Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
A student drives 105.0 mi with an average speed of 61.0 mi/h for exactly 1 hour and 30
minutes for the first part of the trip. What is the distance in miles traveled during this
time?
Answer:
91.5 miles
Explanation:
61 miles per hour so 61(x amount of hours)
so 61 x 1.5 hours is 91.5 miles
A hornet circles around a pop can at increasing speed while flying in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are Group of answer choices down and backwards. down, backwards, and outwards. down and inwards. down and outwards. straight down.
Answer:
down, backwards, and outwards.
Explanation:
For a hornet that is accelerating in flight, this means that there is a net forward motion at a relatively constant vertical height above the ground.
For this flight, the wings beat downwards to counter the weight of the hornet due to gravity, keeping it at that height above the floor.
For the hornet to accelerate forward, there has to be a net backwards force by the wing on the air. This backwards force accelerates tr forward due to the absence of an equal opposing force in the opposite direction save for a little drag.
The wings also beat with forces directed outwards to provide centripetal force to keep the hornet stable. The absence of this would cause it to spiral out of control.
A car travels around an oval racetrack at constant speed. The car is accelerating:________.
A) at all points except B and D.
B) at all points except A, B, C, and D.
C) everywhere, including points A, B, C, and D.
D) nowhere, because it is traveling at constant speed.
2) A moving object on which no forces are acting will continue to move with constant:_________
A) Acceleration
B) speed
C) both of theseD) none of these
Answer:
1A,2D,3B
Explanation:
hope this helps
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a=nb in the figure ).
How fast, relative to an observer, does an Empire ship have to travel for its markings to be confused with those of a Federation ship? Use c for the speed of light in a vacuum.
Express your answer in terms of n and c.
Complete question
The complete question is shown on the first uploaded image
Answer:
The velocity is [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
Explanation:
From the question we are told that
a = nb
The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)
Now this length seen by the observer can be mathematically represented as
[tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]
Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)
So t = a = nb
and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)
i.e h = b
So
[tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]
[tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]
[tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]
[tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]
[tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
You are watching an object that is moving in SHM. When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left. Part A How much farther from this point will the object move before it stops momentarily and then starts to move back to the left
Answer:
2.95m
Explanation:
The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part
But V = w× r; where V is velocity,
w is angular velocity and r is radius.
Also,
a= w2r; where a is linear acceleration
but a = v× r ; by comparing both equations
Hence r = a/v =8.6/2.45 =3.51m
But the horizontal distance of the motion is given by:
X = rcosx ; where x is the angle
X is the distance covered.
We know that the maximum value of cos x is 1 which is 0°
When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:
X = r=3.51m
Meaning the object needs to travel 3.51-0.56=2.95m further.
Note: the acceleration of the motion is constant whether it is swinging towards the left or right.
When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left and the amplitude of motion A = 0.732 m.
What is Amplitude of motion?
The distance between the central and extreme points for a moving particle is known as the amplitude of motion.
The given data to find the amplitude of motion,
Object displaced = 0.560 m
Velocity = 2.45 m/s
Acceleration = 8.60 m/s²
Starting with sine:
x(t) = Asin(ωt)
so that t = 0, x = 0
x(t) = 0.56 m = Asin(ωt)
v(t) = x(t)'= 2.45 m/s = Aωcos(ωt)
a(t) = v(t)'= -8.60 m/s² = -Aω²sin(ωt)
x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)
0.56m / -8.60 m/s² = -1 / ω²
ω² = 15.3571 rad^2/s^2
ω = 3.91881 rad/s
x(t) / v(t) = Asin(ωt) / Aωcos(ωt)
0.560m / 2.45m/s = tan(3.91t) / 3.91rad/s
0.8937= tan(3.91t)
t = 0.176 s
x(0.176) = Asin(3.59×0.176)
0.65 m= Asin(0.631)
A = 0.732 m is the amplitude of motion.
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Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentum (in kg · m2/s) relative to the point (0, 5.0) meters?
