Answer:
Option B is the correct answer.
Explanation:
Shear stress is the ratio of shear force to area.
We have
Shear stress = 3 N/mm² = 3 x 10⁶ N/m²
Area = Area of rectangle = 10 x 10⁻² x d = 0.1d
Shear force = 50000 N
Substituting
[tex]\texttt{Shear stress}=\frac{\texttt{Shear force}}{\texttt{Area}}\\\\3\times 10^6=\frac{50000}{0.1d}\\\\d=0.1667m=16.67cm[/tex]
Width of beam = 16.67 cm
Option B is the correct answer.
To calculate the depth of the beam, shear stress is divided by shear force to get the cross-sectional area, which is then divided by the width to obtain the depth. The depth is found to be 16.67 cm.
Explanation:The subject of the question relates to the calculation of the depth of a rectangular beam based on the given maximum shear force and shear stress. The formula to be used involves shear stress (τ), shear force (V), and the cross-sectional area (A), which is the product of the beam's width and depth (b × d).
Calculation:
Shear Stress (τ) = Shear Force (V) / Area (A)
Given that the maximum shear stress is 3 N/mm² and the shear force is 50000 N, we can rearrange the formula to solve for the cross-sectional area (A):
A = V / τ = 50000 N / 3 N/mm²
Convert 50000 N to mm² by multiplying by 1000 to maintain the correct units:
A = 50000 N × 1000 mm²/N = 50000000 mm²
A = 50000000 mm² / 3 N/mm² = 16666666.67 mm²
Next, we find the depth, remembering the width of the beam is 10 cm (which is 100 mm because 1 cm = 10 mm):
Depth (d) = A / width (b) = 16666666.67 mm² / 100 mm
Depth (d) = 166666.67 mm which is equal to 16.67 cm.
Therefore, the correct answer is (c) 16.67 cm.
A helical gear has 30 teeth and a pitch diameter of 280 mm. Gear module is 8 and pressure angle normal to tooth is 20°. The force normal to the tooth surface is 4500 N. Determine the kW transmitted at 620 rpm.
Answer:
The answer is 30 N a foot.
Explanation:
The power in kW transmitted at 620 rpm is 40.88 kW.
What are gears?A gear is component used to transmit motion. They are used in cars, trucks, machines, bicycles, etc.
A helical gear has 30 teeth and a pitch diameter of 280 mm. Gear module is 8 and pressure angle normal to tooth is 20°. The force normal to the tooth surface is 4500 N. Speed is 620 RPM.
The torque obtained on the gear will be
T = Force x Radius
T = 4500 x (0.280/2)
T = 630 N.m
Power Transmitted P = T ω = 2πNT/60
Plug the values, we get
P = 2π x 620 x 630 /60
P = 40882.8 W
P = 40.88 kW
Thus, the power in kW transmitted at 620 rpm is 40.88 kW.
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A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?
Answer:
13.18 m/s
Explanation:
Let the velocity of sports utility car is
-u as it is moving in opposite direction.
mc = 1200 kg, uc = 31.1 m/s
ms = 2830 kg, us = - u = ?
Using conservation of momentum
mc × uc + ms × us = 0
1200 × 31.1 - 2830 × u = 0
u = 13.18 m/s
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction force will be required?
Answer:
22000 N
Explanation:
Convert velocity to SI units:
98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s
Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.
I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.
Sum of the forces in the radial direction (+x):
∑F = ma
N sin θ + F cos θ = m v² / r
Sum of the forces in the y direction:
∑F = ma
N cos θ - F sin θ - W = 0
To solve the system of equations for F, first solve for N and substitute.
N = (W + F sin θ) / cos θ
Substituting:
((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r
(W + F sin θ) tan θ + F cos θ = m v² / r
W tan θ + F sin θ tan θ + F cos θ = m v² / r
W tan θ + F (sin θ tan θ + cos θ) = m v² / r
W tan θ + F sec θ = m v² / r
F sec θ = m v² / r - W tan θ
F = m v² cos θ / r - W sin θ
F = m (v² cos θ / r - g sin θ)
Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:
F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)
F = 21577 N
Rounding to two sig-figs, you need at least 22000 N of friction force.
A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.
(a) Find the torque the net thrust producesabout the center of the circle.
N·m
(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2
(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2
(a) 24.6 Nm
The torque produced by the net thrust about the center of the circle is given by:
[tex]\tau = F r[/tex]
where
F is the magnitude of the thrust
r is the radius of the wire
Here we have
F = 0.795 N
r = 30.9 m
Therefore, the torque produced is
[tex]\tau = (0.795 N)(30.9 m)=24.6 N m[/tex]
(b) [tex]0.035 rad/s^2[/tex]
The equivalent of Newton's second law for a rotational motion is
[tex]\tau = I \alpha[/tex]
where
[tex]\tau[/tex] is the torque
I is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is
[tex]I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2[/tex]
And so we can solve the previous equation to find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2[/tex]
(c) [tex]1.08 m/s^2[/tex]
The linear acceleration (tangential acceleration) in a rotational motion is given by
[tex]a=\alpha r[/tex]
where in this problem we have
[tex]\alpha = 0.035 rad/s^2[/tex] is the angular acceleration
r = 30.9 m is the radius
Substituting the values, we find
[tex]a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2[/tex]
A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and its initial displacement is s(0) = 9 cm. Find its position function, s(t).
Answer:
The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].
Explanation:
Given that,
Acceleration [tex]a =12t+10[/tex]
Initial velocity [tex]v_{0} = -5\ cm/s[/tex]
Initial displacement [tex]s_{0}=9\ cm[/tex]
We know that,
The acceleration is the rate of change of velocity of the particle.
