The pitch of the helical path followed by an electron in a magnetic field is given by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the electron's velocity, and T is the period of the electron's motion in the magnetic field. By substituting the given values into this formula, we can calculate the pitch.
Explanation:The pitch of a helical path can be determined by considering how the charged particle reacts in a magnetic field. The helical path of an electron in a magnetic field is caused by the perpendicular and parallel components of the electron's velocity. In particular, the parallel component of the velocity, v(cos(theta)), is responsible for the linear movement of the electron along the field lines. This leads to the corkscrew-like helical path, and the pitch of this helix is equivalent to the linear distance moved by the electron in one revolution.
The pitch (p) can be calculated by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the velocity of the electron, and T is the period of circular motion of the electron in the magnetic field. From the provided data, we can first calculate the velocity of the electron using its kinetic energy, 22.5 eV: v = sqrt((2 * KE) / m), where KE is the kinetic energy and m is the mass of the electron (9.11 * 10^-31 kg). Next, we calculate the period using the formula T = 2 * pi * m / (q * B), where q is the charge of the electron (-1.6 * 10^-19 C) and B is the magnetic field intensity (4.55 * 10^4 T).
Finally, we substitute these values into the pitch formula and calculate the pitch of the helical path taken by the electron in the magnetic field.
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The pitch of the helical path taken by the electron is approximately [tex]\( 9.20 \times 10^{-11} \)[/tex] meters.
To find the pitch of the helical path taken by the electron, we need to follow these steps:
1. Calculate the speed of the electron:
The kinetic energy (K.E.) of the electron is given by:
[tex]\[ \text{K.E.} = \frac{1}{2}mv^2 \][/tex]
where [tex]\( m \)[/tex] is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and [tex]\( v \)[/tex] is the speed of the electron.
First, we need to convert the kinetic energy from electron volts (eV) to joules (J):
[tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]
[tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
Now, using the kinetic energy formula:
[tex]\[ \text{K.E.} = \frac{1}{2}mv^[/tex]
To find the pitch of the helical path taken by the electron, we need to determine the components of the electron's velocity and the motion along the magnetic field.
1. Calculate the speed of the electron:
The kinetic energy (KE) of the electron is given by:
[tex]\[ \text{KE} = \frac{1}{2} mv^2 \][/tex]
where m is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and v is the speed of the electron.
First, convert the kinetic energy from electron volts (eV) to joules (J):
[tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]
[tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
Now, solve for the speed \( v \) using the kinetic energy formula:
[tex]\[ \frac{1}{2} mv^2 = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
[tex]\[ v^2 = \frac{2 \times 3.6045 \times 10^{-18}}{9.11 \times 10^{-31}} \][/tex]
[tex]\[ v^2 = 7.9137 \times 10^{12} \][/tex]
[tex]\[ v = \sqrt{7.9137 \times 10^{12}} \approx 2.81 \times 10^6 \, \text{m/s} \][/tex]
2. Determine the components of the velocity:
The angle between the velocity and the magnetic field is [tex]\( 65.5^\circ \)[/tex]. The component of the velocity parallel to the magnetic field [tex](\( v_{\parallel} \))[/tex] and the component perpendicular to the magnetic field[tex](\( v_{\perp} \))[/tex] can be found using trigonometry:
[tex]\[ v_{\parallel} = v \cos(65.5^\circ) \][/tex]
[tex]\[ v_{\perp} = v \sin(65.5^\circ) \][/tex]
Calculate these components:
[tex]\[ v_{\parallel} = 2.81 \times 10^6 \, \text{m/s} \times \cos(65.5^\circ) \approx 2.81 \times 10^6 \times 0.417 \approx 1.17 \times 10^6 \, \text{m/s} \][/tex]
[tex]\[ v_{\perp} = 2.81 \times 10^6 \, \text{m/s} \times \sin(65.5^\circ) \approx 2.81 \times 10^6 \times 0.909 \approx 2.55 \times 10^6 \, \text{m/s} \][/tex]
3. Find the cyclotron frequency [tex](\( f \))[/tex] and the period [tex](\( T \))[/tex] of the electron's circular motion:
The cyclotron frequency is given by:
[tex]\[ f = \frac{qB}{2\pi m} \][/tex]
where q is the charge of the electron [tex](\( 1.602 \times 10^{-19} \) C)[/tex], B is the magnetic field [tex](\( 4.55 \times 10^4 \) T)[/tex], and m is the mass of the electron.
[tex]\[ f = \frac{1.602 \times 10^{-19} \times 4.55 \times 10^4}{2\pi \times 9.11 \times 10^{-31}} \][/tex]
[tex]\[ f \approx \frac{7.29 \times 10^{-15}}{5.72 \times 10^{-31}} \approx 1.27 \times 10^{16} \, \text{Hz} \][/tex]
The period [tex](\( T \))[/tex] is the inverse of the frequency:
[tex]\[ T = \frac{1}{f} \approx \frac{1}{1.27 \times 10^{16}} \approx 7.87 \times 10^{-17} \, \text{s} \][/tex]
4. Calculate the pitch of the helical path:
The pitch is the distance the electron travels parallel to the magnetic field in one period:
[tex]\[ \text{Pitch} = v_{\parallel} \times T \][/tex]
[tex]\[ \text{Pitch} \approx 1.17 \times 10^6 \, \text{m/s} \times 7.87 \times 10^{-17} \, \text{s} \approx 9.20 \times 10^{-11} \, \text{m} \][/tex]
A ladder leans against a non vertical wall with friction that make a angle of 50 degree with ground. how do show the free body diagram. The wall is not vertical
Answer:
As from the given question,
The free body diagram of a ladder leans against a non vertical wall with friction that make a angle of 50 degree with ground is show below.
Friction force can be calculated by formula:
Friction force = μ N
As we know that
At equilibrium,
The sum of the total horizontal force is equal to '0'.
The sum of the total vertical force is equal to '0'.
