Answer:
rush of adrenaline
Explanation:
This is believed to have been possible due to the rush of adrenaline that the individual was experiencing. This has happened a couple times in moments of extreme distress where an individual's body is flooded with adrenaline. This adrenaline allows for super human like strength as well as increased senses. The individual that has lifted the most amount of weight in such a scenario, lifted a total of 3000 lbs. Therefore, it is fair to say that 1200 lb. rock was lifted due to adrenaline.
a ball is thrown upward with a beginning speed of 40m/s. The graph below shows how the speed of the ball changes until it reaches its maximum height.
use the graph to find
a) the time when the ball reaches its maximum height
b) the acceleration of the ball
c) the maximum height the ball went
Answer:
a) 4.0816s
b) -9.8 ms^-1
c) 81.63265m
During a phase change the temperature of a substance remains constant this is because during a phase heat changes the ____ energy of particles in a substance without changing their ____ energy
Answer:
Explanation:
individual and then net
hope that helps I could be wrong about this one though
At what speed, in m/s, would a moving clock lose 1.3ns in 1.0 day according to experimenters on the ground?
Answer:
v=0.14c
Explanation:
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
Answer:
The added mass will mean a longer period of oscillation.
Explanation:
The period of oscillation here is given by the formula;
T = 2π√(m/k)
Where m is mass and k is spring constant
From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.
Thus, the added mass will mean a longer period of oscillation.
Elizabeth has always believed that people's thoughts can help heal them. She wants to help people use positive thinking to positively affect their
illnesses. What type of psychology would be MOST appropriate for Elizabeth to study?
Answer: Family
Explanation:
What fitness component is plank exercises?
Answer:
The plank (also called a front hold, hover, or abdominal bridge) is an isometric core strength exercise that involves maintaining a position similar to a push-up for the maximum possible time.Explanation:
please mark me has brainllestA block is pushed so that it moves up a ramp at constant speed. Identify from choices (a)-(e) below the appropriate description for the work done by the specified force while the block moves from point A to point B. (a) is zero. (b) is less than zero. (c) is greater than zero. (d) could be positive or negative depending on the choice of coordinate systems. (e) cannot be determined.
Answer:
*The work of the Normal (N) y Wy are zero answer a
*The work of the applied force (F1) is positive answer c
*The work of the friction force (fr) is negative, answer b
*The work of the Wy isnegative, answer d
Explanation:
In this exercise it is asked to identify the type of work, unfortunately the diagram cannot be seen, but in the attached we can see the diagram of a body moving upward on an inclined plane, the existing forces are shown.
As the body moves at constant speed the accelerations are zero. Let's look for the job that is defined
W = F. d
W = F d cos θ
where the dot represents the dot product and the bold letters are vectors.
* The work of the Normal (N) and the y component of the weight (Wy) are zero because they are perpendicular to the motion
answer a
* The work of the applied force (F1) is positive because it is in the same direction of motion
W = F1 Δx
answer c
* The work of the friction force (fr) is negative because the force in the displacement have opposite directions
W = -fr Δx
answer b
* the work the x component of the weight (Wx) in this case is negative
answer d
Using the graph, when (in seconds) would you expect the object to have the highest velocity? Explain your answer.
Answer:
blue
Explanation:
nswer the following about two objects, A and B, whose motion produced ihe following ... "A" starts with a greater (t) position. Since they ... since you are moving away from the origin
Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accelerates at 2 m/s2. Instantly the speeder becomes aware that he is being chased and starts to accelerate at 1 m/s2. How much time (in s) passes until the police catch the speeder
Answer:
The time that passes until the police catch the speeder is 82.6204 seconds.
Explanation:
A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.
The position is calculated by the expression:
x = x0 + v0*t + 1/2*a*t²
where:
x0 is the initial position. v0 is the initial velocity. a is the acceleration. t is the time interval in which the motion is studied.First, let’s look at the police car’s equations of motion. In this case:
x0= 0 v0= 50 m/s a= 2 m/s²So: x = 50 m/s*t + 1/2*2 m/s²*t²
Now for the speeder’s car’s equations of motion you know:
x0= 3 km= 3,000 m v0= 55 m/s a= 1 m/s²So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
When the police catch the speeder they are both in the same position. So:
50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
Solving:
0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²
0= 3,000 + 55 *t + 1/2*t² - 50*t - 1*t²
0= 3,000 + 55 *t - 50*t - 1*t² + 1/2*t²
0= 3,000 + 5*t - 1/2*t²
Applying the quadratic formula:
[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]
x1= -72.6209
and x2= 82.6209
Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.
