Answer:
the correct answer is option C which is 50 units.
Explanation:
given,
two vector of magnitude = 30 units and of 70 units
to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}
cos θ value varies from -1 to 1
so, resultant vector
=[tex]\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}[/tex]
a = 30 units and b = 70 units
= [tex]\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}[/tex]
= 40 units to 100 units
hence, the correct answer is option C which is 50 units.
A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h
Final answer:
To find the shortest time for the passing maneuver, we must consider the car's acceleration and deceleration phases and the distance to be traveled. The task involves calculations using the kinematic equations for uniformly accelerated motion.
Explanation:
To calculate the shortest time in which the driver of the car can complete the passing operation, we must consider the car's acceleration, maximum speed, and the distance to be covered. The car must not only reach a point 40 feet in front of the truck but also get back to the speed of 35 mi/h after overtaking, all while not exceeding a speed of 50 mi/h. The entire maneuver consists of accelerating to the maximum speed, maintaining that speed for a portion of the pass, and then decelerating back to 35 mi/h.
First, convert speeds from miles per hour to feet per second:
35 mi/h = 51.33 ft/s (approximately),
50 mi/h = 73.33 ft/s (approximately).
The car needs to cover an initial 40 ft behind the truck and a further 40 ft to be ahead, for a total of 80 ft. We would calculate the time needed to accelerate to 50 mi/h (maximum passing speed), the time spent traveling at this top speed, and then the time to decelerate back to 35 mi/h, all while covering the required distance. The actual calculations involve using the kinematic equations for uniformly accelerated motion; however, without further details on the constraints such as road length, traffic laws, and exact acceleration and deceleration phases, a precise number cannot be produced.
While John is traveling along a straight interstate highway, he notices that the mile marker reads 245 km. John travels until he reaches the 135 km marker and then retraces his path to the 176 km marker. What is John’s resultant displacement from the 245 km marker? Answer in units of km.
Answer:Displacement =69 km
Explanation:
Given
First John notices 245 km marker i.e. he is 245 away from destination.
John travels until he reach 135 km marker i.e. he is 135 km away from destination.
After that he retraces his path to reach 176 km marker
so his resultant displacement is i.e. shortest distance between initial and final point is 245-176 =69 km
An Earth satellite needs to have its orbit changed so the new orbit will be twice as far from the center of Earth as the original orbit. The new orbital period will be twice as long as the original period. O true O false
Answer:
False.
Explanation:
From Kepler's Third Law of plenetary motion, we know that:
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."
Or, as expressed in mathematical terms:
[tex]\frac{a^3}{T^2}=constant[/tex], where a is the semi-major axis of the orbit (the distance from the center), and T is the orbital period of the satellite.
From this expression we can clearly see that if the orbit's semi-major axis is doubled, orbital period will be [tex]\sqrt{8}[/tex] times longer to compensate the variation.
A simple AC generator consists of a single loop of wire rotating 50 times per second within a magnetic field. The loop is a rectangle 5cm by 12 cm. The field strength is 0.21 T. What is the peak output emf?
Answer:
39.56 x 10⁻² V
Explanation:
Peak emf in a rotating coil
= n BAω
Where n is no of turns of coil , B is magnetic field , A is area of coil and ω is angular velocity of rotation of coil
Here n = 1
B = .21 T
A = 5 X 12 X 10⁻⁴ = 60 X 10⁻⁴
ω = 2π X 50 = 100π
Emf = 1 x .21 x 60 x 10⁻⁴ x 100 x 3.14
= 39.56 x 10⁻² V
Suppose you take a 50gram ice cube from the freezer at an initial temperature of -20°C. How much energy would it take to completely vaporize the ice cube? (Hint: think of this process as four separate steps and calculate the energy needed for each one.)
Answer:
The amount of energy required is [tex]152.68\times 10^{3}Joules[/tex]
Explanation:
The energy required to convert the ice to steam is the sum of:
1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.
2) Latent heat required to convert the ice into water.
3) Energy required to raise the temperature of water from 0 degrees to 100 degrees
4) Latent heat required to convert the water at 100 degrees to steam.
The amount of energy required in each process is as under
1) [tex]Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules[/tex]
where
[tex]'S.heat_{ice}[/tex]' is specific heat of ice =[tex]2.05J/^{o}C\cdot gm[/tex]
2) Amount of heat required in phase 2 equals
[tex]Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules[/tex]
3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals
[tex]Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules[/tex]
where
[tex]'S.heat_{water}[/tex]' is specific heat of water=[tex]4.186J/^{o}C\cdot gm[/tex]
4) Amount of heat required in phase 4 equals
[tex]Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\[/tex][tex]\\\\\\\\[/tex]Thus the total heat required equals [tex]Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules[/tex]
It would take 200 kJ of energy to completely vaporize the ice cube.