[tex]a = \dfrac{dv}{dt}[/tex]
The velocity is the rate of change of position of the particle
[tex]v=\dfrac{dx}{dt}[/tex]
We need to calculate the the position
The acceleration is
[tex]a_{t} = 12t+10[/tex]
[tex]\dfrac{dv}{dt} = 12t+10[/tex]
[tex]a_{t}=dv=(12t+10)dt[/tex]
On integration both side
[tex]\int{dv}=\int{(12t+10)}dt[/tex]
[tex]v_{t}=6t^2+10t+C[/tex]
At t = 0
[tex]v_{0}=0+0+C[/tex]
[tex]C=-5[/tex]
Now, On integration again both side
[tex]v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt[/tex]
[tex]s_{t}=2t^{3}+5t^2-5t+C[/tex]
At t = 0
[tex]s_{0}=0+0+0+C[/tex]
[tex]C=9[/tex]
[tex]s_{t}=2t^3+5t^2-5t+9[/tex]
Hence, The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].
To find the position function of a particle given acceleration a(t) = 12t + 10, one integrates twice. The first integral gives the velocity function, the second gives the position or displacement function. These are determined to be v(t) = 6t^2 + 10t - 5 and s(t) = 2t^3 +5t^2 - 5t + 9, respectively.
Explanation:The subject of this problem involves calculating the position function, or displacement, of a particle given an acceleration function, initial velocity, and initial position. In this case, acceleration is given by a(t) = 12t + 10.
To find the velocity function, v(t), you integrate the acceleration function: ∫a(t) dt = ∫(12t + 10) dt = 6t^2 + 10t + C1, where C1 is the constant of integration. We know that v(0) = −5 cm/s, so C1 is -5: the full velocity function is v(t) = 6t^2 + 10t - 5.
We then integrate the velocity function to find the position function, s(t): ∫v(t) dt = ∫(6t^2 + 10t - 5) dt = 2t^3 +5t^2 - 5t + C2. Given s(0) = 9 cm, we find C2 = 9. The position function s(t) is therefore s(t) = 2t^3 +5t^2 - 5t + 9.
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If B is added to C, the result is a vector in the direction of the positive y-axis with a magnitude equal to that of If C = zi + yj, where x=6.13 and y=7.12 what is the magnitude of B?
Answer:c
Explanation:
A heat engine having the maximum possible efficiency has an efficiency of 0.35 when operating between two heat reservoirs. If the temperature of the hot reservoir is 427°C, what is the temperature of the cold reservoir? (5 pts) A) 200 K B) 245 K C) 350 K D) 455 K E) 600 K
Answer:
Option (D)
Explanation:
n = 0.35,
Temperature of hot reservoir,
T1 = 427 C = 700 K
Temperature of cold reservoir, T2 = ?
Use the formula for efficiency of Carnot cycle
n = 1 - T2 / T1
0.35 = 1 - T2 / 700
0.65 = T2 / 700
T2 = 700 × 0.65
T2 = 455 K
An us bomber is flying horizontally at 300 mph at an altitude of 610 m. its target is an iraqi oil tanker crusing 25kph in the same direction and same vertical plane. what horizontal distance behind the tanker must the pilot observe before he releases the shell to score a direct hit
Answer:
=1419.19 meters.
Explanation:
The time it takes for the shell to drop to the tanker from the height, H =1/2gt²
610m=1/2×9.8×t²
t²=(610m×2)/9.8m/s²
t²=124.49s²
t=11.16 s
Therefore, it takes 11.16 seconds for a free fall from a height of 610m
Range= Initial velocity×time taken to hit the tanker.
R=v₁t
Lets change 300 mph to kph.
=300×1.60934 =482.802 kph
Relative velocity=482.802 kph-25 kph
=457.802 kph
Lets change 11.16 seconds to hours.
=11.16/(3600)
=0.0031 hours.
R=v₁t
=457.802 kph × 0.0031 hours.
=1.41918 km
=1.41919 km × 1000m/km
=1419.19 meters.
Distance is a numerical representation of the distance between two objects or locations. The horizontal distance behind the tanker will be 1419.19 m.
What is the distance?Distance is a numerical representation of the distance between two objects or locations. The distance can refer to a physical length or an estimate based on other factors in physics or common use.
The given data in the problem is ;
v₁ is the horizontal flying velocity = 300 m/h
H is the altitude = 610 m
v₂ is he cruise velocity= 25km/h
If t is the time taken the shell drops to the tanker from the height h is found by the formula;
[tex]\rm H =\frac{1}{2} gt^2 \\\\ \rm t=\sqrt{\frac{H}{2g} } \\\\ \rm t=\sqrt{2gh} \\\\ \rm t=\sqrt{2\times 610 \times 9.81} \\\\ \rm t=11.16 \sec[/tex]
The velocity of bomber obtained after unit conversion;
[tex]V_{12}=300\times 1.60 = 482.802[/tex]
Relative velocity is defined as the velocity of an object with respect to the other object.
Relative velocity=482.802 kph-25 kph=457.802 kph
In one hour there are 3600 seconds then the conversion is found by;
On changing 11.16 seconds to hours we found;
[tex](\frac{11.16}{3600} )=0.0031 \ hours.[/tex]
The range is the horizontal distance which is given by ;
Range= Initial velocity×time taken to hit the tanker.
[tex]\rm R= v \times t \\\\ \rm R= 457.8\times 0.0031\\\\ \rm R= 1.41918\ Km \\\\ \rm R=1419.19 m[/tex]
Hence the horizontal distance behind the tanker will be 1419.19 m.
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Global warming will produce rising sea levels partly due to melting ice caps but also due to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC. Note that this calculation is only approximate because ocean warming is not uniform with depth. (answer in ×10^{-3} −3 m)
The change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC is approximately 0.21 meters.
