Answers
Friction:
When an object slips on a surface, an opposing force acts between the tangent planes which acts in the opposite direction of motion. This opposing force is called Friction. Or in other words, Friction is the opposing force that opposes the motion between two surfaces.
The main component of friction are:
Normal Reaction (R):
Suppose a block is placed on a table in the above picture, which is in resting state, then two forces are acting on it at that time.
The first is due to its weight mg which is working from its center of gravity towards the vertical bottom.
The second one is superimposed vertically upwards by the table on the block, called the reaction force (P). This force passes through the center of gravity of the block.
Due to P = mg, the box is in equilibrium position on the table.
Coefficient of friction ( μ ):
The ratio of the force of friction and the reaction force is called the coefficient of friction.
Coefficient of friction, µ = force of friction / reaction force
μ = F / R
The coefficient of friction is volume less and dimensionless.
Its value is between 0 to 1.
Advantage and disadvantage from friction force:
The advantage of the force of friction is that due to friction, we can walk on the earth without slipping. Brakes in all vehicles are due to the force of friction. We can write on the board only because of the force of friction. The disadvantage of this force is that due to friction, some parts of energy are lost in the machines and there is wear and tear on the machines.How to reduce friction:
Using lubricants (oil or grease) in machines. Friction can be reduced by using ball bearings etc. Using a soap solution and powder.
Related Questions
A person standing on a cliff extends their arm past the cliff's edge and throws a stone with velocity 16.0 m/s downward. The stone is 80.0 m above ground when it leaves the person's hand. a) Draw a diagram of the problem and label the known and unknown quantities. b) When will the stone land? c) What is the stone's velocity when it lands? Show your work
Answers
Answer:
b) 6.03 seconds
c) 43.164 m/s
Explanation:
t = Time taken
u = Initial velocity = 16 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v=u+at\\\Rightarrow 0=16-9.81\times t\\\Rightarrow \frac{-16}{-9.81}=t\\\Rightarrow t=1.63 \s[/tex]
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=16\times 1.63+\frac{1}{2}\times -9.81\times 1.63^2\\\Rightarrow s=13.05\ m[/tex]
So, the stone would travel 13.05 m up
So, total height of the stone would fall is 13.05+80 = 93.05 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 93.05=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{93.05\times 2}{9.81}}\\\Rightarrow t=4.4\ s[/tex]
b) The stone will land 1.63+4.4 = 6.03 seconds later
[tex]v=u+at\\\Rightarrow v=0+9.81\times 4.4\\\Rightarrow v=43.164\ m/s[/tex]
c) The stone's velocity when it lands is 43.164 m/s
When a car crashes, its velocity changes from, say 60 mph to 0 mph in a short amount of time. The driver of the car also has the same change in velocity (assuming the driver is strapped in). When an airbag deploys in a crash, the car and driver have the same change in velocity.
How do airbags protect the driver in a crash, then?
Answers
Answer with Explanation:
The force that is exerted on a object is proportional to the rate of change of momentum. Mathematically
[tex]F=\frac{\Delta p}{\Delta t}[/tex]
As we can see from the above equation the force that is exerted on the object is inversely proportional to the time in which this momentum change is brought meaning if the time interval in which the momentum change is brought about is larger smaller the force that acts on the object.
When the driver of a car crashes into a barrier the airbags in the car are deployed instantly as due to inertia of motion the driver continues to move in the original direction of motion thus hitting the air bag in the process. Since the air in the airbag is compressible it increases the time in which this momentum becomes zero, reducing the force that is exerted on the person, thus protecting the person.
Airbags reduce the force on a driver's head by increasing the time over which the head decelerates, thus applying the principle of impulse to protect during a car crash. They achieve this in tandem with seatbelts and car structures designed to crumple and absorb impact forces.
Airbags protect drivers during a crash by utilizing the principle of impulse, which is the product of the net force acting on an object and the time over which the force acts. The airbag increases the time over which the driver's head decelerates to a stop, in turn decreasing the force that acts on the head. By extending the stopping time from a fraction of a second (if hitting the dashboard) to a longer time (momentum is stopped by the airbag), the average force experienced by the driver's head is considerably lessened. This reduction in force reduces the likelihood of severe injuries. The airbag also has vents to deflate during the crash, preventing bounce-back injuries.
Further, the airbag works with seatbelts that may have variable tension to distribute forces across the body more evenly, thus reducing stress on any single part of the body. Modern vehicles include these safety features not just for comfort but for crucial safety improvements, allowing for a longer time of force application during a crash and therefore less force at any instant. The same concept applies to the car's structure with plastic components designed to crumple and prolong collision time, further reducing forces experienced by the passengers.