Explanation:The process of completely vaporizing an ice cube can be broken down into four separate steps, each requiring a certain amount of energy:
Heating the ice cube from -20°C to 0°C: This step requires 4.1 kJ of energy.Melting the ice at 0°C: This step requires 133.6 kJ of energy.Heating the resulting water from 0°C to 37°C: This step requires 61.9 kJ of energy.Vaporizing the water at 37°C: This step requires an additional 0 kJ of energy, since the water is already at its boiling point.In total, it would take 200 kJ of energy to completely vaporize the ice cube.
You are riding in an elevator on the way to the
eighteenthfloor of your dormitory. The elevator is accelerating
upwardwith a = 1.90 m/s2. Beside you is the
boxcontaining your new computer; box and contents have a total
mass28.0 kg. While the elevator is accelerating upward, you
pushhorizontally on the box to slide it at constant speed toward
thedoor. If the coefficient of kinetic friction between the boxand
elevator floor is mk = 0.32, what magnitudeof
force must you apply?
Answer:
F = 104.832 N
Explanation:
given,
upward acceleration of the lift = 1.90 m/s²
mass of box containing new computer = 28 kg.
coefficient of friction = 0.32
magnitude of force = ?
box is moving at constant speed hence acceleration will be zero.
Now force acting due to lift moving upward =
F = μ m ( g + a )
F = 0.32 × 28 × ( 9.8 + 1.9 )
F = 104.832 N
hence, the force applied should be equal to 104.832 N
While you are driving down a country road at dusk with a velocity of 22 m/s, a deer suddenly runs across the road in front of you. You apply the brakes, reducing your velocity to 7 m/s in 5 s as you avoid hitting the deer. What is the acceleration of the car?
Answer:
[tex]a=\frac{v_{f}-v_{o} }{t_{f}-t_{o} } =\frac{(7-22)\frac{m}{s} }{5s-0s} =-3\frac{m}{s^{2} }[/tex]
Explanation:
We know that the initial velocity is 22 m/s
The final velocity after applying the brakes is 7 m/s, and the time that takes it to break is 5 seconds
We know that the concept of acceleration is the change of velocity in time
[tex]a=\frac{v_{f}-v_{o} }{t_{f}-t_{o} } =\frac{(7-22)\frac{m}{s} }{5s-0s} =-3\frac{m}{s^{2} }[/tex]
The answer is negative since the car is slowing down
Two identical cylinders, A and B, contain the same type of gas at the same pressure. Cylinder A has twice as much gas as cylinder B. Which is true? Two identical cylinders, A and B, contain the same type of gas at the same pressure. Cylinder A has twice as much gas as cylinder B. Which is true? TA > TB TA < TB TA = TB Not enough information to make a comparison.
Answer:
Explanation:
Gas law equation for any gas is as follows .
PV = nRT ( for n moles of gas )
Let the pressure , volume , temperature and mole of gas in cylinder A be P,V
T₁ and 2n . Pressure , volume , temperature and mole in cylinder B will be
P , V , T₂ and n.
Applying Gas laws , we get
For gas in cylinder A
PV = 2n R T₁
For gas in cylinder B
PV = n R T₂
Equating these equations , we get
2n R T₁ = n R T₂
2 T₁ = T₂
or,
[tex]2T_A =T_B[/tex]
[tex]T_A[/tex] is less than [tex]T_B[/tex]
Final answer:
Under the conditions of equal pressure and volume, and using the ideal gas law, we determine that the temperatures of the two cylinders of gas must be the same (TA = TB).
Explanation:
Given that the two identical cylinders, A and B, contain the same type of gas at the same pressure, and cylinder A has twice as much gas as cylinder B, we can use the ideal gas law to determine the relationship between their temperatures. The ideal gas law is PV = n ×R× T, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. Since the pressures are equal and the cylinders are identical, volumes are also equal. Thus, because A has twice as much gas as B (nA = 2 x nB), it has twice the number of moles. Holding pressure and volume constant and since R is a constant, both cylinders must have the same temperature for the equation to be balanced. Therefore, we find that TA = TB.
What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
Explanation:
There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.
The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.
While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.
So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.
A student throws a softball horizontally from a dorm window 15.0 m above the ground. Another student standing 10.0 m away catches the ball at a height of 1.50 m above the ground. What is the initial speed of the ball?
Answer:
Speed of ball equals 6.024 m/s.
Explanation:
Let the student throw the ball with a velocity of 'v' m/s horizontally.
Now the time in which the ball travels 10.0 meter horizontally shall be equal to the time in which it travels (15.0-1.50) meters vertically
Hence the time taken to cover a vertical distance of 13.50 meters is obatined using 2 equation of kinematics as
[tex]s=\frac{1}{2}gt^{2}\\\\t=\sqrt{\frac{2s}{g}}\\\\t=\sqrt{\frac{2\times 13.5}{9.81}}\\\\\therefore t=1.66 seconds[/tex]
Since there is no acceleration in horizantal direction we infer that in time of 1.66 seconds the ball travels a distance of 10 meters
Hence the spped of throw is obatines as
[tex]Speed=\frac{Distance}{Time}\\\\v=\frac{10}{1.66}=6.024m/s[/tex]
Final answer:
The initial speed of the softball can be calculated using the height it drops and the horizontal distance it travels. By determining the time of fall for the vertical drop and knowing the horizontal distance, the initial horizontal velocity can be found, approximately 6.02 m/s.
