Suppose you walk 11 m in a direction exactly 24° south of west then you walk 21 m in a direction exactly 39° west of north. 1) How far are you from your starting point, in meters?
2) What is the angle of the compass direction of a line connecting your starting point to your final position measured North of West in degrees?
Answers
Answer:
(1) 42.94 m
(2) [tex]16.02^\circ[/tex]
Explanation:
Let us first draw a figure, for the given question as below:
In the figure, we assume that the person starts walking from point A to travel 11 m exactly [tex]24^\circ[/tex] south of west to point B and from there, it walks 21 m exactly [tex]39^\circ[/tex] west of north to reach point C.
Let us first write the two displacements in the vector form:
[tex]\vec{AB} = (-11\cos 24^\circ\ \hat{i}-11\sin 24^\circ\ \hat{j})\ m =(-10.05\ \hat{i}-4.47\ \hat{j})\ m\\\vec{BC} = (-21\sin 39^\circ\ \hat{i}+21\cos 39^\circ\ \hat{j})\ m =(-31.22\ \hat{i}+16.32\ \hat{j})\ m[/tex]
Now, the vector sum of both these vectors will give us displacement vector from point A to point C.
[tex]\vec{AC}=\vec{AB}+\vec{BC}\\\Rightarrow \vec{AC}=(-10.05\ \hat{i}-4.47\ \hat{j})\ m+(-31.22\ \hat{i}+16.32\ \hat{j})\ m\\\Rightarrow \vec{AC}=(-41.25\ \hat{i}+11.85\ \hat{j})\ m[/tex]
Part (1):
the magnitude of the shortest displacement from the starting point A to point the final position C is given by:
[tex]AC=\sqrt{(-41.25)^2+(11.85)^2}\ m= 42.94\ m[/tex]
Part (2):
As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.
The angle of the shortest line connecting the starting point and the final position measured north of west is given by:
[tex]\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ[/tex]
Related Questions
An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the engines give the plane a constant acceleration of 1.8 m/s^2. The plane then travels a distance of 1500 m down the runway before lifting off. How long does it take from the application of the acceleration until the plane lifts off, becoming airborne?
Answers
Answer:
t=67.7s
Explanation:
From this question we know that:
Vo = 6m/s
a = 1.8 m/s2
D = 1500m
And we also know that:
[tex]X=V_{o}*t + \frac{a*t^{2}}{2}[/tex] Replacing the known values:
[tex]1500=6t+0.9*t^{2}[/tex] Solving for t we get 2 possible answers:
t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:
t = 67.7s
You have a balloon that contains 1 kg of helium that occupies a volume of 4 m Sometime later, the balloon still holds 1 kg of helium, but it now occupies a volume of 2m'. Your friend suggests that the helium can be considered an isolated system in this case. Can they be correct?
Answers
Answer:
Yes! They can be correct
Explanation:
An isolated system is a system where its total energy and mass stay constant.
In this case, even though the volume changed, the mass remains constant (m = 1 kg), so there is no mass exchange, and we must think that the globe is completely closed.
Now, we don't know exactly what happens with energy, but I can give you an example where the total energy of the globe remains constant.
Imagine that the balloon of volume [tex]V_1[/tex] is at a certain height [tex]h_1[/tex], under some pressure, [tex]P_1[/tex]. If you lower the balloon to a height [tex]h_2[/tex], the pressure increases, and from the Boyle's law,
[tex]P_1 V_1 = P_2 V_2[/tex],
[tex]V_2 = V_1 \frac{P_1}{P_2}[/tex], where [tex]P_2 > P_1[/tex],
[tex]V_2 < V_1[/tex], just like the case you state, and there was no exchange of mass or energy related to the inner gas of the balloon, so yes, They can be correct.
A woman stands on a scale in a moving elevator. Her massis
60.0 kg and the combined mass of the elevator and scale is
anadditional 815 kg. Starting from rest, the elevatoraccelerates
upward. During the acceleration, there is tensionof 9410 N in the
hoisting cable. What does the scale readduring the
acceleration?
Answers
Answer:
Explanation:
Combined mass = 815 + 60 = 875 kg
Net weight acting downwards
= 875 x 9.8
= 8575 N
Tension in the string acting upwards
= 9410
Net upward force = 9410 - 8575
= 835 N
Acceleration
Force / mass
a = 835 / 875
a = .954 ms⁻²
Since the elevator is going up with acceleration a
Total reaction force on the woman from the ground
= m ( g + a )
60 ( 9.8 + .954)
= 645.25 N.
Reading of scale = 645.25 N
A 205 kg log is pulled up a ramp by means of a rope that
isparallel to the surface of the ramp. The ramp is inclined at30%
with respect to the horizontal. The coeffecient ofkinetic frictin
between the log and the ramp is 0.900 and the loghas an
acceleration of 0.800 m/s2. Find the tension in therope.
Answers
Answer:2.737 kN
Explanation:
Given
mass of log(m)=205 kg
ramp inclination[tex]=30^{\circ}[/tex]
coefficient of kinetic friction between log and ramp is ([tex]\mu _k[/tex])=0.9
log has an acceleration of 0.8 m/s^2
Let T be the tension in the rope
[tex]T-mgsin\theta -f_r=ma[/tex]
Where [tex]mgsin\theta [/tex]=Sin component of weight
[tex]f_r=friction\ Force=\mu _KN[/tex](Where N is Normal reaction)
[tex]T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma[/tex]
[tex]T=m\left ( gsin\theta +\mu _kcos\theta +a\right )[/tex]
sin30=0.5
cos30=0.866
[tex]T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )[/tex]
[tex]T=205\times 13.35=2.737 kN[/tex]
To find the tension in the rope, you need to first calculate the gravitational and frictional forces on the log on the inclined plane. Then using Newton's second law, set up an equation for net force as the sum of these forces and the tension. Solve for the tension to get the desired result.