An arrow, starting from rest, leaves the bow with a speed of25
m/s. If the average force exerted on the arrow by the bowwere
tripled, all else remaining the same, with what speed wouldthe
arrow leave the bow?
If the force exerted on an arrow by a bow is tripled, the speed of the arrow leaving the bow would also triple, assuming other factors remain constant. Therefore, the arrow would leave the bow with a speed of 75 m/s.
Explanation:If the arrow starting from rest leaves the bow with a speed of 25 m/s due to a certain force, tripling the force would result in a tripling of the acceleration, assuming mass remains constant. From Newton's second law, we understand that Force equals mass times acceleration (F = ma). The acceleration of an object is directly proportional to the force applied, if mass is kept constant. Therefore, if we triple the force, the acceleration would also triple. With a tripling of acceleration, the final speed of the arrow would also triple, as speed is the product of acceleration and time. Therefore, if the force exerted on the arrow by the bow is tripled, the speed with which the arrow leaves the bow would be 75 m/s.
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The speed at which the arrow would leave the bow when the average force exerted on it is tripled is 38.40 m/s.
Explanation:To find the speed at which the arrow would leave the bow if the average force exerted on it were tripled, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the arrow can be calculated using the formula KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass of the arrow, and v is the initial velocity. Since the mass of the arrow remains the same, tripling the force would cause the initial kinetic energy to triple as well. Therefore, the final velocity can be found using the formula KE = 1/2 * m * v^2 and solving for v.
Let's use an example to illustrate this. Suppose the mass of the arrow is 0.5 kg. Initially, the arrow leaves the bow with a speed of 25 m/s. The initial kinetic energy is given by KE = 1/2 * 0.5 kg * (25 m/s)^2 = 156.25 J. If the average force is tripled, the new kinetic energy will be 3 times the initial kinetic energy, which is 3 * 156.25 J = 468.75 J. Plugging this value into the kinetic energy formula and solving for v, we get v = √(2 * 468.75 J / 0.5 kg) = 38.40 m/s.
Therefore, if the average force exerted on the arrow by the bow were tripled, the arrow would leave the bow with a speed of 38.40 m/s.
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Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.115 nm , and their final separation is 0.105 nm . How much electric potential energy was lost in this reaction (in units of eV)?
Answer:
[tex]1.189eV[/tex]
Explanation:
The electric potential energy is the potential energy that results from the Coulomb force and is associated with the configuration of two or more charges. For an electron in the presence of an electric field produced by a proton, the electric potential energy is defined as:
[tex]U=\frac{kq_{e}q_{p}}{r}[/tex]
where [tex]q_{e}[/tex] is the electron charge, [tex]q_{p}[/tex] is the proton charge, r is the separation distance between the charges and k is the coulomb constant.
Knowing this, we can calculate how much electric potential energy was lost:
[tex]\Delta U=U_{f}-U{i}\\\Delta U=\frac{kq_{e}q_{p}}{r_{f}}-\frac{kq_{e}q_{p}}{r_{i}}\\\Delta U=kq_{e}q_{p}(\frac{1}{r_{f}}-\frac{1}{r_{i}})\\\Delta U=(8.99*10^9\frac{Nm^2}{C^2})(-1.60*10^{-19}C)(1.60*10^{-19}C)(\frac{1}{0.105*10^{-9}m}-\frac{1}{0.115*10^{-9}m})\\\Delta U=1.90*10^{-19}J*\frac{6.2415*10^{18}eV}{1J}=1.189eV[/tex]
The change in electric potential energy between an electron and a proton with changing separation can be calculated using the Coulomb potential energy equation and can be expressed in electron volts (eV).
Explanation:The question involves calculating the change in electric potential energy of a system consisting of an electron and a proton as they change separation distance during a reaction. The energy of the system is described by the Coulomb electrostatic potential energy equation Epot = -e2 / (4πε₀r), where e is the charge of the electron, ε₀ is the vacuum permittivity, and r is the separation distance between the electron and proton.
To find the amount of potential energy lost, we calculate the potential energy at the initial separation (0.115 nm) and the final separation (0.105 nm), then take the difference between these values. The energy will be expressed in electron volts (eV), a common unit used to describe energy at subatomic scales.
Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle of 15.8° with the horizontal. Find the separation between the slits. µm
Answer:
2.2 µm
Explanation:
For constructive interference, the expression is:
[tex]d\times sin\theta=m\times \lambda[/tex]
Where, m = 1, 2, .....
d is the distance between the slits.
Given wavelength = 597 nm
Angle, [tex]\theta[/tex] = 15.8°
First bright fringe means , m = 1
So,
[tex]d\times sin\ 15.8^0=1\times \597\ nm[/tex]
[tex]d\times 0.2723=1\times \597\ nm[/tex]
[tex]d=2192.43481\ nm[/tex]
Also,
1 nm = 10⁻⁹ m
1 µm = 10⁻⁶ m
So,
1 nm = 10⁻³ nm
Thus,
Distance between slits ≅ 2.2 µm
Answer:
The separation between the slit is [tex]2.19\mu m[/tex]
Solution:
As per the question:
Wavelength of light, [tex]\lambda = 597 nm = 597\times 10^{-9} m[/tex]
[tex]\theta = 15.8^{\circ}[/tex]
Now, by Young's double slit experiment:
[tex]xsin\theta = n\lambda[/tex]
here,
n = 1
x = slit width
Therefore,
[tex]x = \frac{597\times 10^{-9}}{sin15.8^{\circ}} = 2.19\times 10^{- 6} m[/tex]
[tex]x = 2.19\mu m[/tex]
You plan a trip on which you want to average 91 km/h. You cover the first half of the distance at an average speed of only 49 km/h. What must your average speed be in the second half of the trip to meet your goal? Note that the velocities are based on half the distance, not half the time. km/hIs this reasonable? Yes No
Answer:
637 km/h, not reasonable
Explanation:
Total average speed = 91 km/h
Average speed of first half, v1 = 49 km/h
Let the average speed for the next half is v2.
