The velocity of the swimmer relative to the shore is 2 m/s south by vector addition
What is vector addition ?
Vector addition means putting two or more vectors together. In the addition of vectors, we are adding two or more vectors using the addition operation in order to obtain a new vector that is equal to the sum of the vectors. Vector addition finds its application in physical quantities where vectors are used to represent velocity, displacement, and acceleration.
We can solve this problem by using vector addition.
In fact, we can take north as positive direction, and we can write the following:
vc = - 4m/s is the velocity of the current (which is negative, since it is south)
v s= +2m/s is the velocity of the swimmer (which is positive, since it is north)
Therefore, the velocity of the swimmer relative to the shore is
v = vs + vc = 2+(-4) = - 2m/s
And the negative sign means the direction is south, so the answer is
v = 2 m/s south
The velocity of the swimmer relative to the shore is 2 m/s south by vector addition
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a car carrying a 80-kgkg test dummy crashes into a wall at 28 m/sm/s and is brought to rest in 0.10 ss. part a what is the magnitude of the average force exerted by the seat belt on the dummy? express your answer to two significant figures and include the appropriate units
The magnitude of the average force exerted by the seat belt on the dummy is 224N .
What is an average force ?The average force is the force produced by an object moving over a specific period of time at a given rate of speed, or velocity. This velocity is not instantaneous or precisely measured, as the word "average" indicates.
Briefing:mass of the dummy (m) = 80kg
velocity of the dummy (v) = 28 m/s
time (t) = 10 seconds
Average force exerted (F)
To calculate the average force;
According to the formula;
F = (m × v) ÷ t
Where;
F represents the force exerted
m represents the mass of the dummy
v represents the velocity
t represents the time
F = m * v/t
F = 80 *28/10
F = 224
F = 224 N
The force exerted by the seat belt on the dummy is 224N .
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A freight train moves due north with a speed of 8 m/s. The mass of the train is 6 x 106 kg. How fast would a 1500 kg automobile have to be moving due north to have the same momentum as the train
The freight train has a momentum of mv = (4.5 × 105 kg)(1.4 m/s) = 6.3 ×105 kg-m/s.
An automobile with a mass of 1800 kg would need to go at a speed of 6.3 × 105 kg-m/s ÷ 1800 kg = 350 m/s in order to have a momentum of the same magnitude. (Roughly 800 mph)
A moving body's amount of motion is referred to as momentum. It has units of kg m/s and is theoretically represented as p = m * v.
The momentum equation connects the total force exerted on a fluid element to its acceleration or rate of change of momentum and is a formulation of Newton's Second Law. You're certainly familiar with the formula F = ma, which links applied force and acceleration in solid mechanics analysis.
Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force applied on a particle is equal to the time rate of change of momentum.
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two blocks are released from the top of a building. one falls straight down while the other slides down a smooth ramp. if all friction is ignored, which one is moving faster when it reaches the bottom?
The first and second blocks move at the same speed.
What is the law of conservation of mechanical energy ?
The law of conservation of mechanical energy states that the total mechanical energy of a system is constant if the only forces acting on the system are conservative.
The total mechanical energy of a system is the sum of kinetic and potential energy.
E = KE + PE
This is the total energy, kinetic energy and potential energy of the system.
The kinetic energy of the
system is:
KE = 1/2 x m x v^{2}
This is the mass and velocity of the object.
The gravitational potential energy is:
PE = mgh
Here is the acceleration of gravity and altitude.
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suppose a spring with spring constant 6 n/m is horizontal and has one end attached to a wall and the other end attached to a mass. you want to use the spring to weigh items. you put the spring into motion and find the frequency to be 1.4 hz (cycles per second). what is the mass? assume there is no friction.
When a mass attached to a spring, it will make the spring to produce a harmonic motion with frequency 1.4 Hz. Then the mass is 77.48 gram.
The frequency of a harmonic motion caused by a displaced spring is:
f = 1/2π . √(k/m)
Where:
k = spring constant
m = mass
Parameters given from the problem:
f = 1.4 Hz
k = 6 N/m
Plug those parameters into the formula:
1.4 = 1/2π . √(6/m)
√(6/m) = 8.8
m = 0.07748 kg = 77.48 gram
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mass of bullet 0.25kg, velocity of bullet 230kg, mass of gun 9kg. What is the recoil speed of the gun
The recoil speed of the gun is 6.39m/s
What is recoil speed of the gun ?