Answer:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
Explanation:
In order to calculate the angular momentum of the particle you use the following formula:
[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex] (1)
r is the position vector respect to the point (0 , 5.0), that is:
r = 0m i + 5.0m j (2)
p is the linear momentum vector and it is given by:
[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex] (3)
the direction of p comes from the fat that the particle is moving along the i + j direction.
Then, you use the results of (2) and (3) in the equation (1) and solve for L:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
The angular momentum is -30 kgm^2/s ^k
02
Blue light has a frequency of about 7.5 x 1014 Hz. Calculate the energy, in Joules, of a single photon associated with this frequency
Answer:
49.725× 10^-24J
Explanation:
The Energy associated with a Photon us defined as;
E = hf
Where h is Planck's constant = 6.63× 10^-34m2kg/s
f is the frequency= 7.5 x 10^14 Hz
Hence
E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference of 1.2 x 106 V and then enters a uniform magnetic field whose strength is 2.2 T. The alpha particle moves perpendicular to the field. Calculate (a) the speed of the alpha particle, (b) the magnitude of the magnetic force exerted on it, and (c) the radius of its circular path.
Answer:
a) v = 1.075*10^7 m/s
b) FB = 7.57*10^-12 N
c) r = 10.1 cm
Explanation:
(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:
[tex]K=qV[/tex] (1)
q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C
V: potential difference = 1.2*10^6 V
You replace the values of the parameters in the equation (1):
[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]
The kinetic energy of the particle is also:
[tex]K=\frac{1}{2}mv^2[/tex] (2)
m: mass of the particle = 6.64*10^⁻27 kg
You solve the last equation for v:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]
the sped of the alpha particle is 1.075*10^6 m/s
b) The magnetic force on the particle is given by:
[tex]|F_B|=qvBsin(\theta)[/tex]
B: magnitude of the magnetic field = 2.2 T
The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1
[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]
the force exerted by the magnetic field on the particle is 7.57*10^-12 N
c) The particle describes a circumference with a radius given by:
[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]
the radius of the trajectory of the electron is 10.1 cm
The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m
What is the Magnetic force?
A) We know that the formula for kinetic energy can be expressed as;
K = qV
where;
q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C
V is potential difference = 1.2 × 10⁶ V
K = 3.2 × 10⁻¹⁹ * 1.2 × 10⁶
K = 3.84 × 10⁻¹³ J
Also, formula for kinetic energy is;
K = ¹/₂mv²
where v is speed
Thus;
v = √(2K/m)
v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)
v = 10.75 * 10⁶ m/s
B) The magnetic force is given by the formula;
F_b = qvB
F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)
F_b = 7.57 * 10⁻¹² N
C) The formula to find the radius is;
r = mv/qB
r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)
r = 0.101 m
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A 148 g ball is dropped from a tree 11.0 m above the ground. With what speed would it hit the ground
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy
[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]
[tex]V^2 - U^2 = 2gh[/tex]
[tex]V^2 - 0 = 2gh[/tex]
[tex]V = \sqrt{2 g h}[/tex]
[tex]= \sqrt{2\times9.8\times11}[/tex]
= 14.68m/s
Two students are pushing their stalled car down the street. If the net force exerted on
the car by the students is 1000 N at an angle of 20° below horizontal, the horizontal
component of the force is:
(a) greater than 1000 N.
(b) less than 1000 N.
(c) sum of the pushing force and the weight of the students.
(d) (a) and (b)
(e) (a) and (c)
Answer:
B
Explanation:
Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.
Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly atoms are assembled in this protein synthesis.
Answer:5.35nm
Explanation:
Consider that 1 inch is = 0.0254m
we have,
1m= 1x10^9 nm
While:
0.0254m = 2.54x10^7nm
1/55 (2.54x10^7) = 4.6181 x 10^5nm
1 day= 24 hrs
= (24x60) when calculating in min
= (24x60x60) calculating in seconds we have:
= 8.64x10⁴sec
In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm
Therefore, the amount by which the hair grows in 1 second will be;
= (4.6181 x 10^5)/(8.64x10^4)
= 5.35nm
The rate of growth will be 5.35nm
Chapter 24, Problem 20 GO A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.9 m from his television set. A reporter at the press conference is located 4.1 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Therefore, the distance between politician and TV set is 2536kmExplanation:
Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.