Let the total distance is 2s.
Time taken for first half of the journey, t1 = s / v1 = s / 49
Time take for the next half of the journey, t2 = s / v2
The average speed is defined as the total distance traveled to the total time taken
[tex]91=\frac{2s}{\frac{s}{49}+\frac{s}{v_{2}}}[/tex]
[tex]\frac{s}{49}+\frac{s}{v_{2}}=\frac{2s}{91}[/tex]
[tex]\frac{1}{v_{2}}=\frac{2}{91}-\frac{1}{49}[/tex]
[tex]\frac{1}{v_{2}}=\frac{7}{4459}
v2 = 637 km/h
This is very high speed which is not reasonable.
A uniform plank of length 4.4m and weight 228N
restshorizontally on two supports, with 1.1m of the plank hanging
overthe right support. To what distance x can a person who
weighs449N walk on the overhanging part of the plank before it
begins totip.
Answer:0.558 m
Explanation:
Given
weight of Plank=228 N
weight of man=449 N
Length of plank=4.4 m
let [tex]R_a[/tex] and [tex]R_b[/tex] be the reactions at A & B
and Reaction at A becomes zero when plank is about to rotate at B
Taking moment about B at that instant so that plank is just about to rotate.
[tex]228\times 1.1=449\times x[/tex]
Where x is the maximum distance that man can walk
x=0.558 m
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 11.8 m/s. The magnitude of the gravitational acceleration g = 9.8 m/s2 Take the water surface to be y0 = 0. Choose UPWARD as positive y direction. Keep 2 decimal places in all answers. (a) How high (what maximum height) in meters does his body rise above the water? To solve this part, first note that the body's final velocity at the maximum height is known (implicitly) and identify its value. Then identify the unknown, and chose the appropriate equation (based on the knowns and unknown) to solve for it. After choosing the equation, solving for the unknown, checking units, think about whether the answer is reasonable.(b) How long in seconds is the dolphin in the air? Neglect any effects due to his size or orientation. Note: this is the total time of jumping up from water to hightest point and falling down to water.
Answer:
a) [tex]y_{max}=7.10m[/tex]
b) [tex]t=2.40 s[/tex]
Explanation:
From the exercise, we know the initial velocity, gravitational acceleration and initial position of the dolphin.
[tex]v_{oy}=11.8m/s[/tex]
[tex]y_{o}=0m\\ g=9.8m/s^{2}[/tex]
a) To find maximum height, we know that at that point the dolphin's velocity is 0 and it becomes coming down later.
Knowing that, we need to know how much time does it take the dolphin to reach maximum height.
[tex]v_{y}=v_{oy}+gt[/tex]
[tex]0=11.8m/s-(9.8m/s^{2} )t[/tex]
Solving for t
[tex]t=1.20 s[/tex]
So, the dolphin reach maximum point at 1.20 seconds
Now, using the equation of position we can calculate maximum height.
[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]
[tex]y=0+11.8m/s(1.20s)-\frac{1}{2}(9.8m/s^{2} )(1.20s)=7.10m[/tex]
b) To find how long is the dolphin in the air we need to analyze it's hole motion
At the end of the jump the dolphin return to the water at y=0. So, from the equation of position we have that
[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]
[tex]0=0+11.8t-\frac{1}{2}(9.8)t^{2}[/tex]
What we have here, is a quadratic equation that could be solve using:
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-\frac{1}{2} (9.8)[/tex]
[tex]b=11.8\\c=0[/tex]
[tex]t=0s[/tex] or [tex]t=2.40 s[/tex]
Since the answer can not be 0, the dolphin is 2.40 seconds in the air.
A comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distance traveled by the comet during that time
Answer:
The distance covered by comet is [tex]200,000 km[/tex]
Explanation:
Speed is defined as the rate of change of distance with time. It is given by the equation speed= [tex]\bold{\frac{distance}{time}}[/tex]
Thus distance= [tex]speed*time[/tex]
In this problem it is given that speed of comet= [tex]\frac{50,000km}{hr}[/tex]
time travelled by the comet= 4 hours
Thus distance= [tex]speed*time[/tex]
= [tex]500000*4[/tex]
= [tex]\bold{200,000km}[/tex]
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providing a steady force of 161 N that is parallel to the motion of the sled. The combined mass of the girl and the sled is 87.7 kg, and the coefficient of kinetic friction between the snow and the runners of the sled is 0.151. How much time is required for the sled to travel down a 271-m slope, starting from rest?
Answer:
The sled required 9.96 s to travel down the slope.
Explanation:
Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.
Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).
The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.
The magnitude of the friction force is calculated as follows:
Ff = μ · Fn
where :
μ = coefficient of kinetic friction
Fn = normal force
The normal force has the same magnitude as the y-component of the gravity force:
Fgy = Fg · cos 30º = m · g · cos 30º
Where
m = mass
g = acceleration due to gravity
Then:
Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º
Fgy = 744 N
Then, the magnitude of Fn is also 744 N and the friction force will be:
Ff = μ · Fn = 0.151 · 744 N = 112 N
The x-component of Fg, Fgx, is calculated as follows:
Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N
The resulting force, Fr, will be the sum of all these forces:
Fw + Fgx - Ff = Fr
(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).
Fr = 161 N + 430 N - 112 N = 479 N
With this resulting force, we can calculate the acceleration of the sled:
F = m·a
where:
F = force
m = mass of the object
a = acceleration
Then:
F/m = a
a = 479N/87.7 kg = 5.46 m/s²
The equation for the position of an accelerated object moving in a straight line is as follows:
x = x0 + v0 · t + 1/2 · a · t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.
x = 1/2· a ·t²
Let´s find the time at which the position of the sled is 271 m:
271 m = 1/2 · 5.46 m/s² · t²
2 · 271 m / 5.46 m/s² = t²
t = 9.96 s
The sled required almost 10 s to travel down the slope.