After firing the bullet, the bullet moves forward with a large momentum and the gun moves backward with an equal momentum. Let m and M be the masses of bullet and gun, respectively. If v and V are the velocities of the bullet and gun, respectively after the firing then:-
0= mv+ MV
Therefore V=-mv/M
The negative sign indicates that the gun moves in a direction to that of
the bullet.
m1v1 = m2v2
m1 = mass of bullet
v1 = velocity of bullet
m2 = mass of gun
v2 = velocity of gun
m1v1=m2v2
0.25*230=9*v2
v2=0.25*230/9 = 6.39m/s
The recoil speed of the gun is 6.39m/s
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ake a 40-pf capacitor and charge it to a potential difference of 500 v. then disconnect it from the battery and connect its terminals to those of an uncharged 10-pf capacitor. what are: (a) the
If we rearrange the circuit so that the inductor and capacitor were connected in series,the impedance would be Decreased and the resonant frequency would be unchanged.
What is capacitor ?
Frequency is the rate at which something occurs over a particular period of time or in a given sample.
Sol-An LC circuit is made up of an inductor (L) and a capacitor (C). At the resonance condition of the LC circuit, the inductive reactance XL. becomes equal to the capacitive reactance Xc.
It is defined as-
XL= 2πfL
Xc= 1/2πfC
So when reactances are equal we have-
XL=Xc
2πfL= 1/2πfC
f^2=1/4π^2LC
f=1/2π√LC
The resonance condition is the same for both the parallel and series LC circuit, so the resonance frequency will not change.
The impedance of a parallel LC circuit is:
Z(w)= jL w^2-w^2•/w
In a series LC circuit, when w
, the impedance becomes equal to 0.
Therefore the impedance of a series LC circuit will decrease and the resonant frequency will remain the same.
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why are the rocky inner planets different than the gaseous outer planets? heat from the growing sun allowed rock to mainly form close in gases come from further away from the center of the solar system the rocky inner planets got rocks from the sun later solar wind pushing material away had more effect on metals gas has more momentum and flies further away naturally
The correct answer is option A.
The rocky inner planets are different than the gaseous outer planets because heat from the growing sun allowed rock to mainly form close in.
In our solar system, the inner planets are rocky while the outer planets are in a gaseous state.
Rocky planets are also called terrestrial planets as their outer surface is solid and composed of rocks and metal.
They have large thick gaseous atmospheres which are abundant in gases such as nitrogen, carbon dioxide, and oxygen.
While the outer planets are giant balls of gases that orbit around the sun.
Unlike terrestrial planets, these gaseous planets do not have thick atmospheres and are 2x larger in size than the rocky inner planets.
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a. What is pressure? Write down its Sl unit
Answer:
Pascal (Pa)
Explanation:
the si for pressure is the pascal pa equal to one newton per sqaure meter (N/m^2, or kgxm-^1xs-^2)
[tex] \boxed {\huge \: \sf \red{~~~Answer~~~} \: }[/tex]
Pressure is defined as the physical force exerted on an object. SI unit of pressure is Pascal (pa).[tex] \\ \\ [/tex]
______________________An 85.0 kg man stands on a spring scale in an elevator when it begins to accelerate upwards. If the scale at that moment indicates that his "weight" is 1020 N, what is the acceleration of the elevator?
2.20 m/s^2 is the acceleration of the elevator.
Option c is correct answered .
What does the scale read when the elevator accelerates upward?If you stand on a scale in an elevator accelerating upward, you feel heavier because the elevator's floor presses harder on your feet, and the scale will show a higher reading than when the elevator is at rest. On the other hand, when the elevator accelerates downward, you feel lighter.
How do you find the acceleration of an elevator with weight?support force F = mass x acceleration + weight
For a mass m= kg, the elevator must support its weight = mg = Newtons to hold it up at rest. If the acceleration is a= m/s² then a net force= Newtons is required to accelerate the mass.
Fnet = m a
Fn + (-Fg) = m a
a = 1020 - (85)(9.8) /85
a = 2.20 m/s^2
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When a red light with a wavelength of 670 nm shines on a piece of metal, an electron is ejected. What is the energy of the photon absorbed by the electron to overcome the attractive forces?.