The reporter hears the sound is
4.1 / 343 = 0.01195 s later
The viewer hears the sound from the TV is
2.9 / 343 = 0.00845s
the difference is 0.00845 sec
the question is how far the TV signal can travel in that time.
the distance between politician and TV set is
= 0.00845 * 3*10^8 m
= 2536 km
d = 2536km
Therefore, the distance between politician and TV set is 2536kmA sulfur dioxide molecule has one sulfur
atom and two oxygen atoms. Which is its
correct chemical formula?
A. SO2
C. S2O2
B. (SO)
D. S20
Answer:
a. SO2
Explanation:
Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now
Answer:
1.92 Watt lost
Explanation:
Power rating of each toaster = 600 Watts
Current that flows = 5 Amperes
Wasted power = 1 Watt
Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R
where I = current
and R = Resistance
600 = [tex]5^{2}[/tex] x R
R = 600/25 = 24 Ohms.
According to joules loss due to heating of wire
Power loss P ∝ [tex]I^{2}[/tex]R
imputing values,
1 ∝ [tex]5^{2}[/tex] x 24
1 ∝ 600
to remove the proportionality sign, we introduce a constant k
1 = 600k
k = 1/600 = 0.00167
For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.
The resistance across the wire becomes
1200 = [tex]10^{2}[/tex]R
R = 1200/100 = 12 Ohms
power loss P = k x [tex]I^{2}[/tex]R
P = 0.0016 x [tex]10^{2}[/tex] x 12
P = 1.92 Watt lost
This question involves the concepts of power, current, and resistance.
The power wasted by the wires in the home for two units will be "4 watt".
POWER WASTAGEThe power wasted by the wires can be given in terms of current and resistance by the following formula:
[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]
where,
P₁ = Power wasted for one unit = 1 wattI₁ = current through wires for one unit = 5 AR = Resistance of wires = constantP₂ = Power wasted for two units = ?I₂ = Current through wires for two units = 10 ATherefore,
[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]
P₂ = 4 watt
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A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks distance, D = 130 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is an angle θ = 25°. How wide is the river?
Answer:
The width of the river is [tex]z = 60.62 \ m[/tex]
Explanation:
From the question we are told that
The distance of the base line is D = 130 m
The angle is [tex]\theta = 25^o[/tex]
A diagram illustration the question is shown on the first uploaded image
Applying Trigonometric Rules for Right-angled Triangle,
[tex]tan 25 = \frac{z}{130}[/tex]
Now making z the subject
[tex]z = 130 * tan (25)[/tex]
[tex]z = 60.62 \ m[/tex]
In an RC-circuit, a resistance of R=1.0 "Giga Ohms" is connected to an air-filled circular-parallel-plate capacitor of diameter 12.0 mm with a separation distance of 1.0 mm. What is the time constant of the system?
Answer:
[tex]\tau = 1\ ms[/tex]
Explanation:
First we need to find the capacitance of the capacitor.
The capacitance is given by:
[tex]C = \epsilon_0 * area / distance[/tex]
Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)
The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:
[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]
So the capacitance is:
[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]
[tex]C = 10^{-12}\ F = 1\ pF[/tex]
The time constant of a rc-circuit is given by:
[tex]\tau = RC[/tex]
So we have that:
[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]
What do you call a group of sea turtles?
Answer:
a bale
Explanation:
a bale is a group of turtles
Answer:
A bale or nest
Explanation:
What is the frequency if 140 waves pass in 2 minutes?
Answer:
1.16 Hz
Explanation:
frequency, basically, is the number of wave on 1 second
so, in math we write like this
f = n/t
n = number of waves
t = time to do that (in sec)
f = 140/120 = 7/6 Hz
f = 1.16 Hz
first law of equilibrium
Answer:
For an object to be an equilibrium it must be experiencing no acceleration.
