Answer:
Explanation:
Work done by the spring is negative .
Work done by force F creating displacement d is given by the following expression .
Work = F x d
Both force and displacement are vector quantity .
When direction of force and direction of displacement is same , work is positive . When direction of force and direction of displacement is opposite , work is negative .
When spring is compressed , it exerts a restoring or opposing force in a direction opposite to the direction of displacement of box . Hence here force is opposite to displacement . Restoring force acts opposite to displacement . Hence work done by spring on box is negative .
Using complete sentences and your own words, describe an example of the water cycle in action. Please use at least two of our key terms for today (Water Cycle, Evaporation, Condensation, Precipitation, Crystallization, Run-off, Transpiration, Aquifer) can you please help me i will give brainlyest it says collage but it is middle school pleas help me.
Answer:
Where does the water that forms rain come from? How about fog? Where does that come from? Have you ever been asked these questions by people, especially those kids who keep asking, 'Why?' Has someone ever told you that the water falling as snow has always been here, or that the water we use was once dinosaur blood, or that we are drinking someone's sweat, or, worse yet, drinking someone else's… gulp? How can this be possibl…
Explanation:
A 1570 kg car skidding due north on a level frictionless icy road at 156 km/h collides with a 2245.1 kg car skidding due east at 120 km/h in such a way that the two cars stick together. At what angle (−180◦ ≤ θ ≤ +180◦ ) East of North do the two coupled cars skid off at?
Answer:
θ = 47.75º East of North.
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, if we project it along E-W and N-S axes, the momentum components along these axes must be conserved too.So, for the N-S axis, we can write the following equation:[tex]p_{Northo} = p_{Northf} (1)[/tex]
Since the car moving due east has no speed component along the N-S axis, the initial momentum along this axis is simply:[tex]p_{Northo} = m_{1} * v_{1o} (2)[/tex]
where m₁ = 1570 kg, v₁₀ = 156 km/h
In order to work with the same units, we need to convert the speed in km/h to m/s, as follows:[tex]v_{1o} = 156 km/h *\frac{1000m}{1 km}*\frac{1h}{3600s} = 43.3 m/s (3)[/tex]
Replacing by the values in the left side of (1), we get:[tex]p_{Northo} = m_{1} * v_{1o} = 1570 kg* 43.3 m/s = 67981 kg*m/s (4)[/tex]
Since the collision is inelastic, both cars stick together, so we can write the right side of (1) as follows:[tex]p_{Northf} = (m_{1} + m_{2})* v_{fNorth} = 3815.1 kg* v_{fNorth} (5)[/tex]
From (4) and (5) , we can solve for VfNorth:[tex]V_{fNorth} = \frac{67981kg*m/s}{3815.1kg} = 17.8 m/s (6)[/tex]
We can repeat exactly the same process for the E-W axis:[tex]p_{Easto} = p_{Eastf} (7)[/tex]
Since the car moving due north has no speed component along the E-W axis, the initial momentum along this axis is simply:[tex]p_{Easto} = m_{2} * v_{2o} (8)[/tex]
As we did with v₁₀, we need to convert v₂₀ from km/h to m/s, as follows:[tex]v_{2o} = 120 km/h *\frac{1000m}{1 km}*\frac{1h}{3600s} = 33.3 m/s (9)[/tex]
Replacing by the values in the left side of (7), we get:[tex]p_{Easto} = m_{2} * v_{2o} = 2245.1 kg* 33.3 m/s = 74762 kg*m/s (10)[/tex]
Since the collision is inelastic, both cars stick together, so we can write the right side of (7) as follows:[tex]p_{Eastf} = (m_{1} + m_{2})* v_{fEast} = 3815.1 kg* v_{fEast} (11)[/tex]
From (10) and (11) , we can solve for VfEast:[tex]V_{fEast} = \frac{74762 kg*m/s}{3815.1kg} = 19.6 m/s (12)[/tex]
In order to find the angle East of North of the velocity vector, as we know the values of the horizontal and vertical components, we need just to apply a little bit of trigonometry, as follows:[tex]tg \theta = \frac{V_{fEast}}{V_{fNorth} } = \frac{19.6}{17.8} = 1.1 (13)[/tex]
The angle θ, East of North, is simply tg⁻¹ (1.1):θ = tg⁻¹ (1.1) = 47.75º E of N.