A taxi traveling along a straight section of road starts from rest, accelerating at 2.00 m/s^2 until it reaches a speed of 29.0 m/s. Then the vehicle travels for 87.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. How long is the taxi in motion (in s)? What is the average velocity of the taxi for the motion described? (Enter the magnitude in m/s.)
Answers
Answer:
a) The taxi is 107 s in motion
b) The average velocity is 26.2 m/s
Explanation:
First, the car travels with an acceleration of 2.00 m/s². The equations for position and velocity that apply for the car are:
x = x0 + v0 t + 1/2 a t²
v = v0 + a t
where
x = position at time t
x0 = initial position
v0 = initial speed
t = time
a = acceleration
v = speed
Let´s calculate how much distance and for how long the taxi travels until it reaches a speed of 29.0 m/s:
Using the equation for velocity:
v = v0 + a t
v - v0 / a = t
(29.0 m/s - 0 m/s) / 2 m/s² = t
t = 14.5 s
Then, in the equation for position:
x = x0 + v0 t + 1/2 a t²
x = 0 + 0 + 1/2 * 2.00 m/s² * (14.5 s)²
x = 210 m
Then, the vehicle travels at constant speed for 87 s. The distance traveled will be:
x = v * t
x = 29.0 m/s * 87.0 s = 2.52 x 10³ m
Lastly the car stops (v = 0) in 5 s. In this case, the car has a constant negative acceleration:
Using the equation for velocity:
v = v0 + a t
if v=0 in 5 s, then:
0 = 29.0 m/s + a * 5.00 s
a = -29.0 m/s / 5.00 s
a = -5.80 m/s²
Using now the equation for the position, we can calculate how far has the taxi traveled until it came to stop:
x = x0 + v0 t + 1/2 a t²
x = 0 + 29.0 m/s * 5.00 s -1/2 * 5.80 m/s² * (5.00s)²
x = 72.5 m
a) The taxi has been in motion for:
Total time = 14.5 s + 87.0 s + 5.00s = 107 s
Note that we have always used x0 = 0, then, we have calculated the displacement for each part of the trip.
Adding all the displacements, we will get the total displacement:
Total displacement = 210 m + 2.52 x 10³ m + 72.5 m = 2.80 x 10³ m
Average speed = total displacement / total time
Average speed = 2.80 x 10³ m / 107 s = 26.2 m/s
Abel is not sure why, from h = vi2 sin2 θi 2g , the height the tennis ball reaches is maximum when θi = 90°, and asks Kato to explain. Which of Kato's responses is correct? "When θi = 90°, sin2 θi = sin(2θi) = 1, which is its maximum value, so h is maximum." "When θi = 90°, sin2 θi is maximum, so h is maximum." "When θi = 90°, sin θi = 1, sin2 θi = 2 · 1 = 2, which is its maximum value, so that means h is maximum." "When θi = 90°, sin2 θi is minimum, so h is maximum."
Answers
The correct response from Kato is: 'When θi = 90°, sin^2 θi = sin(2θi) = 1, which is its maximum value, so h is maximum.
Explanation:
When a tennis ball is thrown into the air at an angle, its height can be calculated using the equation:
h = (vi^2 * sin^2(θi)) / (2g)
where:
h is the heightvi is the initial velocityθi is the launch angleg is the acceleration due to gravityThe term sin^2(θi) represents the square of the sine of the launch angle.
When the launch angle is 90°, sin^2(θi) equals 1, which is its maximum value. This means that the height the tennis ball reaches is maximum when the launch angle is 90°.
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Kato's second response, "When θi = 90°, sin^2(θi) is maximum, so h is maximum," is the correct explanation.
Explanation:
To determine the maximum height of a tennis ball, we can use the equation:
h = (vi^2 * sin^2(θi)) / (2g)
where h is the maximum height, vi is the initial velocity, θi is the launch angle, and g is the acceleration due to gravity.
Let's analyze Kato's responses:
"When θi = 90°, sin^2(θi) = sin(2θi) = 1, which is its maximum value, so h is maximum."This response is incorrect. The equation sin^2(θi) = sin(2θi) is not valid. The maximum value of sin^2(θi) is 1, but it does not equal sin(2θi)."When θi = 90°, sin^2(θi) is maximum, so h is maximum."
This response is correct. When the launch angle is 90°, sin^2(θi) equals 1, which is its maximum value. This means that the term (vi^2 * sin^2(θi)) in the equation is maximized, resulting in the maximum height of the tennis ball."When θi = 90°, sin θi = 1, sin^2(θi) = 2 · 1 = 2, which is its maximum value, so that means h is maximum."
This response is incorrect. The value of sin θi is 1 when θi = 90°, but sin^2(θi) is not equal to 2. The maximum value of sin^2(θi) is 1."When θi = 90°, sin^2(θi) is minimum, so h is maximum."