A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?
Answer:
v = 22.54 mph.
Explanation:
Given that,
Distance moved, d = 200 m
Time, t = 19.8 s
We need to find the runner's average speed.
We know that,
1 mile = 1609.34 m
200 m = 0.124 miles
19.8 seconds = 0.0055 h
So,
Speed = distance/time
[tex]v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph[/tex]
So, the runner's average speed is 22.54 mph.
It takes 500 W of power to move an object 96 m in 12 s. What force is being applied to the object?
Answer:
Explanation:
Power, by definition, is the amount of work per unit of time. We arent given work in this question, but we can find it because work is how much force per unit of distance.
[tex]P=\frac{W}{t_f-t_i} =\frac{F*d}{t_f-t_i}[/tex]
Plug in all the values, and its algebra at this point
[tex]500=\frac{F*96}{12}[/tex]
6000 = 96F
F = 62.5 Newtons
The force which is being applied to the object with the power of 500 Watt is 62.5 N.
Power can be defined as the rate of doing work. It is the work done in unit time. The SI unit of power is Watt (W) which is equal to joules per second (J/s). Sometimes, the power of motor vehicles and other machines is given in terms of Horsepower (hp), This unit is approximately equal to 745.7 watts of power.
The power of a system can be calculated as the product of force applied and distance travelled by the system per unit time taken.
Therefore, the force of the object with power 500W can be calculated as:
P = f × d/ t
where, P = Power of the object,
f = Force applied,
d = distance travelled,
t = time taken to cover the distance
f = (P × t)/ d
f = (500 × 12) / 96
f = 6000/ 96
f = 62.5 N
Therefore, the force which is being applied to the object is 62.5N (Newton).
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A train 350 m long is moving on a straight track with a speed of 84.1 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 15.8 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
Answer:
t = 25.0 s
Explanation:
Assuming that the engineer applies the brakes just over the crossing, the train moves exactly 350 m at a constant acceleration, with a final speed (when the last car of the train leaves the crossing) of 15.8km/h.Since we know the initial and final speeds, and the horizontal distance traveled (the length of the train) we can use the following kinematic equation to get the acceleration:[tex]v_{f}^{2} - v_{o}^{2} = 2*a* \Delta x (1)[/tex]
Since we need to find the time in seconds, it is advisable to convert vf and vo to m/s first, as follows:[tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]
[tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]
Replacing (2) and (3) in (1), since Δx =350m, we can solving for a:[tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]
In order to get the time, we can simply use the definition of acceleration, and rearrange terms:[tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]
a ball has a mass of 140g if it thrown with a velocity of 450m/s what is its kenetic energy?
Kinetic energy = mv²
Therfore kinetic energy =14175 joule
pls help ;-; this is the question btw
The answer is "Infrared"
Hope this helps
Answer:
x-rays
Explanation:
A car travels 100 km due East in 2 hours. It then travels 50 km South in 1 hour. What is its average velocity?
The average velocity of the car is 37.27 km/h.
The given parameters;
Initial displacement of the car, x = 100 kmTime of motion, t = 2 hoursFinal displacement of the car, y = 50 kmtime of motion, t = 1 hourThe average velocity of the car is calculated as follows;
[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]
The total displacement of the car is calculated as follows;
[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]
The average velocity of the car is calculated as follows;
[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]
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Donuts are a simple or complex carbohydrate? *
Ok
Answer:
simple carbohydrate
Explanation:
Simple carbs are like cakes, donuts, and candy, etc.