Explanation:
Hope it helps.
Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosmonaut Valdimir tosses a banana at you at a speed of 16 m/s. At exactly the same instant, you fling a scoop of ice cream at Valdimir along exactly the same path. The collision between banana and ice cream produces a banana split 8.2 m from your location 1.4 s after the banana and ice cream were launched.
1. How fast did you toss the ice cream?
2. How far were you from Valdimir when you tossed the ice cream?
Answer:
a
The speed is [tex]s = 5.857 m/s[/tex]
b
The distance is [tex]D = 22.4 \ m[/tex]
Explanation:
From the question we are told that
The speed of the banana is [tex]v = 16 \ m/s[/tex]
The distance from my location is [tex]d = 8.2 \ m[/tex]
The time taken is [tex]t = 1.4 \ s[/tex]
The speed of the ice cream is
[tex]s = \frac{d}{t}[/tex]
substituting values
[tex]s = \frac{8.4}{1.4}[/tex]
[tex]s = 5.857 m/s[/tex]
The distance of separation between i and Valdimir is the same as the distance covered by the banana
So
[tex]D = v * t[/tex]
substituting values
[tex]D = 16 * 1.4[/tex]
[tex]D = 22.4 \ m[/tex]
If a cart of 8 kg mass has a force of 16 newtons exerted on it, what is its acceleration?
Answer:
Explanation:
From Newton's 2nd Law,
F = m×a
Where F is Force
m is mass
a is acceleration
Hence a= F/m
a= 16/8= 2m/s2
A 550 kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s
Answer:
[tex]52.25\times10^4W\\699.1 hp[/tex]
Explanation:
According to the energy conversation:
ΔK=[tex]-f_kd+W[/tex]
ΔK=[tex]K_f-K_i ; K=1/2 mv^2[/tex]
where,
[tex]k_i, k_f[/tex] are initial and final kinetic energy of the system.
[tex]v_i[/tex]= initial velocity of the system
[tex]v_f[/tex]=final velocity of the system
W= total work done on the system
[tex]f_k[/tex]= friction force
d= distance traveled
Given: [tex]v_f[/tex]=110m/s
d=400m
[tex]f_k[/tex]=1200N
[tex]v_i[/tex]=0m/s
t=7.3s
ΔK=[tex]-f_kd+W[/tex]
W= ΔK + [tex]f_kd[/tex]
=[tex]K_f-K_i+f_kd\\[/tex]
[tex]=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500[/tex]
[tex]P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp[/tex]
. A ball weighs 120g on the earth surface,
i) What is its mass on the surface of the moon? 1mk
Answer:
WEIGHT ON MOON IS 0.2004N
Explanation:
mass of the body=120g=[tex]\frac{120}{1000}[/tex]kg=0.12kg (we will convert g into kg)
gravity on moon=1.67m/s²( to find the mass of anybody on another we should know its gravity)
as we know that (from the formula of weight)
weight=mass×gravity
w=mg
w=0.12kg²×1.67m/s²
w=0.2004N
A sound level of 96 dB is how many times as intense as one of 90 dB?
Answer:
A sound level of 96 dB is 4 times as intense as one of 90 dB
Explanation:
The formula of the intensity level of sound in decibels is given as follows:
Intensity Level = 10 log₁₀(I/I₀)
where,
I = Intensity of Sound
I₀ = Reference Intensity Level = 10⁻¹² W/m²
Therefore, for 96 dB sound level:
96 = 10 log₁₀(I₁/10⁻¹²)
log₁₀(I₁/10⁻¹²) = 96/10
I₁/10⁻¹² = 10^9.6
I₁ = (10⁻¹²)(4 x 10⁹)
I₁ = 0.004 W/m²
For 90 dB sound level:
90 = 10 log₁₀(I₂/10⁻¹²)
log₁₀(I₂/10⁻¹²) = 90/10
I₂/10⁻¹² = 10^9
I₂ = (10⁻¹²)(10⁹)
I₂ = 0.001 W/m²
Therefore,
I₁/I₂ = 0.004/0.001
I₁ = 4 I₂
Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.
g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the dragster at the end of 2.1 s, (b) the final velocity of the dragster at the end of of twice this time, or 4.2 s, (c) the displacement of the dragster at the end of 2.1 s, and (d) the displacement of the dragster at the end of twice this time, or 4.2 s?
