Answer:
Impulse is defined as change in momentum of an object divided by time interval.
at t= 0s
initial velocity = u , initial momentum = mu
at some time t .
final velocity = V, final momentum= mv.
now, change in momentum= ( final - initial) = ( mv-mu)
time interval = (t-0) = t
impulse force = (mv-mu)/ ( t)
Ft = (mv-mu) proved .
this law is known as Newton's second law.
A block and tackle pulley system has 3pulley wheels in the lower movable block. Determine the load that can be lifted by an effort of 350N if the efficiency of the system is 80%
A block and tackle pulley design has a velocity ratio 3. <br> Draw a labelled diagram of this system. In your diagram, indicate absolutely the points of application and the directions of the load and effort
How to calculate the load that can be lifted by an effort of 350N if the efficiency of the system is 80%?VR =3
VR = n=3
Efficiency of the system = 80%
Thus , Mechanical asvantage [tex]$=V R \times \eta=$[/tex]80/100×3 =2.4
Man can lift load with effort =350N
Thus,
Load= MA× effort = 2.4×350 = 840 N.
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Driving space is the space around your car_____
vulnerable to collision.
O
To the side
All of the above
To the rear
To the front
in which you are
Answer:
to the rear as there will be an elastic collision
Driving space is the space around your car to the rear vulnerable to collision. the correct answer is option(c).
Driving space indicates that you have a space or a larger area to drive in that is unrestricted. You see only broad, open spaces on the road ahead and there are no limits around your automobile, so you are free to travel forward or change lanes without encountering any obstacles.
Any driver must have room around their car, preferably in the front, back, and on the sides, in order to react to problems as they arise. Your separation from the car in front of you is the most important distance to leave. Drivers' Speeding Patterns Federal and state organizations that research various elements of speeding have taken notice of speeding as the most frequent cause of deadly collisions. The NHTSA has discovered tendencies among drivers who speed recently.
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what is elecric field line
A vector field and a starting position inside the field define the locus as a "field line." We have electric field lines for the electric fields.
What do you mean by electric field line?In general, an electric field line is a curve drawn with each point's tangent pointing in the direction of that point's net field. An arrow on the curve is obviously required to designate the direction of the electric field from the two potential directions provided by a tangent to the curve. A field line is a space curve or a three-dimensional curve.The vector gets shorter as you move away from the origin and always points radially outward because the electric field varies as the inverse of the square of the distance that points from the charge.The density of the lines indicates the size of the field. This indicates that the location with a high density of field lines has a stronger electric field due to the charged substance. The electric field is weaker in the area where the density of these lines is low.The electric field can be depicted by joining these vectors to form a line.The direction of an imaginary line called an electric field line at any place must match the direction of the field there.In general, a field line is a curve drawn so that each point's tangent points in the direction of the net field.To learn more about the electric field, refer to:
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A fisherman and his young son are in a boat on a small pond. Both are wearing life jackets. The son is holding a large helium filled balloon by a string, holding a significant part of his weight. Consider each action below independently and indicate whether the level of the water in the pond, Rises, Falls, is Unchanged or Cannot tell.
The son pops the helium balloon.
The fisherman knocks the tackle box overboard and it sinks to the bottom.
The son finds a cup and bails some water out of the bottom of the boat
The fisherman lowers himself in the water and floats on his back.
The fisherman lowers the anchor and it hangs one foot above the bottom of the pond. (the anchor is initially inside the boat)
The son gets in the water, looses his grip on the string, letting the balloon escape upwards.
(a) The son pops the helium balloon (unchanged)
(b) The fisherman knocks the tackle box overboard and it sinks to the bottom (rises)
(c) The son finds a cup and bails some water out of the bottom of the boat (falls)
(d) The fisherman lowers himself in the water and floats on his back (unchanged)
(e) The fisherman lowers the anchor and it hangs one foot above the bottom of the pond (rises)
(f) The son gets in the water, looses his grip on the string, letting the balloon escape upwards (rises).
Displaced volume of water in the pond
According to Archimedes principle, when a body is partially or completely immersed in a fluid, it experiences an upthrust which is equal to weight of to fluid displaced.
