Answer:
explained
Explanation:
If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.
A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.
Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.
An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.
Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).
In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.
How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s
straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s
relative to her body, then the velocity of the ball is 35 m/s
relative to the stationary, profusely sweating goalkeeper standing in front of the goal.
In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (
The figure shows components of velocity v in horizontal vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.
Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors and
These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.
A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Answer:
The correct answer is - c. Spaceship will have a velocity to the East and will be slowing down.
Explanation:
In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.
As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.
Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks
A=B=50NAngle=theta=35°
We know
[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]
[tex]\\ \sf\longmapsto R=22.4i[/tex]
Resultant using the Law of Sines and the Law of Cosines will be R=95 N
What is force?Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
The Magnitude of two forces =50 N
Angle between the forces = 35
By using the resultant formula
[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]
[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]
[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]
[tex]\rm R=\sqrt{5000+4500}[/tex]
[tex]\rm R=95\ N[/tex]
Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N
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The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle
Answer:
4.38 m/s
Explanation:
Q)what are convex mirrors?
Answer:
A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.
A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
The cation that is reabsorbed from the urine in response to aldosterone
Answer:
If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)
Answer:
[tex]E=35921.96N/C[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.321mm[/tex]
Charge Density [tex]\mu=0.100[/tex]
Distance [tex]d= 5.00 cm[/tex]
Generally the equation for electric field is mathematically given by
[tex]E=\frac{mu}{2\pi E_0r}[/tex]
[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]
[tex]E=35921.96N/C[/tex]
Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day
what do you mean about it
A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
the rate of cooling determines ....... and ......
Answer:
freezing point and melting point
Why we use semiconductor instead of metal in thermopile.
Answer:
Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.
A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.
Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.
The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.
Momentum is conserved, so the total momentum of the system is the same before and after the collision:
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
==>
(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'
==>
-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'
where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.
Kinetic energy is also conserved, so that
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²
==>
(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
==>
1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is
v₁' ≈ -3.11 m/s
v₂' ≈ -0.167 m/s
and take the absolute values to get the magnitudes.
If you want to instead use the masses from the "Required" section, you would end up with
v₁' ≈ -3.18 m/s
v₂' ≈ -0.236 m/s
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
Answer:
Explanation:
Since energy is conserved:
2
mu
2
=
2
mv
2
+mgh
⇒u
2
=v
2
+2gh
⇒(3)
2
=v
2
+2(9.8)(0.5−0.5cos60)
⇒v=2m/s
Acceleration of the simple pendulum is 2.62 m/s².
What is meant by a simple pendulum ?When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.
Here,
Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.
Let T be the tension acting on the string.
As, the bob passes through the angle,
The weight of the bob becomes equal to the vertical component of the tension.
mg = T cos15°
Also, the horizontal component of the tension,
T sin15° = ma
By solving these two equations, we get that,
Acceleration of the simple pendulum,
a = g tan15°
a = 9.8 x 0.267
a = 2.62 m/s²
Hence,
Acceleration of the simple pendulum is 2.62 m/s².
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A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.
Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Solution :
We know,
Distance,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
[tex]$S=ut+0.5(a)(t)^2$[/tex]
For the first 20 ms,
[tex]$S=0+0.5(220)(0.020)^2$[/tex]
S = 0.044 m
In the remaining 30 ms, it has constant velocity.
[tex]$v=u+at$[/tex]
[tex]$v=0+(220)(0.020)[/tex]
v = 4.4 m/s
Therefore,
[tex]$S=ut+0.5(a)(t)^2$[/tex]
[tex]$S'=4.4 \times 0.030[/tex]
S' = 0.132 m
So, the required distance is = S + S'
= 0.044 + 0.132
= 0.176 m
Therefore, the tongue can reach = 0.176 m or 17.6 cm
Answer:
The total distance is 0.176 m.
Explanation:
For t = 0 s to t = 20 ms
initial velocity, u = 0
acceleration, a = 220 m/s^2
time, t = 20 ms
Let the final speed is v.
Use first equation of motion
v = u + at
v = 0 + 220 x 0.02 = 4.4 m/s
Let the distance is s.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]
Now the distance is
s' = v x t
s' = 4.4 x 0.03 = 0.132 m
The total distance is
S = s + s' = 0.044 + 0.132 = 0.176 m
If a jet travels 350 m/s, how far will it travel each second?
Answer:
It will travel 350 meters each second.
Explanation:
The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.
Answer:
5.83 seconds
Explanation:
60 seconds in 1 minute
350 meters per second
350/60
=5.83
What do you understand by moment of inertia and torque?
Word limit 50-60
Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.
Answer:
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire
Answer:
the weight of the wire + 25kg
Explanation:
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
S.I unit for distance =______
(A) m (B)cm
(c) km (d) mm
Answer:
opinion a
Explanation:
the si units of distance is metre (m)
Answer:
A
Explanation:
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
i don't understand this, can someone help please??
Explanation:
N2 + H2 --> NH3
balance them:
N2 + 3 H2 --> 2 NH3
so if 6 moles of N2 react, 12 moles of NH3 will form.
(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )
A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?
Answer:
2
Explanation:
pulling force because of it force
Answer:
5.9 cm
Explanation:
f: frequency of oscillation
frequency of oscillationk: spring constant
frequency of oscillationk: spring constantm: the mass
[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]
in this problem we know,
F= 1.4 Hz
m= 0.26 kg
By re-arranging the formula we get
[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]
The restoring force of the spring is:
F= kx
where
F= 1.2 N
k= 20.1 N/m
x: the displacement of the block
[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
