Answer: 84.63 m
Explanation:
The speed of sound in the air, at [tex]0 \°C[/tex] is [tex]331.5 m/s[/tex], and for each degree Celsius the temperature rises, the speed of the sound increases by [tex]0.6 m/s[/tex]. So, if we estimate the speed of sound at [tex]30 \°C[/tex] it will be [tex]349 m/s[/tex].
On the other hand, the speed of sound [tex]V[/tex] is defined as the distance traveled [tex]d[/tex] in a especific time [tex]t[/tex]:
[tex]V=\frac{d}{t}[/tex]
Where:
[tex]V=349 m/s[/tex] is the speed of sound
[tex]t=\frac{0.485 s}{2}=0.2425 s[/tex] is half the time the sound wave travels since you say "hello", the sound wave hits the building and then returns to you
[tex]d[/tex] is the distance between you and the building
[tex]349 m/s=\frac{d}{0.2425 s}[/tex]
Finding [tex]d[/tex]:
[tex]d=(349 m/s)(0.2425 s)[/tex]
Finally:
[tex]d=84.63 m[/tex]
To find the distance to the building, calculate the speed of sound at 30.6°C, multiply it by the echo time of 0.485 seconds, and divide by 2 to account for the round trip of the sound. Then, convert the distance from meters to feet to get approximately 280.95 feet.
Explanation:To calculate the distance of the building when an echo is heard 0.485 seconds after shouting, we need to consider the speed of sound at the given air temperature of 30.6°C. Since the speed of sound varies with temperature, we use the formula v = v_0 + 0.6t, where v_0 is the speed of sound at 0°C (approximately 331.5 m/s), t is the temperature in Celsius, and v is the speed of sound at t degrees Celsius. At 30.6°C, the speed of sound is v = 331.5 m/s + 0.6(30.6°C), which calculates to roughly 349.86 m/s. Since the echo travels the distance to the building and back, the actual distance to the building is half the total distance the sound wave traveled.
Therefore, the one-way distance to the building is 0.485 s × 349.86 m/s ÷ 2. Converting meters to feet (1 meter = 3.28084 feet), we find that the building is approximately 280.95 feet away.
2.85 A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. The police officer begins to pursue the speeder - first there is a 0.800 s reaction time when the officer has no change in speed, then the officer accelerates at 5.00 m/s2. Including the reaction time, how long does it take for the police car to reach the same position as the speeding car
Answer:
11.1 s
Explanation:
Speed of the police car as given = v = 18 m/s
Speed of the car = V = 42 m/s
Reaction time = t = 0.8 s
Distance traveled by the police car during the reaction time = d₁= 0.8 x 18 = 14.4 m
Distance traveled by speeding car = d₂ =0.8 x 42 = 33.6 m
Acceleration of the police car = a = 5 m/s/s
The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.
Distance traveled by the police car = D = d₁ + v t +0.5 at², according to the kinematic equation.
⇒ D = 14.4 + 18 t + 0.5 (5) t²
⇒ D = 14.4 + 18 t+2.5 t² → (1)
For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).
⇒ 14.4 + 18 t+2.5 t²= 33.6 + 42 t
⇒ 2.5 t² -24 t - 19.2 = 0
⇒ t = 10.3 s
Total time = t +0.8 s
⇒ Time taken for the police car to reach the speeding car = 10.3+0.8= 11.1 s
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 18.5 kg and an initial velocity of v0A = 8.15 m/s, due east. Object B, however, has a mass of mB = 30.5 kg and an initial velocity of v0B = 5.00 m/s, due north. Find the magnitude of the final velocity of the two-object system after the collision.
