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
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.
51. If two cylinders of dissimilar geometry are connected together to form a closed hydrostatic system, why would one cylinder piston travel a greater distance than the other
Answer:
see that for the same pressure the displaced height in each cylinder is different because its diameter is different. Pascal's principle
Explanation:
The pressure on a system is given by the relations
P = ρ g h
P = F / A
where ρ is the density of the liquid, h the height and A the area
The expressions above we see that if for the same height the pressure is the same regardless of the shape of the cylinder.
With the second expression we see that if the system has a different area, the pressure is
P = [tex]\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }[/tex]
where we use subscript 1 for one body and subscript 2 for the other body
F₁ = [tex]\frac{A_{1} }{A_{2} } F_{2}[/tex]
The cylinder displacement is
V = A h
where V is the volume and h the height, in general the liquids are incompressible therefore the displaced volume is constant in the two bodies
V = A₁ h₁ = A₂ h₂
[tex]\frac{A_{1} }{A_2} = \frac{h_2}{h_1}[/tex]
we substitute
F₁ = [tex]\frac{h_2}{h_1}[/tex] F2
From here we see that for the same pressure the displaced height in each cylinder is different because its diameter is different.
If the diameter is the same, the offset height is the same
Why was Cassini launched toward Venus if it is going to explore Saturn?
The gravity of Venus will boost the speed of the space probe
The orbit of Venus is between Earth and Saturn
Venus is needed to reflect the radio signals to the probe
Answer: A. The gravity of Venus will boost the speed of the space probe
Explanation: Hope this helped in time (:
A car is traveling due north at 23.6 m>s. Find the velocity of the car after 7.10 s if its acceleration is (a) 1.30 m>s2 due north, or (b) 1.15 m>s2 due south.
Answer:
a) v = 32.8 m/s
b) v= 15.4 m/s
Explanation:
a)
Applying the definition of acceleration (assumed to be constant), we can write the following expression for the velocity v:[tex]v = v_{o} + a*t (1)[/tex]
where v₀ is the initial velocity and a is the acceleration, being t the
time elapsed.
In the case a) the acceleration and the velocity vectors have both the same direction (due north), so both have the same sign, which means that the car is speeding up.Replacing by the givens v₀ and t in (1), we get:[tex]v = 23.6 m/s + (1.30 m/s2 * 7.10 s) = 32. 8 m/s (2)[/tex]
b)
In this case, the acceleration vector and the velocity vector have opposite directions, so the car slows down, due to both vectors have opposite signs.Replacing by the givens in (1) and taking into account the signs, we get:[tex]v = 23.6 m/s + ((-1.15 m/s2) * 7.10 s)) = 15. 4 m/s (3)[/tex]
4. True or False. Improving your muscular endurance can improve your
power *
Answer:
it's true
Explanation:
A strong body allows you to perform movements and activities that require power without getting tired.
can i eat air? im hungry
Answer:
Yes
Explanation:
need help asap.......................
Answer:
vt3bryhtevy4g24vu5hy4
Which property of a solid measures how resistant the material is to deformation?
A. Elasticity
B. Hardness
C. Plasticity
D. resilience
Answer: the answer is a
Explanation:
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]
a mechanic uses a hydraulic lift to raise a 1,200 kg car 0.50 m off the ground . how much work does the lift do on the car
Answer:
The work done is 5880 J
Explanation:
Recall the formula for work as force times distance, therefore in this case the force is that oppose to the gravitational force on the car(of magnitude m * g), and the distance is 0.5 meters.
Then:
Work = 1200 * 9.8 * 0.5 J = 5880 J
A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the rock after it has fallen halfway to the ground.
Answer:
2.45 J
Explanation:
The following data were obtained from the question:
Mass (m) = 0.5 kg
Height (h) = 1 m
Kinetic energy (KE) =?
Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 1/2 = 0.5 m
Final velocity (v) =?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 0.5)
v² = 9.8
Take the square root of both side
v = √9.8
v = 3.13 m/s
Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:
Mass (m) = 0.5 kg
Velocity (v) = 3.13 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 3.13²
KE = 0.25 × 9.8
KE = 2.45 J
Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J
Which heat related illness requires immediate medical attention
Answer:
Heat stroke
Explanation:
Heat Stroke is the most serious heat-related illness and requires immediate medical attention.
Answer:
Heat Stroke is the most serious heat-related illness and requires immediate medical attention. Symptoms include: confusion, fainting, seizures, very high body temperature and hot, dry skin or profuse sweating. CALL 911 if a coworker shows signs of heat stroke.