Explanation:The physics subject of this problem involves the understanding of Newton's second law, forces in equilibrium, the concept of tension, friction, and forces on inclined planes. Given are the mass of the log (m) = 205 kg, acceleration (a) = 0.8 m/s2, angle of inclination (θ) = 30 degrees, and coefficient of kinetic friction (μ) = 0.900.
Firstly, calculate the force due to gravity along the ramp (fgravity) = m*g*sin(θ), where g = 9.8 m/s2. Secondly, calculate the force due to friction (ffriction) = μ*m*g*cos(θ). The net force (Fnet) pulling the log up the ramp is the difference between the tension force (T), the force due to gravity along the ramp and the frictional force, i.e., Fnet = T - fgravity - ffriction.
Since we also know from Newton’s second law that Fnet = m*a, we can set these equal and solve for the tension: T = m*a + fgravity + ffriction. Plugging in the given values into this equation will provide the tension in the rope.
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A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the xy plane. The 4.80-kg puck has a velocity of 1.00î m/s at one instant. Eight seconds later, its velocity is (6.00î + 6.0ĵ) m/s.(a) Assuming the rocket engine exerts a constant horizontal force, find the components of the force.
Answers
Answer:
[tex]F=(3i+3.6j)\ N[/tex]
Explanation:
It is given that,
Mass of the puck, m = 4.8 kg
Initial velocity of the puck, [tex]u=(1i+0j)\ m/s[/tex]
After 8 seconds, final velocity of the puck, [tex]v=(6i+6j)\ m/s[/tex]
Let the x and y component of force is given by [tex]F_x\ and\ F_y[/tex].
x component of force is given by :
[tex]F_x=m\times \dfrac{v-u}{t}[/tex]
[tex]F_x=4.8\times \dfrac{6-1}{8}[/tex]
[tex]F_x=3\ N[/tex]
y component of force is given by :
[tex]F_y=m\times \dfrac{v-u}{t}[/tex]
[tex]F_y=4.8\times \dfrac{6-0}{8}[/tex]
[tex]F_y=3.6\ N[/tex]
So, the component of the force is [tex]F=(3i+3.6j)\ N[/tex]. Hence, this is the required solution.
The components of the force exerted by the rocket engine on the puck are calculated using Newton's Second Law. The acceleration in each direction (X and Y) is determined by dividing the change in velocity by the change in time. The force is then found by multiplying the mass of the puck by the acceleration, resulting in a force of 3.00î N in the X direction and 3.60ĵ N in the Y direction.
Explanation:The physics behind this problem involves the concept of Newton's Second Law of motion, which states that force equals mass times acceleration (F=ma). To solve this problem, we'll need to find the change in velocity and divide it by the change in time to determine the acceleration. Then, we'll use Newton's Second Law to find the force.
The initial velocity of the puck is 1.00î m/s and the final velocity is (6.00î + 6.0ĵ) m/s. Therefore, the change in velocity (final - initial) in the X direction is 5î m/s and in the Y direction is 6ĵ m/s. The time duration for this change is 8 seconds. So, the acceleration (change in velocity divided by time) is 5/8 = 0.625î m/s² in the X-direction and 6/8=0.75ĵ m/s² in the Y-direction.
Finally, using Newton's Second Law (F=ma), we can find the force in each direction by multiplying the mass of the puck (4.80 kg) by the acceleration. This results in a force of 0.625 * 4.80 = 3.00 N in the X direction (3.00î N) and 0.75 * 4.80 = 3.60 N in the Y direction (3.60ĵ N).
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An airplane weighs 18,000lbs. The local gravitational acceleration g is 32fps^2. What is the mass of the airplane?
Answers
Answer:
Mass of the airplane, m = 562.5 slugs
Explanation:
Given that,
Weight of the airplane, W = 18,000 lbs
The acceleration due to gravity is, [tex]g=32\ fp/s^2[/tex]
Let m is the mass of the airplane. The weight of an object is equal to the product of mass and acceleration due to gravity as :
W = m × g
[tex]m=\dfrac{W}{g}[/tex]
[tex]m=\dfrac{18000}{32}[/tex]
m = 562.5 slugs
So, the mass of the airplane is 562.5 slugs. Hence, this is the required solution.
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in meter rounded to three decimal places? (k = 1/ 40 = 9.00 x 10°N.m2/C2, 1°C = 10 C)
Answers
Answer:
r = 5.335 meters
Explanation:
Given that,
Charge 1, [tex]q_1=-165\ \mu C[/tex]
Charge 2, [tex]q_2=115\ \mu C[/tex]
Force of attraction between two charges, F = 6 N
The force of attraction between two charges is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex], r is the separation between two charges
[tex]r=\sqrt{\dfrac{kq_1q_2}{F}}[/tex]
[tex]r=\sqrt{\dfrac{9\times 10^9\times 165\times 10^{-6}\times 115\times 10^{-6}}{6}}[/tex]
r = 5.335 m
So, the separation between two charges is 5.335 meters. Hence, this is the required solution.
The vertical component of the magnetic induction in the Earth's magnetic field at Hobart is approximately 6×10-5T upward. What electric field is set up in a car travelling on a level surface at 100 km h-1due to this magnetic field? Which side or end of the car is positively charged? Approximately what p.d. is created across a car of typical size?
