Answer:
Part a) Big bang occurred about 13.8 billion years ago. This time is arrived after see the shift in the cosmic background radiation that fills the universe. The result of the big bang was the creation of universe itself. Everything we know space, matter, time was created by the event thus giving birth to the universe and stars, galaxies,e.t.c.
Part b) No string theory is currently not proven as of yet. The basic problem in formulating the theory of everything is to mix 2 completely different theories known as quantum theory of matter and theory of relativity. But this has not been achieved as of yet. There are different theories that try to explain the both quantum and relativistic nature of the matter and string theory is one of the theories that has been proposed to explain the same.The theory is not completely developed as of yet as there are numerous inconsistencies with it such as it proposes 34 dimension's of nature which have not been observed as yet.Also there is an inherent nature of the theory that it deals with quantities with [tex]10^{34}m[/tex] scales of length making it impossible to verify the theory which still is mathematically incomplete.
Part c) Strong Nuclear forces come into action in the nucleus of an atom and holds the nucleon's together. As we know that a pair of protons will repel due to nature of their charge the strong nuclear force holds them together.
Part d) Cosmology is the branch of astronomy that deals with the origin and evolution of the universe. It basically is the study about the universe was formed and how it evolved during different stages.
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
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.
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.
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
Two particles with positive charges q1 and q2 are separated by a distance s. Part A Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero? Express your answer in terms of some or all of the variables s, q1, q2 and k =14πϵ0. If your answer is difficult to enter, consider simplifying it, as it can be made relatively simple with some work.
The solution for the distance x from [tex]\(q_1\)[/tex] along the line connecting the charges where the total electric field is zero is given by: [tex]\[x = \frac{s \cdot (q_1 + q_1\sqrt{q_2})}{q_1 + q_2}\][/tex].
Let's go through the complete solution step by step.
Given:
- Charges: [tex]\(q_1\)[/tex] and [tex]q_2[/tex]
- Distance between charges: s
- Coulomb's constant: [tex]\(k = \frac{1}{4\pi\epsilon_0}\)[/tex]
We want to find the distance (x) from charge [tex]\(q_1\)[/tex] along the line connecting the charges where the total electric field is zero.
The electric field (E) due to a point charge (q) at a distance (r) is given by Coulomb's law:
[tex]\[E = \dfrac{k \cdot q}{r^2}\][/tex]
At a distance x from charge [tex]\(q_1\)[/tex], the electric fields due to [tex]\(q_1\) (\(E_1\))[/tex] and [tex]\(q_2\) (\(E_2\))[/tex] will have magnitudes given by:
[tex]\[E_1 = \dfrac{k \cdot q_1}{x^2}\][/tex]
[tex]\[E_2 = \dfrac{k \cdot q_2}{(s - x)^2}\][/tex]
For the total electric field to be zero, [tex]\(E_1\)[/tex] and [tex]\(E_2\)[/tex] must cancel each other out:
[tex]\[E_1 + E_2 = 0\][/tex]
Substitute the expressions for [tex]\(E_1\)[/tex] and [tex]\(E_2\)[/tex]:
[tex]\[\dfrac{k \cdot q_1}{x^2} + \dfrac{k \cdot q_2}{(s - x)^2} = 0\][/tex]
Cross-multiply and simplify:
[tex]\[q_1 \cdot (s - x)^2 = -q_2 \cdot x^2\][/tex]
Expand and rearrange:
[tex]\[q_1 \cdot (s^2 - 2sx + x^2) = -q_2 \cdot x^2\][/tex]
Solve for x:
[tex]\[q_1 \cdot s^2 - 2q_1 \cdot sx + q_1 \cdot x^2 + q_2 \cdot x^2 = 0\][/tex]
[tex]\[(q_1 + q_2) \cdot x^2 - 2q_1 \cdot sx + q_1 \cdot s^2 = 0\][/tex]
This is a quadratic equation in terms of x. Solve for x using the quadratic formula:
[tex]\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
Where:
[tex]\(a = q_1 + q_2\)[/tex]
[tex]\(b = -2sq_1\)[/tex]
[tex]\(c = q_1s^2\)[/tex]
Calculate x using the quadratic formula:
[tex]\[x = \dfrac{-(-2sq_1) \pm \sqrt{(-2sq_1)^2 - 4(q_1 + q_2)(q_1s^2)}}{2(q_1 + q_2)}\][/tex]
Simplify the expression inside the square root:
[tex]\[x = \dfrac{2sq_1 \pm \sqrt{4s^2q_1^2 - 4(q_1 + q_2)(q_1s^2)}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{2sq_1 \pm \sqrt{4s^2q_1^2 - 4q_1^2s^2 - 4q_2q_1s^2}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{2sq_1 \pm \sqrt{-4q_2q_1s^2}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{2sq_1 \pm 2q_1s\sqrt{-q_2}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{s \cdot (q_1 \pm q_1\sqrt{-q_2})}{q_1 + q_2}\][/tex]
Since [tex]\(q_1\)[/tex] and [tex]\(q_2\)[/tex] are both positive charges, x will be a positive value.