The energy of the photon of light as we can see here is 2.9 * 10-19 J.
What is the energy of the light?We know that light has to do with any of the components of the electromagnetic spectrum. These are the kinds of radiation that are able to pass through vacuum because they do not need a medium for the propagation of the wave. We are now asked to obtain the energy of the photon based in the wavelength that we have been supplied in the question here.
We have that the wavelength of the photon from what we can see in the question have been given as 670 nm. Then;
E = hc/λ
E = energy of the light
h = Plank's constant
c = speed of light
λ = wavelength of the light
It then follows that;
E = 6.6 * 10^-34 * 3 * 10^8/670 * 10^-9
E = 2.9 * 10-19 J
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Calculate the area of the plates of a 1 pF parallel plate capacitor in a vacuum if the separation of the plates is 0.1 mm. [ε0=8.85×10-12C2N-1m-2 ]Select one:6.2 m214.4 m211.3 m218.4 m2
Apply:
C = A e0 / d
Where:
C = capacitance = 1pf = 1 x10^-12 F
A = area
ε0 = 8.85×10^-12
1 pf = 1 x 10^-12 F
A = Cd / ε0 = 1 x 10^-12 ( 1 x 10^-4 ) / 8.85×10^-12 = 1.13 x10 ^-5 m^2
What element has 4 electrons 3 protons and 6 neutrons
The element that has 4 electrons, 3 protons and 6 neutrons is lithium ion.
What is an element?An element is one of the simplest chemical substances that cannot be decomposed in a chemical reaction or by any chemical means and made up of atoms all having the same number of protons.
The atom of an element is made up of three subatomic particles namely;
ProtonElectronNeutronIn a neutral element, the number of protons and electrons are the same i.e. equal. According to this question, an element has 4 electrons, 3 protons and 6 neutrons.
The element in the periodic table with 3 protons is Lithium. This means that the element is lithium ion because it has gained one electron.
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two boxes of the same mass are connected by a string of negligible mass and pulled across a level tabletop with negligible friction by another horizontal string. the force applied by the leading string is f. m m t f a how does the magnitude of the force t exerted by the string connecting the boxes compare to the magnitude f?
The importance of the pressure t exerted via the string connecting the bins evaluate to the value f is T< f.
A force is a power that can trade the movement of an object. A force can encourage an item with mass to exchange its tempo, i.e., to boost up. stress can also be described intuitively as a push or a pull. A strain has both importance and path, making it a vector quantity.
considering both masses together
The acceleration of the system = a
a = f/m+m = f/2m
Now, considering one mass towards left m
T = ma
T = (f/2m)
m = f/2
T = f/2
T<f
Therefore, T must have a magnitude less than f.
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suppose you were hired to build a dam. what features would you look for in a site? be sure to consider the impact on living things as well as the physical characteristics of the site.
I need to get it done right now
Answer: The wall and base
Explanation:
24) A 0.042-kg pet lab mouse sits on a 0.35 kg air-track cart. The cart is at rest, as is a second cart with a mass of 0.25 kg. The lab mouse now jumps to the second cart. After the jump, the 0.35-kg cart has a speed of u1=0.88m/s. What is the speed v2 of the mouse and the 0.25-kg cart? |
The mass of the mouse, m₁=0.042 kg
The mass of the 1st cart, m₂=0.35 kg
The mass of the second cart, m₃=0.25 kg
The speed of the 1st car after the jump, u₁=0.88 m/s
From the law of conservation of momentum, the momentum of a system should always remain the same.
Before the mouse jumps from one cart to another, all the objects were at rest. Therefore the total momentum of the system was zero.
Thus after the mouse jumps, the total momentum of the system should be equal to zero.
Thus the momentum of the second cart and the mouse will be equal in magnitude to the momentum of the first cart but it will be in the opposite direction.
Thus,
[tex]m_2u_1=(m_1+m_3)v_2[/tex]On substituting the known values,
[tex]\begin{gathered} 0.35\times0.88=(0.042+0.25)v_2 \\ \Rightarrow v_2=\frac{0.35\times0.88}{(0.042+0.25)} \\ =1.05\text{ m/s} \end{gathered}[/tex]Thus the speed of the mouse and the second cart after the mouse jumps is 1.05 m/s
27. Beyond "c", the speed of the rocket is;
(A) constant.
(B) continuously increasing.