Answer:
v =4.36 m/s
Explanation:
given,
mass of object A = 18.5 Kg
initial velocity of object A = 8.15 m/s in east
mass of object B = 30.5 kg
initial velocity of object B = 5 m/s
[tex]P = P_A+P_B[/tex]
[tex]P = m_Av_A\widehat{i} + m_B v_B\widehat{j}[/tex]
[tex]P = 18.5\times 8.15 \widehat{i} + 30.5\times 5\widehat{j}[/tex]
[tex]P = 150.775 \widehat{i} + 152.5 \widehat{j}[/tex]
[tex]P = \sqrt{150.775^2+152.5^2}[/tex]
P = 214. 45 N s
velocity after collision is equal to
[tex]v =\dfrac{214.45}{18.5+30.5}[/tex]
v =4.36 m/s
hence, velocity after collision is equal to 4.36 m/s
Answer:
The magnitude of the final velocity of the two-object system is [tex]v=4.37\frac{m}{s}[/tex]
Explanation:
As the Momentum is conserved, we can compare the instant before the collision, and the instant after. Also, we have to take in account the two components of the problem (x-direction and y-direction).
To do that, we put our 0 of coordinates where the collision takes place.
So, for the initial momentum we have that
[tex]p_{ix}=m_{a}v_{0a}+0[/tex]
[tex]p_{iy}=0+m_{b}v_{0b}[/tex]
Now, this is equal to the final momentum (in each coordinate)
[tex]p_{fx}=(m_{a}+m_{b}) v_{fx}[/tex]
[tex]p_{fy}=(m_{a}+m_{b}) v_{fy}[/tex]
So, we equalize each coordinate and get each final velocity
[tex]m_{a}v_{0a}=(m_{a}+m_{b}) v_{fx} \Leftrightarrow v_{fx}=\frac{m_{a}v_{0a}}{(m_{a}+m_{b})}[/tex]
[tex]m_{b}v_{0b}=(m_{a}+m_{b}) v_{fy} \Leftrightarrow v_{fy}=\frac{m_{b}v_{0b}}{(m_{a}+m_{b})}[/tex]
Finally, to calculate the magnitude of the final velocity, we need to calculate
[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}[/tex]
which, replacing with the previous results, is
[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}=(\sqrt{(\frac{18.5*8.15}{49})^{2}+(\frac{30.5*5.00}{49})^{2}})\frac{m}{s}[/tex]
Therefore, the outcome is
[tex]v_{f}=4.37\frac{m}{s}[/tex]
A 7.94-nC charge is located 1.77 m from a 4.14-nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. (b) is the force attractive or repulsive?
Answer:
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
Explanation:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters(m)
Equivalence
1nC= 10⁻⁹C
Data
K=8.99x10⁹N*m²/C²
q₁ = 7.94-nC= 7.94*10⁻⁹C
q₂= 4.14-nC= 4.14 *10⁻⁹C
d= 1.77 m
Magnitude of the electrostatic force that one charge exerts on the other
We apply formula (1):
[tex]F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }[/tex]
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
A cube whose sides are of length = 1.8 m is placed in a uniform electric field of magnitude E = 5.8 ✕ 10^3 N/C so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube (in N · m^2/C)?
Answer:
Zero
Explanation:
Electric flux is defined as the number of electric field lines which passes through any area in the direction of area vector.
The formula for the electric flux is given by
[tex]\phi =\overrightarrow{E}.\overrightarrow{dS}[/tex]
Here, E is the strength of electric field and dS be the area vector.
It is a scalar quantity.
According to the Gauss's theorem, the electric flux passing through any surface is equal to the [tex]\frac{1}{\epsilon _{0}}[/tex] times the total charge enclosed through the surface.
Here, the charge enclose is zero, so the total flux is also zero.
Amy initially 5.0 mi west of the United Center's Michael Jordan statue is running with a constant velocity of 5.0 mi/h due east. Alejandro is initially 4.0 mi east of the statue and is running with a constant velocity of 7.0 mi/h due west. How far are the runners from the statue when they meet?