Answers
Answer:
Explanation:
Magnetic field B = 6 X 10⁻⁵ T.
Width of car = L (Let )
Velocity of car v = 100 km/h
= 27.78 m /s
induced emf across the body ( width ) of the car
= BLv
= 6 X 10⁻⁵ L X 27.78
166.68 X 10⁻⁵ L
Induced electric field across the width
= emf induced / L
E = 166.68 X 10⁻⁵ N/C
We suppose breadth of a typical car = 1.5 m
potential difference induced
= 166.68 x 1.5 x 10⁻⁵
250 x 10⁻⁵ V
= 2.5 milli volt.
The side of the car which is positively charged depends on the direction in which car is moving , whether it is moving towards the north or south.
A diode vacuum tube consists of a cathode and an anode spaced 5-mm apart. If 300 V are applied across the plates. What is the velocity of an electron midway between the electrodes and at the instant of striking the plate, if the electrons are emitted from the cathode with zero velocity?
Answers
Answer:
Explanation:
There is electric field between the plates whose value is given by the following expression
electric field E = V /d where V is potential between the plates and d is distance between them
E = 300 / 5 x 10⁻³
= 60 x 10³ N/c
Force on electron = q E where q is charge on the electron
F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.
Acceleration a = force / mass
a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹
= 10.55 x 10¹⁵ m / s²
For midway , distance travelled
s = 2.5 x 10⁻³ m
s = 1\2 a t²
t = [tex]\sqrt{\frac{2s}{a\\ } }[/tex]
= [tex]\sqrt{\frac{2\times2.5\times10^{-3}}{ 10.55\times10^{15}}[/tex]
t = .474 x 10⁻¹⁸ s
For striking the plate time is calculated as follows
t = [tex][tex]\sqrt{\frac{2\times5\times10^{-3}}{ 10.55\times10^{15}}[/tex][/tex]
t = 0.67 x 10⁻¹⁸ s
The brakes are applied to a moving van, causing it to uniformly slow down. While slowing, it moves a distance of 40.0 m in 7.70 s to a final velocity of 1.80 m/s, at which point the brakes are released. (a) What was its initial speed (in m/s), just before the brakes were applied? m/s (b) What was its acceleration (in m/s^2) while the brakes were applied? (Assume the initial direction of motion is the positive direction. Indicate the direction with the sign of your answer.) m/s^2
Answers
Answer:
a)8.59 m/s
b)-0.8818 m/s²
Explanation:
a) Given the van moved 40 m in 7.70 seconds to a final velocity of 1.80 m/s
Apply the equation for motion;
[tex]d=(\frac{V_i+V_f}{2} )*t[/tex]
where
t=time the object moved
d=displacement of the object
Vi=initial velocity
Vf=final velocity
Given
t=7.70s
Vf=1.80 m/s
d=40m
Vi=?
Substitute values in equation
[tex]40=(\frac{V_i+1.80}{2} )7.70\\\\\\40=\frac{7.70V_i+13.86}{2} \\\\80=7.70V_i+13.86\\\\80-13.86=7.70V_i\\\\66.14=7.70V_i\\\\\frac{66.14}{7.70} =\frac{7.70V_i}{7.70} \\8.59=V_i[/tex]
b)Acceleration is the rate of change in velocity
Apply the formula
Vf=Vi+at
where;
Vf=final velocity of object
Vi=Initial velocity of the object
a=acceleration
t=time the object moved
Substitute values in equation
Given;
Vf=1.80 m/s
Vi=8.59 m/s
t=7.70 s
a=?
Vf=Vi+at
1.80=8.59+7.70a
1.80-8.59=7.70a
-6.79=7.70a
-6.79/7.70=7.70a/7.70
-0.8818=a
The van was slowing down.
When the brakes are applied to a moving van, it travels a distance of 40.0 m in 7.70 s with a final velocity of 1.80 m/s.
a) The initial speed of the van just before the brakes were applied was 8.59 m/s.
b) The acceleration of the van while the brakes were applied was -0.88 m/s².
a) The initial speed of the van can be calculated as follows:
[tex] v_{f} = v_{i} + at [/tex]
Where:
[tex] v_{f}[/tex]: is the final velocity = 1.80 m/s
[tex] v_{i}[/tex]: is the initial velocity =?
a: is the acceleration
t: is the time = 7.70 s
By solving the above equation for a we have:
[tex] a = \frac{v_{f} - v_{i}}{t} [/tex] (1)
Now, we need to use other kinematic equation to find the initial velocity.
[tex] v_{i}^{2} = v_{f}^{2} - 2ad [/tex] (2)
By entering equation (1) into (2) we have:
[tex] v_{i}^{2} = v_{f}^{2} - 2d(\frac{v_{f} - v_{i}}{t}) = (1.80 m/s)^{2} - 2*40.0 m(\frac{1.80 m/s - v_{i}}{7.70 s}) [/tex]
After solving the above equation for [tex]v_{i}[/tex] we get:
[tex] v_{i} = 8.59 m/s [/tex]
Hence, the initial velocity is 8.59 m/s.
b) The acceleration can be calculated with equation (1):
[tex] a = \frac{v_{f} - v_{i}}{t} = \frac{1.80 m/s - 8.59 m/s}{7.70 s} = -0.88 m/s^{2} [/tex]
Then, the acceleration is -0.88 m/s². The minus sign is because the van is decelerating.
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How long in seconds will it take a car traveling at 100 miles/hour to travel the length of an American Football field?
Answers
Answer:
It will take 2.45 seconds.