(C) continuously decreasing.
(D) increasing for a while and constant thereafter.
(E) constant for a while and decreasing thereafter.
Beyond "c", the speed of the rocket is; (D) increasing for a while and constant thereafter.
How quickly does a rocket move?The space shuttle must accelerate from zero to 8,000 metres per second (nearly 18,000 miles per hour) in eight and a half minutes to reach the minimum height necessary to circle the Earth.
The orbital velocity is 7.9 kilometres per second, which translates to more than 20 times the speed of sound. A rocket attaining orbital velocity (1st cosmic velocity) will enter orbit around the Earth (C), whereas a faster rocket would follow an elliptical trajectory (D).
On Earth, air tends to prevent exhaust gases from exiting the engine. This lessens the thrust. However, because there is no atmosphere in space, the exhaust gases may depart considerably more easily and quickly.
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a window-mounted air conditioner removes 3.5 kj from the inside of a home using 1.75 kj work input. how much energy is released outside and what is its coefficient of performance?
The amount of heat released outside is 3.5 kJ and the coefficient of performance is 2.
What is efficiency of a machine?The efficiency of the a machine or any device is the measure of effectiveness of the device. Efficiency describes how a machine converts input energy to output energy without wasting much of the input energy.
Mathematically, efficiency of a machine is given as;
E = output energy / input energy x 100%
The coefficient of performance (COP) of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required.
The energy released outside = 3.5 kJ
COP = 3.5 kJ / 1.75 kJ
COP = 2
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A 0. 20-kg baseball is struck with a force of 100 n from a 0. 94-kg baseball bat. What will be the acceleration of the ball and the bat, in that order, while the bat and ball are in contact?.
The acceleration of the baseball is 500 m/s².
A force is an influence that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull.
The mass of the baseball, m = 0.2 kg
The force acting on the baseball is 100 N.
By Newton's second law of motion,
F = ma
Substituting the values in the above equation,
100 N = 0.2 kg × a
a = 100 / 0.2
a = 500 m/s²
The total mass when the baseball and baseball bat are in contact is:
m = 0.2 + 0.94 = 1.14 kg
Then the acceleration will be:
F = ma
100 = 1.14 × a
a = 87.7 m/s²
The acceleration of the ball after the contact is 500 m/s².
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what is an elliptical galaxy? what is an elliptical galaxy? any galaxy with an elliptical halo when observed at radio wavelengths. a spiral galaxy seen from an angle, giving it an elliptical profile. a galaxy with an elliptical outline and a smooth distribution of brightness (no spiral arms). a spiral galaxy with an elliptically shaped nuclear bulge and the spiral arms starting from the ends of the ellipse.
An elliptical galaxy is a galaxy with an elliptical outline and a smooth distribution of brightness (no spiral arms).
In the field of astronomical physics, we can describe elliptical galaxies as stretched galaxies that have an elliptical outline. This type of galaxy is usually made through old stars and has an even, smooth distribution of light.
Although the elliptical galaxies are not a major part of the universe they make up 10–15% in the Virgo Supercluster.
As the elliptical galaxy is made from broken stars, their size may vary from being a dwarf elliptical to a supergiant.