Explanation:To calculate the change in length of a column of water for a temperature increase of 1.00ºC, we need to use the coefficient of thermal expansion for water. The coefficient of thermal expansion for water is approximately 0.00021 per degree Celsius.
To find the change in length, we multiply the height of the column (1.00 km) by the coefficient of thermal expansion (0.00021) and the temperature increase (1.00ºC).
Change in length = 1.00 km × 0.00021 × 1.00ºC = 0.00021 km = 0.21 m
Therefore, the change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC is approximately 0.21 meters.
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A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slides a distance of 2.7 m in 5.8 s. Find the coefficient of kinetic friction between the block and plane.
Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
A perfect electric dipole feels a net force in a uniform electric field. 1. True 2. False
A perfect electric dipole doesn't feel a net force in a uniform electric field.
To determine whether this statement is true or false, we need to know about the behavior of perfect dipole in an electric field.
How does a perfect dipole behave in uniform electric field?When a perfect dipole is kept in uniform electric field, both the charges of the dipole experience same amount of force.But these forces are not linear, so they collectively rotate the dipole.So, in a uniform electric field, the perfect dipole only experiences torque about its center of axis.How does a perfect dipole behave in non-uniform electric field?When non-uniform electric field is applied, both charges of the perfect dipole experience different amounts of force.So the dipole experiences a torque as well as a net linear force.Thus, we can conclude that a perfect electric dipole doesn't feel a net force in a uniform electric field.
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You notice a very one-dimensional-thinking snail crawling along one rail of a railroad track. Naturally, you remove it to a safer place. But before you do, you observe it as it moves from position xi = -92.9 cm to position xf = -64.9 cm along the rail, as measured from a point where one rail abuts the next. What is the snail\'s displacement Δx in centimeters?
Answer:
28.0 cm
Explanation:
Displacement is final position minus initial position.
Δx = xf − xi
Δx = -64.9 cm − -92.9 cm
Δx = 28.0 cm
The magnitude of displacement is 28 cm.
We have a snail moving on a railway track.
We have to determine snail's displacement.
What is Displacement ?The length of the straight line joining the initial and final position of an object is called its displacement.
According to the question -
Initial Position x[i] = -92.9 cm
Final Position x[f] = -64.9 cm
Since, the motion is one dimensional, the net displacement can be calculated by the formula -
|Δx| = x[f] - x[i] = - 64.9 - (- 92.9) = - 64.9 + 92.9 = |- 28| = 28 cm
Hence, the magnitude of displacement is 28 cm.
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An airplane flying parallel to the ground undergoes two consecutive displacements. The first is 76 km at 39.8◦ west of north, and the second is 156 km at 59.9◦ east of north. What is the magnitude of the plane’s total displacement? Answer in units of km. 020 (part 2 of 2) 10.0
Answer:
162 km
Explanation:
A diagram can be helpful.
Using the law of cosines, we can find the magnitude of the distance (c) to satisfy ...
c^2 = a^2 +b^2 -2ab·cos(C)
where C is the internal angle of the triangle of vectors and resultant. Its value is ...
180° -39.8° -59.9° = 80.3°
Filling in a=76 and b=156, we get ...
c^2 = 76^2 +156^2 -2·76·156·cos(80.3°) ≈ 26116.78
c ≈ √26116.78 ≈ 161.607
The magnitude of the total displacement is about 162 km.
_____
Please note that in the attached diagram North is to the right and East is up. That alteration of directions does not change the angles or the magnitude of the result.
Answer:
Magnitude of total displacement = 162.87 km
Explanation:
Let east be x axis and north be y axis.
The first is 76 km at 39.8◦ west of north.
Displacement 1 = 76 km at 39.8◦ west of north = 76 km at 129.8◦ north of east.
Displacement 1 = 76 cos129.8 i + 76 sin 129.8 j = -48.65 i + 58.39 j
The second is 156 km at 59.9◦ east of north.
Displacement 2 = 156 km at 59.9◦ east of north = 156 km at 31.1◦ north of east.
Displacement 2 = 156 cos31.1 i + 156 sin 31.1 j = 133.58 i + 80.58 j
Total displacement = Displacement 1 + Displacement 2
Total displacement = -48.65 i + 58.39 j + 133.58 i + 80.58 j = 84.93 i + 138.97 j
[tex]\texttt{Magnitude of total displacement =}\sqrt{84.93^2+138.97^2}=162.87km[/tex]
Magnitude of total displacement = 162.87 km
A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.25 m. (a) What is its speed just before striking the sand?
Final answer:
The speed of the meteorite just before striking the sand is 0 m/s.
Explanation:
To find the speed of the meteorite just before striking the sand, we can use the principle of conservation of energy. Let's consider the initial energy when the meteorite is 850 km above the Earth's surface and the final energy when it comes to rest in the sand. Initially, the meteorite has gravitational potential energy and no kinetic energy. Finally, when it comes to rest in the sand, it has no potential energy and only kinetic energy.
Using the equation for gravitational potential energy, PE = mgh, where m is the mass of the meteorite, g is the acceleration due to gravity (9.8 m/s²), and h is the height above the Earth's surface, we can calculate the initial potential energy. PE = (2.0 × 10¹³ kg)(9.8 m/s²)(850,000 m) = 1.61 × 10¹⁹ J.
The final kinetic energy of the meteorite just before striking the sand can be calculated using the equation KE = (1/2)mv², where m is the mass of the meteorite and v is its velocity. We know that the meteorite comes to rest in a distance of 3.25 m, so its final velocity is 0. Using this information, we can solve for the final kinetic energy. 0 = (1/2)(2.0 × 10¹³ kg)v². By rearranging the equation, we find v = 0 m/s.