Hope this helps :D
A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path
Answer:
4units
Explanation:
To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;
D = √(x2-x1)²+(y2-y1)²
Given the coordinate points
(3,5) and (7,5)
Substitute
D = √(7-3)²+(5-5)²
D = √(7-3)²+0²
D = √4²
D = √16
D = 4
Hence the total distance the beam will travel along this path is 4units
A truck travels on a straight road at a velocity of 17 meters per second. Over 20
seconds, it accelerates uniformly to 27 meters per second. What distance did the truck
travel during this acceleration?
Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m
The equations of motion can be used to obtain the distance covered as 440 m.
We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;
v^2 = u^2 + 2as and v = u + at
v = final velocity
u = initial velocity
a = acceleration
s = distance
To obtain the acceleration;
27 = 17 + 20(a)
27 - 17 = 20a
a = 0.5 ms-2
Now, to obtain the distance;
v^2 = u^2 + 2as
v^2 - u^2/as = s
s = (27)^2 - (17)^2/2(0.5)
s = 440 m
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Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the proton at the end of the motion
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, [tex]v_p_i[/tex] = 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
[tex]W =K.E_f - K.E_i[/tex]
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
[tex]K.E_f =EQd + \frac{1}{2}m_pv_p_i^2[/tex]
[tex]m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C[/tex]
[tex]K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\[/tex]
[tex]K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J[/tex]
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Choose the words that make each statement correct.
(i) After being released from rest in a uniform electric field, a pro- ton will move [(a) in the same direction as; (b) opposite the direction of] the electric field to regions of [(c) higher; (d) lower] electric potential.
(ii) After being released from rest in a uniform electric field, an electron will move [(e) in the same direction as; (f) opposite the direction of] the electric field to regions of [(g) higher; (h) lower] electric potential.
Answer:
i). (a) in the same direction as , (d) lower
ii). (f) opposite the direction of, (g) higher
Explanation:
An proton may be defined as a sub atomic particle and it has a positive electrical charge. Its mass is slightly less than that of a neutron. When a proton is placed in an electrical field that is uniformly charged, it is at rest. When the proton first moves out from rest from the uniform electric field, it will move in a direction which is same as that of the electric field and it will move to a region of higher potential.
An electron is defined as the subatomic particle having negative electric charge. When an electron is released form rest from an uniform electric field, it will move in the opposite direction of the uniform electric field and will move to the region of lower electric potential.
What genetic test would you get if there was a specific genetic disease in your family
Emma is working in a shoe test lab measuring the coefficient of friction for tennis shoes on a variety of surfaces. The shoes are pushed against the surface with a force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine has to pull with a force of only 200 N to keep the material moving.
a. What is the coefficient of static friction between the shoe and the material?
b. What is the coefficient of dynamic friction between the shoe and the material?
c. Draw a Free Body Diagram for the above.
Answer:
Explanation:
Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .
a )
Given,
Normal force N = 400 N
Force of friction = 300 N
μ = coefficient of static friction = ?
Putting the values ,
300 = 400 μ
μ = .75
b )
Normal force N = 400 N
Force of friction = 200 N
μ = coefficient of kinetic friction = ?
Putting the values ,
200 = 400 μ
μ = .50
c ) see attached file .
Two trucks are traveling east on I-40 when the Walmart truck (61 mph, 57 ft long) decides to pass the FedEx truck (53 mph, 60 ft long). A 'pass' starts when the front of the Walmart truck is 45 ft behind the rear of the FedEx truck and ends when the rear of the Walmart truck is 45 ft in front of the front of the FedEx truck.
How long does it take for the Walmart truck to pass the FedEx truck?
How far does the Walmart truck travel while passing the FedEx truck?
Answer:
The answer is below
Explanation:
Let us assume that the slower truck which is the FedEx truck travels d mile. The pass started when the Walmart truck was 45 ft behind the FedEx truck and ended when the Walmart truck is 45 ft in front of the FedEx truck.
