Answer:
a) [tex]F_{x} = 58.2 N[/tex]
[tex] F_{y} = 48.9 N [/tex]
b) a = 1.5 m/s²
Explanation:
a) The horizontal and vertical components of Emily's force can be found knowing the angle and the exerted force.
Since the handle is inclined at 40.0° below the horizontal we have:
[tex] F_{x} = |F|*cos(\theta) = 76.0 N*cos(40) = 58.2 N [/tex]
[tex] F_{y} = |F|*sin(\theta) = 76.0 N*sin(40) = 48.9 N [/tex]
b) The acceleration of the car can be calculated as follows:
[tex] F_{x} = ma [/tex]
We used the horizontal component of the force because the cart is moving in that direction.
[tex] a = \frac{F_{x}}{m} = \frac{58.2 N}{38.8 kg} = 1.5 m/s^{2} [/tex]
Hence, the acceleration of the car is 1.5 m/s².
I hope it helps you!
A 5.1 kg textbook is raised a distance of 21.2 cm as a student prepares to leave for school. How much work did the student do on the book ?
Work done on the book is 10.595 Joules .
Work done on the book would be equal to the Potential Energy
Work Done = Potential Energy
W = m g h
where , ' m ' represents mass of the book
' g ' stands for Acceleration due to gravity
' h ' represents the height to which the book is raised
So, according to the given question ,
m = 5.1 kg
g = 9.8 m / s ²
h = 21.2 cm
= 0.212 m
W = m g h
= 5.1 kg x 9.8 m / s ² x 0.212 m
= 10 . 595 Joules
Hence, Work done on the book is 10.595 Joules .
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If there is 100 ml of liquid mercury in a graduated cylinder that weighs 25 grams. What must be the density of the liquid mercury, if the combined mass of the liquid mercury and the graduated cylinder is 1,385.0 grams?
The density of 100 ml of liquid mercury in the graduated cylinder that weighs 25 grams is 13.6 g / ml
ρ = m / V
ρ = Density
m = Mass
V = Volume
V = 100 ml
Mass of cylinder = 25 g
Combined mass of cylinder and mercury = 1385 g
Mass of mercury = 1385 - 25 = 1360 g
ρ = 1360 / 100
ρ = 13.6 g / ml
Density is the amount of molecules present in a given volume of the substance. It is denoted by ρ.
Therefore, the density of 100 ml of liquid mercury in the graduated cylinder that weighs 25 grams is 13.6 g / ml
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100 Points!
A projectile is fired in the earth's gravitational field with a horizontal velocity of v=9.00 m/s. How far does it go in the horizontal direction in 0.550s? Show your work.
B) How far does the projectile go in the vertical direction in 0.550s. Show your work
Answer:
A) 4.95 m
B) 1.48225 m
Explanation:
Constant Acceleration Equations (SUVAT)
[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]
When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.
Consider the horizontal and vertical motion of the projectile separately.
Part AThe horizontal component of velocity is constant, as there is no acceleration horizontally.
Resolving horizontally, taking → as positive:
[tex]u=9.00\quad v=9.00 \quad a=0\quad t=0.550[/tex]
[tex]\begin{aligned}\textsf{Using} \quad s & = \left(\dfrac{u+v}{2}\right)t:\\\\s&= \left(\dfrac{9+9}{2}\right)(0.550)\\s&= (9)(0.550)\\ \implies s&= 4.95\:\sf m\\\end{aligned}[/tex]
Part BAs the projectile is fired horizontally, the vertical component of its initial velocity is zero.
Acceleration due to gravity = 9.8 ms⁻²
Resolving vertically, taking ↓ as positive:
[tex]u=0\quad a=9.8\quad t=0.550[/tex]
[tex]\begin{aligned}\textsf{Using} \quad s & = ut+\dfrac{1}{2}at^2:\\\\s&= (0)(0.550)+\dfrac{1}{2}(9.8)(0.550)^2\\s&= 0+(4.9)(0.3025)\\\implies s&= 1.48225\:\sf m\\\end{aligned}[/tex]
a sailboat moves faster as the wind pick up this is an example Of?