Therefore, the speed of the meteorite just before striking the sand is 0 m/s.
A 140 g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.537 m. The ball makes 2.27 revolutions in a second. What is its centripetal acceleration?
Answer:
109.13 rad/s^2
Explanation:
m = 140 g = 1.4 kg, r = 0.537 m, f = 2.27 rps
The centripetal acceleration is given by
a = r ω^2
a = r x (2 π f)^2
a = 0.537 x ( 2 x 3.14 x 2.27)^2
a = 109.13 rad/s^2
A 7.00 g bullet moving horizontally at 200 m/s strikes and passes through a 150 g tin can sitting on a post. Just after the impact, the can has a horizontal speed of 180 cm/s. What was the bullet’s speed after leaving the can?
Answer:
160 m/s
Explanation:
Momentum is conserved:
mu = mv + MV
(7.00) (200) = (7.00)v + (150) (1.8)
v = 160 m/s
161.43m/s
Explanation:
Using the principle of conservation of linear momentum i.e
Total momentum before impact is equal to total momentum after impact.
=> Momentum of bullet before impact + Momentum of tin before impact
=
Momentum of bullet after impact + Momentum of tin after impact
i.e
[tex]m_{B}[/tex] [tex]u_{B}[/tex] + [tex]m_{T}[/tex] [tex]u_{T}[/tex] = [tex]m_{B}[/tex] [tex]v_{B}[/tex] + [tex]m_{T}[/tex] [tex]v_{T}[/tex]
Where;
[tex]m_{B}[/tex] = mass of bullet = 7.00g = 0.007kg
[tex]m_{T}[/tex] = mass of tin can = 150g = 0.15kg
[tex]u_{B}[/tex] = initial velocity of bullet before impact = 200m/s
[tex]u_{T}[/tex] = initial velocity of tin can before impact = 0m/s (since the can is stationary)
[tex]v_{B}[/tex] = final velocity of the bullet after impact
[tex]v_{T}[/tex] = final velocity of the tin can after impact = 180cm/s = 1.8m/s
Substitute these values into the equation above;
=> [tex]m_{B}[/tex] [tex]u_{B}[/tex] + [tex]m_{T}[/tex] [tex]u_{T}[/tex] = [tex]m_{B}[/tex] [tex]v_{B}[/tex] + [tex]m_{T}[/tex] [tex]v_{T}[/tex]
=> (0.007 x 200) + (0.15 x 0) = (0.007 x [tex]v_{B}[/tex]) + (0.15 x 1.8)
=> 1.4 + 0 = 0.007[tex]v_{B}[/tex] + 0.27
=> 1.4 = 0.007[tex]v_{B}[/tex] + 0.27
=> 0.007[tex]v_{B}[/tex] = 1.4 - 0.27
=> 0.007[tex]v_{B}[/tex] = 1.13
Solve for [tex]v_{B}[/tex]
=> [tex]v_{B}[/tex] = 1.13 / 0.007
=> [tex]v_{B}[/tex] = 161.43m/s
Therefore, the speed of the bullet after impact (leaving the can) is 161.43m/s
An earthquake in San Francisco produces a wave with a period of 0.80 seconds and a wavelength of 40 m., How long will it take for this earthquake to reach Los Angeles if the displacement between these cities is 560 km? A) 3.11 Hrs B) 1.2 hrs C) 42 minutes D) 11.2 seconds
Answer:
A) 3.11 Hrs
Explanation:
Wave speed is given by the formula
[tex]Speed = \frac{wavelength}{Time\: period}[/tex]
now we will have
[tex]speed = \frac{40}{0.80}[/tex]
[tex]v = 50 m/s[/tex]
now the time taken by the wave to move the distance 560 km is given as
[tex]t = \frac{distance}{speed}[/tex]
[tex]t = \frac{560\times 10^3}{50}[/tex]
[tex]t = 11200 s[/tex]
t = 3.11 hours
Describe how the Fermi level changes with donor level, temperature and contact with p and n type semiconductors
Fermi level determines charge carrier population according to the integral over density of states times Fermi Dirac distribution. Therefore, saying that the position of Fermi level is higher (lower), relative to middle of band gap, is just the mathematical description of saying you have more electrons (holes), relative to the intrinsic state, in the system. Dopants in the system increase the concentration of electrons or holes.
A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a) the vector position of the particle at any time t (where t is measured in seconds), (b) the velocity of the particle at any time t, (c) the position of the particle at t = 7.00 s, and (d) the speed of the particle at t = 7.00
a. The particle has position vector
[tex]\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath[/tex]
[tex]\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath[/tex]
b. Its velocity vector is equal to the derivative of its position vector:
[tex]\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath[/tex]
c. At [tex]t=7.00\,\mathrm s[/tex], the particle has position
[tex]\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath[/tex]
[tex]\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m[/tex]
That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of [tex]\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m[/tex] away from the origin in a direction of [tex]\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ[/tex] relative to the positive [tex]x[/tex] axis.