Explanation:
A football field measured 120 yards, that is arround 0.068 miles, the car is moving with a constant speed so the formula we have to apply for this is:
[tex]t=\frac{x}{v}\\where\\x=distance\\v=speed\\[/tex]
[tex]t=\frac{0.068}{100}\\t=0.00068hrs[/tex]
in order to obtain the time in seconds:
[tex]tsec=0.00068hr*(\frac{3600sec}{1hr} )\\tsec=2.45 sec[/tex]
You drop a rock from the top of a building of height h. Your co-experimenter throws a rock from the same spot with a vertically downward speed vo, a time t after you released your rock. The two rocks hit the ground at the same time. Find the expression for the time t, in terms of vo, g, and h.
Answers
Answer:
[tex]t=\sqrt{2h/g}-(1/g)*(\sqrt{v_{o}^2+2gh}-v_{o})[/tex]
Explanation:
First person:
[tex]y(t)=y_{o}-v_{o}t-1/2*g*t^{2}[/tex]
[tex]v_{o}=0[/tex] the rock is dropped
[tex]y_{o}=h[/tex]
[tex]y(t)=h-1/2*g*t^{2}[/tex]
after t1 seconds it hit the ground, y(t)=0
[tex]0=h-1/2*g*t_{1}^{2}[/tex]
[tex]t_{1}=\sqrt{2h/g}[/tex]
Second person:
[tex]y(t)=y_{o}-v_{o}t-1/2*g*t^{2}[/tex]
[tex]v_{o}[/tex] the rock has a initial downward speed
[tex]y_{o}=h[/tex]
[tex]y(t)=h-v_{o}t-1/2*g*t^{2}[/tex]
after t2 seconds it hit the ground, y(t)=0
[tex]0=h-v_{o}t_{2}-1/2*g*t_{2}^{2}[/tex]
[tex]g*t_{2}^{2}+2v_{o}t_{2}-2h=0[/tex]
[tex]t_{2}=(1/2g)*(-2v_{o}+\sqrt{4v_{o}^2+8gh})[/tex]
the time t when the second person throws the rock after the first person release the rock is:
t=t1-t2
[tex]t=\sqrt{2h/g}-(1/g)*(\sqrt{v_{o}^2+2gh}-v_{o})[/tex]
Vector A has a magnitude of 16 m and makes an angle of 44° with the positive x axis. Vector B also has a magnitude of 13 m and is directed along the negative x axis. Find A+B (in meters and degrees)
Find A-B (in meters and degrees)
Answers
Answer with explanation:
The given vectors in are reduced to their componednt form as shown
For vector A it can be written as
[tex]\overrightarrow{v}_{a}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}[/tex]
Similarly vector B can be written as
[tex]\overrightarrow{v}_{b}=-13\widehat{i}[/tex]
Hence The sum and difference is calculated as
[tex]\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=(16cos(44^{o})-13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}+\overrightarrow{v}_{b}=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}+\overrightarrow{v}_{b}|=\sqrt{(-1.49)^{2}+11.11^{2}}=11.21m[/tex]
The direction is given by
[tex]\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{-1.49}=97.64^{o}[/tex]with positive x axis.
Similarly
[tex]\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=(16cos(44^{o})+13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}-\overrightarrow{v}_{b}=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}-\overrightarrow{v}_{b}|=\sqrt{(24.51)^{2}+11.11^{2}}=26.91m[/tex]
The direction is given by
[tex]\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{24.51}=24.38^{o}[/tex]with positive x axis.
A car drives on a highway with a speed of 68mi/hr. What is the speed in km/hr?
Answers
Answer:
Speed in km/hr will be 109.412 km/hr
Explanation:
We have given speed of the car on a highway = 68 mi/hr
We have to find the speed in km/hr
For this first we have to change mi to km
We know that 1 mile = 1.609 km
Speed is the ratio of distance and time
So 68 mi/hr [tex]=68\times 1.609km/hr=109.412km/hr[/tex]
So the speed in km/hr will be 109.412 km /hr
An Olympic runner competing in a long-distance event finishes with a time of 2 hours, 45 minutes, and 35 seconds. The event has a distance of 27.3 miles. What is the average speed of the runner in kilometers per hour?
Answers
Answer:
Average speed of Olympic runner will be 23.08 km/hr
Explanation:
We have given distance d = 27.3 miles
We know that 1 mile = 1.6 km
So 27.3 mile = [tex]=27.3\times 1.6=43.68km[/tex]
Given time = 2 hour 45 minutes and 35 sec
We know that 1 hour = 60 minutes
So 45 minutes [tex]=\frac{45}{60}=0.75hour[/tex]
We also know that 1 hour = 3600 sec
So 1 sec [tex]=\frac{1}{3600}=2.777\times 10^{-4}hour[/tex]
So total time t =[tex]=2+0.75+0.000277=2.750277hour[/tex]
We know that speed [tex]=\frac{distance}{time }=\frac{63.48}{2.750277}=23.08km/hour[/tex]
To find the average speed of the runner in kilometers per hour, convert the time to hours and the distance to kilometers, then divide the distance by the time to get an average speed of approximately 15.932 km/h.
Explanation:The calculation of a runner’s average speed in a long-distance event requires converting the given time and distance into a consistent set of units and then applying the formula for average speed. First, we convert the runner's time into hours and the distance into kilometers. The runner finished with a time of 2 hours, 45 minutes, and 35 seconds, which translates to 2.7597 hours when converted into hours entirely. The distance covered was 27.3 miles, which converts to approximately 43.941 kilometers (using the conversion factor where 1 mile = 1.60934 kilometers).