Why is Pluto considered less of an oddball planet now?
Answer:
The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects
Explanation:
If the evaporator outlet temperature on an r410A system is 50f and the evaporator superheat is 10f, what is the evaporating pressure of the refrigerant in the system
Answer:
So, the evaporating pressure of the R410A = 118 psig
Explanation:
Solution:
For R410A system:
Data Given:
Evaporator Outlet Temperature = 50°F
Evaporator Superheat = 10°F
Required:
Evaporating Pressure in the system = ?
For this, first of all, we need to calculate inlet temperature on R410A system from the given value of outlet temperature.
Evaporator inlet temperature is the difference of outlet temperature and evaporator superheat.
Evaporator inlet temperature = Outlet Temperature - Evaporator Superheat
Evaporator inlet Temperature = 50°F - 10°F
Evaporator inlet Temperature = 40°F
Now, as we have the inlet temperature and the R410A system. We can consult the pressure temperature chart or PT chart, which I have attached and highlighted the value of evaporating pressure for 40°F inlet temperature.
So, the evaporating pressure of the R410A = 118 psig
How long will it take a planet 150 million kilometers from a star
of mass 1.99 x 1030
kg to complete one full orbit?
Answer: 370 days.
Explanation:
It will take the planet 16.69 minutes to complete one full orbit.
Data Given;
r = 150*10^6kmmass of star = 1.99*10^30kgGravitational ForceApplying gravitational and centripetal force
[tex]F = \frac{Gm_1m_2}{r^2}\\ F = \frac{mv^2}{r}\\ \frac{mv^2}{r}=\frac{Gm_1M2}{r^2}\\ v = \frac{2\pi r}{T} \\ \frac{m(2\pi r/T)^2}{r} = \frac{GM_1M_2}{r^2}\\ \frac{4\pi ^2r^2}{T^2} = \frac{GM}{r} \\ T^2 = \frac{4\pi^2 r^3}{Gm}\\ [/tex]
Let's substitute the values into the equation
[tex]T^2 = \frac{4\pi ^2 * (150*10^6)^3}{6.67*10^-11 * 1.99 *10^30}\\ T^2 = 1002799.605 \\ T = 1001.398s\\ T = \frac{1001.398}{60} = 16.69min [/tex]
It will take the planet 16.69 minutes to complete one full orbit.
Learn more on planetary rotation here;
https://brainly.com/question/21222010
Answer quick please thank
1.
A proton has a __________ charge.
positive
negative
neutral
2.
___________ and ___________ are two subatomic particles found in the nucleus of an atom
electrons; neutrons
protons; neutrons
protons; electrons
3.
A(n) ______________ is a subatomic particle that has a negative charge.
proton
electron
neutron
Answer:
positive
proton and neutron
electron
economy?
A
Supply decreases
Price decreases
B.
Pnce increases
Scaroty decreases
c.
Demand increases
Pace increases
DO
Demand increases
Price decreases
Answer:
I. don't. get. this. question
C. Demand increases
Pace increases
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
the following using the energy/work formulae only:
1.The kinetic energy at the top of the cliff
2.The total mechanical energy at the top of the cliff
3.The kinetic energy of the Rock half way down 4.the speed of the Rock half way down
5.The speed of the Rock as it hits the ground
Answer:
this is the answer......
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radial component of the acceleration (in m/s2) on the edge the wheel?