d. The speed of the particle at [tex]t=7.00\,\mathrm s[/tex] is the magnitude of the velocity at this time:
[tex]\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath[/tex]
[tex]\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}[/tex]
Then its speed at this time is
[tex]\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}[/tex]
(a) The vector position of the particle at any time t; is s(t) = (8i)t + (j)t²
(b) the velocity of the particle at any time is v = 8i + (2j)t
(c) the position of the particle at t = 7.00 s is 56i + 49 j at a distance of 74.4 m
(d) the speed of the particle at t = 7.00 is 16.12 m/s
The given parameters;
acceleration of the particle, a = 2 j m/s²
initial velocity of the particle, v = 8 i m/s
(a) The vector position of the particle at any time t;
[tex]s = v_0t + \frac{1}{2} at^2\\\\s(t) = (8 \ i) t \ + (0.5 \times 2 \ j)t^2\\\\s(t) = (8 \ i)t \ + (j)t^2[/tex]
(b) the velocity of the particle at any time
[tex]velocity = \frac{\Delta \ displacement }{\Delta \ time} \\\\s' = v = 8i \ + (2j)t[/tex]
(c) the position of the particle at t = 7.00 s
[tex]s(t) = (8 \ i)t \ + (j)t^2\\\\s(7) = (8 \ i)\times 7 \ \ + \ \ (j)\times 7^2\\\\s(7) = 56i \ \ + \ \ 49j\\\\|s| = \sqrt{(56)^2 + (49)^2} = 74.4 \ m\\\\|s| = 74.4 \ m[/tex]
(d) the speed of the particle at t = 7.00
[tex]v(t) = 8i + (2j)t\\\\v(7) = 8i \ + (2j \times 7)\\\\v(7) = 8i + 14j\\\\|v| = \sqrt{8^2 + 14^2} \\\\|v| = \sqrt{260} \\\\|v| = 16.12 \ m/s[/tex]
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A body of mass 400 g is moving along a smooth surface at a velocity of 1.25 m/s towards the east. It strikes a body of mass 600 g, initially at rest. The 400 g mass then moves at a velocity of 1.00 m/s in a direction 36.9? north of east. (a) what is the easterly component of the total momentum of the system before and after the collision? (b) What is the northerly component of the total momentum of the system before and after the collision? (c) Determine the final velocity of the 600 g body.
(a) 0.5 kg m/s
Before the collision, only the first body is moving, so only the first body contributes to the total momentum.
The first body has
m = 400 g = 0.4 kg (mass)
v = 1.25 m/s (velocity, towards east direction)
So its momentum is
[tex]p_x=mv = (0.4 kg)(1.25 m/s)=0.5 kg m/s[/tex]
And since the body is moving along the east direction, this is also the easterly component of the total momentum before the collision.
(b) Zero
Before the collision, we have:
- The first body moving along the east direction --> so its northerly component is zero
- The second body at rest --> this means that it does not contribute to the momentum, since it is zero
This means that the northerly component of the total momentum before the collision is zero.
(c) 0.5 m/s at 53.1 degrees south of east
The law of conservation of momentum states that each component of the total momentum must be conserved.
- Along the easterly direction:
[tex]p_x = p_{1x} + p_{2x}[/tex]
where
[tex]p_x = 0.5 kg m/s[/tex] is the easterly component of the total momentum
[tex]p_{1x} = m v cos \theta = (0.4 kg)(1.00 m/s) cos 36.9^{\circ} =0.32 kg m/s[/tex] is the easterly component of the momentum of the first body after the collision
[tex]p_{2x}[/tex] is the easterly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision
Solving the equation we find
[tex]p_{2x} = p_x - p_{1x} = 0.5 kg m/s - 0.32 kg m/s = 0.18 kg m/s[/tex]
- Along the northerly direction:
[tex]p_y = p_{1y} + p_{2y}[/tex]
where
[tex]p_y = 0 kg m/s[/tex] is the northerly component of the total momentum
[tex]p_{1y} = m v sin \theta = (0.4 kg)(1.00 m/s) sin 36.9^{\circ} =0.24 kg m/s[/tex] is the northerly component of the momentum of the first body after the collision
[tex]p_{2y}[/tex] is the northerly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision
Solving the equation we find
[tex]p_{2y} = p_y - p_{1y} = 0 - 0.24 kg m/s = -0.24 kg m/s[/tex]
So now we find the momentum of the 600 g body after the collision:
[tex]p_2=\sqrt{p_{2x}^2 + p_{2y}^2}=\sqrt{(0.18)^2+(-0.24)^2}=0.3 kg m/s[/tex]
and so its final speed is
[tex]v=\frac{p_2}{m}=\frac{0.3 kg m/s}{0.6 kg}=0.5 m/s[/tex]
and the direction is
[tex]\theta=tan^{-1} (\frac{p_{2y}}{p_{2x}})=tan^{-1} (\frac{-0.24}{0.18})=-53.1^{\circ}[/tex]
so 53.1 degrees in the south-east direction.
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 44 ft/s2. What is the distance covered before the car comes to a stop? (Round your answer to one decimal place.) ft
Answer:
Distance, d = 61.13 ft
Explanation:
It is given that,
Initial speed of the car, u = 50 mi/h = 73.34 ft/s
Finally, it stops i.e. v = 0
Deceleration of the car, [tex]a=-44\ ft/s^2[/tex]
We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}[/tex]
s = 61.13 ft
So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.
Answer:
The distance covered by car before stopping is 61.13 ft.
Explanation:
Given data:
Initial Speed of car is, [tex]u=50 \;\rm mi/h = 50 \times 1.467 =73.35 \;\rm ft/s[/tex].
Deceleration of car is, [tex]a=-44\;\rm ft/s^{2}[/tex]. (Negative sign shows negative acceleration)
Applying the second kinematic equation of motion as,
[tex]v^{2}=u^{2}+2as[/tex]
Here, s is the distance covered and v is the final speed. Since, car stops finally, v = 0.
Solving as,
[tex]0^{2}=73.35^{2}+2(-44) \times s\\88s =73.35^{2}\\s \approx 61.1 \;\rm ft[/tex]
Thus, distance covered by car before stopping is 61.1 ft.
