Answer:
Force is too large for a person to exert.
Explanation:
m = Mass of child = 14 kg
u = Initial velocity of compartment = 64 mi/h = [tex]\dfrac{64\times 1609.34}{3600}=28.61\ \text{m/s}[/tex]
v = Final velocity of compartment = 0
t = Time taken by the compartment to stop = 0.05 s
Force is given by
[tex]F=ma\\\Rightarrow F=m\dfrac{v-u}{t}\\\Rightarrow F=14\times \dfrac{0-28.61}{0.05}\\\Rightarrow F=-8010.8\ \text{N}[/tex]
The person has to exert a force of [tex]-8010.8\ \text{N}[/tex] in the opposite direction to keep holding the child. This is a huge amount of force and cannot be done by a person's physical strength alone.
____made up of glucose and fructose and found in plants.
1.Xylose
2.Maltose
3.Lactose
4.Sucrose
Answer:
1.Lactose
I Hope its help for you
Have a good day
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.
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the pilot whose weight is 676 N, maintains a constant speed of 2.25 x 102 m/s. At what speed, in m/s, will the pilot experience weightlessness
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
3. What is the acceleration of a cart with a F = 80 N and m = 32 kg? *
2.0 m/s^2
O 2.5 m/s^2
2.8 m/s^2
2.7 m/s^2
Answer:
2.5 m/s^2
Explanation:
Given the following data;
Force = 80N
Mass = 32kg
To find acceleration;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
Making acceleration (a) the subject, we have;
[tex]Acceleration (a) = \frac{F}{m}[/tex]
Substituting into the equation;
[tex]Acceleration (a) = \frac{80}{32}[/tex]
Acceleration = 2.5m/s²
What genetic test would you get if there was a specific genetic disease in your family
Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled with some standard solution,such as aqueous potassium chloride. The conductivity of water is 76 mS m^(-1) at 25 C and the conductivity of 0.100 mol dm^(-3) KCl (aq) is 1.1639 S m^(-1) A cell has a resistance of 33.21ohm when filled with 0.100 mol dm^(-3) KCl (aq) and 300.0 ohm when filled with 0.100 mol dm CHaCOOH (aq). What is the molar conductivity of acetic acid at that concentration and temperature?
Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as = [tex]\frac{Cell constant}{Resistance }[/tex]
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
Determine the molar conductivity of acetic acid
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1
The primary reason for the path of motion of an object being a smooth curve is: Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a the third derivative of parabolas is always zero. b inertia. c tangent direction unit vectors change continuously. d calculus must have continuous derivatives to apply correctly.
Answer:
the correct answer is d
Explanation:
The laws of mechanics are related
F = m a
the acceleration of the body is given by the kinematics
a = [tex]\frac{dv}{dt}[/tex]
v = [tex]\frac{dx}{dt}[/tex]
substituting
a = \frac{d2x}{dt^2}
F = m [tex]\frac{d^2x}{dt^2}[/tex]
Therefore, in order to obtain the force (interaction of a body), continuous curves are needed and derivable from the position and the speed, for which all change in the trajectory of a body must be smooth where smooth is understood to have until the second derived.
Consequently the correct answer is d
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.8 mm above the river, whereas the opposite side is a mere 1.3 mm above the river. The river itself is a raging torrent 53.0 mm wide.
A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?
Answer:
A) 26.5 m/s
B) 33.0 m/s
Explanation:
A)
Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.So, we can apply the definition of average velocity to find this speed as follows:[tex]v_{x} = \frac{\Delta x}{\Delta t} (1)[/tex]
We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).Since the acceleration is constant, we can use the following kinematic equation, as follows:[tex]\Delta y = y_{f} - y_{o} = v_{o} * t + \frac{1}{2} * g *t^{2} (2)[/tex]
if we take the river level as our x-axis, this means that yf = 1.3 m andy₀ = 20.8 m.