The dragster's velocity v at time t with constant acceleration a is
[tex]v=at[/tex]
since it starts at rest.
After 2.1 s, it will attain a velocity of
[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]
or 92.4 m/s.
Doubling the time would double the final velocity,
[tex]v=a(2t)=2at[/tex]
so the velocity would be twice the previous one, 184.8 m/s.
The dragster undergoes a displacement x after time t with acceleration a of
[tex]x=\dfrac12at^2[/tex]
if we take the starting line to be the origin.
After 2.1 s, it will have moved
[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]
or 88 m.
Doubling the time has the effect of quadrupling the displacement, since
[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]
so after 4.2 s it will have moved 352 m.
A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (the normal force exerted on her by the bench) of the person as she passes through the highest point of her motion? ( type in your answer with no units in form xx0)
Answer:
The apparent weight of the person as she pass the highest point is [tex]N = 458.8 \ N[/tex]
Explanation:
From the question we are told that
The radius of the Ferris wheel is [tex]r = 5.0 \ m[/tex]
The period of revolution is [tex]T = 8.0 \ s[/tex]
The weight of the person is [tex]W = 670 \ N[/tex]
Generally the speed of the wheel is mathematically represented as
[tex]v = \frac{2 \pi r}{T }[/tex]
substituting values
[tex]v = \frac{2 * 3.142 * 5}{8 }[/tex]
[tex]v = 3.9 3 \ m/s[/tex]
The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as
[tex]N = mg - \frac{mv^2}{r}[/tex]
Where m is the mass of the person which is mathematically evaluated as
[tex]m = \frac{W}{g}[/tex]
substituting values
[tex]m = \frac{670}{9.8}[/tex]
[tex]m = 68.37 \ kg[/tex]
So
[tex]N = 68.37 * 9.8 - \frac{68.37 * {3.93}^2}{5}[/tex]
[tex]N = 458.8 \ N[/tex]
A 1,269 kg rocket is traveling at 413 m/s with 2,660 kg of fuel on board. If the rocket fuel travels at 1,614 m/s relative to the rocket, what is the rockets final velocity after it uses half of its fuel?
Answer:
About 2104m/s
Explanation:
[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]
Hope this helps!
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K = 3.9 kJ
Explanation:
The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:
[tex] K_{T} = K_{f} - K_{i} [/tex]
The initial kinetic energy is:
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]
Where m₁ is the mass of the object before the explosion and v₁ is its velocity
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]
Now, the final kinetic energy is:
[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]
Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces
Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex] (1)
By conservation of momentum we have:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]
[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]
[tex] v_{2} = \frac{5}{4}v_{1} [/tex] (2)
By entering (2) into (1) we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]
Hence, the kinetic energy added is:
[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]
Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.
I hope it helps you!
Assume that the coefficient of static friction between the board and the box is not known at this point. What is the magnitude of the acceleration of the box in terms of the friction force f?
Answer:
Explanation:
From Newton's second Law of Motion,
F = ma
Ff + F = ma
Where F is the applied force, Ff is the frictional force, a is the acceleration and m is the mass of the object or box.
Magnitude of the acceleration:
a = Ff+F/m
This must act in the direction of F or the box would slide or accelerate off the negative side of the board (taking the direction of F to be positive
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to: A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:__________.
a) 19 s
b) 17 s
c) 21 s
d) 23 s
e) 15 s
Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is
[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]
For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],
[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]
which makes B, approximately 17 s, the correct answer.
The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)
Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:
[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)
Where:
[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.
[tex]\omega[/tex] - Final angular velocity, in radians per second.
[tex]\alpha[/tex] - Angular acceleration, in radians per square second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:
[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 16.667\,s[/tex]
The time interval of angular deceleration is 16.667 seconds. (Answer: B)
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