It is an example of acceleration due to the wind.
Acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration of all other motions.
Acceleration is a vector quantity since it has both a magnitude and a direction. A vector quantity is also velocity. The velocity vector change during a time interval divided by the time interval is the definition of acceleration.
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26. Which of the following is NOT a result of the Earth's precession?
a. the Earth wobbles (like a spinning top) with a period of 26,000 years
b. where the Earth's axis points in the sky changes over the centuries and millennia
c. Polaris will no longer be the North Star in several thousand years
d. the stars twinkle when seen from the surface of planet Earth
e. the signs of the zodiac most astrologers use are no longer in accord with the constellations in which
the Sun is currently found over the course of the year
The Earth's axis' orientation in space gradually shifts due to the precession.
Precession is the name given to this motion, which is characterized by a cyclic wobble in the direction of the Earth's axis of rotation over a period of 25,772 years. The third motion of the Earth to be discovered, following the much more evident daily rotation and yearly revolution, is precession. The axial tilt of the Earth gradually changes over time as a result of precession. The points on the Earth's orbit where the equinoxes and solstices fall will shift over time. As a result, 13,000 years from now, the location of the summer solstice will change to that of the winter solstice. The Earth's axis changes its direction due to precession. In relation to the stars, the celestial poles steadily shift.
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pls pls help !!! <33
Answer: Vx is changing and Vx is greatest near the beginning of the launch.
Explanation: Think of Vx as a bullet fired from a gun. As soon as it is fired it starts to slow down so Vx is changing and at the beginning Vx has its greatest velocity because it can only go slower from there.
answers are B and C. horizontal velocity never changes so A cannot be the answer. that makes B one of the answers. I would also say that the horizontal velocity is greatest near the beginning of the launch as it just starts, since it stays at a constant pace after, so D is not an answer. hope this helps ^_^
Suppose the elevator starts from rest and maintains a constant upward acceleration of 2.90 m/s2 . A bolt in the elevator ceiling 3.00 m above the elevator floor works loose and falls. It falls out the instant the elevator begins to move. a.)How long does it take for the bolt to reach the floor of the elevator? b.)Just as it reaches the floor, how fast is the bolt moving according to an observer in the elevator? c.)Just as it reaches the floor, how fast is the bolt moving according to an observer standing on the floor landings of the building? d.)According to an observer in the elevator, how far has the bolt traveled between the ceiling and floor of the elevator? e.)According to an observer standing on the floor landings of the building, how far has the bolt traveled between the ceiling and floor of the elevator?
The correct answer is 2.26 m.
Initial velocity of elevator u =0
Constant upward acceleration of the elevator a= 2.90 m/s²
Height of the bolt from the floor of the elevator h=3.00 m
Initial velocity of the bolt in the elevator u = 0 m/s
Elevator is moves upward with a constant acceleration 2.90 m/s²
so the net acceleration a' = (a+g) m/s²
=( 2.90 + 9.8) m/s²
a' = 12.7 m/s²
Initial velocity of the bolt u = 0 m/s
Distance traveled by the bolt h = 3.00 m
Substitute these values in the equation
h=ut+ 1/2 ( a )t², we get
3.00 = (0)x+1/2 (12.7)(t^2)
3.00 = 6.35 t²
t²= 3.00/6.35
t = 0.68 s
i.e., after 0.68 s the bolt reaches the floor of the elevator.
