Answer:
The materials used to make electronic components like transistor and integrated, circuit behave as if effective particles known as electron through them, causing electrical properties
A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cable makes angles of 41° and 63° with the horizontal. Calculate the tension in each of the three cables.
Answer:
T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N
Explanation:
This is a balance exercise where we must apply the expressions for translational balance in the two axes
∑ F = 0
Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º
decompose the tension of the two upper cables
cos 41 = T₁ₓ / T1
sin 41 = T₁y / T1
T₁ₓ = T₁ cos 41
T₁y= T₁ sin 41
for cable gold
cos 63 = T₂ / T₂
sin 63 = [tex]T_{2y}[/tex] / T₂
We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.
Let's start by analyzing the point where the traffic light meets the vertical cable
T₃ - W = 0
T₃ = W
T₃ = 200 N
now let's write the equations for the single point of the three wires
X axis
- T₁ₓ + T₂ₓ = 0
T₁ₓ = T₂ₓ
T1 cos 41 = T2 cos 63
T1 = T2 cos 63 / cos 41 (1)
y Axis
[tex]T_{1y}[/tex] + T_{2y} - T3 = 0
T₁ sin 41 + T₂ sin 63 = T₃ (2)
to solve the system we substitute equation 1 in 2
T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W
T₂ (cos 63 tan 41 + sin 63) = W
T₂ = W / (cos 63 tan 41 + sin 63)
We calculate
T₂ = 200 / (cos 63 tan 41 + sin 63)
T₂ = 200 / 1,2856
T₂ = 155.6 N
we substitute in 1
T₁ = T₂ cos 63 / cos 41
T₁ = 155.6 cos63 / cos 41
T₁ = 93.6 N
therefore the tension in each cable is
T₁ = 93.6 N
T₂ = 155.6 N
T₃ = 200 N
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a horizontal shelf of 8 m, then another drop of 3 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.87 kg rock, giving it an initial horizontal velocity that barely clears the shelf below.
1. What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? The acceleration of gravity is 9.8 m/s2 . Consider air friction to be negligible. Answer in units of m/s.
2. How far from the bottom of the second cliff will the projectile land? Answer in units of m.
Answer:
1. v = 6.67 m/s
2. d = 9.54 m
Explanation:
1. To find the horizontal velocity of the rock we need to use the following equation:
[tex] d = v*t \rightarrow v = \frac{d}{t} [/tex]
Where:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
[tex] t = \sqrt{\frac{2d}{g}} [/tex]
Where:
g: is gravity = 9.8 m/s²
[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s [/tex]
Now, the horizontal velocity of the rock is:
[tex] v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s [/tex]
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. To calculate the distance at which the projectile will land, first, we need to find the time:
[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s [/tex]
So, the distance is:
[tex] d = v*t = 6.67 m/s*1.43 s = 9.54 m [/tex]
Therefore, the projectile will land at 9.54 m of the second cliff.
I hope it helps you!
Two boats start together and race across a 50-km-wide lake and back. Boat A goes across at 50 km/h and returns at 50 km/h. Boat B goes across at 25 km/h, and its crew, realizing how far behind it is getting, returns at 75 km/h. Turnaround times are negligible, and the boat that completes the round-trip first wins.
Required:
a. Which boat wins?
b. What is the average velocity of the winning boat?
Answer:
Boat A wins and 50km/hr
Explanation:
To solve this problem we need to calculate the time required for both boats to make the round trip.
For Boat A.
The time required for the first trip =distance/ speed = 50/50 = 1h
For the return trip the time is same since it's the same distance on the same speed.
Hence the time taken for boat A to make the round trip is 2hrs.
For boat B,
The time required for the first trip =distance/ speed = 50/25 = 2h
The time for the return trip is;
=distance/ speed = 50/75 = 2/3h= 2/3×60= 40mins
By comparing both time.
Boat A requires less time hence Boat A wins.
2.The average velocity of the winning boat is the velocity of boat A.
Which is total distance/ total time for the trip.
100km/2hrs = 50km/hr
boat A would win the race and the average velocity of the winning boat would be 50 km/h
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem Two boats start together and race across a 50-km-wide lake and back. Boat A goes across at 50 km/h and returns at 50 km/h. Boat B goes across at 25 km/h, and its crew, realizing how far behind it is getting, returns at 75 km/h.
