
To determine the collision point, we set up equations based on the formula for distance covered under constant acceleration from rest, which is s = 0.5 * a * t2, where s is the distance, a is acceleration, and t is time.
Since they begin at the same time and collide at the same time, we have:
For Sir George: sG = 0.5 * 0.300 * t2
For Sir Alfred: sA = 0.5 * 0.200 * t2
As they cover the 88.0 m together, sG + sA = 88.0. Substituting the equations for sG and sA and solving for t can give us the individual distances they covered, and hence, the point of collision relative to Sir George's starting point.
An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and sticks 16 m above the ground. You may ignore air resistance. a) What was the initial speed of the arrow? b) At what angle above horizontal was the arrow shot?
Answer:
(a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.
Explanation:
Given that,
Horizontal distance = 230 m
Time t = 6 sec
Vertical distance = 16 m
We need to calculate the horizontal component
Using formula of horizontal component
[tex]R =u\cos\theta t [/tex]
Put the value into the formula
[tex]\dfrac{230}{6} = u\cos\theta[/tex]
[tex]u\cos\theta=38.33[/tex].....(I)
We need to calculate the height
Using vertical component
[tex]H=u\sin\theta t-\dfrac{1}{2}gt^2[/tex]
Put the value in the equation
[tex]16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2[/tex]
[tex]u\sin\theta=\dfrac{16+9.8\times18}{6}[/tex]
[tex]u\sin\theta=32.06[/tex].....(II)
Dividing equation (II) and (I)
[tex]\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}[/tex]
[tex]\tan\theta=0.8364[/tex]
[tex]\theta=\tan^{-1}0.8364[/tex]
[tex]\theta=39.90^{\circ}[/tex]
(a). We need to calculate the initial speed
Using equation (I)
[tex]u\cos\theta\times t=38.33[/tex]
Put the value into the formula
[tex]u =\dfrac{230}{6\times\cos39.90}[/tex]
[tex]u=49.96\ m/s[/tex]
(b). We have already calculate the angle.
Hence, (a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.