(b)For the observer in the elevator acceleration
a'=(a+g) m/s²
=( 2.90 + 9.8) m/s²
a'= 12.7 m/s²
Time taken bolt to reach the floor of the elevator t = 0.68 s From the equation v=u+a't we get
v=(0)+(12.7) (0.68)
v = 8.636 m/s
Velocity of the bolt when it reaches the floor of the elevator according to an observer into elevator 8.636 m/s
ii) An observer standing on the floor landings of the buildings: for observer standing on the floor landing, the acceleration of the
bolt
appears a = g m/s²
= 9.8m/s²
Time taken the bolt to reach the floor t = 0.68 s
Initial velocity of the bolt u = 0 m/s
Substitute these values in the equation
v=u+at we get
v=0+(9.8) (0.68 s)
v= 8.636 m/s
Velocity of the bolt when it reaches the floor of the elevator according to
an observer standing on the floor landing = 8.636 m/s
c) i) According to an observer in the elevator in this situation acceleration a= 12.7 m/s²
time t= 0.68 s initial velocity u = 0
substitute these values in the equation 1
s = ut +1/2 at²
We get
s =(0) (0.659)+1/2(12.7 ) (0.68)²
= 2.93 m
Distance traveled by the bolt according to the observer on the elevator 2.93 m
iii)According to an observer standing on the floor landing in this situation
acceleration a=g m/s²
= 9.8m/s²
Time t= 0.68 s
Initial velocity u = 0 m/s Substitute these values in the equation
1 ut + 1/2 at ²,
s=ut+1/2at²
we get
s=(0)+1/2(9.8) (0.68)²
= 2.26 m
Distance traveled by the bolt according to the observer standing on the floor landing 2.26 m.
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HELP! Along the line through the centers of the Moon and the Earth, there is a point where the gravitational forces exerted by the Moon and the Earth, respectively, are equal. We consider an elementary mass located at this point. How can this state be rated? Stable equilibrium, unstable equilibrium, total mechanical energy minimum, total mechanical energy maximum or minimum
We consider an elementary mass located at this point to be in a stable equilibrium.
What is stable equilibrium?An equilibrium is said to be stable if small, externally induced displacements from that state produce forces that tend to oppose the displacement and return the body or particle to the equilibrium state.
Also a body is said to be in a stable equilibrium, if the opposing forces acting on a body are equal.
Examples of a stable equilibriuma weight suspended by a springa brick lying on a level surfacea cone resting on its baseThus, along the line through the centers of the Moon and the Earth, there is a point where the gravitational forces exerted by the Moon and the Earth, respectively, are equal. We consider an elementary mass located at this point to be in a stable equilibrium.
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Purpose
Students will us vector addition on
the Flat-Earth Model.
Theory
Vector addition is easy and useful.
To add (24,0) to (0,-7), the 1st numbers
are added, as are the 2nd, and the order
stays the same: (24,-7). These can be x-
and y-positions, velocities, forces, etc.
Which way are the streaks on the
side windows of a car going 24 MPH if
driving through rain falling at 7 MPH?
That's a vector of (24,-7). The angle is
found by taking the arctangent of the
second 2nd over the 1st: tan¹(-4)=16⁰.
The Flat-Earth Model claims earth
is a horizontal disk. The north pole is at
the center, "Antarctica" is a rim wall at
the edge, and the sun and moon move
in circular paths above the disk. If true,
it should be easy to measure the sun's
altitude and make predictions.
Procedure
1. Find a cardboard box about the size
of a microwave, cut off the top flaps,
and set it on its side outdoors at noon.
2. Cut a coffee-cup-sized hole in the
middle of the upper wall and tape a
sheet of aluminum foil over it. Poke a
small hole in that with a pin or tack.
3 With the sun overhead, even if it is
slightly cloudy, a spot of light should
show on the bottom surface in the box.
Propping the box at a tilted angle may
be necessary, but then don't move it.
Mark the box bottom where the spot is
at noon, and then again at 1PM.
4. Find a ruler and measure the height
from the noon mark up to the foil hole.
Let this be y. Also measure the distance
between the marks. Let this be x. Units
can be inches, centimeters, whatever.
The sun's ray vector in the box is (x,y).
This vector can also be used to find the
sun's angle above the horizon.
Figure 02: The box as described.
5. The sun apparently moves 1000 mi
in an hour. (Call someone in Ft. Worth
and ask them what time of day the sun
is highest in the sky, if skeptical.) The
sun's height in miles is h = 1000-y/x.
6 If the Flat-Earth model is right, the
sun always keeps this height, so after
seven hours, the spot goes 7-x laterally.
Find the predicted sun angle above the
horizon at 7 PM using tan-¹(). Wait
for 7 PM and see if that's nearly right.