Thus, the solution for the distance x from [tex]\(q_1\)[/tex] is given by [tex]\[x = \dfrac{s \cdot (q_1 + q_1\sqrt{q_2})}{q_1 + q_2}\][/tex].
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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
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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Water and iron have quite different specific heats. Iron heats up more than water when the same amount of heat energy is added to it. Which one of these substances do you deduce has the higher specific heat? water
iron
both have the same specific heat
cannot be determined
Answer:
water
Explanation:
The specific heat of an object denotes the amount of energy which is required to raise the objects temperature by 1 unit of temperature and mass of the object is 1 unit.
If the same amount of heat is added to both water and iron the temperature difference in their final and initial temperatures will be different.
The difference in the initial and final temperature of iron will be higher than that of water. This means that the amount of energy which is required to raise the objects temperature by 1 unit of temperature for iron is lower than water.
So, water has higher specific heat
Also
Q = mcΔT
where
Q = Heat
m = Mass
c = Specific heat
ΔT = Change in temperature
[tex]c=\frac{Q}{m\Delta T}[/tex]
This means specific heat is inversely proportional to change in temperature.
So, for water the change in temperature will be lower than iron which means that water will have a higher specific heat.
(a) What is the force of gravity between two 1160 kg cars separated by a distance of 35 m on an interstate highway? (b) How does this force compare with the weight of a car?
Explanation:
(a) Mass of the cars, m₁ = m₂ = 1160 kg
Distance between cars, d = 35 m
The gravitational force between two cars is given by :
[tex]F=G\dfrac{m_1m_2}{d^2}[/tex]
[tex]F=6.67\times 10^{-11}\times \dfrac{(1160)^2}{(35)^2}[/tex]
[tex]F=7.32\times 10^{-8}\ N[/tex]
So, the force of gravity between the cars is [tex]7.32\times 10^{-8}\ N[/tex].
(b)The weight of a car is given by :
[tex]W=mg[/tex]
[tex]W=1160\times 9.8[/tex]
W = 11368 N
On comparing,
[tex]\dfrac{W}{F}=\dfrac{11368}{7.32\times 10^{-8}}[/tex]
[tex]\dfrac{W}{F}=1.55\times 10^{11}[/tex]
[tex]W=1.55\times 10^{11}\times F[/tex]
So, the weight of the car is [tex]1.55\times 10^{11}[/tex] times the force of gravitation between the cars.
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?
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.
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.61 m/s. The stone subsequently falls to the ground, which is 18.1 m below the point where the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g=9.81 m/s^2 for the acceleration due to gravity.
The stone impacts the ground with a speed of approximately [tex]\(17.32 \, \text{m/s}\)[/tex].
To find the impact speed of the stone, we can use the kinematic equation that relates initial velocity [tex](\(v_0\))[/tex], final velocity [tex](\(v\))[/tex], acceleration [tex](\(g\))[/tex], and displacement [tex](\(s\))[/tex]:
[tex]\[v^2 = v_0^2 + 2gs\][/tex]
Where:
- [tex]\(v\)[/tex] is the final velocity (impact speed),
- [tex]\(v_0\)[/tex] is the initial velocity (throwing speed),
- [tex]\(g\)[/tex] is the acceleration due to gravity (9.81 m/s²),
- [tex]\(s\)[/tex] is the displacement (height the stone falls, -18.1 m, as it falls downward).