Explanation:
To solve for the initial speed of the softball, we need to analyze the motion in two dimensions separately: horizontal and vertical.
Horizontal Motion
The horizontal motion can be considered with constant velocity since there's no acceleration in that direction (ignoring air resistance).
Vertical Motion
The vertical motion can be described by the kinematic equations for uniformly accelerated motion (free fall). Given the height difference from the window to the catchpoint (13.5 m) and the acceleration due to gravity (9.81 m/s²), we can calculate the time it takes for the ball to fall this height. The equation we use is:
h = vit + (1/2)at²
Substitute h = 13.5 m, a = 9.81 m/s², and vi = 0 m/s (since the ball is thrown horizontally, the initial vertical speed is 0) to solve for t.
We find that the time t is approximately 1.66 seconds. Using this time and the horizontal distance of 10.0 m, we can now calculate the initial horizontal speed.
vh = d/t
Substituting d = 10.0 m and t = 1.66 s, the initial horizontal speed vh is approximately 6.02 m/s.
Two point charges are fixed on the y axis: a negative point charge q1 = -24 µC at y1 = +0.19 m and a positive point charge q2 at y2 = +0.33 m. A third point charge q = +8.0 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 18 N and points in the +y direction. Determine the magnitude of q2.
Answer:
4.51 * 10^{-5} C
Explanation:
The force between two charge q and q1 is given as
[tex]F = \frac{k*q*q1}{r^2}[/tex]
[tex]= \frac{(9.0 * 10^9)(24 * 10^{-6})(8 * 10^{-6} C)}{(0.19m)^2} [/tex]
= 47.86 N in the +y direction
We need the force between q and q2 to be (47.86 - 18) = 29.86 N in the other direction to get the desired result.
solving for q2,
[tex]q2 = \frac{Fr^2}{(kq)} [/tex]
[tex]= \frac{(29.86)(0.33 m)^2}{(9.0 * 10^9*8*10^{-6} C)}[/tex]
[tex]= 4.51 * 10^{-5} C[/tex]
Final answer:
To determine the magnitude of q2, we can use Coulomb's Law to calculate the net force exerted on charge q. By setting up an equation using the given information, we can find that the magnitude of q2 is approximately 4.84 μC.
Explanation:
To determine the magnitude of q2, we can use Coulomb's Law, which states that the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, we have two charges, q1 and q2, with a distance of 0.33 - 0.19 = 0.14m between them.
The net electrostatic force on charge q can be calculated using the equation F = k * (|q1 * q| / r1^2) + k * (|q2 * q| / r2^2), where k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2).
From the given information, we know that F = 18 N and points in the +y direction, so we can set up the following equation: 18 = k * (|(-24 μC) * (8 μC)| / (0.19m)^2) + k * (|q2 * (8 μC)| / (0.33m)^2).
Solving this equation, we can find the magnitude of q2, which is approximately 4.84 μC.
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to earth. (a) Find the rocket height and velocity when it runs out of fuel. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes into the ground. (d) Find the total elapsed time from launch to ground impact.
Answer:
a) 200m, 100m/s
b) 710.20m
c) -117.98 m/s
d) 26.24 s
Explanation:
To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:
[tex]V=Vo+a*t[/tex] (1)
[tex]X=Xo+Vo*t+\frac{1}{2}*a*t^2\\[/tex] (2)
[tex]V^2=Vo^2+2*a*X[/tex] (3)
where:
Vo is initial velocity
Xo=intial position
V=final velocity
X=displacement
a)
[tex]X=0+0*4+\frac{1}{2}*25*4^2[/tex]
the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.
[tex]X=200m[/tex]
[tex]V=0+25*4[/tex]
[tex]V=100m/s[/tex]
b)
The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8[tex]\frac{m}{s^2}[/tex] because it is going downward.
[tex]0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m[/tex]
c)
In order to find the velocity when it crashes, we can use the formula (3).
the initial velocity is 0 because in that moment is starting to fall.
[tex]V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s[/tex]
the minus sign means that the object is going down.
d)
We can find the total amount of time adding the first 4 second and the time it takes to going down.
to calculate the time we can use the formula (2) setting the reference at 200m:
[tex]-200=0+100*t+\frac{1}{2}*(-9.8)*t^2[/tex]
solving this we have: time taken= 22.24 seconds
total time is:
total=22.24+4=26.24 seconds.
You drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. Your average speed over the 12 km drive will be ____
(A) a greater than 70 km/h.
(B) less than 70 km/h.
(C) exactly 38 km/h.
(D) It cannot be determined from the information given because we must also know directions traveled.
(E) equal to 70 km/h.
Answer:
The average speed is less than 70 km/h.
(B) is correct option.