Answer:
The statue is 1.67 miles west of both the runners
Explanation:
Wherever Amy and Alejandro meet they will have covered a total distance of 5+4 = 9 mi
They will also have run the same amount of time if they started at the same moment = t
Speed of Amy = 5 mi/h
Speed of Alejandro = 7 mi/h
Distance = Speed × Time
Distance travelled by Amy = 5t
Distance travelled by Alejandro = 7t
Total distance run by Amy and Alejandro is
5t+7t = 9
[tex]\\\Rightarrow 12t=9\\\Rightarrow t=\frac{12}{9}\\\Rightarrow t=\frac{4}{3}\ hours[/tex]
Distance travelled by Amy
[tex]5\times t=5\times \frac{4}{3}=\frac{20}{3}\\ =6.67\ miles[/tex]
The distance of Amy from the statue would be
6.67 - 5 = 1.67 miles
So, the statue is 1.67 miles west of both the runners
A ball is thrown vertically upward with a speed of 15.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its highest point? s (c) How long does the ball take to hit the ground after it reaches its highest point? s (d) What is its velocity when it returns to the level from which it started? (Assume the positive direction is upward. Indicate the direction with the sign of your answer.) m/s
Answer:
a) 0.76 m
b) 1.53 s
c) 0.39 s
d) 3.8 m/s
Explanation:
This is a problem in which we need to use the equations pertaining Uniformly Accelerated Motion, as the acceleration during all this process is constant: The gravitational pull on the ball, 9.8 m/s².
To make things easier, we can divide this process in two parts: The first one (A) is from the moment the ball is thrown, until the moment it reaches it highest point and momentarily stops. The second one (B) from the moment it starts descending until it hits the ground.
a) During part A, we use the formula Vf²=Vi² +2*a*x . Where Vf is the final velocity (0 m/s, as the ball stopped in midair), Vi is the initial velocity (15 m/s), a is the acceleration (-9.8 m/s², it has a minus sign, as it goes against the direction of the movement) and x is the distance; thus, we're left with:
0 m/s=(15.0 m/s)²+2*(-9.8 m/s²)*x
We solve for x
x = 0.76 m
b) The formula is Vf=Vi +a*t, where t is the time. We're left with:
0 m/s=15.0 m/s + (-9.8 m/s²)*t
We solve for t
t= 1.53 s
c) Now we focus on part B, and use the formula x=Vi * t + [tex]\frac{at^{2}}{2}[/tex] , with the difference that the Vi is 0 m/s. We already know the value of x in exercise a). Note that a does not have a negative sign, as the direction of movement is opposite to the direction of part A
0.76 m=0 m/s * t +[tex]\frac{9.8\frac{m}{s^{2} } *t^{2} }{2}[/tex]
Solve for t
0.76=4.9t²
t=0.39 s
d) Once again we use the formula Vf=Vi +a*t, using the value of t previously calculated in exercise c).
Vf=0 m/s + 9.8 m/s² * 0,39 s
Vf=3.8 m/s
The ball rises to a height of 11.48 m, takes 1.53 s to reach its highest point, the same amount of time to fall back down, and has a velocity of -15.0 m/s when it returns to the starting level.
The question involves concepts from physics, specifically the kinematics of one-dimensional motion under constant acceleration due to gravity. Here is how we can solve each part of the given problem:
(a) How high does it rise?
To find the maximum height reached by the ball, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
v² = u² + 2as
At the highest point, the final velocity (v) is 0 m/s, the initial velocity (u) is 15.0 m/s (upward), the acceleration due to gravity (a) is -9.8 m/s² (downward), and s represents the height. Solving for s:
0 = (15.0)² + 2(-9.8)s
s = (15.0)² / (2 * 9.8)
s = 11.48 m
(b) How long does it take to reach its highest point?
To find the time (t) it takes for the ball to reach its highest point, we can use the equation:
v = u + at
Since the final velocity at the highest point is 0 m/s:
0 = 15.0 + (-9.8)t
t = 15.0 / 9.8
t = 1.53 s
(c) How long does the ball take to hit the ground after it reaches its highest point?