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A rocket is launched from the surface of the earth with a speed of 9.0x103 m/s. What is the maximum altitude reached by the rocket? (MEarth=5.98x1024 kg, REarth=6.37x106 m)
From the Law of conservation of energy, we know that the sum of the kinetic and potential energy of the rocket is the same at the surface of the Earth and at the maximum altitude. Nevertheless, the kinetic energy of the rocket when it is at the maximum altitude is 0:
[tex]\begin{gathered} K_1+U_1=K_2+U_2 \\ K_2=0 \\ \Rightarrow K_1+U_1=U_2 \end{gathered}[/tex]The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]On the other hand, the gravitational potential energy for big changes in altitude (comparable to the radius of the Earth) is given by the expression:
[tex]U=-\frac{GMm}{r}[/tex]Where M is the mass of the Earth, m is the mass of the rocket, r is the distance from the center of the Earth to the rocket and G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\cdot\frac{m^2}{\operatorname{kg}}[/tex]At the beggining of the movement, the value of r corresponds to the radius of the Earth:
[tex]U_1=-\frac{GMm}{R_E}[/tex]At the end of the movement, the value of r corresponds to the radius of the Earth plus the maximum altitude h:
[tex]U_2=-\frac{GMm}{R_E+h_{}}[/tex]Substitute the expressions for U_1, K_1 and U_2 and simplify the equation by eliminating the factor m:
[tex]\begin{gathered} \frac{1}{2}mv^2-\frac{GMm}{R_E}=-\frac{GMm}{R_E+h} \\ \Rightarrow\frac{1}{2}v^2-\frac{GM}{R_E}=-\frac{GM}{R_E+h} \end{gathered}[/tex]Isolate the term GM/(R_E+h):
[tex]\Rightarrow\frac{GM}{R_E+h}=\frac{GM}{R_E_{}}-\frac{1}{2}v^2[/tex]Divide both sides by the factor GM:
[tex]\Rightarrow\frac{1}{R_E+h}=\frac{1}{R_E}-\frac{v^2}{2GM}[/tex]Take the reciprocal to both sides of the equation:
[tex]\Rightarrow R_E+h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}[/tex]Isolate h:
[tex]h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}-R_E[/tex]Substitute the values of each variable: R_E=6.37x10^6m, M=5.98x10^24kg, G=6.67x10^-11 N*m^2/kg^2, and v=9.0x10^-3 m/s:
[tex]\begin{gathered} h=\frac{1}{\frac{1}{6.37\times10^6m}-\frac{(9.0\times10^3\cdot\frac{m}{s})^2}{2(6.67\times10^{-11}N\cdot\frac{m^2}{kg^2})(5.98\times10^{24}kg)}}-6.37\times10^6m \\ =18.03\times10^6m-6.37\times10^6m \\ =11.7\times10^6m \end{gathered}[/tex]Therefore, the maximum altitude reached by a rocket with an initial speed of 9.0x10^3m is:
[tex]11.7\times10^6m[/tex]a very thin horizontal, 2.00-m long, 5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. at the east end of the beam, a 200-n force pushes downward. at the west end of the beam, a 200-n force pushes upward. what is the angular acceleration of the beam?
The angular acceleration of the beam of length 2 meters and mass 5 kg is 240 radian/s².
The length of the rod is 2 meters, the mass of the rod is 5kg, a force of 200N is applied downward in the east direction while a force of 200N applied upward in west direction.
We know,
That the torque on a body is,
T = Fr
T is torque,
F is the force,
r is the perpendicular distance from the axis of rotation.
So, assuming the centre of the beam as the axis of rotation, the torque on the body is,
T = 200(1)+200(1)
T = 400N-m
We also know,
That torque on the body is,
T = IA
Where I is moment of inertia of the beam,
A is the angular acceleration of the body,
We know, moment of inertia of the beam assuming the centre of the rod as axis of rotation is,
I = MR²/12
M is the mass of the body,
R is the length of the rod,
So,
I = 5(2)²/12
I = 20/12 kg-m²
So, now we can write,
T = 20/12A
As both are value of torque, so,
20/12A = 400
A = 240 radian/s²
The angular acceleration is 240 radian/s².
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To which category does galaxy #2 belong? Why does it belong in this category?
Galaxy #2 belongs to spiral. This is because it has a bulge at the centre, a disk as well as the curved arms. Contains both Young and Old stars - contains gas and dust - contains more red giants and red supergiant's than disk - some regular rotation about centre.
What Is Dark Matter?Dark matter is a form of matter thought to account for approximately 85% of the matter in the world. Dark matter consists of particles that do not absorb, reflect or emit light, so they cannot be detected by observing electromagnetic radiation. Dark matter is material that cannot be directly seen. We know dark matter exists because it affects objects we can directly observe.