Explanation:
Given that,
distance = 6.0 km
Speed = 50 km/h
Speed = 90 km/h
We need to calculate the time in 6.0 km distance
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
Put the value in to the formula
[tex]t=\dfrac{6.0}{50}[/tex]
[tex]t=0.12\ hr[/tex]
We need to calculate the time in another distance
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
Put the value in to the formula
[tex]t=\dfrac{6.0}{90}[/tex]
[tex]t=0.067\ hr[/tex]
We need to calculate the average speed
Using formula of average speed
[tex]v=\dfrac{D}{T}[/tex]
Where, D = total distance
T = total time
Put the value into the formula
[tex]v=\dfrac{12}{0.12+0.067}[/tex]
[tex]v=64.17\ km/hr[/tex]
Hence, The average speed is less than 70 km/h.
The average speed over the 12 km drive is less than 70 km/h since the total time for the trip is 0.187 hours, leading to an average speed of 64.2 km/h.
To calculate the average speed for the entire trip, you must divide the total distance by the total time taken. In the given scenario, you drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. First, calculate the time taken for each portion of the trip.
For the first 6.0 km at 50 km/h: time = distance/speed = 6.0 km / 50 km/h = 0.12 hours.For the second 6.0 km at 90 km/h: time = distance/speed = 6.0 km / 90 km/h = 0.067 hours.Now, add the times together and divide the total distance by the total time to find the average speed.
Total time = 0.12 hours + 0.067 hours = 0.187 hours.Average speed = Total distance / Total time = 12 km / 0.187 hours = 64.2 km/h.Therefore, the average speed over the 12 km drive is less than 70 km/h, which corresponds to option (B).
A 75.0 kg stunt man jumps from a balcony and falls 24.0
mbefore colliding with a pile of mattresses. If the mattresses
arecompressed 1.15 m before he is brought to rest, what is the
averageforce exerted by the mattresses on
thestuntman?
Answer:
The force that acts on the man equals 15354.75 newtons.
Explanation:
After falling through a distance of 24.0 meters the speed of the stunt man upon hitting the mattress can be obtained using third equation of kinematics as
[tex]v^{2}=u^{2}+2gh\\\\\therefore v=\sqrt{2gh}\\\\v=\sqrt{2\times 9.81\times 24}\\\\v=21.7m/s[/tex] ( u=0 since the man falls from rest)
Now since the man is decelerated through a distance of 1.15 meters thus the acceleration produced can be obtained from third equation of kinematics as
[tex]v^{2}=u^{2}+2as\\\\0=u^{2}=2as\\\\a=\frac{-u^{2}}{2s}\\\\a=-\frac{(21.7^{2}}{2\times 1.15}=-204.73m/s^{2}[/tex]
Now by newton's second law the force that produced deceleration of the calculated magnitude is obtained as
[tex]F=mass\times acceleration\\\\F=75.0\times -204.73=-15354.75Newtons[/tex]
The negative sign indicates that the direction of force is opposite of motion.
A computer-controlled ball launcher on a ledge at height h = 8 m above level ground ejects a ball at initial speed v_sub_0 = 8.2 m/s at an angle of θ_sub_1 = 35 degrees above the horizontal. The computer then commands the launcher to change its launch angle to a specific value θ_sub_2 below the horizontal, and to launch a second ball exactly 1.00 seconds after the first ball at a different speed from the first one. Assume the balls are launched from the same height, move in the same plane, and that air resistance can be ignored. The launch of the second ball is arranged such that the second ball collides with the first ball at a height h/2 above the level ground. This requires the second ball to have a specific initial speed and launch angle. What is the required initial speed?
Answer:
21 m/s
Explanation:
For the first ball, in the x direction:
x = x₀ + v₀ t + ½ at²
x = 0 + (8.2 cos 35) t + ½ (0) t²
x = 6.72t
In the y direction:
y = y₀ + v₀ t + ½ at²
y = 8 + (8.2 sin 35) t + ½ (-9.8) t²
y = 8 + 4.70t − 4.9t²
When y = 4:
4 = 8 + 4.70t − 4.9t²
4.9t² − 4.70t − 4 = 0
Solve for t with quadratic formula:
t = [ 4.70 ± √((-4.70)² − 4(4.9)(-4)) ] / 9.8
t = (4.70 ± 10.0) / 9.8
t = 1.50
Therefore:
x = 6.72t
x = 10.1
Now, for the second ball in the x direction:
x = x₀ + v₀ t + ½ at²
x = 0 + (v₀ cos (-θ)) (t − 1) + ½ (0) (t − 1)²
x = v₀ cos θ (t − 1)
And in the y direction:
y = y₀ + v₀ t + ½ at²
y = 8 + (v₀ sin (-θ)) (t − 1) + ½ (-9.8) (t − 1)²
y = 8 − v₀ sin θ (t − 1) − 4.9(t − 1)²
When t = 1.50, x = 10.1 and y = 4:
10.1 = v₀ cos θ (1.50 − 1)
v₀ cos θ = 20.1
4 = 8 − v₀ sin θ (1.50 − 1) − 4.9(1.50 − 1)²
4 = 6.76 − 0.50 v₀ sin θ
v₀ sin θ = 5.49
Using Pythagorean theorem:
v₀² = (v₀ cos θ)² + (v₀ sin θ)²
v₀² = (20.1)² + (5.49)²
v₀ = 20.8
Rounded to two significant figures, the required initial speed is 21 m/s.