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How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their original distance of separation? 1. Reduces to one quarter of original value 2. Doubles 3. Doesn’t change 4. Reduces to one half of original value 5. Quadruple
According to Coulomb's Law, the magnitude of the electric force between two charged particles is inversely proportional to the square of the distance between them. Therefore, if the distance between the two particles is halved, the electric force becomes four times as strong, i.e., it quadruples.
Explanation:The phenomenon you're asking about is governed by Coulomb's Law. This law states that the magnitude of the electric (or Coulomb) force between two charged particles is directly proportional to the product of the magnitudes of their charges, and inversely proportional to the square of the distance between them.
So, if the distance between these two charges is halved, the electric force between them would be affected by the square of that change in distance. Therefore, instead of being half as strong, the electric force becomes four times as strong, i.e., it quadruples. Hence, the correct answer to your question is: it quadruples (option 5). The inverse square law, which this is an example of, is a fundamental principle in physics that applies to other forces too, like gravity.
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A battery is rated at 12 V and 160 A-h. How much energy does the battery store? What is the cost of this energy at $0. 15/kWh?
Answer:
The cost of energy is 0.29$.
Explanation:
Given that,
Voltage V= 12 V
Charge Q= 160 A-h
We need to calculate the energy
Using formula of energy
[tex]E = Pt[/tex]
[tex]E =VIt[/tex]
Where, E = energy
I= current
V = Potential
t = time
Put the value into the formula
[tex]E = 12\times160\times3600[/tex]
[tex]E =6.9\times10^{6} J[/tex]
The cost of this energy at 0.15/kwh
[tex]Total\ cost = Energy\times 0.15 kwh[/tex]
Here, Energy = power x time
In units,
[tex]J=watt\times s[/tex]
But , Energy in 1 hour is:
[tex]E = 1920\ J[/tex]
So, The cost of this energy is
[tex] Total\ cost =1920\dfrac{\times0.15}{1000}[/tex]
[tex]total\ cost = 0.29\$[/tex]
Hence, The cost of energy is 0.29$.
For a battery rated at 12V and 160 A-h, the total energy stored is about 6912000 Joules or 1.92 kWh. The cost of this energy, at a rate of $0.15 per kWh, would therefore be $0.288.
Explanation:To find out the amount of energy stored in the battery, we first understand that the battery's rated capacity is given as Ampere-hours (A-h). It is a measure of the electric charge that a battery can deliver over time. We can find the energy stored in this battery by multiplying the charge by the battery's voltage. With a rating of 12 V and 160 A-h, the total energy would be calculated as Energy (E) = Voltage (V) x Charge (Q), where Q (the charge) is 160 A-h, which must be converted to seconds by multiplying by 3600 (the number of seconds in one hour). So, Q = 160 A-hr x 3600 = 576000 Coulombs. Once we substitute these values into the formula, the total energy stored would be E = 12 V * 576000 C = 6912000 J.
Now, to find the cost of this energy, we need to convert this energy (which is in Joules) to kWh by dividing by 3.6 million (as 1 kWh = 3.6 million Joules), leading to E = 6912000 J / 3600000 = 1.92 kWh. The cost of this energy would then be calculated by multiplying the energy by the cost per unit of energy, here given as $0.15 per kWh. Therefore, Cost = 1.92 kWh * $0.15/kWh = $0.288.
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A 1200 kg car traveling north at 10 m/s is rear-ended by a 2000 kg truck traveling at 30 m/s. What is the total momentum before and after the collision?
Answer:
The total momentum before and after collision is 72000 kg-m/s.
Explanation:
Given that,
Mass of car = 1200 kg
Velocity of car = 10 m/s
Mass of truck = 2000 kg
Velocity of truck = 30 m/s
Using conservation of momentum
The total momentum before the collision is equal to the total momentum after collision.
[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V[/tex]
Where, [tex]m_{1}[/tex]=mass of car
[tex]v_{1}[/tex] =velocity of car
[tex]m_{1}[/tex]=mass of truck
[tex]v_{1}[/tex] =velocity of truck
Put the value into the formula
[tex]1200\times10+2000\times30=(1200+2000)V[/tex]
[tex]V=\dfrac{1200\times10+2000\times30}{(1200+2000)}[/tex]
[tex]V = 22.5\ m/s[/tex]
Now, The total momentum before collision is
[tex]P=m_{1}v_{1}+m_{2}v_{2}[/tex]
[tex]P=1200\times10+2000\times30[/tex]
[tex]P=72000\ kg-m/s[/tex]
The total momentum after collision is
[tex]P=(m_{1}+m_{2})v_{2}[/tex]
[tex]P=(1200+2000)\times22.5[/tex]
[tex]P= 72000 kg-m/s[/tex]
Hence, The total momentum before and after collision is 72000 kg-m/s.
If you weigh 665 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 18.0 km ? Take the mass of the sun to be ms = 1.99×10^30 kg , the gravitational constant to be G = 6.67×10^−11 N⋅m2/kg^2 , and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2 .
Your weight on a neutron star can be calculated based on your given weight on Earth using the Universal Law of Gravitation equation. This allows us to determine the gravitational force or weight for any celestial body given we know its mass and radius.
Explanation:The topic at hand is fundamentally related to Physics since it pertains to the gravitational effects of a massive body. From the question, we are given your weight on Earth which is 665 N. The weight on Earth can be related to the weight of the neutron star by the equation of universal gravitation:
W = mg = GMm/r²
Where:
M = mass of the star or planet or rather the celestial bodym = mass of the object or the person (in this case)G = universal gravitational constantr = distance from the center of the celestial body (radius)Given that the diameter of the neutron star is 18km, its radius becomes 9km = 9000m. Substituting the given values into the formula yields a theoretical weight on a neutron star.