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Find the coefficient of static friction between the ramp and object with the following
Horizontal length of ramp: 75cm
Vertical length of ramp: 19cm
mass of object: 128g
The coefficient of static friction between the ramp of 75 cm horizontal length and a vertical length of 19 cm and object with a mass of 128 g is 0.25
Since the ramp forms a right angled triangle,
tan θ = Opposite side / Adjacent side
Opposite side = Vertical length of ramp = 19 cm
Adjacent side = Horizontal length of ramp = 75 cm
tan θ = 19 / 75
θ = [tex]tan^{-1}[/tex] ( 0.25 )
θ = 14°
m = 128 g = 0.128 kg
g = 9.8 m / s²
W = m g
W = 0.128 * 9.8
W = 1.25 N
Since the object is in static condition,
∑ [tex]F_{y}[/tex] = 0
N - [tex]W_{y}[/tex] = 0
N - W cos θ = 0
N = 1.25 * cos 14°
N = 1.25 * 0.97
N = 1.21 N
F[tex]_{s}[/tex] = μ[tex]_{s}[/tex] N
F[tex]_{s}[/tex] = Static frictional force
μ[tex]_{s}[/tex] = Co-efficient of static friction
N = Normal force
Since the object is in static condition,
∑ [tex]F_{x}[/tex] = 0
F[tex]_{s}[/tex] - [tex]W_{x}[/tex] = 0
F[tex]_{s}[/tex] = W sin θ
F[tex]_{s}[/tex] = 1.25 * sin 14°
F[tex]_{s}[/tex] = 1.25 * 0.24
F[tex]_{s}[/tex] = 0.3 N
μ[tex]_{s}[/tex] = F[tex]_{s}[/tex] / N
μ[tex]_{s}[/tex] = 0.3 / 1.21
μ[tex]_{s}[/tex] = 0.25
Therefore, the coefficient of static friction between the ramp and object is 0.25
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I need help with question 7, I just need the answer you don’t have to explain.
Explanation
Step 1
free body diagram
so, for m1
m1= 10 kg
so,
[tex]\begin{gathered} \sum ^{forces}_{\text{ y}}=T_1-mg=ma \\ T_1=m_1(a+g)\Rightarrow equation(1) \\ T_1=10(a+g) \end{gathered}[/tex]and for m2
[tex]\begin{gathered} \sum ^{forces}_{\text{ x}}=m_2g\sin (37)-T_1=m_2a \\ \text{solve for T}_1 \\ T_1=m_2g\sin (37)-m_2a \\ \text{replace} \\ T_1=(3.6\text{ kg)(9.8 }\frac{m}{s^2})\sin 37-3.6a \\ T_1=21.23\text{ -}3.6a\Rightarrow eq(2) \end{gathered}[/tex]Step 2
now, replace in equaiotn (1) and solve for a
[tex]\begin{gathered} T_1=m_1(a+g)\Rightarrow equation(1) \\ 21.23\text{ -}3.6a=10(a+g) \\ 21.23\text{ -}3.6a=10a+10g \\ -7.2a=10a-98.1 \\ -17.2a=-98.1 \\ a=-\frac{98.1}{-17.2} \\ a=5.703\text{ }\frac{m}{s^2} \end{gathered}[/tex]finally, replace in equation (2) to find Tension
[tex]\begin{gathered} T_1=21.23\text{ -}3.6a \\ T_1=21.23\text{ -}3.6(5.703) \\ T_2=3.891\text{N} \end{gathered}[/tex]I hope this helps you
In 2 - 3 sentences, explain the difference in Kinetic energy and Potential energy.
The main difference between Kinetic energy and Potential energy is that kinetic energy refers to movement while potential energy refers to storage.
What are Kinetic energy and Potential energy?Kinetic energy is a type of energy in motion or energy in movement such as a turbine (mechanical energy), while Potential energy refers to the energy that is stored to be used when required (e.g. chemical bonds of foods).
Therefore, with this data, we can see that Kinetic energy is used as movement, while Potential energy is stored to be used in the future.
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Slim Jim, continually maintaining his svelte body, lifts a 70 kg barbell 1.4m above the grounda) How much energy did the barbell have when it was on the ground?b) How much energy does it have after being lifted 1.4m? What kind of energy does it have afterbeing lifted? Where did it come from?c) How much work did Jim do to the lift the object?dIf he lifted it in 1.5s how much power did he use?
Given,
The mass of the barbell, m=70 kg
The height to which Slim Jim lifts the barbell, h=1.4 m
a)
When the barbell was on the ground, it will have zero kinetic energy as it has no velocity. And if assume the height of ground as zero meters, then its potential energy is also zero.
Thus when the barbell was on the ground, its energy was zero joules.
b)
The energy of the barbell when it is at a height of h is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=70\times9.8\times1.4 \\ =960.4\text{ J} \end{gathered}[/tex]Thus the energy that the barbell has after being lifted 1.4 m is 960.4 J
This energy is the energy stored in the barbell due to its position. Thus the energy stored in the barbell is the potential energy.