Turnaround times are negligible, and the boat that completes the round-trip first wins
Average velocity of the boat A = ( 50 + 50 ) / 2
= 50 km /h
Thus, boat A would win the race and the average velocity of the winning boat would be 50 km/h
Learn more about Velocity from here,
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What is the radius of a nucleus of an atom?
Answer:
i think its 1-10 ´ 10-15 m
Explanation:
It is found that nuclear radii range from 1-10 ´ 10-15 m. This radius is much smaller than that of the atom, which is typically 10-10 m. Thus, the nucleus occupies an extremely small volume inside the atom. The nuclei of some atoms are spherical, while others are stretched or flattened into deformed shapes.
Chris and Sue are wearing harnesses and are hanging from the ceiling by means of ropes attached to them. They are face to face and push off against each other. Sue has a mass of 52 kg and Chris has a mass of 78 kg. Following the push, Sue reaches a height of 0.65 meters above her starting point. How high does Chris reach?
Answer:
h = 0.288m
Explanation:
Assume
[tex]v_1[/tex] = Speed of Sue
[tex]v_2[/tex] = Speed of Chris immediately after the push
Sue's KE = [tex]\frac{1}{2} mv_1\ ^2 = 26 v_1 \ ^2[/tex]
now she swings this is converted into gravitation at PE of
mg n = 52 × 9.8 × 0.65
= 331.24
[tex]26v_{1}\ ^2[/tex] = 331.24
So, [tex]v_1 = 3.569[/tex]
They started at rest by conservation of momentum in case of push off the magnitude of sue momentum and it is equal to the magnitude of Chris momentum in the opposite or inverse direction
[tex]m_1v_1 = m_2v_2[/tex]
[tex]52 \times 3.569 = 78 \times v_2[/tex]
[tex]v_2 = 2.380[/tex]
Chris kt = [tex]\frac{1}{2} \times 78 \times 2.380^2[/tex]
= 220.827
220.827 = mgh
So, h = 0.288m
A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant rate of 11.0cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf induced in the loop after exactly time 4.00s has passed since the circumference of the loop started to decrease AND find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
emf = 0.02525 V
induced current with a counterclockwise direction
Explanation:
The emf is given by the following formula:
[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}[/tex][tex]\ \ =-B\frac{A_2-A_1}{t_2-t_1}[/tex] (1)
ФB: magnetic flux = BA
B: magnitude of the magnetic field = 1.00T
A2: final area of the loop; A1: initial area
t2: final time, t1: initial time
You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.
In t = 4 s the final circumference will be:
[tex]c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm[/tex]
To find the areas A1 and A2 you calculate the radius:
[tex]r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm[/tex]
r1 = 0.261 m
r2 = 0.190 m
Then, the areas A1 and A2 are:
[tex]A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2[/tex]
Finally, the emf induced, by using the equation (1), is:
[tex]emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV[/tex]
The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.
A box is being pulled to the right. What is the magnitude of the Kinect frictional force?
The so-called "force" of friction always acts exactly opposite to the direction in which an object is moving. So for this question, since the box is moving to the right, the "force" of friction is acting toward the left.
There's not enough information given in the question to determine its magnitude.
with one example each , state and explain five outputs derieved from a business policy
ISC 152
Describe the
two major sources of energy and
give two
examples
each
HEYA!!
Here's your Answer:
Energy is the power we use for transportation, for heat and light in our homes and for the manufacture of all kinds of products. There are two sources of energy: renewable and non-renewable energy
RENEWABLE ENERGY: Renewable energy, often referred to as clean energy, comes from natural sources or processes that are constantly replenished. For example, (sunlight or wind keep shining and blowing), even if their availability depends on time and weather.
NON-RENEWABLE ENERGY:Non-renewable energy comes from sources that will run out or will not be replenished in our lifetimes—or even in many, many lifetimes.
Most non-renewable energy sources are (fossil fuels: coal, petroleum, and natural gas).
HOPE IT HELPS!!!