At the same time, due to in the vertical direction the car has no initial velocity, this means that v₀ = 0.Replacing by the values in (2) , and solving for t:[tex]t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} } = 2 s (3)[/tex]
If we choose t₀ =0 ⇒ Δt = t = 2 sReplacing Δx and Δt in (1):[tex]v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s (4)[/tex]
B)
When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:[tex]v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s (5)[/tex]
Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:[tex]v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}} = 33.0 m/s (6)[/tex]
The (kinetic, radiant) theory explains the motion of particles in matter.
A)Kinetic
B)Radiant
Answer:
radiant
Explanation:
Answer:
kinetic
Explanation:
this is because the kinetic theory is when energy is used in motion.
Can I get help on this question I’m scared to get it wrong .
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.
Learn more:
brainly.com/question/20658056
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.
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:
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
Learn more about the acceleration: https://brainly.com/question/12134554
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.07 m/s. The stone subsequently falls to the ground, which is 19.3 m below the point were the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g = 9.81 m/s^2 for the acceleration due to gravity.
Answer:
the stone hits the gound with a speed of 20.7 m/s
Explanation:
Becuase gravity is constant we know that the initial upward velocity will be equal to the downward velocity when the stone has returned to its intal location.
diffraction of light is the __________ of light as it passes through the Edges of a barrier or a slit.
1, reflections
2, refraction
3,bending
4,absorbing
Answer:
3 bending
Explanation:
A student weighing 5.4 × 102 newtons takes 15 seconds to run up a hill. The top of the hill is 10 meters vertically above her starting point. What power does the student develop during her run?
Answer:
P = 360 Watts
Explanation:
Given that,
The weight of a student, [tex]F=5.4\times 10^2\ N[/tex]
It takes 15 seconds to run up a hill.
The top of the hill is 10 meters vertically above her starting point.
We need to find the power develop during her run. We know that te power developed is given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{5.4\times 10^2\times 10}{15}\\\\P=360\ W[/tex]
So, the power develop during her run is 360 W.
The bond order for a single covalent bond is.
A. two
B. four
C. one
D. three
Answer:
I think it should be C, which is one
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the
power consumed will be-
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
50 W
Explanation:
Case 1
Power = V * I
100 = 220 * I
I = [tex]\frac{100}{220}[/tex] A
Case 2
P = V * I
P = 110 * [tex]\frac{100}{220}[/tex]
P = 50 W
I think the answer is 50 W
Hope it helps
A 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
A virtual image larger than the object can be produced by
1) concave mitor 2) convex mutor 3)plane mirror
concave lens
O1
Answer:
concave mirror
Explanation:
Why This is Correct Because, concave mirrors only form a virtual image when a object is larger than the other. when the object is produced between the focus, object and the image object, it becomes a virtual image.
Answer:
The answer is concave mirror
How long would it take for a car to travel a distance of 220 kilometers if it is traveling at a velocity of 55 km/hr South?
Your answer:
220 hours
12,100 hours
4 hours
0.25 hours
Carousel conveyors are used for storage and order picking for small parts. The conveyorsrotate clockwise or counterclockwise, as necessary, to position storage bins at the storageand retrieval point. The conveyors are closely spaced, such that the operators travel timebetween conveyors is negligible. The conveyor rotation time for each item equals 1 minute;the time required for the operator to retrieve an item after the conveyor stops rotatingequals 0.25 minute. How many carousel conveyors can one operator tend without creatingidle time on the part of the conveyors
Answer:
the number of carousel conveyors that an operator can operate without any idle time is 5
Explanation:
Given the data in the question;
first we express the equation for number of carousel conveyors that can be operated by an operator;
n' = [tex]\frac{(a + t)}{( a + b)}[/tex]
where a is the concurrent activity time ( 0.25 minute )
b is the independent operator activity time
t is the independent machine activity time( 1 )
Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.