F = ρVg
V = F/ρg
where;
V is volume of water displacedρ is density of the immersed objectg is acceleration due to gravityWhen an object with a significant weight is dropped into the pond, the level of water in the pond will rise and vice versa.
The son pops the helium balloonThe water level in the pond will be Unchanged, since the weight of the balloon is almost insignificant.
Fisherman knocks the tackle box overboard and it sinks to the bottomThe water level will rise because the weight of the tackle box is significant.
Son finds a cup and bails some water out of the bottom of the boatThe water level will fall since some volume of water is being removed.
Fisherman lowers himself in the water and floats on his backThe water level will be unchanged, since nothing new is added into the pond.
Fisherman lowers the anchor and it hangs one foot above the bottom of the pondThe water level will rise because the anchor will displace some volume of water upwards when it is dropped inside the pond.
Son gets in the water, looses his grip on the string, letting the balloon escape upwards.The water level will rise, because the balloon will displace the water upwards when it escape upwards.
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An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a length of l = 8.77 m. The end of the boom is supported by another rope which is horizontal and attached to the wall as shown in the figure.
Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.)
The the tension in the horizontal rope is 7,019.4 N.
What is tension?
Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod.
Tension can also be described as the action-reaction pair of forces acting at each end of said elements.
The unit of tension is Newton (same unit as force, since tension is also a force).
Tension in the horizontal ropeThe tension in the horizontal rope is calculated as follows;
Apply the principle of torque;
T(L/2) cosθ = Mg (L/2) sinθ + mg L sinθ
T = (M + 2m)g sinθ/cosθ
T = (M + 2m)g tanθ
let θ = 66⁰ (this value should be given in the question)
T = (M + 2m)g tanθ
T = (74.9 + 2 x 122)(9.8)(tan 66)
T = 7,019.4 N
Thus, the the tension in the horizontal rope is 7,019.4 N.
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An electron is moving the east with a speed of 5.0 × 106 m/s. There is an electric field of
3.0 kV/m in the east direction. What will be the electrons speed after it has moved 1.00 cm?
The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
What is Electric field?Electric field is the physical field that surrounds a charge.
How to find final velocity of the electron when it moves some distance in a certain electric field?From Newton's second law, the acceleration the electron will bea=F/m=qE/m
where q= charge of electronE= electric field
m= mass of electron
=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)
=10¹⁵×0.526m/s²
The kinematics equation v²=v0²+2a(Δx)where v=final velocity of the electronv0=initial velocity of the electron =5×10⁶m/s
a=acceleration of the electron =10¹⁵×0.526m/s²
Δx=distance moved by the electron in east direction =1cm=10^-2m
Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2=25×10¹²+10.52×10¹²
=35.52×10¹²
Now velocity of electron=5.95×10⁶m/s.Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
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Why is glass classified the way it is?
A. The bulb lights up when the switch is closed.
B. Free electrons can carry charge.
C. Delocalized electrons create a current.
D. Not enough charge carriers.
The glass is classified the way it is because of option(b)i.e, Free electrons can carry a charge.
An amorphous solid is a glass. Glass shows all the mechanical properties of a solid even though its atomic-scale structure resembles that of a supercooled liquid. Transparency, heat resistance, pressure and breakage resistance, and chemical resistance are among glass' primary properties. 90% of all manufactured glass is soda-lime glass, making it the most prevalent and least priced type.
High electronegativity and tightly bonded electrons are found in the oxygen atoms that make up glass. This is so that glass can conduct electricity, which requires a significant amount of energy to release the electrons from the oxygen atom. As a result, the glass serves as an insulator at room temperature.
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**please look at attached photo for proper understanding* Inside a cathode ray tube, an electron is in the presence of a uniform electric field with a magnitude of 295 N/C.
What is the magnitude of the acceleration of the electron (in m/s2)?
_____m/s2
The electron is initially at rest. What is its speed (in m/s) after 8.50 ✕ 10−9 s?
________m/s
The magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².
The speed of the electron at the given time is 4.41 x 10⁵ m/s.