This response is incorrect. When the launch angle is 90°, sin^2(θi) is maximum, not minimum.
Therefore, Kato's second response, "When θi = 90°, sin^2(θi) is maximum, so h is maximum," is the correct explanation.
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A baseball catcher is performing a stunt for a television commercial. He will catch a baseball (mass 145 g) dropped from a height of 66.0 m above his glove. His glove stops the ball in 0.0118 s. What is the force (in N) exerted by his glove on the ball? (Indicate the direction with the sign of your answer. Assume the baseball is traveling in the negative direction.)
Answers
Answer:
Explanation:
The ball is going down with velocity. It must have momentum . It is stopped by
catcher so that its momentum becomes zero . There is change in momentum . So force is applied on the ball by the gloves .
The rate of change of momentum gives the magnitude of force. This force must be in upward direction to stop the ball. So force is in positive direction .
Let us measure the force applied on the ball .
Final velocity after the fall by 66 m
V = √ 2gh
= √ 2x9.8 x 66
35.97 m /s
Momentum = m v
0.145 x 35.97
= 5.2156 kgms⁻¹
Change in momentum
= 5.2156 - 0
= 5.2156
Rate of change of momentum
= Change of momentum / time = 5.2156 / .0118
= 442 N
if R be the force exerted by gloves to stop the ball
R - mg represents the net force which stops the ball so
R - mg = 442
R = 442 + mg
= 442 + .145 x 908
443.43 N
According to Newton's second law of motion rate of change of momentum is equal to the applied force. The force exerted by the glove of the player on the ball will be 497.4 N.
What is Newton's second law of motion?According to Newton's second law of motion rate of change of momentum is equal to the applied force.
Momentum is given by the product of mass and velocity. it is denoted by P.it leads to the impulsive force.
P = mv
ΔP = mΔv
[tex] \rm{ F = \frac{\delP}{\delt} } [/tex]
[tex]V = \sqrt{2gh} [/tex]
[tex]\rm{V = \sqrt{2\times9.81\times60} [/tex]
v = 34.31 m/sec
ΔV = v -u
ΔV = 34.31-0
ΔV = 34.31 m /sec.
Δt = 0.0118 sec
[tex]\rm{\frac{\delv}{\delt} } [/tex]= 34.31
F = [tex]m\rm{\frac{\delv}{\delt} }[/tex]
F = [tex] 0.145\times3430 [/tex]
F = 497.4 N
Thus the force exerted by the glove of the player on the ball will be 497.4 N.
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A charge 5.00 nC is placed at the origin of an xy-coordinate system, and a charge -1.97 nC is placed on the positive x-axis at x = 4.00 cm . A third particle, of charge 6.05 nC is now placed at the point x = 4.00 cm , y= 3.02 cm . Q: Find the y-component of the total force exerted on the third charge by the other two
Answers
Answer:
The total force exerted on the Y axis is: -52.07μC
Explanation:
This is an electrostatic problem, so we will use the formulas from the Coulomb's law:
[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]
We are interested only of the effect of the force on the Y axis. We can notice that the charge placed on the x=4cm will exers a force only on the Y axis so:
[tex]Fy1=9*10^9*\frac{6.05*10^{-9}*(-1.97)*10^{-9}}{(3.02*10^{-2})^2}\\[/tex]
Fy1=-117.61μC
For the charge placed on the origin we have to calculate the distance and the angle:
[tex]r=\sqrt{(4*10^{-2}m)^2 +(3.02*10^{-2}m)^2} \\r=5cm=0.05m[/tex]
we can find the angle with:
[tex]alpha = arctg(\frac{3.02cm}{4cm})=37^o[/tex]
The for the Force on Y axis is:
[tex]Fy2=9*10^9*\frac{6.05*10^{-9}*(5)*10^{-9}}{(3.02*10^{-2})^2}*sin(37^o)\\[/tex]
Fy2=65.54μC
The total force exerted on the Y axis is:
Fy=Fy1+Fy2=-52.07μC
A helium-neon laser beam has a wavelength in air of 633 nm. It takes 1.38 ns for the light to travel through 32.0 cm of an unknown liquid. What is the wavelength of the laser beam in the liquid?