Explanation:
I HOPE THIS HELPS * pls mark me brainliest }
A force of 6.7 N acts on a 30 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.
Answer:
(a) 0.748 J
(b) 2.245 J
(c) 3.74 J
(d) 4.482 W
Explanation:
(a) Work done W = Force × distance
W = F×d,
Where d = 1/2(at²)
Therefore,
W =1/2(F×at²)................ Equation 1
Where a = acceleration, t = time.
But,
a = F/m...................... Equation 2
Where m = mass.
Substitute equation 1 into equation 2
W = 1/2(F²t²/m)................. Equation 3
Given: F = 6.7 N, t = 1 s, m = 30 kg
Substitute into equation 3
W₁ = 1/2(6.7²×1²/30)
W = 0.748 J.
(b) Similarly,
The work done in the second seconds is
Where t₂ = 2 s
W₂ = 1/2(F²t₂²/m)- W₁
W = 1/2(6.7²×2²/30)-0.748
W = 2.245 J
(c) The work done in the third seconds is
Where t₃ = 3 s
W₃ = 1/2(F²t₃²/m)-(W₂+W₃)
W = 1/2(6.7²×3²/30)-(2.993)
W = 3.74 J.
(d) P = Fv ............... Equation 4
Where v = velocity.
and,
v = at..................... Equation 5
Substitute equation 5 into equation 4
P = Fat................... Equation 6
Given: F = 6.7 N, a = 6.7/30 = 0.223 m/s², t = 3 s
Substitute into equation 6
P = 6.7×0.223×3
P = 4.482 W.
What are two ways electromagnetic waves are used in a home computer scanner?
A.
An image is produced when a silicon chip releases electrons by the photoelectric effect.
B.
Visible light is reflected off an object and then captured by a charge-coupled device.
C.
X-rays shine on objects, and the rays either pass through or are absorbed by the objects.
D.
Red light shines on a black-and-white object, and light is reflected according to the pattern of black and white.
Two ways electromagnetic waves are used in a home computer scanner are:
An image is produced when a silicon chip releases electrons by the photoelectric effect.Visible light is reflected off an object and then captured by a charge-coupled device.What is electromagnetic wave?In terms of science, electromagnetic radiation (EMR) is made up of electromagnetic (EM) field waves that travel over space while carrying electromagnetic radiant energy. It consists of X-rays, gamma rays, microwaves, infrared, visible light, ultraviolet, and radio waves. These waves are all a component of the electromagnetic spectrum.
Electromagnetic waves, which are synchronised oscillations of the electric and magnetic fields, are the traditional form of electromagnetic radiation. The electromagnetic spectrum is created at various wavelengths depending on the oscillation frequency. Electromagnetic waves move at the speed of light in a vacuum.
Hence, option (A) and (B) are correct.
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what does correlation mean
a mutual relationship or connection between two or more things.
What are similarities between homogeneous mixture and
heterogeneous mixture
Answer:
they both are the mixture of substance
Answer:
a homogeneous mixture is a uniform mixture whose components appear to be in a single state while a heterogeneous mixture is non uniform mixture whose components remain separate
Explanation:
example of homogeneous_ water and ethanol
heterogeneous_ water and chalk
A person standing on a building ledge throws a ball vertically from a launch position 55 m above the ground. It takes 2.0 s for the ball to hit the ground.
With what initial speed was the ball thrown?
Answer:
Vo = 17.69 [m/s]
Explanation:
To solve this problem we must use two equations of kinematics.
[tex]v_{f}^{2} =v_{o}^{2} +2*g*h\\v_{f}=v_{o}+g*t[/tex]
where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 55 [m]
t = time = 2 [s]
Now we replace the gravity acceleration into the second equation:
[tex]v_{f}=v_{o}+9.81*2\\v_{f}=v_{o}+19.62[/tex]
And then into the first equation:
[tex](v_{o}+19.62)^{2}=v_{o}^{2}+2*9.81*55\\v_{o} ^{2}+2*v_{o}*19.62+384.94=v_{o}^{2} + 1079.1\\39.24*v_{0}=694.16\\v_{o}=17.69[m/s][/tex]
The initial speed at which the ball is thrown upward is 17.7 m/s.
According to the question the initial position of the ball, y = 55 m, and the final position of the ball is y' = 0 m. We have assumed upward direction as positive direction and downward as negative direction.
so the total displacement:
d = y'-y = 0 - 55
d = -55 m
now applying the second equation of motion:
d = ut - (1/2)gt²
where t = 2s ( given ) and g = 9.8 m/s².