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.4 m/s at an angle of 24° above the horizontal. It is released 0.70 m above the floor. Q1: What horizontal distance does the ball cover before bouncing? please show how to solve for t in the problem.
Answer:
The horizontal distance is 2.41 mts
Explanation:
For this problem, we will use the formulas of parabolic motion.
[tex]Y=Yo+Voy*t+\frac{1}{2}*a*t^2\\Vx=V*cos(\alpha)\\Vy=V*sin(\alpha)[/tex]
We need to find the time of the whole movement (t), for that we will use the first formula:
we need the initial velocity for that:
[tex]Vy=4.4*sin(24^o)\\Vy=1.79m/s[/tex]
so:
[tex]0=0.70+1.79*t+\frac{1}{2}*(-9.8)*t^2[/tex]
now we have a quadratic function, solving this we obtain two values of time:
t1=0.60sec
t2=-0.234sec
the obvious value is 0.60sec, we cannot use a negative time.
Now we are focusing on finding the horizontal distance.
the movement on X is a constant velocity motion, so:
[tex]x=Vx*t[/tex]
[tex]Vx=4.4*cos(24^o)=4.02m/s\\[/tex]
so:
[tex]x=4.02*(0.60)=2.41m[/tex]
A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 86 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.0 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 18 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation is of the form y(x, t) = ym sin(kx ± ωt + φ), what are (c) ym, (d) k, (e) ω, (f) φ, and (g) the correct choice of sign in front of ω?
Final answer:
The frequency of the wave is 1075 Hz, and the wavelength is 0.08 m. The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave with specific parameters. The amplitude is 0.04 m, the wave number is 25π m⁻¹, the angular frequency is 6755π rad/s, and the phase angle can be either +π or -π. The choice of sign in front of ω determines the direction of wave propagation.
Explanation:
To find the frequency of the wave, we can use the formula f = v/λ, where f is the frequency, v is the speed of the wave, and λ is the wavelength. In this case, the speed of the wave is given as 86 m/s. Since the wave is sinusoidal, the wavelength is equal to twice the amplitude of the transverse displacement of the string particle at x = 0. So, the wavelength is equal to 2 times 4.0 cm, which is 8.0 cm or 0.08 m.
Substituting these values into the formula, we get f = 86 m/s / 0.08 m = 1075 Hz. Therefore, the frequency of the wave is 1075 Hz.
To find the wavelength of the wave, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of the wave, and f is the frequency. Substituting the given values, we get λ = 86 m/s / 1075 Hz = 0.08 m.
Therefore, the wavelength of the wave is 0.08 m.
The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave. In this equation, ym is the amplitude of the wave, k is the wave number, ω is the angular frequency, and φ is the phase angle. The choice of the sign in front of ω determines the direction of wave propagation. If it is positive, the wave propagates in the positive x-direction, and if it is negative, the wave propagates in the negative x-direction.
In this case, the amplitude of the wave is given as 4.0 cm or 0.04 m. The wave number can be calculated using the formula k = 2π/λ, where k is the wave number and λ is the wavelength. Substituting the given wavelength of 0.08 m, we get k = 2π/0.08 m = 25π m⁻¹.
The angular frequency can be calculated using the formula ω = 2πf, where ω is the angular frequency and f is the frequency. Substituting the given frequency of 1075 Hz, we get ω = 2π × 1075 Hz = 6755π rad/s.
The phase angle is given as ± π. The choice between +π and -π determines the phase of the wave at x = 0 and t = 0.
Therefore, (c) ym = 0.04 m, (d) k = 25π m⁻¹, (e) ω = 6755π rad/s, (f) φ = ±π, and (g) the correct choice of sign in front of ω depends on the desired direction of wave propagation.
A 3730-kg SUV can climb up a 600-m high hill in one minute (60 seconds). How much is its power output in horse-powers? 245 hp
490 hp
800 hp
373 hp
600 hp
Answer:
490hp
Explanation:
Power is energy per unit of time, an in this case the energy needed is the gravitational potential energy, so we have:
[tex]P=E/t=mgh/t=(3730kg)(9.8m/s)(600m)/60s=365540W[/tex]
Since all units are in S.I. we get our result in Watts (Joules/s). To convert to horse power (imperial), we need to know that:
[tex]745.7 W = 1 hp[/tex]
Which obviously means:
[tex]\frac{1 hp}{745.7 W} = 1[/tex]
So we can write:
[tex]P=365540W=365540W(\frac{1 hp_m}{745.7 W})=490hp[/tex]
There is also metric horse power ([tex]735.5 W = 1 hp[/tex]), and using this value we get [tex]P=497hp[/tex], although both results are close.