So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item
so
we substitute
0.25min for a, 1 for t and 0min for b
n' = [tex]\frac{(0.25min + 1min)}{( 0.25min+ 0 min)}[/tex]
n' = 1.25 min / 0.25
n' - 5
Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5
rocket fuel contains 50j of energy in its chemical store. the rocket has a mass of 1kg. what is the maximum speed the rocket could reach
acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.63 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.3 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. hardwood floor magnitude m/s2 hardwood floor duration ms carpeted floor magnitude m/s2 carpeted floor duration ms g
From what does oil form?
A. marine organisms
B. terrestrial plants
C. dinosaurs
D. lava or magma
Answer: marine organisms
Explanation:
i just took the test
One of the fastest pitches ever thrown in Major League Baseball was by Aroldis Chapman and had a velocity of 105.1 miles/hour. How many seconds did it take this pitch to travel the 60 feet and 6 inches from the pitcher's mound to home plate
Answer:
t = 0.39 s
Explanation:
Assuming that the ball is launched horizontally, once in the air, if we neglect the resistance of the air, the ball moves at a constant speed, equal to the initial velocity, in this case, 105.1 mi/hr.In order to find time in seconds, it is advisable to convert the speed in mi/hr to m/s, as follows:[tex]105.1 mi/hr * (\frac{1hr}{3600s})*\frac{1609m}{1mi} = 47.0 m/s (1)[/tex]
In the same way, it's advisable to convert 60' 6'' (60.5') to m, as follows:[tex]60.5 ft * \frac{0.3058m}{1ft} = 18.5 m (2)[/tex]
Applying the definition of average velocity, we can find the time traveled by the ball from pitcher's mound to home plate, as follows:[tex]t = \frac{18.5m}{47m/s} = 0.39 s (3)[/tex]
to see if the original results are Which career field is an applied science?
geology
biotechnology
physics
chemistry
Answer:
it is chemistry
Explanation:
The discharge of a pump is 3 m above the inlet. Water enters at a pressure of 138 kPa and leaves at a pressure of 1380 kPa. The specific volume of the water is 0.001 m3/kg. If there is no heat transfer and no change in kinetic or internal energy, what is the work per unit mass
Answer:
The answer is "[tex]1.271 \ \frac{KJ}{kg}\\[/tex]"
Explanation:
[tex]\Delta e_{mech} =\frac{P_2-P_1}{P} + \frac{v_{2}^2-v_{1}^2}{2}+g(z_2-z_1)\\\\\Delta e_{mech} =\frac{ 1380 -138 \times 1000 }{1000} + 0+g(3-0)\\\\P = \frac{1}{v}= \frac{1}{0.001} = 1000 \frac{kg}{m} \\\\ \Delta e_{mech} =1242 +9.81(3)= 1271.43 \frac{J}{kg} \\\\\text{work per unit pass}= 1.271 \ \frac{KJ}{kg}\\[/tex]
Jack D. Ripper flipped out after missing a Must-Do-It question for the third time on his Minds On Physics assignment. Outraged by the futility of his efforts, he flings a 4.0-gram pencil across the room. The pencil lodges into a 221.0-gram Sponge Bob doll which is at rest on a countertop. Once in motion, the pencil/doll combination slide a distance of 11.9 cm across the countertop before stopping. The coefficient of friction between the doll and the countertop is 0.325. Determine the speed at which the pencil is moving prior to striking Sponge Bob.
Answer:
Explanation:
Let the velocity after the 4 gram pencil strikes is v .
kinetic energy of the combination = 1/2 m v²
= .5 x ( 4 + 221 ) x 10⁻³ x v² = work done by friction
friction force acting on the combination = 225 x10⁻³x .325 x 9.8 = .7166 N
work done by friction
= .7166 x .119 = .085 J
.5 x 225 x 10⁻³ v² = .085
v² = .085 / .1125 = .7555
v = .8692 m = 86.92 cm /s
Velocity of combination after collision = 86.92 cm /s
Let velocity of pencil before collision be V
Applying law of conservation of momentum at the time of collision ,
4 x V = 225 x 86.92
V = 4889.25 cm / s
= 48.9 m /s .