A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100 seconds, and walked back to the research building for 1000 seconds. To find out Ellie’s average velocity for each case while running, resting, and walking, begin by plotting a graph between position and time. 1.List the velocity from greatest to least among running, resting, and walking
2.List the speed from greatest to least among running, resting, and walking.
Answer:
1. Running velocity (5 m/s)
2. Resting velocity (0 m/s)
3. Walking velocity (-1 m/s)
1. Running speed (5 m/s)
2. Walking speed (1 m/s)
3. Resting speed (0 m/s)
Explanation:
Attached you will find the plot of position vs time of Ellie´s movement.
The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.
So, the velocity is calculated as follows:
v = Δx / Δt
where
Δx = final position - initial position
Δt = elapsed time
1) The average velocity of Ellie while running is:
v = 1000 m - 0 m / 200 s = 5 m/s
While resting:
v = 0 m - 0 m / 100 s = 0 m/s
And while walking back:
v = 0 m - 1000 m / 1000 s = - 1 m/s
Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.
Comparing with the graphic, the velocity is the slope of the function position(t).
Then:
1. Running velocity (5 m/s)
2. Resting velocity (0 m/s)
3. Walking velocity (-1 m/s)
2) The speed is the distance traveled over time:
Running speed = 1000 m / 200 s = 5m /s
Resting speed = 0 m / 100 s = 0 m/s
Walking speed = 1000 m/ 1000 s = 1 m/s
Then:
1. Running speed (5 m/s)
2. Walking speed (1 m/s)
3. Resting speed (0 m/s)
It takes 15 min to drive 6.0 mi in a straight line to the local hospital. It takes 10 min to go the last 3.0 mi, 2.0 min to go the last mile, and only 30 s (0.50 min) to go the last 0.50 mi. What is the average speed of the trip? Take the positive xx ‑direction to be from the starting point toward the hospital.
Final answer:
The average speed for the trip to the hospital, which covers a distance of 6.0 miles in 15 minutes, is 24 miles per hour (mph).
Explanation:
To calculate the average speed of the trip to the hospital, you take the total distance traveled and divide it by the total time taken. The student has specified that it takes 15 minutes to drive 6.0 miles in total. Therefore, the average speed can be calculated as follows:
Speed = Total Distance / Total Time
Speed = 6.0 miles / 15 minutes
Speed = 0.4 miles per minute
To convert this to miles per hour (mph), multiply by 60 (since there are 60 minutes in an hour):
Speed = 0.4 miles/minute × 60 minutes/hour
Speed = 24 mph.
Therefore, the average speed for the entire trip is 24 mph.
Calculate the rotational inertia of a wheel that has akinetic
energy of 24,400 J when rotating at 566 rev/min.
Answer:
J = 14.4 kg*m^2
Explanation:
Assuming that the wheel is not moving anywhere, and the kinetic energy is only due to rotation:
Ek = 1/2 * J * w^2
J = 2 * Ek / (w^2)
We need the angular speed in rad / s
566 rev/min * (1 min/ 60 s) * (2π rad / rev) = 58.22 rad/s
Then:
J = 2 * 24400 / (58.22^2) = 14.4 kg*m^2
What does it mean when we say that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side?
Answer:
It means at a diverging boundary when magma spreads to both sides it is almost identical in both the side.
Explanation:
At a diverging boundary when magma spreads to both sides it is almost identical in both the side.
If you take a picture of one side of the boundry and the spreading of sea floor and place it before a mirror you can see the image is identical to the picture of other side.
Therefore, the meaning of saying, each side of the sea floor away from the mid ocean ridge is a mirror image of the other side is explained above.
The statement that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side refers to the symmetric nature of seafloor spreading. Molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust, creating symmetric magnetic striping patterns. This is akin to the bilaterally symmetric structure observed in certain organisms.
Explanation:When we say that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side, we are referencing the symmetric nature of seafloor spreading. Just as a mirror image reflects an object exactly, so does the seafloor on one side of the mid-ocean ridge reflect the seafloor on the opposite side. This is because molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust. This new crust then moves away from the ridge due to tectonic forces, and the process repeats, creating a pattern of symmetrical magnetic striping on the seafloor.
This 'mirroring' effect is similar to the bilaterally symmetric structure seen in certain organisms, where a plane cut from the front to back of the organism produces distinct left and right sides that are mirror images of each other. You can see this symmetry in the images of the Moon provided – two different sides, yet mirroring similar physical traits.
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If the distance between the Earth and the Sun were increased by a factor of 2.51, by what factor would the strength of the force between them change? [Hint: Use Newton's Law of Universal Gravitation, and give your answer to 2 decimal places only]
Answer:
Force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times
Explanation:
According to law of gravitation we know that force between two object is given by
[tex]F=\frac{GM_1M_2}{R^2}[/tex], here G is universal gravitational constant [tex]M_1[/tex] is mass of one object and [tex]M_2[/tex] is mass of other object , and R is the distance between them
From the relation we can see that force is inversely proportional to square of distance between them
So force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times
Final answer:
The strength of the gravitational force between the Earth and the Sun would decrease by a factor of approximately 6.30 if the distance between them were increased by a factor of 2.51, according to the inverse square law of universal gravitation.
Explanation:
If the distance between the Earth and the Sun were increased by a factor of 2.51, the strength of the force between them would change according to Newton's law of universal gravitation. This law states that the force is inversely proportional to the square of the distance between two masses. Therefore, if the distance increases by a factor of 2.51, the force of attraction would decrease by a factor of (2.51)2.
To calculate the specific factor by which the force decreases, we square 2.51 which gives us 6.3001. The gravitational force would thus be reduced by this factor. Since the law involves an inverse square relationship, as the distance increases by a factor of 2.51, the strength of the gravitational force decreases by a factor of approximately 6.30.