Answer:
[tex]a=368.97\ m/s^2[/tex]
Explanation:
Given that,
Initial angular velocity, [tex]\omega=0[/tex]
Acceleration of the wheel, [tex]\alpha =7\ rad/s^2[/tex]
Rotation, [tex]\theta=14\ rotation=14\times 2\pi =87.96\ rad[/tex]
Let t is the time. Using second equation of kinematics can be calculated using time.
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s[/tex]
Let [tex]\omega_f[/tex] is the final angular velocity and a is the radial component of acceleration.
[tex]\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s[/tex]
Radial component of acceleration,
[tex]a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2[/tex]
So, the required acceleration on the edge of the wheel is [tex]368.97\ m/s^2[/tex].
The radial component of the acceleration (in m/s2) on the edge the wheel is 369.45 m/s²
Using ω² = ω₀² + 2αθ, we find the final angular speed after 14 revolutions, ω where
ω₀ = initial angular speed = 0 rad/s (since the wheel starts from rest), α = angular acceleration = 7 rad/s and θ = number of revolutions = 14 = 14 rev × 2π rad/rev = 87.965 rad.So, substituting the values of the variables into the equation, we have
ω² = ω₀² + 2αθ,
ω² = (0 rad/s)² + 2 × 7 rad/s² × 87.965 rad.
ω² = 0 rad²/s² + 1231.504 rad²/s²
ω² = 1231.504 rad²/s²
ω = √(1231.504 rad²/s²)
ω = 35.09 rad/s
We know that the radial acceleration a = rω² where
r = radius of wheel = diameter of wheel/2 = 60 cm/2 = 30 cm = 0.30 m and ω = final angular acceleration of wheel = 35.09 rad/s.So, substituting the values of the variables into the equation, we have
a = rω²
a = 0.30 m × (35.09 rad/s)²
a = 0.30 m × 1231.504 rad²/s²
a = 369.45 m/s²
So, the radial component of the acceleration (in m/s2) on the edge the wheel is 369.45 m/s².
Learn more about radial acceleration here:
https://brainly.com/question/25243603
A 0.5 kg apple falls from a tree. When the apple is 1.5 meters above the ground it has a velocity of 2.4 m/s. What is the apple's total mechanical energy
Answer:
8.7975 Joules
Explanation:
Step one:
Given data
mass m= 0.5kg
height h= 1.5m
velocity v= 2.4m/s
The potential energy PE= mgh
The kinetic energy KE= 1/2mv^2
Step two
PE=0.5*9.81*1.5
PE=7.3575 Joules
KE= 1/2*0.5*2.4^2
KE=0.5*0.5*5.76
KE=1.44 Joules
The total mechanical energy is
PE+KE
=7.3575+1.44
=8.7975 Joules
When an electromagnetic wave such as light interacts with a solid and polarizes the electron clouds within the atomic structure of that solid, what are the (2) primary consequences of this interaction
Answer:
1. Absorption or Emission of the light
2. Light induced changes in the matter
Explanation:
When an electromagnetic wave such as light interacts with solid, two consequences are for sure:
1. Absorption or Emission of the light
2. Light induced changes in the matter
When light travels through the solid, the intensity of light decreases as a result of addition of light energy to the body to which it interacts. If the medium or body to which light interacts is low in absorbing due to its atomic structure inside then light passing through it will show it. On the contrary, if a material is high in absorbing, very less intensive light will travel out.
Moreover, there will ionization of the atoms inside the medium to which light interacts. As light carries energy and when it interacts with atoms of the body, atoms gets energy and excited or de-excited accordingly.
Hence, above are the two primary consequences of this interaction.
HELPPPP if this object is accelerating to the right at 2.5 m/s2, arrow C
7.0 N
65 N
27 N
23 N
V (V)
1 (A)
0
135
9
3
4
1
30
2
2
2
2.