Substituting the known values:
[tex]\[v^2 = (7.61 \, \text{m/s})^2 + 2 \times (9.81 \, \text{m/s}^2) \times (-18.1 \, \text{m})\][/tex]
[tex]\[v^2 = 58.0321 - 2 \times 9.81 \times 18.1\][/tex]
[tex]\[v^2 \approx 58.0321 - 357.801\][/tex]
[tex]\[v^2 \approx -299.7689\][/tex]
Since the stone is impacting the ground, we consider the positive root:
[tex]\[v \approx \sqrt{299.7689}\][/tex]
[tex]\[v \approx 17.32 \, \text{m/s}\][/tex]
Therefore, the stone impacts the ground with a speed of approximately [tex]\(17.32 \, \text{m/s}\)[/tex].
A car drives on a highway with a speed of 68mi/hr. What is the speed in km/hr?
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
At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later, the cyclist is 21.0 m due north of the flagpole, going due east with a speed of 13.0 m/s. For the cyclist in this 21.0 s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration? (Give all directions as positive angles relative to due east, where positive is meaured going counterclockwise.)
Answer:
Part a)
[tex]d = 21\sqrt2 = 29.7 m[/tex]
Part b)
Direction is 45 degree North of West
Part c)
[tex]v_{avg} = 1.41 m/s[/tex]
Part d)
direction of velocity will be 45 Degree North of West
Part e)
[tex]a = 0.875 m/s^2[/tex]
Part f)
[tex]\theta = 45 degree[/tex] North of East
Explanation:
Initial position of the cyclist is given as
[tex]r_1 = 21.0 m[/tex] due East
final position of the cyclist after t = 21.0 s
[tex]r_2 = 21.0 m[/tex] due North
Part a)
for displacement we can find the change in the position of the cyclist
so we have
[tex]d = r_2 - r_1[/tex]
[tex]d = 21\hat j - 21\hat i[/tex]
so magnitude of the displacement is given as
[tex]d = 21\sqrt2 = 29.7 m[/tex]
Part b)
direction of the displacement is given as
[tex]\theta = tan^{-1}\frac{y}{x}[/tex]
[tex]\theta = tan^{-1}\frac{21}{-21}[/tex]
so it is 45 degree North of West
Part c)
For average velocity we know that it is defined as the ratio of displacement and time
so here the magnitude of average velocity is defined as
[tex]v_{avg} = \frac{\Delta x}{t}[/tex]
[tex]v_{avg} = \frac{29.7}{21}[/tex]
[tex]v_{avg} = 1.41 m/s[/tex]
Part d)
As we know that average velocity direction is always same as that of average displacement direction
so here direction of displacement will be 45 Degree North of West
Part e)
Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East
So we have
change in velocity of the cyclist is given as
[tex]\Delta v = v_f - v_i[/tex]
[tex]\Delta v = 13\hat i - (-13\hat j)[/tex]
now average acceleration is given as
[tex]a = \frac{\Delta v}{\Delta t}[/tex]
[tex]a = \frac{13\hat i + 13\hat j}{21}[/tex]
so the magnitude of acceleration is given as
[tex]a = \frac{13\sqrt2}{21} = 0.875 m/s^2[/tex]
Part f)
direction of acceleration is given as
[tex]\theta = tan^{-1}\frac{y}{x}[/tex]
[tex]\theta = tan^{-1}\frac{13}{13}[/tex]
[tex]\theta = 45 degree[/tex] North of East
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.
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.
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.
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 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.
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.
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree. At some instant the little creature moves with a velocity of -1.03 m/s. Then, 2.47 s later, it moves at the velocity 1.51 m/s. What is the chipmunk\'s average acceleration during the 2.47-s time interval?
Answer:
a = 1.02834008 m/s2
Explanation:
given data:
initial velocity u = -1.03 m/s
time t = 2.47 s later
final velocity v = 1.51 m/s
average acceleration is given as a
[tex]a = \frac{(v - u)}{t}[/tex]
putting all value to get required value of acceleration:
[tex]= \frac{(1.51 -(- 1.03))}{2.47}[/tex]
[tex]= \frac{1.51+1.03}{2.47}[/tex]
= 1.02834008 m/s2
If y = 0.02 sin (20x – 400t) (SI units), the wave number is
Answer:
Wave number, [tex]k=20\ m^{-1}[/tex]
Explanation:
The given equation of wave is :
[tex]y=0.02\ sin(20x-400t)[/tex]............(1)
The general equation of the wave is given by :
[tex]y=A\ sin(kx-\omega t)[/tex]..............(2)
k is the wave number of the wave
[tex]\omega[/tex] is the angular frequency
On comparing equation (1) and (2) :
[tex]k=20\ m^{-1}[/tex]
[tex]\omega=400\ rad[/tex]
So, the wave number of the wave is [tex]20\ m^{-1}[/tex]. Hence, this is the required solution.