Answers
Answer:
Wavelength = 489.52 nm
Explanation:
Given that the wavelength of the light = 633 nm
The time taken by the light in unknown liquid = 1.38 ns
Also,
1 ns = 10⁻⁹ s
So, t = 1.38 × 10⁻⁹ s
Also,
Distance = 32.0 cm = 0.32 m ( 1 cm = 0.01 m)
So, speed of the light in the liquid = Distance / Time = 0.32 / 1.38 × 10⁻⁹ m/s = 2.32 × 10⁸ m/s
Frequency of the light does not change when light travels from one medium to another. So,
[tex]\frac {V_{air}}{\lambda_{air}}=\frac {V_{liquid}}{\lambda_{liquid}}[/tex]
[tex]{V_{air}}=3\times 10^8\ m/s[/tex]
[tex]{\lambda_{air}}=633\ nm[/tex]
[tex]{V_{liquid}}=2.32\times 10^8\ m/s[/tex]
[tex]{\lambda_{liquid}}=?\ nm[/tex]
So,
[tex]\frac {3\times 10^8\ m/s}{633\ nm}=\frac {2.32\times 10^8\ m/s}{\lambda_{liquid}}[/tex]
Wavelength = 489.52 nm
Wavelength is the distance between two points of the two consecutive waves.
The wavelength of the laser beam in the liquid is 489.52 nm.
What is the wavelength of the wave?Wavelength is the distance between two points of the two consecutive waves.
Given information-
The helium-neon laser beam has a wavelength in air of 633 nm.
The time taken by the helium laser beam to to travel through 32.0 cm or 0.32 m of an unknown liquid is 1.38 ns or [tex]1.38\times10^{-9}\rm s[/tex].
Speed of the light is the ratio of distance traveled by it in the time taken. Thus the speed of the given light is,
[tex]v=\dfrac{0.32}{1.38}\\v=2.32\times10^8 \rm m/s[/tex]
Frequency of the wave is the ratio of speed and wavelength of the wave.
As the frequency of the wave is equal for each medium. Thus,
[tex]f=\dfrac{v_{liq}}{\lambda_{liq}} =\dfrac{v_{air}}{\lambda_{air}}[/tex]
As the speed of the air is [tex]3\times10^8[/tex] m/s.
Thus put the values in the above ratio as,
[tex]\dfrac{2.32\times10^8}{\lambda_{liq}} =\dfrac{3\times10^8}{633}\\\lambda_{liq}=489.52 \rm nm[/tex]
Thus the wavelength of the laser beam in the liquid is 489.52 nm.
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A boy pulls with a 20 N force, at a 20 degree incline. What part of the force moves the wagon?
Answers
Answer:
horizontal direction force move wagon at 18.79 N
Explanation:
given data
force F = 20 N
angle = 20 degree
to find out
What part of the force moves the wagon
solution
we know here as per attach figure
boy pull a wagon at force 20 N at angle 20 degree
so there are 2 component
x in horizontal direction i.e F cos20
and y in vertical direction i.e F sin20
so we can say
horizontal direction force is move the wagon that is
horizontal direction force = F cos 20
horizontal direction force = 20× cos20
horizontal direction force = 18.79 N
so horizontal direction force move wagon at 18.79 N
Some homes that use baseboard heating use copper tubing. Hot water runs through and heats the copper tubing, which in turn heats aluminum fins. It is actually the aluminum fins that heat the air rising through the fins. How much energy would it take to heat a section of the copper tubing that weighs about 715.0 g , from 13.65 ∘C to 26.60 ∘C ? Copper has a specific heat of 0.3850 J/(g⋅∘C).
Answers
Answer:
19 degrees is the answer
Explanation:
A freight train has a mass of 1.5 X 10^7 kg. If the
locomotivecan exert a constant pull of 7.5 X1^5 N, how long does it
take toincrease the speed of the train from rest to 80 km/h?
Answers
Answer:
t=444.4s
Explanation:
m=1.5*10^7 kg
F=7.5*10^5 N
v=80km/h*(1h/3600s)*(1000m/1km)=22.22m/s
Second Newton's Law:
F=ma
a=F/m=7.5*10^5/(1.5*10^7)=0.05m/s^2
Kinematics equation:
vf=vo+at=at
vo: initial velocity equal zero
t=vf/a=22.22/0.05=444.4s
To calculate the time it takes to increase the speed of a train from rest to 80 km/h, you can use Newton's second law of motion.
Explanation:To calculate the time it takes to increase the speed of the train from rest to 80 km/h, we can use Newton's second law of motion. First, we need to calculate the force required to accelerate the train. Given the mass of the train, 1.5 x 10^7 kg, and the acceleration, we can use the formula: force = mass x acceleration
force = (1.5 x 10^7 kg) x (80 km/h to m/s conversion)
Next, we can use the formula: force = mass x acceleration to find the time it takes to accelerate the train:
time = force / (7.5 x 10^15 N)
Plugging in the values, we can calculate the time it takes to increase the speed of the train from rest to 80 km/h.
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An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.2 × 10^-7 C/m^2, and the plates are separated by a distance of 1.3 × 10^-2 m. How fast is the electron moving just before it reaches the positive plate?