Final answer:
To calculate the SUV's power output, we find the work done against gravity and divide it by the time taken to get power in watts. This is then converted to horsepowers leading to the answer of 490 hp. The correct answer is b) 490 hp.
Explanation:
To find the power output in horsepowers of the 3730-kg SUV climbing a 600-meter hill in 60 seconds, first we calculate the work done against gravity, which is equal to the force due to gravity (weight of the SUV) multiplied by the height of the hill:
Work = Weight × Height = (mass × gravity) × height
Then, we convert the weight to newtons by multiplying the mass (kg) by the acceleration due to gravity (9.81 m/s2, approximately), and multiply that result by the height (in meters).
Power in watts is calculated by dividing the work done by the time taken in seconds:
Power (W) = Work done (Joules) / Time (seconds)
To convert watts to horsepowers, we use the conversion rate where 1 horsepower is equivalent to 746 watts:
Power (hp) = Power (W) / 746
After plugging in the values, the SUV's power output in horsepowers would be:
Power (W) = (3730 kg × 9.81 m/s2) × 600 m / 60 s = 365,130 W
Power (hp) = 365,130 W / 746 = 489.43 hp, which we round to 490 hp.
Thus, the correct answer from the given options is 490 hp.
The average American uses electrical energy at the rate of about 1.5 kW. Solar energy reaches Earth’s surface at an average rate of about 300 watts on every square meter (a value that accounts for nights and clouds). What fraction of the United State’s land area would have to be covered with 20% efficient solar cells to provide all of our electricity?
Answer: the fraction is [tex]0.88*10^{-6}[/tex]
Explanation:
Hi!
The land area per capita in USA is 28,310 square km. Then the solar power per capita is (rounding some numbers):
[tex]P_s = 300 \frac{W}{m^2} (28,310) km^2 = 3*10^2*2.83*10^4*10^6 W = 8.49*10^{12} W = 8.49*10^9 kW[/tex]
If we take 20% of this power, the fraction k to have 1.5 kW is:
[tex]1.5 kW = k(0.2*8.49 * 10^9 kW)[/tex]
[tex]k = \frac{1.5}{1.7}*10^{-6} = 0.88 *10^{-9} [/tex]
Final answer:
To answer the question, we need to calculate the total electrical energy consumption in the U.S., understand the solar energy per square meter, consider the 20% efficiency of solar cells, and determine the needed land area for solar panels to meet the country's electrical energy needs.
Explanation:
The question involves calculating what fraction of the United States' land area would need to be covered with 20% efficient solar cells to provide all of its electricity, given that the average American uses electricity at the rate of about 1.5 kW and solar energy reaches the Earth's surface at an average rate of about 300 watts per square meter. To find this fraction, we first need to consider the total electrical energy consumption of the United States and then assess how much of this demand can be met by the solar energy available on a given area of land, taking into consideration the efficiency of the solar cells.
The key steps include calculating the total power used by Americans in watts, understanding the energy provided by the sun per square meter, adjusting for the efficiency of solar conversion, and finally, determining the land area required. Specifically, a 20% conversion efficiency means that only a fifth of the solar energy striking the solar cells is converted into usable electrical energy. With these considerations, the solution entails a mix of energy requirement calculations and solar energy potential assessments to identify the required land area for solar panels.
X-rays are scattered from a target at an angle of 37.2° with the direction of the incident beam. Find the wavelength shift of the scattered x-rays.
Answer:
[tex]\Delta \lambda=4.94\times 10^{-13}\ m[/tex]
Explanation:
It is given that,
X-rays are scattered from a target at an angle of, [tex]\theta=37.2^{\circ}[/tex]
We need to find the wavelength shift of the scattered x-rays. The shift in wavelength is given by :
[tex]\Delta \lambda=\dfrac{h}{mc}(1-cos\ \theta)[/tex]
h is the Planck's constant
m is the mass of electron
c is the speed of light
[tex]\Delta \lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}(1-cos(37.2))[/tex]
[tex]\Delta \lambda=4.94\times 10^{-13}\ m[/tex]
So, the wavelength shift of the scattered x- rays is [tex]4.94\times 10^{-13}\ m[/tex]. Hence, this is the required solution.
James is running toward the base of a tall canyon wall at 2 m/s while screaming at 355 Hz. The wind is blowing the opposite direction at 3 m/s. What beat frequency does he hear from his own echo off the wall? Use 350 m/s for the speed of sound.
Answer:
[tex]f_{ecco} = 360 Hz[/tex]
Explanation:
The change of frecuency of sound due to the movement of the source is colled Doppler Effect.