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?
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 girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later. How fast was the pebble going when it hit the water?
A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water [tex]2.00\ s[/tex] later. The pebble was going at a speed of [tex]19.62\ m/s[/tex] when it hit the water.
Use kinematic equations to solve this problem. The pebble is dropped from rest, so its initial velocity is 0 m/s. The time it takes for the pebble to hit the water is given as 2.00 seconds.
The speed of the pebble when it hits the water:
Given:
Initial velocity [tex](v_o) = 0 m/s[/tex]
Time [tex](t) = 2.00\ seconds[/tex]
Acceleration due to gravity [tex](a) = -9.81\ m/s^2[/tex]
Use the kinematic equation:
[tex]v = v_o + a \times t\\v = 0 + (-9.81) \times (2.00)[/tex]
Calculate the numerical value of v:
[tex]v = -19.62\ m/s[/tex]
So, the pebble was going at a speed of [tex]19.62\ m/s[/tex] when it hit the water.
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The pebble was traveling at a speed of 19.6 m/s when it hit the water.
Explanation:In order to determine the speed at which the pebble hit the water, we can use the equations of motion. Assuming the acceleration due to gravity is 9.8 m/s^2 and neglecting air resistance, we can use the equation:
s = ut + (1/2)at^2
Where s is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration. In this case, the distance traveled is the height of the cliff, the initial velocity is 0 (since the pebble is dropped), t is 2.00 s, and the acceleration is 9.8 m/s^2. Plugging in these values, we can solve for the initial velocity:
s = 0 * (2.00) + (1/2) * 9.8 * (2.00)^2
s = 19.6 m
Therefore, the pebble was traveling at a speed of 19.6 m/s when it hit the water.
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.)
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 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.
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
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?
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 driver of a car going 75 km/h suddenly sees the lights of a barrier 35 m ahead. It takes the driver 0.60 seconds before he applies the brakes, and the average acceleration during braking is -8.5 m/s?. (A) Does the car hit the barrier? (B) What is the maximum speed at which the car could be moving and not hit the barrier 35 meters ahead?
Answer:
A) The car will hit the barrier
b) 19.82 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Distance = Speed × Time
⇒Distance = (75/3.6)×0.6
⇒Distance = 12.5 m
Distance traveled by the car during the reaction time is 12.5 m
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-\frac{75}{3.6}^2}{2\times -8.5}\\\Rightarrow s=25.53\ m[/tex]
Total distance traveled is 12.5+25.53 = 38.03 m
So, the car will hit the barrier
Distance = Speed × Time
⇒d = u0.6
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-u^2}{2\times -8.5}\\\Rightarrow s=\frac{u^2}{17}[/tex]
d + s = 35
[tex]\\\Rightarrow u0.6+\frac{u^2}{17}=35\\\Rightarrow \frac{u^2}{17}+0.6u-35=0[/tex]
[tex]10u^2+102u-5950=0\\\Rightarrow u=\frac{-51+\sqrt{62101}}{10},\:u=-\frac{51+\sqrt{62101}}{10}\\\Rightarrow u=19.82, -30.02[/tex]
Hence, the maximum velocity by which the car could be moving and not hit the barrier 35 meters ahead is 19.82 m/s.
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.
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]
How many miles will a car drive on 25 L of gasoline if the car average mileage is 60 km/gal. provide answer with correct significant figures.
Answer:
Distance traveled by car in 25 L IS 396.15 km
Explanation:
We have given average millage of car = 60 km /gal
Means car travel 60 km in 1 gallon
The amount of gasoline = 25 L
We know that 1 L = 0.2641 gallon
So [tex]25L=25\times 0.2641=6.6025gallon[/tex]
As the car travels 60 km in 1 gallon
So traveled distance by car in 6.6025 gallon of gasoline = 60×6.6025 = 396.15 km
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.