Acceleration of the electronThe magnitude of the acceleration of the electron is calculated as follows;
Force on the electron, F = ma = Eq
ma = Eq
a = Eq/m
where;
E is electric fieldq is charge of electronm is mass of electrona = (295 x 1.6 x 10⁻¹⁹) / (9.1 x 10⁻³¹)
a = 5.187 x 10¹³ m/s²
Speed of the electronThe speed of the electron at the given time is calculated as follows;
v = at
v = ( 5.187 x 10¹³)(8.5 x 10⁻⁹)
v = 4.41 x 10⁵ m/s
Thus, the magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².
The speed of the electron at the given time is 4.41 x 10⁵ m/s.
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If little herbert does something uncon
sciously he will probably
If little herbert does something unconsciously he will probably, he's definitely very unware of he has done
Doing something unconscious may have some implications which may not be favourable.
What is unconscious?Unconscious simply means a state of someone being absent minded of what is happening to him or her at a particular period of time
So therefore, if little herbert does something unconsciously he will probably, he's definitely very unware of he has done
Complete question:
What happens if little herbert does something unconsciously he will probably
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The focal length of a diverging lens is negative. If f = −26 cm or a particular diverging lens, where will the image be formed of an object located 63 cm to the left of the lens on the optical axis?
.... cm to the left of the lens
....What is the magnification of the image?
The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.
Location of the imageThe image distance formed by the lens is calculated as follows;
1/v + 1/u = -1/f
where;
v is the image distanceu is the object distancef is the focal length of the lens1/v = -1/f - 1/u
1/v = -(-1/26) - 1/63
1/v = 1/26 - 1/63
1/v = 0.022588
v = 1/0.022588
v = 44.27 cm
What is magnification of lens?The magnification of a lens is defined as the ratio of the height of an image to the height of an object.
Magnification of the image formedThe magnification of the image is calculated as follows;
Magnification = image distance/object distance
M = 44.27/63
M = 0.7
Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.
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How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km?
1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
What is Speed ?Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s
In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.
What are the parameters to consider ?
The parameters are;
The distance S = 3.85 × [tex]10^{5}[/tex] kmThe Speed of Light C = 3 × [tex]10^{8}[/tex] m/sThe time taken t = ?Speed = distance S ÷ Time t
Convert kilometer to meter by multiplying it by 1000
C = S/t
3 × [tex]10^{8}[/tex] = 3.85 × [tex]10^{8}[/tex] / t
Make t the subject of formula
t = 3.85 × [tex]10^{8}[/tex] / 3 × [tex]10^{8}[/tex]
t = 1.2833
t = 1.3 s
Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
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A 940-kg sports car collides into the rear end of a 2500-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.
What was the speed sports car at impact?
Express your answer to two significant figures and include the appropriate units.
The speed of the sport car at the time of impact is 6.61 m/s.
What is the frictional force of the two cars?Frictional force of the two cars = coefficient of kinetic frictin × mass × acceleration of gravity
= 0.8 × (2500+940) × 9.8
= 26970N
What is the acceleration of the skidded cars?As per Newton's second law of motion, force = mass × accelerationAcceleration= force / mass= 26,970/3440
= 7.8 m/s²
What is the velocity of the sport car at the time of impact?As per Newton's equation of motion, V² - U² = 2aSHere, V = 0 m/s, a= -7.8 m/s², S= 2.8 mSo, 0²-U²= 2×(-7.8)×2.8=> U = √43.68
= 6.61 m/s
Thus, we can conclude that the speed of the sport car at the time of impact is 6.61 m/s.
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In the diagram, q1 = +6.39*10^-9 C and
q2 = +3.22*10^-9 C. What is the electric
field at point P? Include a + or - sign to
indicate the direction.
P
91
0.424 m-
0.636 m
-
92
(Remember, E points away from + charges,
and toward charges.)
(Unit = N/C)
E =+823.12N/C is the electric field at point P
Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field. in the diagram, q1 = +6.39*10^-9 C and
q2 = +3.22*10^-9 C hence E = 823.12 N/C
A region of space surrounding andd P Include a + or - sign toindicate the direction.P910.424 m-0.636 m- electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.
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A force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.
A). What is the potential energy of the spring when it is stretched 0.700 m?
B). What is its potential energy when it is compressed 8.00 cm?