3
4
1
5
0
6
5
6
50
3
Answer:
6755727gvbu7euyeue77377365353663636
Explanation:
ghfjfkjfjfn has a long way is a good relationship is a different person than the difference is that the answer is 8AM is a good relationship with the other 4AM in a certain way is a different type and have different effects and feelings and the difference is the difference is that you can do it for a long period and the same way you know what to smome but you can get to the claire family a lot of times and the difference between the two is that the person you want and you know it would have different opinions on the subject and then use the difference between the two and a half hour and the difference is that the answer is yes but if you're not going anywhere in the past two and your life you can do that with your 6 or an 6th year old girl who is a good things go to
a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cliff
Answer:
The time taken by the projectile to hit the ground is 6.85 sec.
Explanation:
Given that,
Vertical height of cliff = 230 m
Distance = 300 m
Suppose, determine the time taken by the projectile to hit the ground.
We need to calculate the time
Using second equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, s = vertical height of cliff
u = initial vertical velocity
g = acceleration due to gravity
Put the value in the equation
[tex]230=0+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]t=\sqrt{\dfrac{230\times2}{9.8}}[/tex]
[tex]t=6.85 sec[/tex]
Hence, The time taken by the projectile to hit the ground is 6.85 sec.
Needddd helppppppp!!!
Answer:
2/9 times as strong.
Explanation:
From the question given above, the following assumptions were made:
Initial mass of 1st planet (M₁ ) = M
Initial mass of 2nd planet (m₁ ) = m
Initial distance apart (r₁) = r
Initial Force of attraction (F₁) = F
Final mass of 1st planet (M₂) = 2M
Final mass of 1st planet (m₂) = constant = m
Final distance apart (r₂) = 3r
Final force of attraction (F₂) =?
Next, we shall obtain an expression to determine the new force. This can be obtained as follow:
F = GMm / r²
Cross multiply
Fr² = GMm
Divide both side by Mn
G = Fr² / Mm
Since G is constant, then we have
F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂
Finally, we shall determine the new force as follow:
Initial mass of 1st planet (M₁ ) = M
Initial mass of 2nd planet (m₁ ) = m
Initial distance apart (r₁) = r
Initial Force of attraction (F₁) = F
Final mass of 1st planet (M₂) = 2M
Final mass of 1st planet (m₂) = constant = m
Final distance apart (r₂) = 3r
Final force of attraction (F₂) =?
F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂
Fr² / Mm = F₂ × (3r)² / 2M × m
Fr² / Mm = F₂ × 9r² / 2Mm
Cross multiply
Fr² × 2Mm = F₂ × 9r² × Mm
Divide both side by 9r² × Mm
F₂ = Fr² × 2Mm / 9r² × Mm
F₂ = F × 2 / 9
F₂ = 2/9 F
Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.
An object is moving with an initial velocity of 3.3m/s it is subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration
Answer:
218m
Explanation:
Given parameters:
Initial velocity = 3.3m/s
acceleration = 3.7m/s²
time = 10s
Unknown:
How far will it travel during the time of acceleration = ?
Solution:
We use of the kinematics equations to solve this problem;
S = ut + [tex]\frac{1}{2}[/tex] at²
S is the distance
u is the initial velocity
t is the time
a is the acceleration
So;
S = (3.3x10) + ([tex]\frac{1}{2}[/tex] x 3.7 x 10²) = 218m
A sample of 0.500 kg of ice is held at a temperature of 0°C. How much energy
must be added to completely turn the ice into liquid water? (The latent heat of
vaporization for water is 2260 kJ/kg; the latent heat of fusion for water is 333
kJ/kg.)
A. 1130 kJ
B. 666 kJ
C. 167 kJ
D. 4520 kJ
Answer:
C. 167 kJ
Explanation:
The minimum amount of heat require to complete turn the ice into liquid water is equal to the latent heat of the ice, that is, the amount of heat needed by the ice to turn into water. This amount is calculated by:
[tex]Q = m\cdot h_{f}[/tex] (1)
Where:
[tex]m[/tex] - Mass of ice, measured in kilograms.