Analysis
Please answer each of the following
in Canvas using complete sentences:
1. How high up (h) is the sun using the
Flat-Earth Model?
2. At what angle above the horizon is
the sun at 10 PM using this model?
3. What is the diagonal distance to the
sun at 10 PM by this model when its
position vector is (10000,h) miles?
4. Did this lab disprove all unpopular
scientific viewpoints? Why or why not?
* I’ve already found my sun’s ray vector in the box which was (4.25, 17.25). *
On the day of the summer solstice, at noon, Eratosthenes was aware of a deep well on the Nile close to Aswan where the sun shone all the way to the bottom. Because the sun was not overhead in Alexandria, the length of that shadow was not zero.
We believe that the light rays from the sun that reach us are all traveling in the same direction; they are parallel, as the sun is thought to be extremely far away. No shadow would be formed by a vertical stick at point A, a "moderate" shadow by a stick at point B, and a very long shadow by a stick at point C. A radius of the earth leading to point A is shown on the diagram, and another leading to point B. A little geometry reveals that the angle between those two radii and the difference between the shadow angles are the same. In our example, the shadows indicate that during our journey from Florida to Pennsylvania, we had rotated the earth by about 10 degrees.
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Someone pls tell me what the answer is
The distance of the ball that would land which was launched horizontally at a speed (s) of 3.4 m/s from a table that is 7.8 m tall (d) will land is 4.29m
Consider the ball moving horizontally off the table. When it moves, gravitational force acts on it pulling the ball downwards. In addition to that objects that are falling down would accelerate quickly i.e., let’s take an object falling at the speed of x, after time t it would travel at 2x, after 2t it would be 4x.
We need to find time to calculate distance by using speed equation.
For calculating time, let’s use the freefall formula,
d = 1/2 gt²
t = √(2d) / g
t = √2 (7.8)/ 9.8
t = 1.262 s
To calculate distance, we could use speed equation as we have both Time and Velocity
s = d / t
d = s t
d = 1.262 * 3.4
d = 4.29 m
Thus, the distance of the ball that would land is 4.29 m
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convert these into proper Vector notation Westward velocity of 42 km/h
Postion 6.5 measured in m that is North of the refrence point.
Downward acceleration measured in m/s2 that has a magnitude of 1.9.
Proper vector notations are V = 42i km/hr, r = 6.5j m, and a = -1.9j m/s².
In physics, a vector is a quantity with both magnitude and direction. It is typically represented by an arrow whose length is proportional to the magnitude of the quantity and whose direction is the same as that of the quantity.
A) Let us assume that the west direction is the positive X-axis.
In vector notation, the X-axis is denoted by "i". So, the westward velocity of 42 km/hr will be written as
V = 42i km/hr
B) Let us assume that the North direction is the positive Y- axis.
In vector notation, the Y-axis is denoted by "j". Also, the position vector is represented by the letter 'r'. So position 6.5 measured in m, that is North of the reference point will be written as
r = 6.5j m
C) Let us assmue thaat the downward direction is negative Y-axis.
In vector notation, the Y- axis is denoted by "j". so, the downward acceleration measured in m/s2 that has a magnitude of 1.9 will be written as
a = 1.9 × (-j) m/s²
a = -1.9j m/s²
So, proper vector notations are V = 42i km/hr, r = 6.5j m, and a = -1.9j m/s².
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In camping, you drop a stone from the 1st spot to the lake. The splash is heard 4 seconds later. After walking, you drop a stone again at 2nd spot at the same lake. This time the splash is heard 12 seconds later. The height of the 2nd spot is ___ times the 1st spot.
A-3
B-12
C-6
D-9
The height of the 2nd spot is 3 times the height of the 1st spot (Option A)
How to determine the height of the first spotWe'll begin by obtaining the height of the first spot. This can be obtained as follow:
Time (t) = 4 seconds Velocity of sound (v) = 343 m/sHeight of first spot (x) =?v = 2x / t
343 = 2x / 4
Cross multiply
2x = 343 × 4
2x = 1372
Divide by 2
x = 1372 / 2
Height of first spot = 686 m
How to determine the height of the second spotTime (t) = 12 seconds Velocity of sound (v) = 343 m/sHeight of second spot (x) =?v = 2x / t
343 = 2x / 12
Cross multiply
2x = 343 × 12
2x = 4116
Divide by 2
x = 4116 / 2
Height of second spot = 2058 m
How to compare 2nd height with 1st height Height of 1st spot = 686 mHeight of 2nd spot = 2058 mComparison =?