-55 = 2u - 0.5×9.8×4
-55 = 2u - 19.6
u = -17.7 m/s
the negative sign indicated that the initial velocity is opposite to the direction of displacement.
This means the initial velocity is upward as it should be.
Therefore, the initial speed of the ball is 17.7m/s
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What happens to the molecules of a substance when it changes phase?
When it comes to phase shifts or phase change the greater the intermolecular interactions are the closer the molecules are to one another.
What is phase change?A phase change occurs when matter transitions from one state to another (solid, liquid, gas, plasma). When enough energy is provided to the system (or when a significant quantity is removed), as well as when the pressure on the system is adjusted, these changes occur.Any two phases of matter can undergo phase transitions.All phase shifts are accompanied by an energy shift.Isothermal phase shifts occur in all cases.Temperature changes can cause substances to change phase often. Most substances are solid at low temperatures; as the temperature rises, they become liquid; and at even higher temperatures, they become gaseous.To Know more about Phase change here
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A spherical balloon with a 36cm diameter is being deflated. Its volume V is a function of its radius r according to
Question:
A spherical balloon with a 36 cm diameter is being deflated. Its volume V is a function of its radius r according to [tex]V(r) = \frac{4}{3}\pi r^3.[/tex]
As it's deflating, it is much easier to measure its radius than its volume. Suppose the radius of the balloon is [tex]r(t) = 18 - 18e^{-0.24t}[/tex] cm at t seconds.
Determine the value of V'(4) (accurate to 3 decimal places). Write a complete sentence including the units of both 4 and V'(4) in this context. Determine the value of r(4) (accurate to 3 decimal places). Write a complete sentence including the units of both 4 and r(4) in this context.Answer:
[tex]V'(4) = 201.143[/tex]
[tex]r(4) = 11.106[/tex]
Explanation:
Given
[tex]V(r) = \frac{4}{3}\pi r^3.[/tex]
[tex]r(t) = 18 - 18e^{-0.24}[/tex]
Solving (a): V'(4)
First, we determine V('r)
[tex]V(r) = \frac{4}{3}\pi r^3.[/tex]
Differentiate w.r.t r
[tex]V'(r) = 3 * \frac{4}{3}\pi r^{3-1}[/tex]
[tex]V'(r) = 3 * \frac{4}{3}\pi r^2[/tex]
[tex]V'(r) = 4\pi r^2[/tex]
Substitute 4 for r and take [tex]\pi = \frac{22}{7}[/tex]
[tex]V'(4) = 4 * \frac{22}{7} * 4^2[/tex]
[tex]V'(4) = 4 * \frac{22}{7} * 16[/tex]
[tex]V'(4) = \frac{4 * 22* 16}{7}[/tex]
[tex]V'(4) = \frac{1408}{7}[/tex]
[tex]V'(4) = 201.143[/tex]
This means that the volume of the balloon when the radius is deflated to 4 seconds is 201.143 cm^3
Solving (b): r(4)
Substitute 4 for t in [tex]r(t) = 18 - 18e^{-0.24t}[/tex]
[tex]r(4) = 18 - 18e^{-0.24*4}[/tex]
[tex]r(4) = 18 - 18e^{-0.96}[/tex]
[tex]r(4) = 18 - 18*0.383[/tex]
[tex]r(4) = 11.106[/tex]
This means that the radius of the balloon when at 4 seconds of deflation is 11.106 cm
What may result when the energy that builds up at plate boundaries is released because the plates suddenly overcome the force of friction?
volcano
hot spot
earthquake
trench
A young athlete has a mass of 42 kg. On a day when there is no wind she runs a 100m race in 14.2
Answer:
7 m/s approx
Explanation:
Step one:
Given data
Mass m= 42kg
distance covered= 100m
time taken = 14.2 seconds
Required
The speed of the athlete
Step two:
We know that speed= distance/time
Substitute
Speed= 100/14.2
Speed= 7.04 m/s
Speed= 7 m/s approx
For each of the situations described below, the object considered is undergoing some changes. Among the possible changes you should consider are:
(Q) The object is absorbing or giving off heat.
(T) The object's temperature is changing.
(E) The object's thermal energy is changing.
(W) The object is doing mechanical work or having work done on it.
(C) The objects chemical energy is changing.