[tex]h_{f}[/tex] - Latent heat of fusion, measured in kilojoules per kilogram.
[tex]Q[/tex] - Latent heat, measured in kilojoules.
If we know that [tex]m = 0.500\,kg[/tex] and [tex]h_{f} = 333\,\frac{kJ}{kg}[/tex], the latent heat of ice is:
[tex]Q = (0.500\,kg)\cdot \left(333\,\frac{kJ}{kg} \right)[/tex]
[tex]Q = 166.5\,kJ[/tex]
Therefore, the correct answer is C.
Electronic configuration of.
1)Fe.
2)Fe++
3)Fe+++
Answer:
The electronic configuration of Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6 and Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5. Fe2+ contains 2 fewer electrons compared to the electronic configuration of Fe.
Which is an example of current electricity?
What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2100 kg car (a large car) resting on the slave cylinder
Complete Question:
What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2100 kg car (a large car) resting on the slave cylinder ? The master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter
Answer:
[tex]F_1 = 142.92N[/tex]
Explanation:
Given
[tex]m = 2100kg[/tex] --- mass
[tex]D_1 = 2.00\ cm[/tex] --- diameter of the large cylinder
[tex]D_2 = 24.0\ cm[/tex] --- diameter of the slave cylinder
To do this, we apply Archimedes' principle of buoyancy which implies that:
[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2}[/tex]
Where
[tex]F_1 = Force\ on\ the\ master\ cylinder[/tex]
[tex]F_2 = Force\ on\ the\ slave\ cylinder[/tex]
[tex]A_1 = Area\ of\ the\ master\ cylinder[/tex]
[tex]F_2 = Area\ of\ the\ small\ cylinder[/tex]
Calculating the area of the master cylinder.
[tex]A_1 = \pi r_1^2[/tex]
[tex]r_1 = \frac{1}{2}D_1 = \frac{1}{2} * 2.00cm = 1.00cm[/tex]
[tex]A_1 = \pi* 1^2[/tex]
[tex]A_1 = \pi * 1[/tex]
[tex]A_1 = \pi[/tex]
Calculating the area of the slave cylinder.
[tex]A_2 = \pi r_2^2[/tex]
[tex]r_2 = \frac{1}{2}D_2 = \frac{1}{2} * 24.00cm = 12.00cm[/tex]
[tex]A_2 = \pi* 12^2[/tex]
[tex]A_2 = \pi* 144[/tex]
[tex]A_2 = 144\pi[/tex]
Substitute these values in:
[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2}[/tex]
[tex]\frac{F_1}{\pi} = \frac{F_2}{144\pi}[/tex]
Multiply both sides by [tex]\pi[/tex]
[tex]\pi * \frac{F_1}{\pi} = \frac{F_2}{144\pi} * \pi[/tex]
[tex]F_1 = \frac{F_2}{144}[/tex]
The force exerted on the slave cylinder (F2) is calculated as:
[tex]F_2 = mg[/tex]
[tex]F_2 = 2100 * 9.8[/tex]
[tex]F_2 = 20580[/tex]
Substitute 20580 for F2 in [tex]F_1 = \frac{F_2}{144}[/tex]
[tex]F_1 = \frac{20580}{144}[/tex]
[tex]F_1 = 142.92N[/tex]
Hence, the force exerted on the master cylinder is approximately 142.92N
How long does it take a 1.51 × 104 W steam engine to do 8.72 × 106 J of work? Round your answer to three significant figures
Place the lunar phases in the correct order. !QUICK! (I WILL GIVE BRAINLEST!!!!!)
Answer:
Here:
Explanation:
These eight phases are, in order, new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, third quarter and waning crescent. The cycle repeats once a month (every 29.5 days).
With a wooden ruler, you measure the length of a rectangular piece of sheet metal to be 14 mm. With micrometer calipers, you measure the width of the rectangle to be 4.98 mm. Give your answers to the following questions using the correct number of significant figures.