The additional distance covered by the truck Walmart = 45 ft + 60 ft (length of FedEx truck) + 45 ft = 150 ft
1 mile = 5280 ft
150 ft = 150 ft * 1 mile/5280 ft = 0.0284 mile
Therefore the Walmart truck covered an additional 0.0284 mile. Hence the distance covered by the walmart truck = d + 0.0284
Let us say it took t hours for the pass to be completed. Hence:
For FedEx; 53 mph = d / t. t =d / 53
For Walmart; 61 mph = (d + 0.0284) / t. t = (d + 0.0284) / 61
Since it took the trucks the same time. Hence:
d / 53 = (d + 0.0284) / 61
61d = 53d + 1.506
8d = 1.506
d = 1.506/8
d = 0.1882 mile = 993.75 ft
The FedEx truck traveled 993.75 mile while the Walmart truck traveled 1143.75 ft (993.75 + 150 ft)
The time (t) = d / 53 = 0.1882 / 53
t=0.00355 h = 12.8 seconds
Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a current of 7.68 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 65.0o relative to a magnetic field whose magnitude is 0.59 T. The magnitude of the net magnetic force experienced by the two-wire unit is 4.11 N. What is the current I
Answer:
[tex]4.77\ \text{A}[/tex]
Explanation:
F = Magnetic force = 4.11 N
[tex]I_n[/tex] = Net current
[tex]I_2[/tex] = Current in one of the wires = 7.68 A
B = Magnetic field = 0.59 T
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]65^{\circ}[/tex]
[tex]l[/tex] = Length of wires = 2.64 m
[tex]I[/tex] = Current in the other wire
Magnetic force is given by
[tex]F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}[/tex]
Net current is given by
[tex]I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}[/tex]
The current I is [tex]4.77\ \text{A}[/tex].
A sports car of mass m has the same kinetic energy as an SUV with a mass 3m as each is driven along the same road. Which vehicle, if either, has the larger momentum and what is the difference in their momenta, if any
Answer:
Explanation:
Kinetic energy ( KE ) = 1/2 m v²
= m²v² / 2 m = p² / 2m where p is momentum
KE = p² / 2m
p² = 2m KE
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
p₁ = √ ( 2 m x KE )
p₂ = √ ( 6 m x KE )
p₂ - p₁ = √ ( 6 m x KE ) - √ ( 2 m x KE )
= √KE m ( √6 - √2 )
Kinetic energy ( K.E )
[tex]= \frac{1}{2} m v^2\\\\= \frac{m^2 v^2}{2 m} \\\\= \frac{p^2}{2m}[/tex]
where p is momentum
[tex]K.E =\frac{p^2}{2m}\\\\p^2 = 2m. KE[/tex]
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
[tex]p_1 =\sqrt{(2m*K.E)}\\\\p_2 = \sqrt{(6m*K.E)} \\\\p_2 - p_1 = \sqrt{(6m*K.E)} -\sqrt{(2M*K.E} \\\\p_2 - p_1 = \sqrt{K.E m(\sqrt{6}-\sqrt{2}) }[/tex]
The difference is shown above.
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A north magnetic pole is facing another north magnetic pole with a distance xx. If the distance between the poles becomes 12x12x, what happens to the magnitude of the field energy between them
Answer:
Explanation:
When two north magnetic poles are placed close to each other , they will repel each other . If distance between them is increased due to mutual repulsion from x to 12 x , work is done by the magnetic field . This results in decrease of magnetic field energy .
Hence , when the distance between the poles becomes 12x from x , the magnitude of the field energy between them decreases .
What is the momentum of a 100-kilogram fullback carrying a football on a play at a velocity of 3.5 m/sec.
Answer:
100 Kg * 3.5 m/sec = 350 Kg-m/s
Explanation:
Momentum= F= Δ(mv) with m= mass, v= velocity, and Δ the change in mass and velocity
In this problem you are given all the factors you need to solve the equation you simply just plug in your mass (1,000 Kg) and Velocity (3.5m/s) and multiply them by each other to get your answer
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
What is the maximum height achieved if a 0.400 kg mass is thrown straight upward with an initial speed of 40.0 m⋅s−1? Ignore the effect of air resistance
The maximum height : 81.63 m
Further explanationGiven
0.4 kg mass
vo = initial speed = 40 m/s
Required
the maximum height
Solution
We can use the law of conservation energy(ME=PE+KE) or use parabolic motion
For parabolic motion :
h max = (vo²sin²θ)/2g
θ = 90°(straight upward)
Input the value :
h max = (40²sin²90°)/2 x 9.8
h max = 81.63 m
Connective Tissue in a tendon is