Comparison = Height of 2nd spot / Height of 1st spot
Height of 2nd spot / Height of 1st spot = 2058 / 686
Height of 2nd spot / Height of 1st spot = 3
Cross multiply
Height of 2nd spot = 3 × Height of 1st spot
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Summarize the article in a paragraph .
Words are too small for me -_-
Estimate the car's velocity at 4.0 s . Express your answer to two significant figures and include the appropriate units.
A car is stopped for a traffic signal. When the light turns green,
the car accelerates, increasing its speed from 0 to 5.20 m/s in
0.832 s. What are the magnitudes of (a) the linear impulse and
(b) the average total force experienced by a 70.0-kg passenger
in the car during the time the car accelerates?
The impulse of the car is 365kgm/s and average force experienced is 437.5N.
acceleration is defined as the change in velocity per unit time.
average force is defined as the force experienced by a body while the acceleration is constant.
given:
initial velocity (u)=0m/s
final velocity(v)=5.20m/s
time(t)=0.832s
mass (m)=70 kg
a)
first we will find acceleration
a=5.2/0.832
=6.25m/s²
now average force experienced (F)= mass*acceleration
F=70×6.25F
=437.5N
now impulse (I) is given by F×t=437.5×0832I
=365kg×m/s
therefore the impulse is 365kg×m/s
b)
Now we will find the average force applied.
Average force experienced is 437.5N
therefore,
The impulse of the car is 365kgm/s and average force experienced is 437.5N.
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The power in a lightbulb is given by the equation P=I^2 R, where I is the current flowing through the lightbulb and R is the resistance of the lightbulb. What is the current in a circuit that has a resistance of 25.0 Ω and a power of 30.0 W? A 0.830 A B 1,20 A C 0.910 A D 1.09 A
Answer:
Explanation:
The correct option will be the option (d) that is 1.09 A.
According to the equation given in the question:
That is; P=i^2R where I= current, R= resistance
P=30 W
R=25 ohm
30=i^2*25
I^2=30/25
I^2=1.2
I=1.09A
So, the correct answer of the given question will be 1.09A
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An electric stove is rated 1200,
240v-calculate?
(a) The current drawn is 5A.
(b) The resistance of the filament is 48Ω.
(a) The formula for power is:
P = VI
therefore, current drawn is;
[tex]I = \frac{P}{V}[/tex]
[tex]I = \frac{1200}{240}[/tex]
I = 5A
Therefore, the current drawn is 5A.
(b) Using, ohm's law:
V = IR
therefore, resistance of the filament is:
[tex]R = \frac{V}{I}[/tex]
[tex]R = \frac{240}{5}[/tex]
R = 48Ω
Therefore, the resistance of the filament is 48Ω.
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Direction of the vector A
The direction of vector A which is parallel but opposite to the vector I is is 305°
Direction of vector I = 125°
Vector I makes a parallel line with vector A. Consider vector A and I to be collinear. Collinear lines makes up an angle of 180°
[tex]Angle_{A}[/tex] + [tex]Angle_{I}[/tex] = 180°
[tex]Angle_{A}[/tex] + 125° = 180°
[tex]Angle_{A}[/tex] = 55°
Direction of vector A = 360° - 55° = 305°
The direction of a vector is the angle, the vector creates with positive x-axis in counter clockwise direction.
Therefore, the direction of vector A is 305°
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A baseball player throws a ball horizontally.
Which statement best describes the ball's
motion after it is thrown? [Neglect the effect of
friction.]
The vertical speed of the ball increases and its horizontal speed remains constant as usual because air friction is being neglected.