What is:
a. the area of the rectangle.
b. the ratio of the rectangle’s width to its length.
c. the perimeter of the rectangle.
d. the difference between the length and the width.
e. the ratio of the length to the width.
Answer:
Follows are the solution to the given question:
Explanation:
The rectangular part has a length of [tex]14 \ mm[/tex] and its rectangular part has a width of [tex]4.98 \ mm[/tex].
In option A
Calculating the area of the rectangular throgh the given piece:
[tex]\to A_R = WL=(14 mm) (4.98 mm) =69.72 \ mm^2[/tex]
In option B
Calculating the ratio of rectangle's width which is rectangle's length:
[tex]\to R_{WL}=\frac{W}{L}= \frac{4.98 \ mm}{14 \ mm} = 0.3557[/tex]
So, the ratio of rectangle's width to rectangle's length is 0.3557 .
In option C
Calculating the Perimeter of the rectangle:
[tex]\to P_R=2(W+L)=2(14 \ mm+ 4.98 \ mm)= 2(18.98) = 37.96 \ mm[/tex]
In option D
Calculating the difference between length and width:
[tex]\to D_{LW} = L- W = 14\ mm -4.98 \ mm =9.02 \ mm[/tex]
In option E
Calculating the ratio of length to width:
[tex]\to R_{LW}=\frac{L}{W} =\frac{14\ mm}{4.98 \ mm} = 2.811[/tex]
On a horizontal surface is located
an object of mass 20 kg and it is observed that
To set it in motion, a
80 N force, whereas if the object is
in motion, only one
40 N force, to keep it at speed
constant. Determine the values of the coefficients
static and kinetic friction, respectively.
(g = 10 m / s2).
A) 0.32; 0.11
B) 0.36; 0.15
C) 0.40; 0.2
D) 0.45; 0.22
By Newton's second law, the net vertical force acting on the object is 0, so that
n - w = 0
where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that
80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is
40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: µ and ν are the Greek letters mu and nu)
what are the SI units of thermal conductivity?
Answer:
The SI unit of thermal conductivity is watts per meter-kelvin (W/(m⋅K)).
Explanation:
hope this will help u
Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a speed of 4.85 106 m/s
Answer:
8.0995×10^-21 kgms^-1Explanation:
Mass of proton :
[tex]m_P=1.67\times 10^-^2^7\:kg\\[/tex]
Speed of Proton:
[tex]v_P=4.85\times 10^6[/tex]
Linear Momentum of a particle having mass (m) and velocity (v) :
[tex]-> p =m->v\:\:\: (1)[/tex]
Magnitude of momentum :
[tex]p=mv\:\:\: (2)[/tex]
Frome equation (2), magnitude of linear momentum of the proton :
[tex]p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1[/tex]
Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a 1000 kg mass vehicle. (a) For a low speed test at 2.5 rnls, compute the energy in the vehicle just prior to impacl If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 em
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
An endangered species is a species that has died out and no individuals are left. True or false
Answer:
the answer is false... ....
Answer:
False
Explanation:
A species that has died our ans has no individual's left would be considered extict.
In downtown Chicago, the east-west blocks are 400 ft long while the north-south blocks are 280 ft long. Because of the many one-way streets, it can be challenging to get around. Veronica starts at the corner of Dearborn and Ohio Streets. She drives four blocks north to Superior, two blocks east to Wabash, then a block south to get to her destination at Wabash and Huron.
Required:
What is the straight-line distance from her starting point?
Answer:
The answer is "1160 ft".
Explanation:
Using the Pythagoras theorem:
[tex]\to a= 280 \times 3= 840\\\\\to b= 400 \times 2= 800\\\\[/tex]
[tex]\bold{\to y^2= a^2+b^2}\\\\[/tex]
[tex]=840^2 +800^2\\\\= 705600+640000\\\\=1345600\\\\=1160 \ ft\\[/tex]
[tex]\to y=1160 \ ft[/tex]