Answers
Answer:[tex]1.066\times 10^7 m/s[/tex]
Explanation:
Given
Charge per unit area on each plate([tex]\sigma [/tex])=[tex]2.2\times 10^{-7}[/tex]
Plate separation(y)=0.013 m
and velocity is given by
[tex]v^2-u^2=2ay[/tex]
where a=acceleration is given by
[tex]a=\frac{F}{m}=\frac{eE}{m}[/tex]
e=charge on electron
E=electric field
m=mass of electron
[tex]E=\frac{\sigma }{\epsilon _0}[/tex]
[tex]a=\frac{e\sigma }{m\epsilon _0}[/tex]
substituting values
[tex]v=sqrt{\frac{2e\sigma y}{m\epsilon _0}}[/tex]
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 2.2\times 10^{-7}\times 0.013}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}[/tex]
[tex]v=1.066\times 10^7 m/s[/tex]
The velocity of an electron just before it reaches the positive plate of a parallel plate capacitor can be calculated by equating the converted potential energy to kinetic energy, and then solving for velocity using the electric field, the charge and mass of the electron.
Explanation:Calculating the Final Velocity of an Electron in a Parallel Plate Capacitor
To find out how fast an electron is moving just before it reaches the positive plate of a parallel plate capacitor, we can use the concepts of electric fields and potential energy. The electric field E between the plates can be found using the charge per unit area σ (sigma) and the vacuum permittivity ε₀ (epsilon nought), given by E = σ / ε₀. Once the electric field is known, we can determine the force on the electron as F = qE, where q is the charge of the electron.
Since the electron starts from rest, the potential energy at the negative plate is converted entirely into kinetic energy just before it hits the positive plate. We can use the electron's charge and the potential difference (V) between the plates to find this energy: qV. We know that the kinetic energy is ½mv², where m is the mass of the electron and v is the final velocity.
Setting the potential energy equal to the kinetic energy, we get qV = ½mv². Solving for the final velocity, v, we find that v = √(2qV/m). Since the potential difference V can be calculated as the product of electric field E and the distance d between the plates, we can substitute V with Ed. The final equation for velocity is therefore v = √(2qEd/m). This will give us the velocity of the electron just before it reaches the positive plate.
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the ball along the incline is constant at 0.25g, directed down the incline. If the ball is re- leased with a speed of 4 m /s, determine the distance s it moves up the incline before reversing its direc- tion and the total time t required for the ball to re- turn to the child’s hand.
Answers
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
In the return stroke of a typical lightning bolt, a current of 7.01×10^4 A (70.1 kA). exists for 20 μs. How much charge is transferred in this event ?
Answers
Answer:
Charge, q = 1.402 C
Explanation:
Given that,
Current from lightning bolt, [tex]I=7.01\times 10^4\ A[/tex]
Time, t = [tex]t=20\ \mu s=2\times 10^{-5}\ s[/tex]
Let q is the charge transferred in this event. We know that the total charge divided by time is called current. Mathematically, it is given by :
[tex]I=\dfrac{q}{t}[/tex]
[tex]q=I\times t[/tex]
[tex]q=7.01\times 10^4\times 2\times 10^{-5}[/tex]
q = 1.402 C
So, 1.402 coulomb of charge is transferred in this event. Hence, this is the required solution.
An electrostatic paint sprayer has a potential of 25.0 kV that repels paint droplets onto a grounded object. What charge must a 0.63 mg drop of paint have to arrive at the object with a speed of 9 m/s? Give answer in terms of 10^-10 C.
Answers
Answer:
10⁻⁶ C
Explanation:
Let the charge be Q
Gain of energy in electric field
= potential x charge
= 25 x 10³ x Q
Kinetic energy of droplet
= .5 x .63 x 10⁻³ x 9 x 9
= 25.515 x 10⁻³ J
So , equating the energy gained
25 x 10³ x Q = 25.515 x 10⁻³
Q = 10⁻⁶ C
A skateboarder rolls off a horizontal ledge that is 1.32 m high, and lands 1.88 m from the base of the ledge. What was his initial velocity? (Unit = m/s)
Answers
Answer:
initial velocity is 3.62 m/s
Explanation:
given data
high = 1.32 m
length = 1.88 m
to find out
initial velocity
solution
we consider here top height point a and b point at ground and c point at distance 1.88 m away from b on ground
and x is horizontal component and y is vertical component
so at point A initial velocity is Va = Vx i
and at point c velocity = Vc = Vx i + Vy j
first we calculate time taken when it come down by distance formula
distance = 1/2 ×gt² ..............1
1.32 = 1/2 ×(9.8)t²
t = 0.519 sec
so velocity x = distance / time
velocity x = [tex]\frac{1.88}{0.519}[/tex]
velocity x = 3.622 m/s
so initial velocity is 3.62 m/s
Answer:
3.62 m/s
Explanation:
h = 1.32 m
d = 1.88 m
Let u be the initial horizontal velocity and the time taken by the board to reach the ground is t.