By definition, air resistance refers to the forces that work against an object's relative motion as it travels through the atmosphere. These drag forces cause the object to move more slowly by acting in opposition to the speed of the incoming flow. Since drag is the part of the net aerodynamic force operating in the opposite direction to the direction of the movement, it differs from other resistance forces in that it directly depends on velocity.
Fluid friction includes friction in the air. These friction forces differ from the conventional model of surface friction in that they rely on velocity. Only rare examples can be dealt with analytically since the velocity dependence may be extremely intricate. Air resistance is roughly proportional to velocity and can be represented in the form of very low speeds for microscopic particles.
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If a bat hits a ball with 85 N of force, how much force does the ball exert on the bat?
If a bat hits a ball with 85 N of force the bat will also exert the same force
85N.
You are probably already aware of the force the ball applies to the wall when it is thrown against it. Similar to how the wall exerts a force on the ball, the ball bounces off the wall. Similar to how the earth drags you down with gravity. You might not be aware of it, but you are also pushing the planet with the same amount of power. The third law of Newton is a result of this amazing event.
Newton's Third Law states that if object A exerts a force on object B, then object B must apply a force on object A that is equal in magnitude and directed in the opposite direction.
An object interacts with another object when a push or pull is applied to it. A product of interaction is force. Frictional force, for example, falls into the category of contact force. Non-contact force, on the other hand, includes gravitational force. Newton's third rule of motion explains that when two bodies contact, they exert force on one another. This force is referred to as an action and reaction pair.
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The gravitational force between two objects is 3600 N.
What will be the gravitational
force between the objects the
mass of each object is reduced to one-third of its original mass?
O 400 N
O 1200 N
O 10,800 N
O 32,400 N
The gravitational force between the two objects whose mass is reduced to one-third of its original mass is 400 N
We know that,
F = G[tex]m_{1}[/tex][tex]m_{2}[/tex] / r²
F = Gravitational force
G = Gravitational constant
[tex]m_{1}[/tex], [tex]m_{2}[/tex] = Mass of two objects
r = Distance between two objects
G = 3600 N
[tex]m_{1}'[/tex] = [tex]m_{1}[/tex] / 3
[tex]m_{2}'[/tex] = [tex]m_{2}[/tex] / 3
F' = G [tex]m_{1}'[/tex] [tex]m_{2}'[/tex] / r² = G[tex]m_{1}[/tex][tex]m_{2}[/tex] / 9 r²
F' = F / 9
F' = 3600 / 9
F' = 400 N
Gravitational force is the force of attraction between two objects due to the presence of gravity.
Therefore, the gravitational force between the two objects whose mass is reduced to one-third of its original mass is 400 N
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What Happens When Two Objects Are Charged By Rubbing Or Peeling?
a. One object has only one type of charge. The other object has only the other type of charge.
b. Both objects have the same type of charge.
c. Both objects have both types of charge, but there are different amounts of each type on each object. (One object has more negative charges than positive and the other object has more positive charges than negative charges.)
When Two Objects Are Charged By Rubbing Or Peeling, both objects have both types of charge, but there are different amounts of each type on each object. (One object has more negative charges than positive chrages and the other object has more positive charges than negative charges); option C.
What happens when two objects are charged by rubbing or peeling?Charges are the inherent properties of the particles of matter which determines how the atoms will react when placed in an electric filed.
There are two types of charges in an atom:
positive charges which are present on the protonsnegative charges that which are present on the electrosThe electrons in an atom can be easily transferred from one atom to another, hence objects can be charged by rubbing or peeling one against the other.
When when two objects are rubbed or peeled against each other, electrons move from one object to the other.
The object that acquires the electrons become negatively charged whereas the object that loses the electrons become positively charged. However, both objects have both electrons and protons present in them.
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An airplane acceleration is 3m/s^2. The minimum take-off speed is 102 m/s. The minimum runway length is ____ meter.
The airplane starts form rest, then accelerates by rate of [tex]3m/s^2[/tex] until it reaches 102m/s at the end of the runway. so we have:
[tex]Vo=0m/s\\Vf=102m/s\\a=3m/s^2\\\\[/tex]
From the equations of uniformly accelerated motion (since the airplane's acceleration is uniform) [tex]'Vf^2=Vo^2+2aS' \\[/tex] is compatible to find the minimum runway length (S).