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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The route followed by a hiker consists of three displacement vectors A with arrow, B with arrow, and C with arrow. Vector A with arrow is along a measured trail and is 1550 m in a direction 25.0° north of east. Vector B with arrow is not along a measured trail, but the hiker uses a compass and knows that the direction is 41.0° east of south. Similarly, the direction of vector vector C is 20.0° north of west. The hiker ends up back where she started, so the resultant displacement is zero, or A with arrow + B with arrow + C with arrow = 0. Find the magnitudes of vector B with arrow and vector C with arrow.
Answer:
[tex]D_{B}=1173.98m\\D_{C}=675.29m[/tex]
Explanation:
If we express all of the cordinates in their rectangular form we get:
A = (1404.77 , 655.06) m
[tex]B = A + ( -D_{B} *sin(41) , -D_{B} * cos(41) )[/tex]
[tex]C = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )[/tex]
Since we need C to be (0,0) we stablish that:
[tex]C = (0,0) = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )[/tex]
That way we make an equation system from both X and Y coordinates:
[tex]A_{x} + B_{x} + C_{x} = 0[/tex]
[tex]A_{y} + B_{y} + C_{y} = 0[/tex]
Replacing values:
[tex]1404.77 - D_{B}*sin(41) - D_{C}*cos(20) = 0[/tex]
[tex]655.06 - D_{B}*cos(41) + D_{C}*sin(20) = 0[/tex]
With this system we can solve for both Db and Dc and get the answers to the question:
[tex]D_{B}=1173.98m[/tex]
[tex]D_{C}=675.29m[/tex]
Two 1.0 kg masses are 1.0 m apart (center to center) on a frictionless table. Each has a +10μC of charge. a) What is the magnitude of the electric force on one of the masses? b) What is the initial acceleration of one of the masses if it is released and allowed to move?
Answer:
(a). The electric force is 0.9 N.
(b). The acceleration is 0.9 m/s²
Explanation:
Given that,
Mass = 1.0 kg
Distance = 1.0 m
Charge = 10 μC
(a). We need to calculate the electric force
Using formula of electric force
[tex]F = \dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F=\dfrac{9\times10^{9}\times10\times10^{-6}\times10\times10^{-6}}{1.0^2}[/tex]
[tex]F=0.9\ N[/tex]
(b). We need to calculate the acceleration
Using newton's second law
[tex]F = ma[/tex]
[tex]a =\dfrac{F}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{0.9}{1.0}[/tex]
[tex]a=0.9\ m/s^2[/tex]
Hence, (a). The electric force is 0.9 N.
(b). The acceleration is 0.9 m/s²
By applying Coulomb's law and Newton's second law, we find that the magnitude of the electric force on one of the masses is 0.0899 N and the initial acceleration of one of the masses is 0.0899 m/s^2.
Explanation:This question pertains to Coulomb's law, which deals with the force between two charges. Coulomb's law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The equation used for this law is F = k(q1*q2)/r², where F represents the force, q1 and q2 represent the charges, r represents the distance between the charges and k is Coulomb's constant (8.99 × 10⁹ N×m²/C²).
a) To find the magnitude of the electric force on one of the masses, we would use the Coulomb's law equation: F = k(q1×q2)/r², where q1 and q2 are both +10μC (which need to be converted to C by multiplying by 10⁻⁶), and r is 1.0m. This gives us F = (8.99 × 10⁹ N×m²/C²) × ((10 × 10⁻⁶ C)²) / (1.0m)² = 0.0899 N.
b) To find the initial acceleration of one of the masses, we would use Newton's second law (F = ma), where F is the force (0.0899 N from the previous calculation), m is the mass (1.0 kg), and a is acceleration. So a = F/m = 0.0899 N / 1.0 kg = 0.0899 m/s².
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The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this field? (Enter the magnitude.) m/s2
Answer:
Answer:
6.68 x 10^16 m/s^2
Explanation:
Electric field, E = 3.8 x 10^5 N/C
charge of electron, q = 1.6 x 10^-19 C
mass of electron, m = 9.1 x 10^-31 kg
Let a be the acceleration of the electron.
The force due to electric field on electron is
F = q E
where q be the charge of electron and E be the electric field
F = 1.6 x 10^-19 x 3.8 x 10^5
F = 6.08 x 10^-14 N
According to Newton's second law
Force = mass x acceleration
6.08 x 10^-14 = 9.1 x 10^-31 x a
a = 6.68 x 10^16 m/s^2
Explanation:
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.
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]
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?
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
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)
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.