14. High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross-sectional area of 4.9 x 10-4 m2 . What is the resistance of ten kilometers of this wire
Answer:
The resistance of the aluminium wire is 0.541 ohms
Explanation:
Given:
Cross sectional area of the aluminum wire, A = 4.9 x 10-4 m2
Length of the wire, L = 10 km = 10,000 m
Resistance of the wire = (pL) / A
Where:
p is resistivity of aluminium wire = 2.65 x 10^-8 ohm-m
L is length of the wire
A is cross sectional area of the wire
Substitute these values and solve for resistance of the wire.
Resistance = (2.65 x 10^-8 x 10,000) / (4.9 x 10^-4)
Resistance = 0.541 ohms
Therefore, the resistance of the aluminium wire is 0.541 ohms
Which best describes earth's magnetic field lines?
Answer:
The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.
Explanation:
Answer: the field lines go out of earth near the north pole, enter earth in the south pole, and are not aligned with the geographic poles.
Explanation: just took the test on edge 2020.
A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 60 i + 64 k, with speed measured in feet per second. The spin of the ball results in a southward acceleration of 6 ft/s2, so the acceleration vector is a = −6 j − 32 k. Where does the ball land? (Round your answers to one decimal place.) ft from the origin at an angle of ° from the eastern direction toward the south. With what speed does the ball hit the ground? (Round your answer to one decimal place.)
The ball's position in the air at time t is given by the vector,
p(t) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ²
and its velocity is given by
v(t) = (60 i + 64 k) + (-6 j - 32 k) t
The ball is in the air for as long as it takes for the vertical (k) component of the position vector to reach 0, so we solve,
64 t - 32/2 t ² = 0 ==> t = 0 OR t = 4
and so the ball is in the air for 4 s.
After this time, the ball has position vector
p(4) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ² = 240 i - 48 j
which has magnitude
||p(4)|| = √(240² + (-48)²) = 48 √26 ≈ 244.8 ft
in a direction θ in the x,y plane from the positive x axis such that
tanθ = -48/240 = -1/5 ==> θ = -arctan(1/5) ≈ -11.3º
or 11.3º South of East.
The ball hits the ground with speed
||v(4)|| = ||60 i - 24 j - 64 k|| = √(60² + (-24)² + (-64)²) = 4 √517 ≈ 91.0 ft/s
kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is
r = 244.8 ft, tea = 21.8º from East to South
v = 91.0 ft / s
given parameters
Initial velocity v = (60 i + 64 k) ft / sBody acceleration a = (-6 j - 32 k) ft / s²
to find
where it reaches the groundground speed
Kinematics allows finding the position, velocity and acceleration of the body, in this case we have a problem in three dimensions, where they establish a Cartesian coordinate system, a method to solve this exercise is to solve each component independently
a) The acceleration of gravity acts on the z axis, so let's find the time it takes to reach the ground, if the initial vertical velocity is v_{oz} = 64 ft/s and the acceleration is a_z = g = -32 ft / s², we assume that the ball leaves the ground (z₀ = 0)
z = z₀ + v_{oz} t + ½ a_z t²
when reaching the ground its height of zero and
0 = 0 + v_{oz} t + ½ a_z t²
t (v_{oz) + ½ a_z t) = 0
t (64 - 16 t) = 0
the solusion of this squadron is
y = 0
t = 4 s
the first time is when it leaves and the second time is for when it reaches the ground, therefore the flight time is t = 4s
with this time we find the displacement is each exercise
X axis
in this axis there is no acceleration, so we use the uniform motion relationships
vₓ = x / t
x = vₓ t
x = 60 4
x = 240 ft
Y Axis
on this axis there is an acceleration of a_y = -6 ft/s and an initial velocity v_{oy} = 0
we use the kinematic relation
y = v_{oy} t + ½ a_y t²
y = 0 - ½ 6 4²
y = - 48 ft
let's use the Pythagoras theorem to find the position
r² = (x -x₀) ² + (y -y₀) ² + (z -z₀) ²
r² = (240-0) ² + (-48-0) ² + (0-0) ²
r = 244.8 ft
We use trigonometry to find the direction
tan θ = y / x
θ = tan⁻¹ [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ [tex]\frac{96}{240}[/tex]
θ = -21.8º
This angle is measured clockwise from the x axis, it can also be read
θ = 21.8º from East to South
b) Let's look for the speed when we hit the ground
X axis
vₓ = v_{ox} + aₓ t
vₓ = 60 - 0
vₓ = 60 ft / s
Y Axis
v_y = v_{oy} + a_y t
v_y = 0 - 6 4
v_y = -24 ft / s²
Z axis
v_z = v_{oz} + a_z t
v_z = 64 -32 4
v_z = -64 ft / s
With the Pytagoras theorem find the modulus of this speed is
v² = vx² + vy² + vz²
v² = 60² + 24 ² + 64²
v = 91.0 ft / s
In conclusion, the kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is
a) r = 244.8 ft, θ = 21.8º from East to South
b) v = 91.0 ft / s
learn more about kinematics here:
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As part of an exercise program, a woman walks south at a speed of 2.00 m/s for 60.0 minutes. She then turns around and walks north a distance 3000 m in 25.0 minutes (a) What is the woman's average velocity during her entire motion?