To find the average speed, we divide the total distance in kilometers by the total time in hours:
Average speed = Total distance / Total time
This results in an average speed of approximately 15.932 kilometers per hour (km/h).
What current is required in the windings of a long solenoid that has 420 turns uniformly distributed over a length of 0.575 m in order to produce inside the solenoid a magnetic field of magnitude 4.22×10−5 T? The permeablity of free space is 1.25664 × 10 Tm/A.
Answers
Answer:
The current in the solenoid=0.463 Ampere
Explanation:
Given:
Number of turns , N=420
Length of the solenoid=0.575 m
Magnitude of Magnetic Field,[tex]B=4.22\times10^{-5}\ \rm T[/tex]
We know that the magnitude of the magnetic Field inside the solenoid is
[tex]B=\mu_0 ni[/tex]
Where
[tex]\mu_0[/tex] is the permeability of free space.n is the number of turns per unit lengthi is the currentAccording to question
[tex]B=\mu_0 ni\\\\4.22\times10^{-5}=1.24664\times10^{-7}\times \dfrac{420}{0.575}\times i\\=0.463\ \rm Amp[/tex]
Hence the current is calculated.
A block is placed on a plane inclined at 37.0° to the horizontal. The coefficient of kinetic friction is 0.40. If the block is released, what is its acceleration? include a diagram.
Answers
Answer:
2.77 m/s^2
Explanation:
For simplicity, we will say that the x-axis is parallel to the inclined plane and the y-axis is perpendicular to the inclined plane. The force that the surface applies on the block N will be perpendicular to the plane. The force of friction can be expressed like this:
[tex]Fr = u*N[/tex]
Where u is the coefficient of kinetic friction.
The x-component and y-component of the weight force will be:
[tex]W_y = W*cos(37)\\W_x = W*sin(37)[/tex]
In the y-axis, the acceleration of the block is 0. Then:
[tex]N - W_y = 0\\N = W_y = W*cos(37) = m*g*cos(37)[/tex]
In the X-axis, the summation of forces would be:
[tex]W_x - Fr = m*a\\m*g*sin(37) - u*N = m*g*sin(37) - u*m*g*cos(37) = ma\\a = \frac{mg(sin(37)-u*cos(37))}{m} = 9.81m/s^2*(sin(37)-0.4*cos(37))=2.77m/s^2[/tex]
Which is true concerning the acceleration due to gravity? A. It decreases with increasing altitude. B. It is different for different objects in free fall. C. It is a fundamental quantity. D. It is a universal constant. E. all of these
Answers
Answer:
The correct answer is option 'a': It decreases with increase in altitude
Explanation:
Acceleration due to gravity is the acceleration that a body is subjected to when it is freely dropped from a height from surface of any planet, ignoring the resistance that the object may face in it's motion such as drag due to any fluid.. The acceleration due to gravity is same for all the objects and is independent of their masses, it only depends on the mass of the planet and the radius of the planet on which the object is dropped. it's values varies with:
1) Depth from surface of planet.
2)Height from surface of planet.
3) Latitude of the object.
Hence it neither is a fundamental quantity nor an universal constant.
The variation of acceleration due to gravity with height can be mathematically written as:
[tex]g(h)=g_{surface}(1-\frac{2h}{R_{planet}})[/tex]
where,
R is the radius of the planet
[tex]g_{surface}[/tex] is value of acceleration due to gravity at surface.
hence we can see that upon increase in altitude the value of 'g' goes on decreasing.
A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a stop in 20 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)
Answers
Answer:
Part a) Force on car = 2833.84 Newtons
Part b) Time to stop the car = 3.8 seconds
Part c) Factor for stopping distance is 4.
Part d) Factor for stopping time is 1.
Explanation:
The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as
[tex]v^2=u^2+2as[/tex]
where
v = final speed of the car ( = 0 in our case since the car stops)
u = initial speed of the car = 38 km/hr =[tex]\frac{38\times 1000}{3600}=10.56m/s[/tex]
a = deceleration produces
s = distance in which the car stops
Applying the given values we get
[tex]0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2[/tex]
Now the force can be obtained using newton's second law as
[tex]Force=\frac{Weight}{g}\times a[/tex]
Applying values we get
[tex]Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons[/tex]
The negative direction indicates that the force is opposite to the motion of the object.
Part b)
The time required to stop the car can be found using the first equation of kinematics as
[tex]v=u+at[/tex] with symbols having the same meanings
Applying values we get
[tex]0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds[/tex]
Part c)
From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.
Part d)
From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.
A projectile is launched from ground level with an initial velocity of v 0 feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s equals negative 16 t squared plus v 0 t. Find the time(s) that the projectile will (a) reach a height of 192 ft and (b) return to the ground when v 0equals128 feet per second.
Answers
Answer:
a) At times t = 2 s and t = 6 s, the projectile will be at a height of 192 ft.
b) The projectile will return to the ground at t = 8 s.
Explanation:
a) The height of the projectile is given by this equation:
s = -16·t² + 128 f/s·t (see attached figure)
If the height is 192 ft, then:
192 ft = 16·t² + 128 ft/s· t
0 = -16·t² + 128 ft/s·t - 192 ft
Solving the quadratic equation:
t = 2 and t = 6
At times t = 2 s and t = 6 s, the projectile will be at a height of 192 ft.
b) When the projectile return to the ground, s = 0. Then:
0 = -16·t² + 128 ft/s·t
0 =t(-16·t + 128 ft/s)
t = 0 is the initial point, when the projectile is launched.
-16·t + 128 ft/s = 0
t = -128 ft/s / -16 ft/s² = 8 s
The projectile will return to the ground at t = 8 s.