As James (the source) is running toward the wall, the frecuency reaching the wall (so the eco sound) will be higher than the source. In this case the frecuency at the wall will be:
[tex]f_{2} = f_{1} (\frac{v}{v-v_{s} } )[/tex]
where [tex]v_{s}[/tex] is the speed of source, 2 m/s
and [tex]v[/tex] is the speed of sound, given that we have wind movind the air in the opposite direction respect to the wall, the speed of sound would be:
[tex]v = 350 \frac{m}{s} - 3 \frac{m}{s} = 347 \frac{m}{s}[/tex]
Replacing the values: [tex]f_{2} = 357 Hz[/tex]
Now the wall becames the new source, and James (the observer is aproaching the source), for an observer aproaching the source the new frecuency will be:
[tex]f_{3} = f_{2} (1 + \frac{v_{s}}{v} )[/tex]
Now the waves are traveling in the direction of wind, so the velocity of sound will be:
[tex]v = 350 \frac{m}{s} + 3 \frac{m}{s} = 353 \frac{m}{s}[/tex]
Replacing:
[tex]f_{3} = 360 Hz[/tex]
Objects that are thrown upwards as they gain height their speed goes: a. Increasing
b. Staying the same
c. Decreasing
d. Duplicating the initial
Answer:
The correct option is 'c': Decreases.
Explanation:
As we know that when we throw a ball upwards we imoart a kinetic energy to the bsll. As the ball moves upwards the kinetic energy of the ball starts to decrease as the ball slows down as it moves upwards as work has to be done for movement against gravity. This work done against gravity is stored as potential energy of the ball.
Mathematically as we throw the ball upward's with velocity 'u' it's initial kinetic energy equals [tex]E_{initial}=\frac{1}{2}mu^{2}[/tex]
As the ball attain's a height 'h' it's total energy is sum of the potential and the remaining kinetic energy as follows
[tex]E(h)=mgh+\frac{1}{2}mv^{2}[/tex]
Equating both the energies we get
[tex]\frac{1}{2}mu^{2}=mgh+\frac{1}{2}mv^{2}\\\\\\therefore v=\sqrt{u^{2}-2gh}[/tex]
Hence we can see as the height increases the velocity decreases.
Given a particle that has the acceleration a(t) = 4e3 meters, a. Find the displacement at 2 seconds, assuming an initial position of O & from rest. b. Find velocity at 1 second assuming starting from rest.
Answer: Hi!
The acceleration of our particle is then a = 4*[tex]10^{3[/tex] meters per second square. Also we know that we start at X₀ = O m (i am not sure if you are using a a 0 or a O, but you will see that in this problem it doesn't matter, se i will use a constant O, that can be ani real number as the initial position) and V₀ = 0 m/s.
Ok, we have the acceleration, so if we want the velocity, we must integrate the acceleration over time.
V(t) = ∫ 4*[tex]10^{3[/tex]dt = 4*[tex]10^{3[/tex]*t + V₀
where we added the constant of integration, who is the inicial velocity, that we already know that is zero.
and for the position, we must integrate again.
X(t) = ∫ 4*[tex]10^{3[/tex]*tdt = 4*[tex]10^{3[/tex]*[tex]t^{2}[/tex]/2 + X₀
ok, in the part a) we want to know the displacement at 2 seconds.
that is X(2s) - X(0s) = (2000*4 + O) - O meters = 8000 meters.
in the b part we want to know the velocity at 1 second; that is:
V(1s) = 4000*1 m/s = 4000 meters per second.
A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in the car's momentum?
Answer:
Change in momentum is [tex]2.667\times 10^{4} kg.m/s[/tex]
Solution:
The momentum of any body is the product of its mass and the velocity associated with the body and is generally given by:
[tex]\vec{p} = m\vec{v}[/tex]
Now, as per the question:
Mass of the car, M = 1500 kg
The velocity in the east direction, [tex]v\hat{i} = 40\hat{i} km/h[/tex]
The velocity in the north direction, [tex]v\hat{j} = 50\hat{j} km/h[/tex]
Now, the momentum of the car in the east direction:
[tex]p\hat{i} = mv\hat{i} = 1500\times 40\hat{i} = 60000\hat{i} kg.km/h[/tex]
Now, the momentum of the car in the north direction:
[tex]p\hat{j} = mv\hat{j} = 1500\times 50\hat{j} = 75000\hat{j} kg.km/h[/tex]
Change in momentum is given by:
[tex]\Delta p = p\hat{i} - p\hat{j} = 60000\hat{i} - 75000\hat{j}[/tex]
Now,
[tex]|\Delta p| = |60000\hat{i} - 75000\hat{j}|[/tex]
[tex]|\Delta p| = \sqrt{60000^{2} + 75000^{2}}[/tex]
[tex]|\Delta p| = 96046 kg.km/hr = \frac{96046}{3.6} = 2.667\times 10^{4} kg.m/s[/tex]
(Since, [tex]1kg.km/h = \frac{1}{3.6} kg.m/s[/tex])
At a certain instant, a particle is moving in the direction with momentum 18 kg·m/s. During the next 0.5 s, a constant force <-4, 12, 0> N acts on the particle. What is the momentum of the particle at the end of this 0.5 s interval?
Answer:
24.325 kg m/s
Explanation:
Initial momentum, pi = 18 kg m/s
F = < -4, 12, 0>
t = 0.5 s
Let the final momentum is pf.
The magnitude of force is
[tex]F=\sqrt{(-4)^{2}+12^{2}+0^{2}}=12.65 N[/tex]
According to the Newton's second law, the rate of change of momentum is equal to the force.