For each of the two situations, identify which of the five changes is taking place, indicating the applicable letters Q T U W C (or none) as appropriate while providing your reasoning.
a. An ice cube sits in the open air and is melting.
b. A cylinder with a piston on top contains a compressed gas and is sitting on a thermal reservoir (a large iron block that can provide or take thermal energy as the system needs). After everything has come to thermal equilibrium, the piston is moved upward somewhat (very slowly). The object to be considered is the gas in the cylinder.
Answer: a. (Q), (T), (E)
b. (Q), (T), (E), (W)
Explanation: Thermal Energy and Temperature are closely related: when the temperature rises causing atoms or molecules to move, thermal energy is produced. Thermal energy is the energy within the system.
Mechanical Work is the amount of energy transferred due to an applied force.
Chemical Energy is the energy contained in the bonds of chemical structures of the molecules released when a chemical reaction happens.
Given the explanations, let's analyse the situations:
a. For an ice cube to be melting, it has to be absorbing heat, which means its thermal energy is changing and, consequently, so does its temperature;
b. First, the gas inside the cylinder reaches a thermal equilibrium, which means its thermal energy and temperature changed. Since there were exchange of heat to reach the equilibrium, the gas absorbed or gave off heat. After the equilibrium, when the piston starts to be moved, the energy of the pressure is transferred to the gas, so mechanical work had been done.
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has force constant 815 N/m . The coefficient of kinetic friction between the floor and the block is μk=0.40. The block and spring are released from rest and the block slides along the floor.
Required:
What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)
Answer:
v = 0.41 m/s
Explanation:
In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.So, we can write the following general equation, taking the initial and final values of the energies:[tex]\Delta K + \Delta U = W_{ffr} (1)[/tex]
Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.⇒ Kf = 1/2*m*vf² (2)The change in the potential energy, can be written as follows:[tex]\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)[/tex]
where k = force constant = 815 N/m
xf = final displacement of the block = 0.01 m (taking as x=0 the position
for the spring at equilibrium)
x₀ = initial displacement of the block = 0.03 m
Regarding the work done by the force of friction, it can be written as follows:[tex]W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)[/tex]
where μk = coefficient of kinettic friction, Fn = normal force, and Δx =
horizontal displacement.
Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:Fn = Fg= m*g (5)Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:[tex]\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)[/tex]
[tex]\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)[/tex]
Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:[tex]\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)[/tex]
[tex]v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)[/tex]The driver of a 1000 kg car traveling on the interstate at 35 m/s slams on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the brakes are applied, a constant friction force of 800 N acts on the car. Ignore air resistance, at what minimum distance should the brakes be applied to avoid collision with the other vehicle
Answer:
765.625 m
Explanation:
When brakes are applied , work is done by frictional force to reduce the kinetic energy of the car . To stop the car , the work done by brake must equalize the kinetic energy of car .
1/2 m v² = F d
m is mass of the car , v its velocity , F is frictional force and d is displacement of car .
1/2 x 1000 x 35² = 800 x d
d = 765.625 m
a car accelerate at 9 m/s squared. Assuming the car starts from rest how far will it travel in 10 seconds
Answer:
450m
Explanation:
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 + t + \frac{1}{2} a {t}^{2} \\ \frac{1}{2 } \times 9 \times {10}^{2} = 450m[/tex]
The greenhouse effect is
2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the collision Object B is moving at 15m/s.What is the velocity of Object A AFTER the collision?
Answer:
Explanation:
We shall apply law of conservation of momentum .
Momentum before collision = momentum after collision .
Momentum before collision = 400 kg m/s
Momentum after collision = 5 x v + 11 x 15
where v is velocity of A after the collision .
5 x v + 11 x 15 = 400
5 v = 400 - 165
5v = 235
v = 47 m /s .
Which property of a solid measures how resistant the material is to deformation?
Answer: the answer would be b in k12
Explanation:
Every experiment most control for variables that are not being tested that might also affect the dependent variable. True or false
Answer: Scientist use an experiment to look for cause and effect relationships in nature. They design an experiment so that changes to one Variable (any factor, trait or ... Independent Variable = what you are testing or what you are in control of or what ... A part of the experiment that is not being tested and is used for comparison
Explanation:
Estimate the weight of a 1000kg car that is accelerating at 3 m/s/s.
Answer:
W = 9800 N
Explanation:
Given that,
Mass of a car, m = 1000 kg
Acceleration of the car, a = 3 m/s²
We need to find the weight of the car. Weight of an object is given by the product of mass and acceleration due to gravity on the Earth.
W = mg
Put all the values,
W = 1000 kg × 9.8 m/s²
= 9800 N
So, the weight of the car is 9800 N.