In searching the bottom of a pool at night, a watchman shinesa
narrow beam of light from his flashlight, 1.3 m above the
waterlevel, onto the surface of the water at a point 2.7 m from the
edgeof the pool (Figure 23-50). Where does the spot of light hit
thebottom of the pool, measured from the wall beneath his foot, if
thepool is 2.1 m deep?
Answer:
The spot of light hits the bottom of the pool at 4.634 m from the wall beneath the watchman's feet.
Explanation:
Use the diagram attached below this answer to see the notation we will use.
For this case, we're trying to find x and we have:
h=1.3 m
b=2.1 m
a=2.7 m
We also know Snell's law for refraction:
[tex]n_{1} sin\theta_{1}=n_{2}sin\theta_{2}[/tex]
n is the refractive index of each substance (in this case, air and water), which are:
[tex]n_{air}=1[/tex]
[tex]n_{water}=1.33[/tex]
Triangle theory says that [tex]\theta_{1}=\beta[/tex] and:
[tex]tan\beta=\frac{a}{h}[/tex]
[tex]\beta=arc tan(\frac{a}{h})=arctan(\frac{2.7m}{1.3m})=64.29[/tex]
Using Snell's law:
[tex]\theta_{2}=arcsin(\frac{n_{1}sin\theta{1}}{n_{2}})=arcsin(\frac{1sin(64.29)}{1.33})=42.644[/tex]
Using triangle theory:
[tex]tan\theta_{2}=\frac{(x-a)}{b}[/tex]
[tex]x=b*tan\theta_{2}+a=2.1m*tan(42.644)+2.7m=4.634m[/tex]
Answer:
The distance of spot of light from his feet equals 3.425 meters.
Explanation:
The situation is represented in the attached figure below
The angle of incidence is computed as
[tex]\theta _i=tan^{-1}(\frac{1.3}{2.7})\\\\\therefore \theta _i=25.71^{o}[/tex]
Now by Snell's law we have
[tex]n_{i}sin(\theta _i)=n_{r}sin(\theta _r)[/tex]
where
[tex]n_{i},n_{r}[/tex] are the refractive indices of the incident and the refracting medium respectively
[tex]\theta _i,\theta _r[/tex] are the angle of incidence and the angle of refraction respectively
Thus using the Snell's relation we have
[tex]1.0\times sin(25.71)=1.33\times sin(\theta _r)\\\\\therefore sin(\theta _r)=\frac{sin(25.71}{1.33}=0.326\\\\\therefore \theta _r=sin^{-1}(0.326)=19.04^{o}[/tex]
from the attached figure we can see
[tex]tan(\theta _r)=\frac{L_{2}}{H}=\frac{L_{2}}{2.1}\\\\\therefore L_{2}=2.1\times tan(19.04)=0.725m[/tex]
Thus distance of spot on the pool bed from his feet equals [tex]2.7+0.725=3.425m[/tex]
Convert the following binary numbers to a decimal: 1001, 10101, 1010001, and 1010.1010.
Answer:
1) 1001
=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 1 = (9)₁₀
2) 10101
=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰
= 16 + 4 + 1 = (21)₁₀
3) 1010001
=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 64 + 16 + 1 = (81)₁₀
4) 1010.1010
= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴
= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀
Answer:
1) 1001
=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 1 = (9)₁₀
2) 10101
=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰
= 16 + 4 + 1 = (21)₁₀
3) 1010001
=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 64 + 16 + 1 = (81)₁₀
4) 1010.1010
= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴
= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀
Explanation:
A police car at rest, passed by a speeder traveling at a constant 120 km/h, takes off in hot pursuit. The police officer catches up to the speeder in 750 m, while maintaining a constant acceleration. Calculate (a) how long it took the police car to overtake the speeder, (b) the required police acceleration, and (c) the velocity of the police car at the moment it reaches the speeder.
The police car takes approximately 30 seconds to reach the speeder, requires an acceleration of about 1.67 m/s², and its velocity at the moment it reaches the speeder is approximately 50 m/s.
Explanation:This problem is a two-body pursuit scenario. The speeder's motion can be described by x = Ut, while the police car's motion is represented by the equation x = 1/2at². The speeder is moving at a constant speed of 120 km/h which is equal to 33.33 m/s.
(a) Time for the police car to overtake the speeder:
To find the time, we equate the two equations (since they both cover the same distance of 750m) and solve for 't'. Thus, 750m = 33.33m/s * t = 1/2 * a * t². By solving this equation, we get two values of 't', out of which the realistic answer is t = 30 seconds.
(b) Required acceleration of the police car:
Plugging the time into the equation for the police car, we get 750m = 1/2 * a * (30s)². Solving for 'a', we find the required acceleration to be approximately 1.67 m/s².
(c) Velocity of the police car at the moment it reaches the speeder:
Since the police car starts from rest and maintains a constant acceleration, the final velocity can be calculated using the equation v = at. Hence, at the moment it reaches the speeder, the police car's velocity would be approximately 50 m/s.
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The alternating current which crosses an apparatus of 600 W has a maximum value of 2.5 A. What is efficient voltage between its demarcations? A. 140 V
B. 240 V
C. 340 V
D. Impossible to find without knowing the resistance of the apparatus
Answer: Option (b) is correct.
Explanation:
Since we know that,
P = VI
where;
P = power
V= Voltage
I = Current
Since it's given that,
P = 600W
I = 2.5 A
equating these values in the above equation, we get;
V = [tex]\frac{600}{2.5}[/tex]
V = 240 V
What is the repulsive force between two pith balls that are 9.00 cm apart and have equal charges of -28.0 nC?
Answer:
Force, [tex]F=8.71\times 10^{-4}\ N[/tex]
Explanation:
Given that,
Charges on pith balls, [tex]q_1=q_2=-28\ nC=-28\times 10^{-9}\ C[/tex]
Distance between balls, d = 9 cm = 0.09 m
Let F is the repulsive force between two pith balls. We know that the repulsive force between two charges is given by :
[tex]F=k\dfrac{q_1^2}{d^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{(-28\times 10^{-9})^2}{(0.09)^2}[/tex]
F = 0.000871 N
or
[tex]F=8.71\times 10^{-4}\ N[/tex]
So, the repulsive force between the pith balls is [tex]8.71\times 10^{-4}\ N[/tex]. Hence, this is the required solution.