Answer:
0.824m/s
Explanation:
To calculate the average velocity we have to find the total distance and the total time
We have to find the distance and time in each motions
FIRST MOTION
The values given are
Speed= 2m/s , t = 60minutes
The time has to be converted to seconds. 60×60 = 3600seconds
Distance= speed×time
= 2× 3600
= 7200m
In the first motion the distance is 7200m and the time is 3600seconds
SECOND MOTION
The values given are
Distance= 3000m
Time= 25 mins to seconds
= 25×60
= 1500 seconds
In the second motion distance is 3000m and the time is 1500 seconds
The total distance can be calculated by applying the formular ( d1-d2) since she moved in an opposite direction
Total distance= 7200-3000
= 4200m
The total time (t1+t2) = 1500+3600
= 5100 seconds
Therefore, average velocity is calculated by applying the formular
Total distance/ Total time
= 4200/5100
= 0.824m/s South
Hence the average velocity is 0.824m/s South.
Answer:
The woman's average velocity during her entire motion is 2 m/s
Explanation:
Given;
initial speed of the woman, u = 2.00 m/s
initial time taken, t₁ = 60 minutes = 3600 seconds
initial displacement of the woman, x₁ = ?
final displacement of the woman, x₂ = 3000 m north
final time taken , t₂ = 25.0 minutes = 1500 seconds
The woman's average velocity during her entire motion:
initial displacement of the woman, x₁ = u x t₁ = 2.00 m/s x 3600 seconds
= 7200 m South
[tex]Average \ velocity = \frac{\delta X}{\delta t} = \frac{X_1-X_2}{t_1-t_2} \\\\V_{avg.} = \frac{7200-3000}{3600-1500} = \frac{4200}{2100} = 2 \ m/s[/tex]
Therefore, the woman's average velocity during her entire motion is 2 m/s
To Da
The radius of a circular track is 7 m. What will be the speed of a scooter, if it takes 20 seconds to complete one round of the
track?
published yet
04
published yet
A: 2.2 m/s
published yet
B: 22 m/s
02
published yet
C: 0.22 m/s
D: 022m/s
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Answer:
A
Explanation:
Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.
Answer:
i see they did already i am just supporting them
Answer:
A
Explanation:
Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.
A plane travels 2.5 KM at an angle of 35 degrees to the ground, then changes direction and travels 5.2 km at an angle of 22 degrees to the ground. What is the magnitude and direction of the planes total displacement??
Answer:
7.7 km 26°
Explanation:
The total x component is:
x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87
The total y component is:
y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38
The magnitude is:
d = √(x² + y²)
d = 7.7 km
The direction is:
θ = atan(y/x)
θ = 26°
18. Compared to its weight on Earth, a 5 kg object on the moon will weigh A. the same amount. B. less. C. more.
Answer:
B. less
Explanation:
acceleration due to gravity on Earth, g = 9.8 m/s²
acceleration due to gravity on Moon, g = 1.6 m/s²
Given mass of the object as, m = 5 kg
Weight of an object is given as, W = mg
Weight of the object on Earth, W = 5 x 9.8 = 49 N
Weight of the object on Moon, W = 5 x 1.6 = 8 N
Therefore, the object weighs less on the moon compared to its weight on Earth.
The correct option is "B. less"
A natural force of attraction exerted by the earth upon objects, that pulls
objects towards earth's center is called
A natural force of attraction exerted by the earth upon objects, that pulls objects towards earth's center is called Gravitational force .