Use second equation of motion in vertical direction
[tex]s=ut+0.5at^{2}[/tex]
1.32 = 0 + 0.5 x 9.8 x t^2
t = 0.52 second
The horizontal distance = horizontal velocity x time
1.88 = u x 0.52
u = 3.62 m/s
Thus, the nitial speed of the skate board is 3.62 m/s.
A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.60 g/cm3 at one end to 18.5 g/cm3 at the other end. This linearly increasing density can be described using the equation ???? = B + Cx. (a) Find the constants B and C required for this rod, assuming the low-density end is placed at x = 0 cm and the high-density end is at x = 12 cm. (b) The mass of the rod can be found using:m=dV=Adx=(B+Cx)Adx
Answers
Answer:
(a) The constants required describing the rod's density are B=2.6 and C=1.325.
(b) The mass of the road can be found using [tex]A\int_0^{12}\left(B+Cx)dx[/tex]
Explanation:
(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by
[tex]2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}[/tex]
recalling that the coordinate x is measured in centimeters.
(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:
[tex]m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6[/tex],
hence, the mass of the rod is 126.6 g.
The density of an alloy is 5000 kg/m3. What is the density in g/cm 0.005 @ 0.500 5.000 50.000
Answers
Answer:
5 g/cm³
Explanation:
Density of an alloy = 5000 kg/m³
The Density of a material is the ratio of the mass by the volume.
The units here are of the metric system
Mass is the resistance a body has which opposes motion when force is applied.
Volume is the amount of material in an object
Convert kg/m³ to g/cm³
1 kg = 1000 g
1 m³ = 10⁻⁶ cm
[tex]5000\ kg/m^3=5000\times 1000\times 10^{-6}=5\ g/cm^3[/tex]
∴ 5000 kg/m³ = 5 g/cm³
.In the Highscope curriculum _____and _____ are active partners in shaping the educational experience
A. Curriculum, lesson plan
B. Teachers, students
C. Directors, teachers
D. Community members, teachers
Answers
Answer:
D. Community members, teachers
Explanation:
In the HighScope Curriculum teachers work in collaboration with family members, thus encouraging greater learning in students. They do this by providing information about the curriculum, inviting family members to participate in the activities carried out in the classroom, workshops for parents are also held. This allows discussing children's progress and sharing ideas to extend classroom learning to home.
Calculate that electric field that will apply a force of 25,000 N to an electron.
Answers
Answer:
[tex]E=1.56\times 10^{23}\ N/C[/tex]
Explanation:
Given that,
Electric force applied to the electron, [tex]F=25000\ N[/tex]
Charge on electron, [tex]q=1.6\times 10^{-19}\ C[/tex]
We need to find the electric force acting on the electron. The electric field is given by :
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{25000}{1.6\times 10^{-19}}[/tex]
[tex]E=1.56\times 10^{23}\ N/C[/tex]
So, the electric field acting on the electron is [tex]1.56\times 10^{23}\ N/C[/tex]. Hence, this is the required solution
Why does an astronaut in a spacecraft orbiting Earthexperience
a feeling of weightlessness?
Answers
Answer:
Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.
while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.
Both spacecraft and the astronauts both are in a free-fall condition.
A uniform continuous line charge with net positive charge Q and length L lies on the x-axis from −L2 to +L2. This problem asks about the electric field at a point on the +y-axis: (0,a). continuous line charge
(1) What is the direction of the x-component of the electric field at (0,a)?
(A) +x(to the right)
(B) -x (to the left)
(C) zero (there is no net horizontal component of the E-field)
Answers
Answer:
(C) zero (there is no net horizontal component of the E-field)
Explanation:
If we subdivide the bar into small pieces, each piece (dx) contains a charge (dq), the electric field of each piece is equivalent to the field of a punctual electric charge, and has a direction as shown in the attached figure. For each piece (dx) in the negative axis there is another symmetric piece (dx) in the positive axis, and as we see in the figure for symmetry the sum of their electric fields gives a resultant in the Y axis (because its components in X are cancelled by symmetry).
Then the resultant of the electric field will be only in Y.
(C) zero (there is no net horizontal component of the E-field)
The x-component of the electric field at point (0,a) due to a uniform continuous line charge on the x-axis is zero, due to the symmetrical distribution of charge and corresponding cancellation of horizontal electric field components.
Explanation:The student is asking about the direction of the electric field at a point on the positive y-axis due to a uniform continuous line charge distributed along the x-axis. To find the direction of the x-component of the electric field at the point (0,a), we can consider the symmetry of the charge distribution. For any small element of charge on the positive side of the x-axis, there is an identical element of charge on the negative side at the same distance from the origin. The electric fields produced by these two elements at point (0,a) on the y-axis will have the same magnitude but opposite x-components. These x-components will cancel each other out, resulting in a net x-component of the electric field being zero. Therefore, the correct answer to the student's question is (C) zero (there is no net horizontal component of the electric field).