A light source of wavelength λ illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 2.00 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 6.00 eV. What is the work function of the metal?
Answer:
[tex]W=2eV[/tex]
Explanation:
Energy from a light source (photons):
E=h*c/λ
Photoelectric effect and Work function (W):
E=W+Ek
Ek: maximum kinetic energy from photoelectrons
then:
h*c/λ=W+Ek
Case 1:
[tex]h*c/lambda_{1}=W+Ek_{1}[/tex] (1)
Case 2:
[tex]h*c/lambda_{2}=W+Ek_{2}[/tex]
but [tex]lambda_{2}=lambda_{1}/2[/tex]
[tex]2h*c/lambda_{1}=W+Ek_{2}[/tex] (2)
If we divide (2) by (1):
[tex]2=\frac{W+Ek_{2}}{W+Ek_{1}}[/tex]
[tex]W=Ek_{2}-2Ek{1}=2eV[/tex]
A cannon tilted up at a 29.0° angle fires a cannon ball at 81.0 m/s from atop a 22.0 m -high fortress wall. What is the ball's impact speed on the ground below?
Answer:
The speed of the ball when it hits the ground is 83.4 m/s
Explanation:
Please see the attached figure for a description of the problem.
The vector velocity at time "t" can be written as follows:
[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]
where:
v0 : module of the initial velocity vector.
α: launching angle.
g: acceleration due to gravity.
t: time
To calculate the impact speed, we can use this equation once we know the time at which the ball hits the ground. For this, we can use the equation for position:
[tex]r = ( x0 + v0*t*cos\alpha ; y0 + v0*t*sin\alpha + 1/2 g*t^{2})[/tex]
where:
r= vector position
x0 = horizontal initial position
y0 = vertical initial position
The problem gives us information about the vertical displacement. If we take the base of the wall as the center of the reference system, at the time at which the ball hits the ground, the module of the vertical component of the vector position, r, is 0 m (see figure). The initial vertical position is 22 m.
if ry is the vertical component of the vector r at final time:
[tex]ry = (0; y0 + v0*t*sin\alpha +1/2 g*t^{2})[/tex]
then, the module of ry is (see figure):
module ry =[tex]0 = 22m + v0*t*sin\alpha + 1/2*g*t^{2}[/tex]
Let´s replace with the given data:
[tex]0 = 22m + 39.3 m/s*t -4.9 m/s^{2}*t^{2}[/tex]
Solving the quadratic equation:
t = -0.5 and t = 8.5 s
At 8.5 s after firing, the ball hits the ground.
Now, we can find the module of the velocity vector when the ball hits the ground:
[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]
at time t = 8.5s
[tex]v = (81.0m/s * cos\alpha ; 81.0m/s * sin\alpha - 9.8 m/s^{2} *8.5s)[/tex]
v = (70.8 m/s; -44.0 m/s)
module of v = [tex]\sqrt{(70.8m/s)^{2} + (-44.0m/s)^{2}}[/tex] = 83.4 m/s
A sodium atom will absorb light with a wavelength near 589 nm if the light is within 10 MHz of the resonant frequency. The atomic mass of sodium is 23. (i) Calculate the number of "yellow" photons of wavelength 2 = 589 nm that must be absorbed to stop a sodium atom initially at room temperature (V-600 m/s). [7 marks] (ii) What is the minimum time needed to cool a sodium atom?
Answer:
i)20369 photons
ii) 40 ps
Explanation:
Momentum of one Sodium atom:
[tex]P=m*v =600m/s*23amu*\frac{1 kg}{6.02*10^{23}amu}\\P=2.29*10^{-23}kgm/s[/tex]
In other to stop it, it must absorb the same momentum in photons:
[tex]P=2.29*10^{-23}kgm/s=n_{photons}*\frac{h_{planck}}{\lambda}\\=n*\frac{6.63*10^{-34}}{589*10^{-9}} \\==>n=20369 photons[/tex]
Now, for the minimun time, we use the speed of light and the wavelength. For the n photons:
[tex]t=n*T=n*\frac{\lambda}{c} =20369*\frac{589nm}{3*10^{8}m/s}=4*10^{-11} second=40 ps[/tex]
The heaviest flying bird is the trumpeter swan, which weighsin
at about 38 pounds at its heaviest. What is the weight
innnewtons?
Answer:169.1 N
Explanation:
Given
bird weigh 38 pounds
and we know
1 Pound is equal to 0.453592 kg
Thus 38 pounds is equal to 17.236 kg
Thus the weight of bird is [tex]17.236\times 9.8=169.09 N \approx 169.1 N[/tex]
Answer:
The weight in newtons is 169.1.
Explanation:
This question can be solved as a simple rule of three problem.
We have that each pound has 4.45 newtons. So, how many newtons are there in 38 pounds?
So
1 pound - 4.45N
38 pounds - xN
[tex]x = 38*4.45[/tex]
[tex]x = 169.1N[/tex]
The weight in newtons is 169.1.
Vector A is 3 m long and vector B is 4 m long. The length
ofthe sum of the vectors must be:
A. 5 m
B. 7 m
C. 12 m
D. some value from 1 m to 7 m.
Answer:d
Explanation:
Given
Magnitude of Vector A is 3 m
and Magnitude of vector B is 4 m
So the maximum value of resultant can be 7 m when both are at an angle of [tex]0^{\circ}[/tex]
and its minimum value can be 1 m when both are at angle of [tex]180 ^{\circ}[/tex]
So the resultant must lie between 1 m to 7 m
The length of the sum of the vectors A and B with no direction given must be some value from 1 m to 7 m. Thus, the correct option is D.