The time for the ball to fall back down is the same as the time taken to reach the highest point, so this is also 1.53 s.
(d) What is its velocity when it returns to the level from which it started?
The velocity on returning to the starting level will be the same magnitude as the initial velocity but in the opposite direction, so it will be -15.0 m/s, with the negative sign indicating the downward direction.
A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light Years thick. If this Galaxy contains 1,078 Billion stars, estimate the average distance between the stars in this galaxy. Hint: calculate the average volume per star in cubic Light Years, and then estimate the approximate linear dimension across such a volume. (Indicate your answer to one decimal place.)
Answer:
Approximate linear dimension is 2 light years.
Explanation:
Radius of the spiral galaxy r = 62000 LY
Thickness of the galaxy h = 700 LY
Volume of the galaxy = πr²h
= (3.14)(62000)²(700)
= (3.14)(62)²(7)(10)⁸
= 84568×10⁸
= [tex]8.45\times 10^{12}[/tex] (LY)³
Since galaxy contains number of stars = 1078 billion stars ≈ [tex]1.078\times 10^{12}[/tex]
Now volume covered by each star of the galaxy = [tex]\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}[/tex]
= [tex]\frac{8.45\times 10^{12} }{1.078\times 10^{12}}[/tex]
= 7.839 Light Years
Now the linear dimension across the volume
= [tex](\text{Average volume per star})^{\frac{1}{3}}[/tex]
= [tex](7.839)^{\frac{1}{3}}[/tex]
= 1.99 LY
≈ 2 Light Years
Therefore, approximate linear dimension is 2 light years.
The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.
Answer:
E = 1.655 x 10⁷ N/C towards the filament
Explanation:
Electric field due to a line charge is given by the expression
E = [tex][tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex][/tex]
where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is
8.85 x 10⁻¹².
Putting the given values in the equation given above
E = [tex]\frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}[/tex]
E = 1.655 x 10⁷ N/C
Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a distance of 45.0 m in which to stop it.
Answer:
4.44 kN in the opposite direction of acceleration.
Explanation:
Given that, the initial speed of the car is, [tex]u=20m/s[/tex]
And the mass of the car is, [tex]m=1000 kg[/tex]
The total distance covered by the car before stop, [tex]s=45m[/tex]
And the final speed of the car is, [tex]u=0m/s[/tex]
Now initial kinetic energy is,
[tex]KE_{i}=\frac{1}{2}mu^{2}[/tex]
Substitute the value of u and m in the above equation, we get
[tex]KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J[/tex]
Now final kinetic energy is,
[tex]KE_{f}=\frac{1}{2}mv^{2}[/tex]
Substitute the value of v and m in the above equation, we get
[tex]KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J[/tex]
Now applying work energy theorem.
Work done= change in kinetic energy
Therefore,
[tex]F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN[/tex]
Here, the force is negative because the force and acceleration in the opposite direction.
What wavelength photon is required to excite a hydrogen atom from the n = 1 state to the n = 3 state?
Answer:
The wavelength required is 102.9 nm.
Explanation:
The energy levels for the hydrogen atom are
[tex]E(n) = \frac{-13.6 \ eV}{n^2}[/tex]
So, for a transition from the first level to the third level we got
[tex]\Delta E = E(3) - E (1)[/tex]
[tex]\Delta E = \frac{-13.6 \ eV}{ 3 ^2} - \frac{-13.6 \ eV}{1^2}[/tex]
[tex]\Delta E = \frac{-13.6 \ eV}{ 9} - \frac{-13.6 \ eV}{1}[/tex]
[tex]\Delta E = \frac{8}{9} 13.6 \ eV[/tex]
[tex]\Delta E = 12.09 \ eV[/tex]
[tex]\Delta E = 12.09 \ eV * \frac{1.6 \ 10^-19 \ Joules}{1 \ eV}[/tex]
[tex]\Delta E = 1.93 \ 10^-18 \ Joules[/tex]
So we need a photon with this energy.