[tex]F = \frac{p_{f}-p_{1}}{t}[/tex]
[tex]12.65= \frac{p_{f}-18}}{0.5}[/tex]
pf - 18 = 6.325
pf = 24.325 kg m/s
Thus, the momentum of body after 0.5 s is 24.325 kg m/s.
Imagine that you are standing 5 feet from, and looking directly toward, a brass ball (a convex mirror) 1 foot in diameter. Describe the image you world see in the ball (a) Determine the image distance. (b) Is this a real or virtual image?
Answer:
virtual the image is virtual because image formed viruvirualconvex mirror
A 4000-kg truck traveling with a velocity of 20 m/s due south collides head-on with a 1350-kg car traveling with a velocity of 10 m/s due north. The two vehicles stick together after the collision. What is the momentum of each vehicle prior to the collision?
Answer:Momentum of Truck[tex]=4000\times 20=80,000 kg-m/s[/tex]
Momentum of car[tex]=1350\times 10=13,500 kg-m/s[/tex]
Explanation:
Given
mass of truck(m)=4000 kg
Velocity of Truck is ([tex]V_T[/tex])=[tex]-20\hat{j}[/tex]
mass of car ([tex]m_c[/tex])=1350 kg
Velocity of car[tex](V_c)=10 \hat{j}[/tex]
Conserving momentum
[tex]4000\times \left ( -20\right )+1350\left ( 10\right )=5350v[/tex]
[tex]v=\frac{66,500}{5350}=12.42 m/s[/tex]
Momentum of Truck[tex]=4000\times 20=80,000 kg-m/s[/tex]
Momentum of car[tex]=1350\times 10=13,500 kg-m/s[/tex]
Answer:
1) Momentum of truck before collision is [tex]\overrightarrow{p_{1}}=-80000kgm/s[/tex]
2) Momentum of car before collision is [tex]\overrightarrow{p_{2}}=13500kgm/s[/tex]
Explanation:
The momentum of an object with mass 'm' travelling with speed 'v' is mathematically given by
[tex]\overrightarrow{p}=mass\times \overrightarrow{v}[/tex]
In this problem we shall assume that direction's coincide with the Cartesian axis for simplicity
Thus for Truck we have
Mass = 4000 kg
Velocity = [tex]-20\widehat{j}[/tex]m/s
Thus momentum of truck becomes
[tex]\overrightarrow{p_{1}}=4000\times -20\overrightarrow{j}=-80000kgm/s[/tex]
Similarly for car we have
Mass = 1350 kg
Velocity = [tex]10\widehat{j}[/tex]m/s
Thus momentum of car becomes
[tex]\overrightarrow{p_{2}}=1350\times 10\overrightarrow{j}=13500kgm/s[/tex]
The magnitude of a negative of a vector is negative. O True O False
Answer:
The given statement is false.
Explanation:
For any negative vector
[tex]\overrightarrow{r}=-x\widehat{i}-y\widehat{j}[/tex]
The magnitude of the vector is given by
[tex]|r|=\sqrt{(-x)^{2}+(-y)^{2}}\\\\|r|=\sqrt{x^2+y^2}[/tex]
As we know that square root of any quantity cannot be negative thus we conclude that the right hand term in the above expression cannot be negative hence we conclude that magnitude of any vector cannot be negative.
If the price of gasoline at a particular station in Europe is 5 euros per liter. An American student in Europe is allowed to use 40 euros to buy gasoline. Knowing that 4 quarts make a gallon, and that 1 liter is about 1 US liquid quart, about how many gallons can the student buy?
Answer:
Number of gallons =2 gallon
Explanation:
given data:
rate of gasoline ineurope = 5 euro per liter
total money to buy gasoline = 40 euro
total gasoline an american can buy in europe [tex]= \frac{40}{5}[/tex]
= 8 litres of gasoline
As given in the question 1 ltr is 1 quarts therefore
Total no. of quarts is 8 quarts
As from question 4 quarts is equal to one gallon, hence
Number of gallons[tex]= \frac{8}{4} = 2 gallon[/tex]
Two 1.1 kg masses are 1 m apart (center to center) on a frictionless table. Each has +10 JC of charge. What is the initial acceleration (in m/s2) of this massif it is released and allowed to move?
Answer:
acceleration = 0.8181 m/s²
Explanation:
given data
mass = 1.1 kg
apart d = 1 m
charge q = 10 μC
to find out
What is the initial acceleration
solution
we know that acceleration is
acceleration = [tex]\frac{force}{mass}[/tex] .................1
here force = [tex]k \frac{q1q2}{r^2}[/tex]
here q1 q2 is charge and r is distance and Coulomb constant k = 9 × [tex]10^{9}[/tex] Nm²/C²
force = [tex]9*10^{9} \frac{(10*10^{-6})^2}{1^2}[/tex]
force = 0.9 N
so from equation 1
acceleration = [tex]\frac{0.9}{1.1}[/tex]
acceleration = 0.8181 m/s²