You are pushing a cart across the room, and the cart has wheels at the front and the back. Your hands are placed on top of the cart at the center (left to right) of the top edge, pushing horizontally. There is friction between the wheels and the floor. Is the normal force between the floor and the front wheels greater than, smaller than, or equal to the normal force between the floor and the rear wheels? a) The force on the front wheels must be smaller than the force on the rear wheels. b) The force on the front wheels must be equal to the force on the rear wheels. c) The force on the front wheels must be greater than the force on the rear wheels.
Answer:
The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)
If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.
The force on the front and rear wheels can be determined by finding the
force acting under equilibrium conditions.
The normal reaction between the floor and the front wheel is larger. The
correct option is c) The force on the front wheels must be greater than the
force on the rear wheels.
Reasons:
The forces acting on the wheels are;
Vertical forces acting on the cart:
The weight of the cart acting on the floor
The normal reaction of the floor on the wheels
The horizontal acting on the cart:
Pushing force acting horizontally
Friction force acting in the reverse direction
At equilibrium, the cart does not tip over
Sum of moment about the rear wheel = [tex]\mathbf{W \times \dfrac{d}{2} + F \times h - N_{front} \times d} = 0[/tex]
Where;
h = The height of the cart
d = depth of the cart
[tex]N_{front}[/tex] = The normal reaction at the front wheels
Therefore;
[tex]\displaystyle N_{front} = \frac{W \times \dfrac{d}{2} + F \times h}{d} = \mathbf{ \frac{W}{2} + \frac{F \times h}{d}}[/tex]
Sum of moment about the front wheel = [tex]\mathbf{ F \times h + N_{rear} \times d - W \times \dfrac{d}{2} } = 0[/tex]
Therefore;
[tex]\displaystyle N_{rear} = \frac{W \times \dfrac{d}{2} - F \times h}{d} = \mathbf{ \frac{W}{2} - \frac{F \times h}{d}}[/tex]
Which gives;
[tex]\displaystyle N_{front} = \mathbf{ \displaystyle N_{rear} + 2 \times \frac{F \times h}{d}}[/tex]
[tex]\displaystyle 2 \times \frac{F \times h}{d} > 0[/tex]
Therefore;
[tex]\displaystyle N_{front} > \displaystyle N_{rear}[/tex]
The correct option is c) The force on the front wheels must be greater than
the force on the rear wheels.
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Adiabatic Equation of State Derive the adiabatic equation of state (2.3.19) using particle conservation (2.3.7) and energy conservation (2.3.21), by assuming that the heat flow vector q and all collision terms in these equations are zero.
p=Cn^{\gamma } Equation 19
\frac{\partial n}{\partial t}+\bigtriangledown (nu)=G-L Equation 7
\bigtriangledown (\frac{3}{2}pu)=\frac{\partial }{\partial t}(\frac{3}{2}p)\mid c Equation 22
Answer:
Explanation:
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A 58.0 kg snow skier is on the top of 351 m high hill. After she has gone down a vertical distance of 142 m, what is her mechanical energy? Explain your answer (CER)!
Answer:
Explanation:
Initially skier is at a height of 351 m . Her kinetic enery will be nil because she is at rest . Her potential energy will be calculated as follows
potential energy = mgh where m is mass , h is height and g is acceleration due to gravity
potential energy = 58 x 9.8 x 351
= 199508 .4 J
Total mehanical energy = potential energy + kinetic energy
= 199508.4 J
According to conservation of mechanical energy , at the height of 142 m also total mechanical energy will be same . At this height some potential energy will be converted into kinetic energy but total of potential and kinetic energy will be same.
Calculate the speed of a wave on a guitar string which has a mass of 55 g and a length of 30 cm if the tension in the string is 200 N
Answer:
33 m/s
Explanation:
v = √(T / (m/L))
where v is the velocity,
T is the tension,
and m/L is the mass per length.
v = √(200 N / (0.055 kg / 0.30 m))
v = 33 m/s
An object on a number line moved from x = 15 cm to x = 165 cm and then
moved back to x = 25 cm, all in a time of 100 seconds.
What was the average velocity of the object?
Answer:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:
[tex]v_{avg}=\frac{x_{all}}{t}[/tex]
x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s
[tex]v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}[/tex]
hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
Answer:
The average velocity of the object is 0.1cm/s
Explanation:
Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds
The average velocity is calculated as thus.