The energy of a photon its given by
[tex]E = h \nu = h \frac{c}{\lambda}[/tex]
So, the wavelength will be
[tex]\lambda = \frac{h c}{E}[/tex]
[tex]\lambda = \frac{6.62 \ 10^{-34} \ \frac{m^2 kg}{s} \ * 3.00 \ 10^8 \ \frac{m}{s}}{1.93 \ 10^-18 \ Joules}[/tex]
[tex]\lambda = 10.29 \ 10^{-8} m[/tex]
[tex]\lambda = 1.029 \ 10^{-7} m[/tex]
[tex]\lambda = 102.9 \ nm[/tex]
A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is 1. After how many seconds does the ball strike the ground?
Answer:
[tex]t=6.96s[/tex]
Explanation:
From this exercise, our knowable variables are hight and initial velocity
[tex]v_{oy}=96ft/s[/tex]
[tex]y_{o}=112ft[/tex]
To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
[tex]0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]
Solving for t using quadratic formula
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-\frac{1}{2} (32.2)\\b=96\\c=112[/tex]
[tex]t=-0.999s[/tex] or [tex]t=6.96s[/tex]
Since time can't be negative the answer is t=6.96s
For each of the motions described below, determine the algebraic sign (+, -, or 0) of the velocity and acceleration of the object at the time specified. For all of the motions, the positive y axis is upward.
Part A
An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped.
Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .
Part B
A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball immediately after the ball leaves the child's hand?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .
Part C
A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball at the very top of the ball's motion (i.e., the point of maximum height)?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .
Part A: -,+
The elevator is moving downward, this is what determines the direction of the velocity, as it will follow the direction of the movement. As we are told that the positive direction is upward, then the velocity has negative direction. Also, after the button is pressed, the elevator starts to stop, in other words, its velocity starts to decreased. This means that the acceleration has an opposite direction to the velocity, therefore, its sign is +.
Part B: +, -
The ball is moving upward, and as said before, this is what determines the direction of the velocity, as it will follow the direction of the movement. Then, velocity has a + sign.
Also, after the ball is thrown, there is no other force other than gravity, which will oppose to the movement of the ball, trying to make it come back to the ground. This means that the acceleration has an opposite direction to the velocity, in other words, it's directed downward, therefore, its sign is -.
Part C: 0, -
The acceleration of the ball since it was thrown until it fell to the ground will always be the gravity, which will always go downward (-).
After being thrown, the ball's velocity will start to decrease because of gravity. When its velocity has turned to 0, the ball will have reached maximum height . At this point it will start to fall again, accelerated by gravity. But at the very top, the velocity of the ball is 0.
If a negative charge is placed in an electric field, what direction will it be accelerated? a) In the direction of the field.
b) In the opposite direction of the field.
c) Perpendicular to the direction of the field.
d) It will not be accelerated.
e) None of the above.
Answer:
option B
Explanation:
the correct answer is option B
when negative charge is placed in electric field then direction will be accelerated in opposite direction of the field.
charged electric particle produce electric field and they exert force on other charged particle.
when a charged particle is positive it will accelerate in the direction of the electric field if the charge is negative then particle will accelerate in opposite direction of electric field.
A closed system consisting of 4 lb of gas undergoes a process in which the relation between pressure and volume is pVn = constant. The process begins with p1 = 15 psi, v1 = 1.25 ft3/lb and ends with p2 = 53 psi and v2 = 0.5 ft3/lb. Determine: a) the volume in ft3 occupied by the gas at states 1 and 2, and b) the value of n.
Answer:
V1=5ft3
V2=2ft3
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5ft3
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= 2ft3
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
How do resistors in parallel affect the total resistance?