What is Vector sum?Vector addition is the property or operation of addition of two or more vectors together into a vector sum. The sign used with the vectors depends upon the direction in which they move. If the vectors are moving in the same direction then they will be added and if they are moving in opposite directions then they are substracted.
Given, Magnitude of Vector A is 3 m and Magnitude of vector B is 4 m.
So, in this case the maximum value of resultant vector can be 7 m when both are moving in the same direction with an angle of 0° and it has minimum value which can be 1 m when both are at angle of 180° in opposite direction.
So, the resultant vector must lie between 1 m to 7 m.
Therefore, the correct option is D.
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How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 4.5 mi (23,760 ft) away and the air temperature is 74.0°F?
Answer:
t = 20.96 seconds
Explanation:
given,
lightning strikes at = 4.5 mi( 23,760 ft)
air temperature =74.0°F
74.0°F = 23.3 °C
speed of the sound at 23.3 °C
V = 331.5 + 0.6 × T
V = 331.5 + 0.6 × 23.3
= 345.48 m/s
distance given = 4.5 mile
1 mile = 1609.4
4.5 mile = 4.5 × 1609.4 = 7242.3 m
time taken =
[tex]t =\dfrac{d}{v}[/tex]
[tex]t =\dfrac{7242.3}{345.48}[/tex]
t = 20.96 seconds
hence, time taken by the sound to reach by the observer is 20.96 sec.
Final answer:
To estimate the time elapsed between seeing lightning and hearing thunder, one must know the speed of sound, which varies with temperature, and the distance to the lightning strike. At a distance of 4.5 miles and an air temperature of 74.0°F, approximately 21 seconds will elapse.
Explanation:
Calculating the Time Difference Between Lightning and Thunder
The question involves the relationship between the speed of light and sound to determine how long it will take for the sound of thunder to reach an observer after the lightning is seen. Since light travels at approximately 300,000 kilometers per second (3 × 108 meters per second), lightning is seen almost instantaneously. However, the sound of thunder, which is caused by the rapid expansion of superheated air due to a lightning strike, travels much slower.
At a temperature of 74.0°F, we need to convert this to Celsius since the speed of sound in air is given by the formula v = 331.4 m/s + 0.6T, where T is in Celsius. The temperature in Celsius can be found through T(°C) = (T(°F) - 32) × 5/9, giving us T(°C) = (74 - 32) × 5/9 = 23.33°C.
The speed of sound at this temperature is v = 331.4 m/s + 0.6 × 23.33 = 345.4 m/s. To find the time it takes for the sound to travel 4.5 miles (or 23,760 feet), we need to convert the distance to meters since the speed of sound is given in meters per second. We have 1 mile = 1,609.34 meters, so 4.5 miles = 7,242 meters approximately. Finally, we calculate the time as time = distance/speed = 7,242 m / 345.4 m/s ≈ 20.97 s.
Therefore, approximately 21 seconds will elapse between seeing the lightning and hearing the thunder if the lightning strikes 4.5 miles away and the air temperature is 74.0°F.
If the mass of an object is measured to be 53.5 ± 0.1 g and its volume is measured to be 22.30 ± 0.05 cm^3 , what is the density? Report the density uncertainty in both forms: as a percentage and as an absolute number with units.
Final answer:
The density of the object is 2.40 g/cm³ with an absolute uncertainty of 0.01 g/cm³, which corresponds to a relative uncertainty of approximately 0.411%.
Explanation:
To calculate the density of an object, you divide the mass by the volume. Given that the mass is 53.5 ± 0.1 g and the volume is 22.30 ± 0.05 cm³, the density can be computed as follows:
Density = Mass/Volume = 53.5 g / 22.30 cm³ = 2.3991 g/cm³
However, when it comes to reporting this value, we need to match the significant figures to the least number in any of the measurements used, which in this case would be three significant figures. Thus, the density is reported as 2.40 g/cm³.
To calculate the uncertainty in density, you combine the relative uncertainties of mass and volume by simply adding them because we're dividing the two quantities. The relative uncertainty of mass is (0.1 g / 53.5 g) * 100% ≈ 0.187%, and the relative uncertainty of volume is (0.05 cm³ / 22.30 cm³) * 100% ≈ 0.224%. Adding these gives us the total uncertainty:
Total relative uncertainty = 0.187% + 0.224% ≈ 0.411%
To find the absolute uncertainty in density, you multiply the total relative uncertainty by the calculated density:
Absolute uncertainty = 0.411% * 2.3991 g/cm³ ≈ 0.00986 g/cm³
Rounded to match the significant figures of the calculated density, this would be 0.01 g/cm³. So, the density reported with its absolute uncertainty is 2.40 ± 0.01 g/cm³.
A speed trap is set up with two pressure-activated
stripsplaced across a highway, 110 m apart. A car is speeding
alongat 33m/s, while the speed limit is 21 m/s. At the instant
thecar activates the first strip, the driver begins slowingdown.
What minimum deceleration is needed in order that theaverage speed
is within the limit by the time the car crosses thesecond
marker?
The question is asking to calculate the required minimum deceleration of a speeding car to fit the speed limit when it crosses the end of a measured stretch of the road. The deceleration required can be calculated using the motion equations of physics by first calculating the time it takes to travel the distance between the two strips and then using this time to calculate the required deceleration.
Explanation:The subject area of this question is kinematics, which is a branch of physics that deals with the concepts of distance, displacement, speed, velocity, and acceleration. The problem is asking for the minimum deceleration the car needs to have in order to not exceed the speed limit when it reaches the second strip. To solve this, you would use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is acceleration (in this case deceleration, so it will be negative), and t is time. But first, we need to figure out the time. The time it takes to travel between the two strips can be calculated by t = d/v. The distance d is given as 110m, and the average speed v we want is 21 m/s (the speed limit).
Once we have the time, we can substitute the values into the first equation and solve for a. The values are v = 21 m/s (the speed we want to have in the end), v0 = 33 m/s (the initial speed), and we've calculated t from the previous step. For deceleration, as the speed is decreasing, a will be a negative number.