The eye of a hurricane passes over Grand Bahama Island in a direction 60.0° north of west with a speed of 43.0 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 23.0 km/h. How far from Grand Bahama is the hurricane 4.80 h after it passes over the island?
Answers
Answer:
The hurricane will be 166.15km away from the island.
Explanation:
We can solve this problem with trigonometry and the formulas of constant velocity motion.
we need to find the displacement on the X and Y axis.
the speed on the X axis on the first trame of movemont is defined as:
[tex]Vx=V*cos(ang)[/tex]
we will take the reference to the left of the X axis as positive, so:
[tex]Vx=43*cos(60)\\Vx=21.5 km/h[/tex]
So the displacement on X is:
[tex]dx=Vx*t\\dx=21.5km/h*3h\\dx=64.5km[/tex]
we will do the same for the Y axis:
[tex]Vy=V*sin(ang)\\Vy=43sin(60)\\Vy=37.24km/h[/tex]
the Y displacement is:
[tex]dy=Vy*t\\dy=37.24km/h*3h\\dy=111.72km[/tex]
Then on the second trame of the movement we only have displacement on the Y axis.
[tex]dy2=Vy2*t\\dy2=23km/h*1.8h\\dy2=41.4km[/tex]
now we need the total displacement on the X and Y axis:
[tex]Y=dy+dy2\\Y=153.12km\\X=64.5km[/tex]
The magnitud of the total displacement will be:
[tex]D=\sqrt{(X)^2+(Y)^2}\\D=\sqrt{(64.5)^2+(153.12^2}\\ \\D=166.15km[/tex]
Final answer:
The distance of the hurricane from Grand Bahama 4.80 hours after passing over the island is 170.4 km.
Explanation:
To find the distance of the hurricane from Grand Bahama 4.80 hours after passing over the island, we need to calculate the distance it has traveled during that time.
First, let's calculate the distance traveled during the initial three hours when the hurricane was moving at a speed of 43.0 km/h. The distance traveled is given by:
Distance = Speed × Time = 43.0 km/h × 3 h = 129.0 km
Now, let's calculate the distance traveled during the remaining 1.80 hours when the hurricane was moving at a speed of 23.0 km/h. The distance traveled is given by:
Distance = Speed × Time = 23.0 km/h × 1.80 h = 41.4 km
Therefore, the total distance traveled by the hurricane is 129.0 km + 41.4 km = 170.4 km.
A rocket ship at rest in space gives a short blast of its engine, firing 50 kg of exhaust gas out the back end with an average velocity of 400 m/s. What is the change in momentum of the rocket during this blast?
Answers
Answer:
20,000 Ns
Explanation:
mass of exhaust gases, m = 50 kg
velocity of exhaust gases, v = 400 m/s
The momentum of a body is defined as the measurement of motion of body. mathematically, it is defined as the product of mass off the body an its velocity.
change in momentum of rocket = final momentum - initial momentum
= m x v - 0
= 50 x 400
= 20,000 Ns
Final answer:
The change in momentum of the rocket (also known as impulse) during the blast is 20,000 kg·m/s. This is calculated by multiplying the mass of the exhaust gas (50 kg) by its average velocity (400 m/s).
Explanation:
The change in momentum of the rocket during the blast (also known as impulse) can be calculated using the conservation of momentum. The momentum of the exhaust gas expelled will be equal in magnitude and opposite in direction to the change in momentum of the rocket.
To calculate the change in momentum of the rocket, we use the formula:
Change in momentum = mass of exhaust × velocity of exhaust.
In this case, the mass of the exhaust gas is 50 kg, and the average velocity at which it is expelled is 400 m/s:
Change in momentum = 50 kg × 400 m/s = 20000 kg·m/s.
This is the momentum gained by the rocket in the opposite direction, due to Newton's third law of motion.
A particle leaves the origin with an initial velocity v⃗ =(2.40 m/s)xˆv→=(2.40 m/s)x^ , and moves with constant acceleration a⃗ =(−1.90 m/s2)xˆ+(3.20 m/s2)yˆa→=(−1.90 m/s2)x^+(3.20 m/s2)y^ . (a) How far does the particle move in the x direction before turning around? (b) What is the particle’s velocity at this time? (c) Plot the particle’s position at t=0.500 st=0.500 s , 1.00 s, 1.50 s, and 2.00 s. Use these results to sketch position versus time for the particle.
Answers
Answer:
distance stop 1.52m,
velocity 4.0 m/s y^
Explanation:
The movement of the particle is two-dimensional since it has acceleration in the x and y axes, the way to solve it is by working each axis independently.
a) At the point where the particle begins to return its velocity must be zero (Vfx = 0)
Vfₓ = V₀ₓ + aₓ t
t = - V₀ₓ/aₓ
t = - 2.4/(-1.9)
t= 1.26 s
At this time the particle stops, let's find his position
X1 = V₀ₓ t + ½ aₓ t²
X1= 2.4 1.26 + ½ (-1.9) 1.26²
X1= 1.52 m
At this point the particle begins its return
b) The velocity has component x and y
As a section, the X axis x Vₓ = 0 m/s is stopped, but has a speed on the y axis
Vfy= Voy + ay t
Vfy= 0 + 3.2 1.26
Vfy = 4.0 m/s
the velocity is
V = (0 x^ + 4.0 y^) m/s
c) In order to make the graph we create a table of the position x and y for each time, let's start by writing the equations
X = V₀ₓ t+ ½ aₓ t²
Y = Voy t + ½ ay t²
X= 2.4 t + ½ (-1.9) t²
Y= 0 + ½ 3.2 t²
X= 2.4 t – 0.95 t²
Y= 1.6 t²
With these equations we build the table to graph, for clarity we are going to make two distance graph with time, one for the x axis and another for the y axis
Chart to graph
Time (s) x(m) y(m)
0 0 0
0.5 0.960 0.4
1 1.45 1.6
1.50 1.46 3.6
2.00 1.00 6.4
In the problem, we have a particle moving with constant acceleration. We can find the distance before it turns around using the equations of motion, and the velocity at that time is zero. To plot its position over time, again we have to use the equations of motion for each coordinate.