Answer:
399j and 456j
Explanation:
(a)spring energy= force * distance
spring energy=570 * 0.700= 399 joules
(b) convert 8.00 cm to m we have 0.08m
E =f * s
E=570 *0.08= 456j
The potential energy of the spring when it is stretched 0.700 m would be 199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.
What is the spring constant?The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.
The mathematical relation for calculating the spring constant is as follows
F = - Kx
As given in the problem, if a force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.
k = 570 / 0.7
k = 814.28 N / m
The potential energy stored in spring when it is stretched 0.700 m
E = 1/2 kx²
=0.5×814.28×0.7²
= 199.5 Joules
The potential energy stored in spring when it is stretched 8 cm
E = 1/2 kx²
=0.5×814.28×0.08²
=2.60 Joules
Thus, the potential energy of the spring when it is stretched 0.700 m would be 199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.
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*please refer to photo attached* The figure below shows a small, charged sphere, with a charge of q = +44.0 nC, that moves a distance of d = 0.189 m from point A to point B in the presence of a uniform electric field E of magnitude 300 N/C, pointing right.
What is the magnitude (in N) and direction of the electric force on the sphere?
magnitude_______N
Direction?
- toward the right
- toward the left
- the magnitude is zero
(b) What is the work (in J) done on the sphere by the electric force as it moves from A to B?
__________J
(c) What is the change of the electric potential energy (in J) as the sphere moves from A to B? (The system consists of the sphere and all its surroundings.)
PEB − PEA = ______J
(d) What is the potential difference (in V) between A and B?
VB − VA = ________V
(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
(d) The potential difference between A and B is 56.7 V.
Electric force on the sphere
The electric force on the sphere is calculated as follows;
F = Eq
where;
E is electric fieldq is the chargeF = 300 x (44 x 10⁻⁹)
F = 1.32 x 10⁻⁵ N
The direction of the force is towards the right.
Work done on the sphereW = Fd
W = 1.32 x 10⁻⁵ N x 0.189 m
W = 2.5 x 10⁻⁶ J
Change of the electric potential energyThe change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.
Potential difference between A and BVB − VA = Ed
VB − VA = 300 N/C x 0.189 m
VB − VA = 56.7 V
Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
The potential difference between A and B is 56.7 V.
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Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for the function in the figure?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The value of parameter C for the function in the figure is 2.
What is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
f(x) = Acos(x - C)
where;
A is amplitude of the waveC is phase difference of the waveWhat is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
From the blue colored graph; at y = 1, x = -2 cm
1 = cos(2 - C)
(2 - C) = cos^(1)
(2 - C) = 0
C = 2
Thus, the value of parameter C for the function in the figure is 2.
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A 92-kg fullback is running at 3.6 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west.
a)Calculate the original momentum of the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the momentum is to the east, and negative value if the direction of the momentum is to the west.
b)Calculate the impulse exerted on the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
c)Calculate the impulse exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
d)Calculate the average force exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the force is to the east, and negative value if the direction of the force is to the west.
a) The original momentum of the fullback is 331.2 kgm/s.
b) The impulse exerted on the fullback is - 331.2 kgm/s
c) The impulse exerted on the tackler is 331.2 kgm/s
d) The average force exerted on the tackler is 389.64 N
Given:Mass of the fullback, m = 92 kg
Initial velocity of the fullback , u = 3.6 m/s
Time of the motion of the fullback , t = 0.85s
The original momentum of the fullback ;
[tex]P_{i} = mv\\P_{i} = (92) (3.6)\\P_{i} = 331.2 kg m/s[/tex]
The impulse exerted on the full-back;
J = ΔP = [tex]m v_f - m v_i[/tex]
J = [tex]m ( v_f - v_i)[/tex]
J = 92 ( 0- 3.6)
J = - 331.2 kgm/s
The impulse exerted on the tackler;
[tex]J _1 = - J _2\\J_2 = - J_1\\J_2 = -( - 331.2)\\J_2 = 331.2 kgm/s[/tex]
The average force exerted on the tackler;
F= Δmv/t
[tex]F =\frac{331.2}{0.85} \\F = 389.64 N[/tex]
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