Average Velocity = ∆D/t
Where ∆D represent the displacement.
The displacement is calculated as follows.
∆D = End point - Start Point.
From the question, the end and start point are 25cm and 15cm respectively.
Hence,
∆D = 25cm - 15cm
∆D = 10cm.
t = 100 seconds
So, Average Velocity = 10cm/100s
Average Velocity = 0.1cm/s
Hence, the average velocity of the object is 0.1cm/s
A projectile is fired with an initial speed of 37.7 m/s at an angle of 41.2° above the horizontal on a long flat firing range. Determine the maximum height reached by the projectile.
Answer:
h = 31.46 m
Explanation:
We have,
Initial speed of a projectile is 37.7 m/s
It was projected at an angle of 41.2° above the horizontal on a long flat firing range.
It is required to find the maximum height reached by the projectile. The formula used to find it is given by :
[tex]h=\dfrac{u^2\sin^2\theta}{2g}[/tex]
Plugging all the known values,
[tex]h=\dfrac{(37.7)^2\times \sin^2(41.2)}{2\times 9.8}\\\\h=31.46\ m[/tex]
So, the maximum height reached by the projectile is 31.46 m.
A ball is fired from a toy gun with an initial speed of 7.00 m/s with the gun aimed at 300 upward with respect to the horizontal. The point of firing is 1.32 m above the horizontal floor. What is the horizontal component of the ball's initial velocity?
Answer:
6.0621 m/s
Explanation:
we all know that horizontal component of the velocity in projectile motion is
= u cosα
u= initial speed of 7.00 m/s of the gun.
α = angle of the initial velocity with the horizontal = 30°
therefore, horizontal component of the ball's initial velocity = 7×cos30°
[tex]7\times\frac{\sqrt{3} }{2} \\=6.0621 \text{m/s}[/tex]
The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algebra, find an equation for an object's speed v after falling a vertical distance h. Do this by equating KE to its change of PE.
Answer:
[tex]v = \sqrt{20h} [/tex]
Explanation:
The potential energy (PE) we are looking here is gravitational potential energy (GPE).
GPE= mgh,
where m is the mass of an object,
g is the gravitational field strength
h is the height of the object
KE= ½mv²,
where m is the mass and v is the velocity
loss in GPE= gain in KE
mgh= ½mv²
gh= ½v² (divide by m throughout)
Assuming that the object is on earth, then g= 10N/Kg
½v²= 10h (substitute g=10)
v²= 20h (×2 on both sides)
v= √20h (square root both sides)
An atom with 19 protons and 18 neutrons is an
Two loudspeakers are located 2.65 m apart on an outdoor stage. A listener is 19.1 m from one and 20.1 m from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range (20 Hz–20 kHz). The speed of sound in the air is 343 m/s. What are the three lowest frequencies that give minimum signal (destructive interference) at the listener's location?
Answer:
171.5Hz,514.5Hz and 857.5Hz
Explanation:
We are given that
Distance between two loudspeaker,d=2.65 m
Distance of listener from one end=19.1 m
Distance of listener from other end=20.1 m
[tex]\Delta L=20.1-19.1=1[/tex]m
Speed of sound,v=343m/s
For destructive interference
[tex]f_{min,n}=\frac{(n-0.5)v}{\Delta L}[/tex]
Using the formula and substitute n=1
[tex]f_{min,1}=\frac{(1-0.5)\times 343}{1}=171.5Hz[/tex]
For n=2
[tex]f_{min,2}=\frac{(2-0.5)\times 343}{1}=514.5Hz[/tex]
For n=3
[tex]f_{min,3}=(3-0.5)\times 343=857.5Hz[/tex]
Hence, the three lowest frequencies that give minimum signal (destructive interference) at the listener's location is given by
171.5Hz,514.5Hz and 857.5Hz
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t2. (a) What is the maximum height reached by the ball? ft (b) What is the velocity of the ball when it is 384 ft above the ground on its way up? (Consider up to be the positive direction.) ft/s What is the velocity of the ball when it is 384 ft above the ground on its way down?