Slim Jim has to do some work to lift the barbell to the given height. This work done will be stored in the barbell in the form of potential energy. That is, the energy of the barbell is supplied to it from Slim Jim through the work.
c)
All the work done by Jim will be stored in the barbell in the form of potential energy. Thus, the work done by Jim is equal to the potential energy of the barbel.
Therefore, the work done by Jim is 960.4 J
d)
Given,
The time interval, t=1.5 s
The power is given by,
[tex]P=\frac{W}{t}[/tex]Where W is the work done by Jim.
On substituting the known values,
[tex]\begin{gathered} P=\frac{960.4}{1.5} \\ =640.27\text{ W} \end{gathered}[/tex]Thus the power used by Jim is 640.27 W
7 The distance-time graph for a motorcyclist riding
off from rest is shown in Figure 1.2.18.
a Describe the motion.
b
Calculate how far the motorbike moves in
30 seconds.
c Calculate the speed.
Answer:
a) Uniform rectilinear motion
b) 600 m
c) 20 m/s
Explanation:
a) average velocity is described as the change of position divided by the time of travel, so v= ∆x/∆t. This is a linear function (y = mx+ q) which is the same structure of the equation needed to calculate average velocity (x2= x1 + vt). If you do the ratio between a distance and the corresponding time you'll see that the average velocity is the same, which is typical of this kind of motion.
b) At the the time of 30 seconds the corresponding distance on the y axis is 600 m.
c) Again, v=∆x/∆t and we can rewrite is as v=x2-x1/t. If you take the same data we've used before you can write v= 600m-0/ 30s= 20 m/s. The average velocity is 20 m/s.
Please help me with this, it's a Kinematics Equation 2 problem!
The distance covered by the object is given as,
[tex]d=ut+\frac{1}{2}at^2[/tex]Plug in the known values,
[tex]\begin{gathered} d=(0\text{ m/s)(9 s)+}\frac{1}{2}(2m/s^2)(9s)^2 \\ =0\text{ m+}81\text{ m} \\ =81\text{ m} \end{gathered}[/tex]Thus, the distance covered by the object is 81 m.
a group of students are provided with three objects all of the same mass and radius. the objects include a solid cylinder, a thin hoop (or cylindrical shell), and a solid sphere. the students are asked to predict which will get to the bottom of a ramp first if all three are released together from the same distance up the ramp. which prediction is correct if the objects are listed in order from fastest to slowest?
Sphere, cylinder, hoop is the correct prediction. Moment of inertia is defined as the quantity represented by a body that resists an angular acceleration that is the sum of the product of each particle's mass times the square of its distance from the axis of rotation.
Ball moment of inertia Is = 2/5 M r²
Cylinder Ic = ½ M r²
Tire Ih = M r²
Substitute each to calculate time
Ball
ts = (d / √2gy) √ ( 1 + 2/5 Mr² / mr²)
ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)
ts = (d / √ 2gy) 1.1
Cylinder
tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)
tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5
tc = (d / √ 2gy ) 1.2
tires
th = (d / √2gy) √ (1 + mr² / mr²)
th = (d / √2gy) √(1 + 1) = (d / √ 2gy ) √ 2
th = ( d / √ 2gy) 1.41
The result is that the bullet hits the fastest and the body with the most tires. So the correct answer is
ts < . tc < th
ball, cylinder tire
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a 2 900-kg truck moving at 11.0 m/s strikes a car waiting at a traffic light, hooking bumpers. the two continue to move together at 7.00 m/s. what was the mass of the struck car?
The mass of the car that was stuck is 1657 Kg.
What is the mass of the struck car?We know that we have to use the laws of the conservation of the linear momentum here. It is known that the momentum before collision must be equal to the total momentum that we have after collision.
In this case, we have;
m1u1 + m2u2 = m1v1 + m2v2
m1 = mass of the truck
m2 = mass of the car
u1 and v1 = the initial and final velocities of the truck
u2 and v2 = initial and final velocities of the car
Again we are told that the two continue to move together hence v1 + v2 = v
m1u1 + m2u2 = (m1 + m2) v
(2900 * 11.0) + 0 = (2900 + m2) 7
31900 = 20300 + 7m2
31900 - 20300 = 7m2
m2 = 31900 - 20300 /7
m2 = 1657 Kg
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