A 1500 kg vehicle is moving on a dry pavement where the coefficient of friction is 0.80. If the vehicle is traveling at 25 m/s, what minimum distance is required to stop it when the brakes are applied. a. 40 m
b. 55 m
c. 72 m
d. 64 m
Answers
Answer:
The minimum distance is 40 m.
(a) is correct option.
Explanation:
Given that,
Mass of vehicle = 1500 kg
Coefficient friction = 0.80
Speed = 25 m/s
We need to calculate the acceleration
Using frictional force
[tex]f=\mu mg[/tex]
Put the value into the formula
[tex]ma=\mu mg[/tex]
[tex]a = \mu g[/tex]
We need to calculate the distance
Using equation of motion
[tex]v^2=u^2+2as[/tex]
Put the value into the formula
[tex]0=25^2-2times\mu g\times s[/tex]
[tex]s=\dfrac{25^2}{2\times0.80\times9.8}[/tex]
[tex]s=39.8 = 40 m[/tex]
Hence, The minimum distance is 40 m.
A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. The distance the wagon travels is: a) 32 m b) 1.7m c) 0.39 m d) 3.4 m
Answers
Answer:
Option d)
Solution:
As per the question:
Work done by farm hand, [tex]W_{FH} = 972J[/tex]
Force exerted, F' = 310 N
Angle, [tex]\theta = 23^{\circ}[/tex]
Now,
The component of force acting horizontally is F'cos[tex]\theta[/tex]
Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.
Thus
[tex]W_{FH} = \vec{F'}.\vec{d}[/tex]
[tex]972 = 310\times dcos23^{\circ}[/tex]
d = 3.406 m = 3.4 m
The distance the wagon travels is 3.4 m. So the correct option is d.
Explanation:To find the distance the wagon travels, we need to use the work-energy principle. Since the force is applied at an angle, we can find the horizontal component of the force by multiplying the force by the cosine of the angle. The work done is equal to the force multiplied by the distance, so we can rearrange the equation to solve for distance. Therefore, the distance the wagon travels is equal to the work done divided by the force component, or 972 J / (310 N * cos(23°)) = 3.4 m.
Learn more about Work done here:https://brainly.com/question/35917320
#SPJ11
An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (average), calculate the velocity (magnitude) of the plane relative to the ground.B) Calculate the velocity (direction) of the plane relative to the ground.C) Calculate how far from its intended position will it be after 11 min if the pilot takes no corrective action.
Answers
Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position
Explanation:
Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where
see fig 1
Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then
90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,
this because we have an isosceles triangle, then the cathetus length is
hypotenuse/[tex]\sqrt{2}[/tex]
using Pythagoras, here the hypotenuse is 90, then the cathetus are of length
90/[tex]\sqrt{2}[/tex] km/h= 63,6396 km/h.
Now the total speed of the plane is
690km/h south + 63,6396 km/h north +63,6396 km/h east,
this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.
then using again Pythagoras we get the magnitude of the total speed it is
[tex]\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h [/tex],
the direction is calculated with respect to the south using trigonometry, we know the
sin x= cathetus opposed / hypotenuse,
then
x= [tex]sin^{-1}'frac{63,6396}{629,5851}[/tex]=5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form
626,6396 km/h south + 63,6396km/h east.
Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get
Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,
since 11 min=11/60 hours=0.1833 hours.
Then the total detour from the expected position, the one it should have without the influence of the wind, we get
Xdetour=[/tex]\sqrt{2* 11,6672x^{2} }[/tex] = 16,5km at 45 degrees from east pointing north
The situation is sketched as follows see fig 2
A bug lands on a windshielf wiper. Explain why the bug is more likely to be dislodged when the wipers are turned on at the high rather than low setting.
Answers
Answer and Explanation:
The motion of the viper is circular and it completes a semi- circle forming an arc and then retraces its path back.
Thus the force experienced by the bug is centripetal force thus centripetal acceleration and in the absence of this force the bug will get dislodged.
This centripetal force is mainly provided by the static friction between the blades and the bug.
When the wipers are moving with high velocity when turned on, larger centripetal force is required to keep the bug moving with the wiper on the arc than at low setting.
Thus there are more chances for the bug to be dislodged at higher setting than at low setting.
A 24 cm candle is placed 6 m in front of a thin diverging lens with a focal length magnitude of 3 m. Find the image's height and classify it as real or virtual, upright or inverted, enlarged or reduced.
Answers
Answer:
Image is virtual and formed on the same side as the object. 2 m from the lens.