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We find this acceleration to be approximately 4.09×[tex]10^{12} m/s^2[/tex]. Thus, your weight would be around 2.77×[tex]10^{14}[/tex] N on the neutron star.
To calculate your weight on the surface of a neutron star, you need to find the acceleration due to gravity on that neutron star using Newton's Universal Law of Gravitation.
The formula for gravitational force is:
F = G x (m1 x m2) /[tex]R^2[/tex]
Where G is the gravitational constant, m1 and m2 are the masses, and R is the radius of the object.
Given:
Mass of the neutron star, M = 1.99×[tex]10^{30}[/tex] kg
Radius, R = 18.0 km = 18,000 meters
Gravitational constant, G = 6.67×[tex]10^{-11} N⋅m^2/kg^2[/tex]The acceleration due to gravity on the neutron star, g_s, can be calculated as:
g_s = G x (M / [tex]R^2[/tex])
Substituting in the given values:
g_s = (6.67×[tex]10^{-11}[/tex] N⋅[tex]m^2/kg^2[/tex] x 1.99×[tex]10^{30}[/tex] kg) / [tex](18,000 m)^2[/tex]
g_s ≈ 4.09×[tex]10^{12}[/tex] m/[tex]s^2[/tex]
Your weight on Earth is given as 665 N. Since weight (W) is mass (m) times the acceleration due to gravity (g):
m = W / g = 665 N / 9.810 m/[tex]s^2[/tex] ≈ 67.8 kg
So, your weight on the neutron star would be:
Weight on neutron star = mass x g_s = 67.8 kg x 4.09×[tex]10^{12} m/s^2[/tex] ≈ 2.77×[tex]10^{14}[/tex] N
A certain particle at rest has a lifetime of 2.52 mu s. what must be the speed of the particle for its lifetime observed to be 8.32 mu s? 0.985c 0.953c 0.913c 0.302c None of the above
Answer:
0.953c
Explanation:
T₀ = lifetime of the particle at rest = 2.52 μs
T = Observed lifetime of the particle while in motion = 8.32 μs
v = speed of particle
c = speed of light
Using the formula for time dilation
[tex]T = \frac{T_{o}}{\sqrt{1-(\frac{v}{c})^{2}}}[/tex]
inserting the values
[tex]8.32 = \frac{2.52}{\sqrt{1-(\frac{v}{c})^{2}}}[/tex]
v = 0.953c
Final answer:
To calculate the speed at which a particle's lifetime appears dilated to 8.32 μs, one uses the time dilation formula from special relativity. The calculated speed is approximately 0.995c, which does not match the provided choices. A careful review of the calculation and rounding effects might lead to one of the options being the nearest correct answer.
Explanation:
To find the required speed for the particle's lifetime to be observed as 8.32 μs, we need to apply the concept of time dilation from Einstein's theory of special relativity. Time dilation is given by the equation τ' = τ / √(1 - v²/c²), where τ' is the dilated lifetime, τ is the proper lifetime (lifetime at rest), v is the velocity of the particle, and c is the speed of light.
The problem gives us τ' = 8.32 μs and τ = 2.52 μs. We substitute these values into the equation and solve for v:
8.32 μs = 2.52 μs / √(1 - v²/(3.00 × 10⁸ m/s)²)
We first square both sides, rearrange the equation, and then take the square root to find v:
(8.32 μs / 2.52 μs)² = 1 / (1 - v²/c²)
(3.30)² = 1 / (1 - v²/c²)
10.89 = 1 + v²/c²
v²/c² = 10.89 - 1
v²/c² = 9.89
v = √(9.89) ∑ c
v ≈ 0.995c
Thus the required speed of the particle for its lifetime to be observed as 8.32 μs is approximately 0.995c. However, this speed is not listed among the multiple-choice options, suggesting either a calculation error or that none of the provided answers are correct. It's important to review the calculation carefully and consider rounding effects that might lead to one of the provided answers being the nearest correct choice.
A 81 kg man lying on a surface of negligible friction shoves a 79 g stone away from himself, giving it a speed of 4.5 m/s. What speed does the man acquire as a result?
Answer:
The speed of the man is 4.54 m/s.
Explanation:
Given that,
Mass of man=8100 g
Mass of stone = 79 g
Speed = 4.5 m/s
We need to calculate the speed of the man
Using momentum of conservation
[tex](m_{1}+m_{2})V=m_{1}v_{1}+m_{2}v_{2}[/tex]
Where,
[tex]m_{1}[/tex]=mass of man
[tex]m_{2}[/tex]=mass of stone
[tex]v_{1}[/tex]=velocity of man
[tex]v_{2}[/tex]=velocity of stone
Put the value in the equation
As the stone is away from the man
So, the speed of stone is zero
[tex]8179\times4.5=8100\times v+0[/tex]
[tex]v=\dfrac{8179\times4.5}{8100}[/tex]
[tex]v = 4.54\ m/s[/tex]
Hence, The speed of the man is 4.54 m/s.
How far from a converging lens with a focal length of 16 cm should an object be placed to produce a real image which is the same size as the object? Express your answer using two significant figures.
Answer:
32 cm
Explanation:
f = focal length of the converging lens = 16 cm
Since the lens produce the image with same size as object, magnification is given as
m = magnification = - 1
p = distance of the object from the lens
q = distance of the image from the lens
magnification is given as
m = - q/p
- 1 = - q/p
q = p eq-1
Using the lens equation, we get
1/p + 1/q = 1/f
using eq-1
1/p + 1/p = 1/16
p = 32 cm
To create a real image with the same size as the object using a converging lens, the object should be placed at a distance equal to twice the focal length of the lens.
Given the focal length (f) of the lens is 16 cm, the object should be placed at a distance of 2 * 16 cm = 32 cm from the lens.
