Answer:
The car stops after 32.58 m.
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 20 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -6 m/s²
Time taken by the car to stop
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20}{-6}\\\Rightarrow t=3.33\ s[/tex]
Total Time taken by the car to stop is 0.5+3.33 = 3.83 s
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 3.83+\frac{1}{2}\times -6\times 3.83^2\\\Rightarrow s=32.58\ m[/tex]
The car stops after 32.58 m.
Distance between car and obstacle is 50-32.58 = 17.42 m
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
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
The density of an alloy is 5000 kg/m3. What is the density in g/cm 0.005 @ 0.500 5.000 50.000
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
What is the electric field in N/C) at a point where the force on a -2.1 x 10-5 C charge is (4.31 -6.89) x 10-6 N?
Answer: 0.123 N/C
Explanation: In order to solve this question we have to use the electric Force on a particle produced by an electric field which is given by:
F=q*E
so E=F/q= -2,58* 10^-6/-2.1*10^-5= 0.123 N/C
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.
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 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
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.
If a marathon runner averages 8.83 mi/h, how long does it take him or her to run a 26.22-mi marathon? Express your answers in h, min and s.
Answer:
t=2.97h
t= 10692s
t= 178.2 min
Explanation:
We propose the following ratio:
[tex]\frac{t}{26.22mi} =\frac{1h}{8.83mi}[/tex]
[tex]t=\frac{1h*26.22mi}{8.883mi}[/tex]
t=2.97h
Equivalences
1h=3600s
1h=60 min
Calculation of t in minutes (min) and seconds(s):
t=2.97h*3600s/h= 10692s
t=2.97h*60min/h= 178.2 min
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).
Answer:
19 degrees is the answer
Explanation:
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
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 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.
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.
Calculate that electric field that will apply a force of 25,000 N to an electron.
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
A 2450-Hz sound source and an observer are moving away from each other. The source is moving at 15.0m/s, and the observer is moving at 25.0m/s. Assuming the speed of sound to be 343 m/s, what is the frequency heard by the observer?
Answer:
[tex]f'=2176.256983Hz[/tex]
Explanation:
The relationship between observed frequency f' and the emitted frequency f is given by the doppler effect equation. In this case the observer and the source are moving in opposite direction away from each other, so:
[tex]f'=\frac{c-v_0}{c+v_s} f[/tex]
Where:
[tex]c=Speed-of-the-sound-waves=343m/s[/tex]
[tex]v_0=Velocity -of-observer=25m/s[/tex]
[tex]v_s=Velocity -of-source=15m/s[/tex]
[tex]f=Emitted -frequency=2450Hz[/tex]
[tex]f'=Observed-frequency[/tex]
Evaluating the data in the equation:
[tex]f'=\frac{343-25}{343+15}*2450=2176.256983Hz[/tex]
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.
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
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
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.
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.
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
Why does an astronaut in a spacecraft orbiting Earthexperience
a feeling of weightlessness?
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 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?
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 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.)
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 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)
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.
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 ?
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.
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)
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 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
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.
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A hot-air balloon is descending at a rate of 1.9 m/s when a passenger drops a camera. If the camera is 47 m above the ground when it is dropped, how much time does it take for the camera to reach the ground? If the camera is 47 m above the ground when it is dropped, what is its velocity just before it lands? Let upward be the positive direction for this problem.
Answer:
The camera lands in t=2.91s with a velocity:
[tex]v=-30.45m/s[/tex]
Explanation:
The initial velocity of the camera is the same as the hot-air ballon:
[tex]v_{o}=-1.9m/s[/tex]
[tex]y_{o}=47m[/tex]
Kinematics equation:
[tex]v(t)=v_{o}-g*t[/tex]
[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]
when the camera lands, y=0:
[tex]0=47-1.9t-4.91*t^{2}[/tex]
We solve this equation to find t:
t1=-3.29s This solution have not sense in our physical point of view
t2=2.91s
So, the camera lands in t=2.91s
We replace this value in v(t):
[tex]v=v_{o}-g*t=-1.9-9.81*2.91=-30.45m/s[/tex]
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."
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."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 boy pulls with a 20 N force, at a 20 degree incline. What part of the force moves the wagon?
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
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
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 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?
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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A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±Q. Let's consider how the electric field changes if one of these variables is changed while the others are held constant. Part A What is the ratio E(final)/E(initial) of the final to initial electric field strengths if Q is doubled?
Answer:
E(final)/E(initial)=2
Explanation:
Applying the law of gauss to two parallel plates with charge density equal σ:
[tex]E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\[/tex]
So, if the charge is doubled the Electric field is doubled too
E(final)/E(initial)=2
A hanging magnesium wire with diameter 2.1 mm (2.1 × 10-3 m) is initially 1.15 m long. When a 104 kg mass is hung from it, the wire stretches an amount 8.46 mm. A mole of magnesium has a mass of 24 grams, and its density is 1.74 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic force.
The effective spring stiffness of the interatomic force within a magnesium wire, when subject to stretching by a 104 kg mass, is approximately 120451.067 N/m. This value is calculated using Hooke's Law, which relates the force applied to the extension of the spring (wire) and the spring's stiffness.
Explanation:To find the effective spring stiffness of the interatomic force in a magnesium wire, we can use Hooke's Law, which states that the force F needed to extend or compress a spring by some distance x is proportional to that distance. This is mathematically represented as F = kx, where k is the spring stiffness (or force constant), and x is the extension or compression of the spring (in this case, the wire).
First, we calculate the force applied by the mass hung from the magnesium wire. The force due to gravity is F = mg, where m is the mass (104 kg), and g is the acceleration due to gravity (approximately 9.8 m/s2).
F = 104 kg * 9.8 m/s2 = 1019.2 N.
The elongation of the wire (x) is given as 8.46 mm, which is 8.46 * 10-3 m.
Using Hooke's Law, we can solve for the spring stiffness, k, by rearranging the equation: k = F / x.
k = 1019.2 N / (8.46 * 10-3 m) = 120451.067 N/m (approximately).
This calculation gives us the effective spring stiffness of the interatomic force within the magnesium wire when a 104 kg mass is hung from it, causing it to stretch by 8.46 mm.
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?
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.
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?
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.