Answer:
a) h_max = 400ft
b) v = 1024 ft/s
Explanation:
The general equation of a motion is:
[tex]s(t)=v_ot-\frac{1}{2}gt^2[/tex] (1)
You have the following equation of motion is given by:
[tex]s(t)=160t-16t^2[/tex] (2)
You compare both equations (1) and (2) and you obtain:
vo: initial velocity = 160ft/s
g: gravitational acceleration = 32ft/s
a) The maxim height reached by the ball is given by:
[tex]h_{max}=\frac{v_o^2}{2g}=\frac{(160ft/s)^2}{64ft/s^2}=400ft[/tex]
b) The velocity is given by:
[tex]v^2=v_o^2-2gh\\\\v=\sqrt{(160ft/s)^2-2(32ft/s^2)(384ft)}=1024ft/s[/tex]
Answer:
a) Smax = 400 ft
b) v = 32 ft/s
c) v = - 32 ft/s
Explanation:
(a)
The function given for the height of ball is:
s = 160 t - 16 t²
Therefore, in order to find the time to reach maximum height (in-flexion point), we must take the derivative with respect to t and set it equal to zero:
Therefore,
160 - 32 t = 0
32 t = 160
t = 160/32
t = 5 sec
Therefore, maximum distance is covered at a time interval of 5 sec.
Smax = (160)(5) - (16)(5)²
Smax = 800 - 400
Smax = 400 ft
b)
First we calculate the time at which ball covers 384 ft
384 = 160 t - 16 t²
16 t² - 160 t + 384 = 0
Solving Quadratic Equation:
Either:
t = 6 sec
Or:
t = 4 sec
Since, the time to reach maximum height is 5 sec. Therefore, t < 5 sec
Therefore,
t = 4 sec
Now, we find velocity at 4 sec by taking derivative od s with respect to t at 4 sec:
v = 160 - 32 t
v = 160 - (32)(4)
v = 32 ft/s
c)
Since, t = 6 s > 5 s
The second value of t = 6 sec must correspond to the instant when the ball is 384 ft above the ground while traveling downward.
Hence, velocity at that time will be:
v = 160 - (32)(6)
v = -32 ft/s
Negative sign due to downward motion.
A bike with tires of radius
0.330 m speeds up from rest
to 5.33 m/s in 6.27's. Through
what angle do the wheels turn
in that time?
(Unit = rad)
Answer:
The wheels turn 101.27 radians in that time.
Explanation:
First, we need to find the angular velocity of the tires. We use the following formula for this purpose:
Δv = rω
ω = Δv/r
where,
ω = Angular Velocity of Tires = ?
Δv = change in linear velocity = 5.33 m/s - 0 m/s = 5.33 m/s
r = radius of tire = 0.33 m
Therefore,
ω = (5.33 m/s)/(0.33 m)
ω = 16.15 rad/s
Now, the angular displacement covered by tires can be found out by using the general formula of angular velocity:
ω = θ/t
θ = ωt
where,
θ = angular displacement = ?
t = time = 6.27 s
Therefore,
θ = (16.15 rad/s)(6.27 s)
θ = 101.27 radians
The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.6 V and b = -4.90 V/m. (a) Determine the potential at x = 0. 10.6 Correct: Your answer is correct. V Determine the potential at x = 3.00 m. V (b) Determine the magnitude and direction of the electric field at x = 0. magnitude 4.9 Correct: Your answer is correct. V/m direction +x Correct: Your answer is correct. Determine the magnitude and direction of the electric field at x = 3.00 m. magnitude V/m direction
Answer:
a) 10.6V
b) E = 4.9V/m, +x direction
c) E = 4.9V/m, +x direction
Explanation:
You have the following function:
[tex]V=a+bx[/tex] (1)
for the potential in a region between x=0 and x=6.00 m
a = 10.6 V
b = -4.90V/m
[tex]V=10.6V-4.90\frac{V}{m}x[/tex]
a) for x=0 you obtain for V:
[tex]V=10.6V-4.90\frac{V}{m}(0m)=10.6V[/tex]
b) The relation between the potential difference and the electric field can be written as:
[tex]E=-\frac{dV}{dx}=-b=-(-4.9V/m)=4.9V/m[/tex] (2)
the direction is +x
c) The electric field is the same for any value of x between x=0 and x=6m.
Hence,
E = 4.9V/m, +x direction