[tex]Vf^2=Vo^2+2aS\\Vf^2-Vo^2=2aS\\S=Vf^2-Vo^2/2a\\[/tex]
[tex]S=((102m/s)^2-(0m/s)^2)/2m/s^2\\S=(10404m^2/s^2-0m^2/s^2)/2m/s^2\\S=(10404m^2/s^2)/2m/s^2\\S=5202m[/tex]
The magnitude of the impulse on a ball bouncing off a wall is equal?
The impulse is equal to the rate of change of momentum.
What is impulse?The term impulse refers to the product of force and time. Let us recall that from Newton's law, impulse is proportional to the change in the momentum of the ball.
We can write from Newtons law;
F. t = Mv - Mu
F = force
t = time
M = mass
v = final velocity
u = initial velocity
Thus, the impulse is equal to the rate of change of momentum.
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Trenton is a middle-aged man with Down syndrome. Compared to unaffected adults his age, Trenton is at greater risk of developing ______.
Answer:
because
Explanation:
ok just because
Girl runs 40m due south in 40 seconds,he then return to North,20m in 10 seconds,, calculate
1. average speed
2.average velocity
3.change in velocity
4.acceleration
An object’s weight is equal to the force of gravity on it. The Moon has a gravity of 1/6 that of earth’s, while Jupiter has a gravity of 2.64 times that of the earth how much more would a 5.00 kg turtle weigh on Jupiter than the moon?
Answer:
See below
Explanation:
On Earth , turtle weight = 5 * 9.81 =49 N
Moon = 490.5 * 1/6 = 8.18 N
Jupiter 490.5 * 2.64 = 129.4 N
129.4 - 8.18 N = 121.2 N more on Jupiter than moon
I 'm driving down the street at 15 m/s, it takes me 20 minutes to get to my destination. What was the distance to my destination?
The distance to the destination is 18,000 meters or 18 kilometers. the formula for calculating the distance is speed × Time.
Using the formula:
Distance = Speed × Time
Speed = 15 m/s (meters per second)
Time = 20 minutes
First, we need to convert the time from minutes to seconds, as the speed is given in meters per second.
1 minute = 60 seconds
Time in seconds = 20 minutes × 60 seconds/minute = 1200 seconds
Now, we can calculate the distance:
Distance = 15 m/s × 1200 seconds
Distance = 18,000 meters
So, the distance to the destination is 18,000 meters, or 18 kilometers (since 1 kilometer = 1000 meters).
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A charge of 10 coulomb passes through a cross-section of a conductor in 20 seconds. What is the magnitude of current flowing through it?
The magnitude of current flowing through the conductor is 0.5A.
The charge flowing through the cross-section of the conductor is q = 10C.The time taken by the charge to flow is t = 20s.A stream of charged particles, such as electrons or ions, travelling through an electrical conductor or a vacuum is known as an electric current. The net rate of electric charge flowing through a surface or into a control volume is how it is calculated.The equation for the current flowing in the circuit is I = q/t. Where I is the current in the circuit.Substitute the values of q and t,I = 10C/20sI = 1/2AI = 0.5AThus, the current flowing through the conductor is 0.5 A.To learn more about current, visit :
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An object is dropped from an airplane and takes 72 seconds to hit the ground what is the magnitude of its Vfy
The magnitude of its Vfy i.e Vertical componet is mathematically given as
V=705.6 m/s
This is further explained below.
What is the magnitude of its Vfy?Generally, At this point, we need to do an analysis of the vertical motion of the item, which may be thought of as a free fall motion with a constant acceleration.
g = 9.8
As a result, we are able to employ the suvat equation that is as follows:
v = u +at
As a result, the following expression may be used to determine the magnitude of the object's vertical velocity after t = 72 seconds:
v = 0 + (9.8) (72)
V=705.6 m/s
In conclusion, the magnitude
V=705.6 m/s
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