Explanation:The problem given relates to motion in two dimensions specifically linear motion with constant acceleration. Let's address each part of the question.
Since the acceleration in the x-direction is negative, the particle will eventually stop and then begin to move in the negative x direction. To find the time at which it stops, we use the first equation of motion: final velocity = initial velocity + (acceleration * time). Here, the final velocity becomes 0 when it stops, and thus, time = initial velocity / magnitude of acceleration. Once we know the time, we can find the distance traveled in the x-direction before turning around by using the second equation of motion: displacement = initial velocity*time + 0.5*acceleration*(time^2). The velocity at the time of turn-around would be 0, because the term 'turning around' means the point where it just stops in the x direction and is about to move in the opposite direction. To plot the particle position at different time points, we would have to use the equations of motion separately for x and y direction and find the positions at time points given. Learn more about Motion with constant acceleration here:
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A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway worker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?
Answers
Answer:22.76 m/s
Explanation:
Given
Train length(L)=75 m
Front of train after travelling 125 m is 18 m/s
Time taken by the front of train to cover 125 m
[tex]v^2-u^2=2as[/tex]
[tex]18^2-0=2\times a\times 125[/tex]
[tex]a=1.296 m/s^2[/tex]
Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel a distance of 125+75=200 m
[tex]v^2-u^2=2as[/tex]
[tex]v^2=2\times 1.296\times 200[/tex]
[tex]v=\sqrt{518.4}=22.76 m/s[/tex]
A hot-air balloon is descending at a rate of 2.1 m/s when a passenger drops a camera. If the camera is 42 m above the ground when it is dropped, how long does it take for the camera to reach the ground?
Express your answer using two significant figures.
If the camera is 42 m above the ground when it is dropped, what is its velocity just before it lands? Let upward be the positive direction for this problem.
Express your answer using two significant figures.
Answers
Answer:
a) 2.7s
b) 29 m/s
Explanation:
The equation for the velocity and position of a free fall are the following
[tex]v=v_{0}-gt[/tex] -(1)
[tex]x=x_{0}+v_{0}t-gt^{2}/2[/tex] - (2)
Since the hot-air ballon is descending at 2.1m/s and the camera is dropped at 42 m above the ground:
[tex]v_{0}=-2.1m/s[/tex]
[tex]x_{0}=42m[/tex]
To calculate the time which it takes to reach the ground we use eq(2) with x=0, and look for the positive solution of t:
[tex]t = \frac{1}{84}(2.1\pm\sqrt{2.1^{2} - 4\times42\times9.81/2} )[/tex]
t = 2.71996
Rounding to two significant figures:
t = 2.7 s
Now we calculate the velocity the camera had just before it lands using eq(1) with t=2.7s
[tex]v=-2.1-9.81*(2.71996)[/tex]
v = -28.782 m/s
Rounding to two significant figures:
v = -29 m/s
where the minus sign indicates the downwards direction
A model rocket rises with constant acceleration to a height of 3.1 m, at which point its speed is 28.0 m/s. a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket's acceleration?
c. Find the height of the rocket 0.10 s after launch.
d. Find the speed of the rocket 0.10 s after launch.
Answers
Explanation:
It is given that,
Height, h = 3.1 m
Initial speed of the rocket, u = 0
Final speed of the rocket, v = 28 m/s
(b) Let a is the acceleration of the rocket. Using the formula as :
[tex]a=\dfrac{v^2-u^2}{2h}[/tex]
[tex]a=\dfrac{(28)^2}{2\times 3.1}[/tex]
[tex]a=126.45\ m/s^2[/tex]
(a) Let t is the time taken to reach by the rocket to reach to a height of h. So,
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{28\ m/s}{126.45\ m/s^2}[/tex]
t = 0.22 seconds
(c) At t = 0.1 seconds, height of the rocket is given by :
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
[tex]h=\dfrac{1}{2}\times 126.45\times (0.1)^2[/tex]
h = 0.63 meters
(d) Let v' is the speed of the rocket 0.10 s after launch.
So, [tex]v'=u+at[/tex]
[tex]v'=0+126.45\times 0.1[/tex]
v' = 12.64 m/s
Hence, this is the required solution.
What gravitational force does the moon produce on the Earth is their centers are 3.88 x 10^8 m apart and the moon has a mass of 7.34 x 10^22 kg?
Answers
The gravitational force that the moon produces on the Earth, when their centers are 3.88 x 10^8 m apart and the moon has a mass of 7.34 x 10^22 kg, can be calculated using Newton's Law of Universal Gravitation. The force comes out to be approximately 1.82 * 10^20 N.
Explanation:To calculate the gravitational force exerted by the moon on the earth, we would use Newton's Law of Universal Gravitation. This law states that every point mass attracts every other point mass by a force pointing along the line intersecting both points. The force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between the point masses.
The law is mathematically expressed as: F = G * (m1 * m2)/d² , where F is the force of gravity between the two bodies, G is the gravitational constant (approximately 6.67 * 10^-11 N * m²/kg²), m1 and m2 are the masses of the two bodies, and d is the distance between the centers of the two bodies.