1. The lens formula is given by:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens.
- v is the image distance.
- u is the object distance.
2. In this case, we want a real image of the same size as the object, which means that the magnification (M) is equal to 1. Magnification is given by:
M = -v/u
Since M = 1, we have v = -u.
3. For a real image formed by a converging lens, v is negative. So, we have v = -u = -2f.
4. Substituting the focal length (f = 16 cm) into the equation, we find that u = -2 * 16 cm = -32 cm. The negative sign indicates that the object is placed on the same side as the object. Therefore, the object should be placed 32 cm from the lens to produce the desired image.
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A 2 kg ball travellng to the right with a speed of 4 m/s collidees with a 5 kg ball traveling to the left with a speed of 3 m/s. Take right to be the positive direction. What is the total momentum of the two balls before they collide? What is the total momentum of the two balls after they collide?
Explanation:
Momentum before collision:
(2 kg) (4 m/s) + (5 kg) (-3 m/s) = -3 kg m/s
No external forces act on the balls, so momentum is conserved. Therefore, momentum after collision is also -3 kg m/s.
The total momentum of the two balls before the collision is -7 kg*m/s. The total momentum of the two balls after the collision is also -7 kg*m/s.
Explanation:To find the total momentum before the collision, we need to calculate the individual momentum of each ball and then add them together. To compute an object's momentum, multiply its mass by its velocity.
For the first ball, the momentum is 2 kg * 4 m/s (since it is travelling to the right) = 8 kg*m/s. For the second ball, the momentum is 5 kg * (-3 m/s) (since it is travelling to the left) = -15 kg*m/s. Adding these two momenta together, we get a total momentum of 8 kg*m/s + (-15 kg*m/s) = -7 kg*m/s.
After the collision, if the balls stick together and move as one, the speed of the combined mass can be calculated using the principle of conservation of momentum. Since momentum is conserved in a collision, the total momentum after the collision will be the same as before. In this case, since the total momentum before the collision was -7 kg*m/s, the total momentum after the collision will also be -7 kg*m/s.
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deals with the concepts that are important in this problem. A grasshopper makes four jumps. The displacement vectors are (1) 31.0 cm, due west; (2) 32.0 cm, 34.0 ° south of west; (3) 16.0 cm, 57.0 ° south of east; and (4) 16.0 cm, 75.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.
To find the magnitude and direction of the resultant displacement of a grasshopper's jumps, calculate the components using trigonometry and then determine the magnitude and angle of the resultant vector.
Explanation:The magnitude of the resultant displacement: First, calculate the x and y components of each jump using trigonometry. Then sum up all the x components and y components separately to find the resultant components. Finally, use these components to calculate the magnitude.
The direction of the resultant displacement: Use inverse trigonometric functions to find the angle the resultant displacement makes with respect to due west.
Resultant displacement: (a) Magnitude ≈ 17.09 cm, (b) Direction ≈ 116.88° south of west.
To find the resultant displacement, we can first break down each displacement vector into its horizontal (x) and vertical (y) components. Then we'll sum up all the x-components and y-components separately to get the total horizontal and vertical displacement. Finally, we'll use these values to find the magnitude and direction of the resultant displacement.
Let's start by breaking down each displacement vector:
(1) 31.0 cm due west:
- Horizontal component (x): -31.0 cm
- Vertical component (y): 0 cm
(2) 32.0 cm, 34.0° south of west:
- Horizontal component (x): 32.0 cm * cos(34.0°)
- Vertical component (y): -32.0 cm * sin(34.0°)
(3) 16.0 cm, 57.0° south of east:
- Horizontal component (x): 16.0 cm * cos(57.0°)
- Vertical component (y): -16.0 cm * sin(57.0°)
(4) 16.0 cm, 75.0° north of east:
- Horizontal component (x): 16.0 cm * cos(75.0°)
- Vertical component (y): 16.0 cm * sin(75.0°)
Now let's calculate the components:
For (2):
- Horizontal component (x): 32.0 cm * cos(34.0°) ≈ 26.61 cm
- Vertical component (y): -32.0 cm * sin(34.0°) ≈ -17.09 cm
For (3):
- Horizontal component (x): 16.0 cm * cos(57.0°) ≈ 7.96 cm
- Vertical component (y): -16.0 cm * sin(57.0°) ≈ -13.57 cm
For (4):
- Horizontal component (x): 16.0 cm * cos(75.0°) ≈ 4.14 cm
- Vertical component (y): 16.0 cm * sin(75.0°) ≈ 15.43 cm
Now, let's sum up all the x-components and y-components:
Total horizontal displacement (x):
= -31.0 cm + 26.61 cm + 7.96 cm + 4.14 cm
= 7.71 cm
Total vertical displacement (y):
= 0 cm - 17.09 cm - 13.57 cm + 15.43 cm
= -15.23 cm
Now, to find the magnitude (R) of the resultant displacement:
R = sqrt((Total horizontal displacement)^2 + (Total vertical displacement)^2)
= sqrt((7.71 cm)^2 + (-15.23 cm)^2)
≈ sqrt(59.44 cm^2 + 232.52 cm^2)
≈ sqrt(291.96 cm^2)
≈ 17.09 cm
Now, to find the direction of the resultant displacement:
θ = atan(Total vertical displacement / Total horizontal displacement)
= atan(-15.23 cm / 7.71 cm)
≈ atan(-1.975)
This gives us an angle of approximately -63.12° with respect to the horizontal. Since the angle is negative, we'll add 180° to it to find the angle with respect to due west:
θ = -63.12° + 180°
≈ 116.88°
So, the magnitude of the resultant displacement is approximately 17.09 cm, and its direction is approximately 116.88° south of west.