The size of the image is 7.97 cm
Image is upright as the magnification is positive and smaller than the object.
Explanation:
u = Object distance = 6 m
v = Image distance
f = Focal length = -3 m (concave lens)
[tex]h_u[/tex]= Object height = 24 cm
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-3}-\frac{1}{6}\\\Rightarrow \frac{1}{v}=\frac{-1}{2} \\\Rightarrow v=\frac{-2}{1}=-2\ m[/tex]
Image is virtual and formed on the same side as the object. 2 m from the lens.
Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-2}{6}\\\Rightarrow m=0.33[/tex]
[tex]m=\frac{h_v}{h_u}\\\Rightarrow 0.33=\frac{h_v}{0.24}\\\Rightarrow h_v=0.33\times 0.24=0.0797\ m[/tex]
The size of the image is 7.97 cm
Image is upright as the magnification is positive and smaller than the object.
The radii of curvature of a biconvex lens are 4 cm and 15 cm. The lens is in air, its index of refraction is 1.5. An object is at 1 m before the front surface of the lens. Calculate the distance of the image from the back surface of the lens.
Answers
Answer:
-1.19 m
Explanation:
R1 = + 4 cm
R2 = - 15 cm
n = 1.5
distance of object, u = - 1 m
let the focal length of the lens is f and the distance of image is v.
use lens makers formula to find the focal length of the lens
[tex]\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]
By substituting the values, we get
[tex]\frac{1}{f}=\left ( 1.5-1 \right )\left ( \frac{1}{4}+\frac{1}{15} \right )[/tex]
[tex]\frac{1}{f}=\frac{19}{120}[/tex] .... (1)
By using the lens equation
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]\frac{19}{120}=\frac{1}{v}+\frac{1}{1}[/tex] from equation (1)
[tex]\frac{1}{v}=\frac{19-120}{120}[/tex]
v = -1.19 m
The gravitational force of attraction between two 850,000 kg rocks that are 2.40 meters apart is ____ N.
Answers
Answer:
gravitational force between two objects is 8.37 N
Explanation:
given data
mass of object = 85000 kg
distance between two object r = 2.40 m
to find out
gravitational force between two objects
solution
we know that gravitational constant G is 6.67 × [tex]10^{-11}[/tex] m³/ s²-kg
so
gravitational force between two objects formula is
F = [tex]G*\frac{m1m2}{r^2}[/tex]
here E is gravitational force and G is gravitational constant put here all value and m1 and m2 are mass of object
F = [tex]6.67*10^{-11}*\frac{85000^2}{2.40^2}[/tex]
F = 8.37 N
gravitational force between two objects is 8.37 N
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 40.0 m west to a bench where you sit and watch the sunrise. It takes you 28.0 s to walk from your house to the windmill and then 42.0 s to walk from the windmill to the bench. For the entire trip from your front door to the bench, what is your average velocity?
For the entire trip from your front door to the bench, what is your average speed?
Answers
Final answer:
The average velocity for the trip from the front door to the bench is 0.143 m/s due east, and the average speed for the same trip is 1.286 m/s.
Explanation:
To calculate the average velocity for the entire trip, we need to consider the displacement (final position relative to the starting position) and the total time taken for the trip. The displacement is 10.0 m east (50.0 m due east - 40.0 m back west), and the total time is 28.0 s + 42.0 s, which equals 70.0 s. Therefore, the average velocity is displacement divided by time, calculated as follows:
Average Velocity = Displacement/Total Time = 10.0 m / 70.0 s = 0.143 m/s due east.
To calculate the average speed, we consider the total distance traveled and the total time. The distance is 50.0 m east + 40.0 m west = 90.0 m, and the time is the same 70.0 s. Thus, the average speed is:
Average Speed = Total Distance/Total Time = 90.0 m / 70.0 s = 1.286 m/s.
A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your answer to three significant figures and include the appropriate units
Answers
Answer:
The kangaroo was 1.164s in the air before returning to Earth
Explanation:
For this we are going to use the equation of distance for an uniformly accelerated movement, that is:
[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]
Where:
x = Final distance
xo = Initial point
Vo = Initial velocity
a = Acceleration
t = time
We have the following values:
x = 1.66m
xo = 0m (the kangaroo starts from the floor)
Vo = 0 m/s (each jump starts from the floor and from a resting position)
a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)
t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.
Now replace the values in the equation
[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]
[tex]1.66 = 0 + 0t + \frac{1}{2}9.8t^2[/tex]
[tex]1.66 = 4.9t^2[/tex]
[tex]\frac{1.66}{4.9} = t^2[/tex]
[tex]\sqrt{0.339} = t\\ t = 0.582s[/tex]
It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is
t = 0.582s + 0.582s
t = 1.164s
The kangaroo was 1.164s in the air before returning to Earth