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What is the difference in energy between a beta particle at rest and one traveling 0.35c?
Answer:
They have a difference in energy of 35 eV.
Explanation:
The energy at rest of a particle is given by:
[tex]E_{R} = m_{0}c^2[/tex] (1)
Where [tex]m_{0}[/tex] is the mass of the particle at rest and c is the speed of light.
Beta particles are high energy and high velocity electrons or positrons ejected from the nucleus of an atom as a consequence of a radioactive decay. Either if the beta particle is an electron¹ or a positron² it will have the same mass.
Hence, the mass of the beta particle at rest in equation (1) will be equal to the mass of an electron:
[tex]m_{e} = 9.1095x10^{-31} Kg[/tex]
Replacing the values of [tex]m_{e}[/tex] and c in equation (1) it is gotten:
[tex]E_{R} = (9.1095x10^{-31} Kg)(3.00x10^{8} m/s)^{2}[/tex]
[tex]E_{R} = 8.19x10^{-14} Kg.m^{2}/s^{2}[/tex]
But [tex]1 J = Kg.m^{2}/s^{2}[/tex], therefore:
[tex]E_{R} = 8.19x10^{-14} J[/tex]
It is better to express the rest energy in electronvolts (eV):
[tex]1eV = 1.60x10^{-19} J[/tex]
[tex]8.19x10^{-14} J . \frac{1 eV}{1.60x10^{-19} J}[/tex] ⇒ [tex]511.875 eV[/tex]
[tex]E_{R} = 511.875 eV[/tex]
So the energy of the beta particle at rest is 511.875 eV.
Case for the one traveling at 0.35c:
Since it is traveling at 35% of the speed of light it is necessary to express equation (1) in a relativistic way, that can be done adding the Lorentz factor to it:
[tex]E = \frac{m_{0}c^{2}}{sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex] (2)
Where v is the velocity of the particle (for this case 0.35c).
[tex]E = \frac{511.875 eV}{sqrt{1-\frac{(0.35c)^{2}}{c^{2}}}}[/tex]
[tex]E = \frac{511.875 eV}{sqrt{1-\frac{0.1225c^{2}}{c^{2}}}}[/tex]
[tex]E = \frac{511.875 eV}{sqrt{1-0.1225}}[/tex]
[tex]E = \over{511.875 eV}{sqrt{0.8775}}[/tex]
[tex]E = \over{511.875 eV}{0.936}[/tex]
[tex]E = 546.875 eV[/tex]
The difference in energy between the two particles can be determined using the relativistic form of the kinetic energy:
[tex]K = E – E_{R}[/tex] (3)
Where E is the energy of the particle traveling at 0.35c and [tex]E_{R}[/tex] is the energy of the beta particle at rest.
[tex]K = 546.875 eV – 511.875 eV[/tex]
[tex]K = 35 eV[/tex]
They have a difference in energy of 35 eV.
Key terms:
¹Electron: Fundamental particle of negative electric charge.
²Positron: Is an electron with positive electric charge (similar to an electron in all its properties except in electric charge and magnetic moment).
While standing on the roof of a building, a child tosses a tennis ball with an initial speed of 13 m/s at an angle of 35° below the horizontal. The ball lands on the ground 2.6 s later. How tall, in meters, is the building?
Answer:
building height is 52.52 m
Explanation:
given data
initial speed v = 13 m/s
angle = 35°
time = 2.6 s
to find out
how tall is building
solution
we consider here h is height of building
so
initial velocity at angle 35 is express as
u = v sin(θ) ...........1
u = 13 sin35 = 7.45 m/s
so
by distance formula
h = ut + 1/2 at² ...........2
h = 7.45 ( 2.6) + 1/2 × (9.81) × (2.6)²
h = 52.52 m
so building height is 52.52 m
Final answer:
Using the formula for vertical displacement in projectile motion, with an initial velocity of 13 m/s downward and the ball being in the air for 2.6 seconds, the building's height is found to be approximately 52.5 meters.
Explanation:
To solve for the height of the building in the given scenario, we can use the vertical component of the projectile motion. We know that the child throws the tennis ball with an initial speed of 13 m/s at an angle of 35° below the horizontal, and it lands after 2.6 seconds. We can use the formula for the vertical motion under constant acceleration (gravity in this case) to find the height:
Equation for vertical displacement:
s = ut + ½at2
s is the vertical displacement (height of the building), u is the initial vertical velocity, a is the acceleration due to gravity (9.8 m/s2), and t is the time the ball is in the air.
First, we calculate the initial vertical velocity component (u):
u = vinitial × sin(θ)
u = 13 m/s × sin(35°) = 13 m/s × 0.5736 ≈ 7.457 m/s (downward)
Since the initial velocity is downward and we need upward to be positive, we will treat it as negative:
u = -7.457 m/s
Next, we can calculate the height of the building.
s = (-7.457 m/s)(2.6 s) + ½(-9.8 m/s2)(2.6 s)2
s = -19.3886 m - 33.118 m
s = -52.5066 m
Since the displacement is in the negative direction (downward), the height of the building is 52.5 meters (we must take the absolute value of the displacement to get the height).
Your school science club has devised a special event for homecoming. You’ve attached a rocket to the rear of a small car that has been decorated in the green and gold school colors. The rocket provides a constant acceleration for 9.0 s. As the rocket shuts off, a parachute opens and slows the car at a rate of 5.0 m/s2 . The car passes the judges’ box in the center of the grandstand, 990 m from the starting line, exactly 12 s after you fire the rocket. What is the car’s speed as it passes the judges
Answer:
V=120m/s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X (2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem you must divide the problem into two parts 1 and 2, when the rocket accelerates (1), and when the rocket decelerates (2).
Then you raise equations 3 and 1 in both parts.
finally you use algebraic methods to find the value of speed
I attach the complete procedure