Thus, the gravitational force (F) between Earth and the Moon can be calculated as follows: F = (6.67 * 10^-11 N * m²/kg²) * (7.34 * 10^22 kg * 5.97 * 10^24 kg) / (3.88 * 10^8 m)^2. Calculating this gives a force of approximately 1.82 * 10^20 N. This is the gravitational force that the moon exerts on the Earth.
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We calculated that the gravitational force between the Moon and the Earth is approximately 1.95 ×[tex]10^{20}[/tex] N.
Gravitational Force Between the Moon and the Earth
To calculate the gravitational force between the Moon and the Earth, we use Newton's Universal Law of Gravitation. According to this law, the force of gravity ( F ) between two masses ( m1 and m2 ) is given by:
F = G x (m1 x m2) / r²
where:
G is the gravitational constant, approximately 6.67 × [tex]10^{-11}[/tex] N(m²/kg²)
m1 and m2 are the masses of the two objects
r is the distance between the centers of the two masses
Here, we have:
m1 (mass of the Earth) = 5.98 ×[tex]10^{24}[/tex]kg
m2 (mass of the Moon) = 7.34 × [tex]10^{22}[/tex] kg
r (distance between centers) = 3.88 × [tex]10^8[/tex] m
Now, we substitute these values into our formula:
F = (6.67 × [tex]10^{-11}[/tex] N(m²/kg²)) x (5.98 × [tex]10^{24}[/tex]kg x 7.34 × [tex]10^{22}[/tex] kg) / (3.88 × [tex]10^8[/tex] m)²
First, we calculate the numerator:
(6.67 × [tex]10^{-11}[/tex]) x (5.98 × [tex]10^{24}[/tex] x 7.34 × [tex]10^{22}[/tex]) = 2.92 × [tex]10^{37}[/tex] N m²/kg²
Next, we calculate the denominator:
(3.88 × [tex]10^8[/tex])² = 1.50 × [tex]10^{17}[/tex] m²
Finally, we divide the numerator by the denominator to get:
F = 2.92 × [tex]10^{37}[/tex] N m²/kg² / 1.50 × [tex]10^{17}[/tex] m² ≈ 1.95 × [tex]10^{20}[/tex] N
Therefore, the gravitational force that the Moon produces on the Earth is approximately 1.95 × [tex]10^{20}[/tex] N .
The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A (20, 15,0)m to the point B (0,07) m. What is the work done on the particle? The work done is given by F. f, where is the resultant force (here F = Fi + F2) and is the displacement.
Answers
Answer:
-600 J
Explanation:
F₁ = 8i +29 j + 32k
F₂ = 48 i - 59 j - 22 k
F = F₁ +F₂ = 8i +29 j + 32k +48 i - 59 j - 22 k
F = 56i - 30 j + 10 k
displacement d = ( 0 - 20 )i + ( 0 - 15 )j + ( 7 -0) k
d = - 20 i - 15 j + 7 k
Work Done = F dot product d
F . d = - 56 x 20 - 30 x - 15 + 10 x 7
= - 1120 +450 + 70
= -600 J
A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person boxing through the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?
Answers
Answer:
4.14 m
Explanation:
In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .
Let in this last leg , u be the initial velocity.
s = ut + 1/2 g t²
2 = .2 u + .5 x 9.8 x .04
u = 9.02 m /s .
Let v be the final velocity in this leg
v² = u² + 2 g s
v² = (9.02)² + 2 x 9.8 x 2
= 81.36 +39.2
v = 10.97 m / s
Now consider the whole height from where the ball dropped . Let it be h.
Initial velocity u = 0
v² = u² +2gh
(10.97 )² = 2 x 9.8 h
h = 6.14 m
Height from window
= 6.14 - 2m
= 4.14 m
Water is poured into a bowl at a constant rate of 17.0 cm^3/s. The bowl has a circular cross section, but does not have a uniform diameter. (That is, different horizontal cross sections taken at different heights of the bowl have different diameters.) As the water fills the bowl, the water level reaches a point where the diameter of the bowl is
d1 = 1.45 cm.
What is the rate (in cm/s) at which the water level rises at this diameter?
Answers
Answer:
10.29 cm/s
Explanation:
Discharge in to the bowl = 17.0 cm³/s
Diameter of the bowl, d₁ = 1.45 cm
Now,
Rate at which water level rise at its diameter = [tex]\frac{\textup{Discharge}}{\textup{Area of cross-section}}[/tex]
also,
Area of cross-section = [tex]\frac{\pi}{\textup{4}}\times1.45^2[/tex]
or
Area of cross-section = 1.651 cm²
Therefore,
Rate at which water level rise at its diameter = [tex]\frac{\textup{17}}{\textup{1.651}}[/tex]
or
Rate at which water level rise at its diameter = 10.29 cm/s
The rate at which the water level rises at a diameter of 1.45 cm is 8.44 cm/s.
Explanation:To find the rate at which the water level rises at this diameter, we can use the concept of continuity equation. The continuity equation states that the flow rate of a fluid is constant at all points in a pipe or container.
Therefore, the rate at which the water level rises at a particular diameter can be found by dividing the flow rate by the cross-sectional area at that diameter.
Rate of water level rise = Flow rate / Cross-sectional area
Substituting the given values, we get:
Rate of water level rise = 17.0 cm³/s / (π * (d1/2)²)
Rate of water level rise = 17.0 cm³/s / (π * (1.45 cm/2)²)
Rate of water level rise = 8.44 cm/s