Answer:
They're going to increase the total resistance as [tex]R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}[/tex]
Explanation:
If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then
[tex]I = I_1 + I_2 + ... + I_N[/tex]
where
[tex]I_i = \frac{V_i}{R_i}[/tex]
but
[tex]V_i = V_j = V[/tex] for [tex]i,j= 1, 2,..., N[/tex]
so
[tex]I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}[/tex]
where
[tex]R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1} [/tex]
You drive a car 690 ft to the east, then 380 ft to the north. a) What is the magnitude of your displacement?
b) What is the direction of your displacement?
Answer:displacement =787.71 m
Explanation:
Given
Driver driver the car 690 ft to the east
then turn 380 ft to the north
(a)magnitude of acceleration is [tex]=\sqrt{380^2+690^2}=\sqrt{620500}[/tex]
displacement=787.71 m
(b)direction of displacement
[tex]tan\theta =\frac{380}{690}=0.5507[/tex]
[tex]\theta =28.84^{\circ}[/tex] with east direction
The magnitude of the displacement is approximately 775 ft and its direction is roughly 29 degrees north of east.
Explanation:Your total displacement after driving a car 690 ft to the east and then 380 ft to the north is calculated using the Pythagorean theorem, which states that the magnitude of the hypotenuse (displacement) of a right triangle (formed by the eastwards and northwards journeys of the car) can be found by sqrt((eastwards travel)^2 + (northwards travel)^2). Applying the theorem, we obtain sqrt((690 ft)^2 + (380 ft)^2) = 775 ft approximately.
The direction of the displacement can be found using the tangent of the angle, which is the ratio of the opposite (northwards) to the adjacent side (eastwards). Applying the inverse tangent function, we get tan^-1(380/690), which gives us approximately 29 degrees. Therefore, the direction of the displacement is 29 degrees north of east.
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A free positive charge released in an electric field will:
accelerate in the direction opposite the electric field.
accelerate along a circular path.
accelerate in the direction in which the electric field is pointing.
remain at rest.
accelerate in a direction perpendicular to the electric field.
Answer:
accelerate in the direction in which the electric field is pointing.
Explanation:
The positive charge feels a force in the same direction as the electric field
F=Eq
F and E are vectors, q is a scalar
(if it were a negative charge the force would be in the opposite direction)
that force will produce an acceleration in the same direction, that acceleration will cause the particle to move in the same direction, ie the direction of the electric field.
A free positive charge in an electric field will accelerate in the direction in which the electric field is pointing due to electrostatic forces.
Explanation:In the context of physics and electric fields, a free positive charge released in an electric field will accelerate in the direction in which the electric field is pointing. This is because electric field lines are drawn from positive to negative, indicating the direction that a positive test charge would move. Therefore, when a positive charge is placed in an electric field, it is repelled from the positive source charge and attracted to the negative source charge, causing it to accelerate along the electric field lines in the direction they are pointing.
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Take the following measured gauge (or gage) pressures and convert them to absolute pressures in both kPa and psi units for an ambient pressure equal to 98.10 kPa a) pgage 152 kPa, b) Pgage=67.5 Torr, c) pvac 0.10 bar, d) pvac 0.84 atm
Answer:
Explanation:
Given
ambient Pressure =98.10 kPa
(a)gauge pressure 152 kPa
we know
Absolute pressure=gauge pressure+Vacuum Pressure
[tex]P_{abs}[/tex]=152+98.10=250.1 kPa or 36.27 psi
(b)[tex]P_{gauge}[/tex]=67.5 Torr or 8.99 kpa
as 1 Torr is 0.133 kPa
[tex]P_{abs}[/tex]=8.99+98.10=107.09 kPa or 15.53 psi
(c)[tex]P_{vaccum}[/tex]=0.1 bar or 10 kPa
Thus absolute pressure=98.10-10=88.10 kPa or 12.77 psi
as 1 kPa is equal to 0.145 psi
(d)[tex]P_{vaccum}[/tex]=0.84 atm or 85.113 kPa
as 1 atm is equal to 101.325 kPa
[tex]P_{abs}[/tex]=98.10-85.11=12.99 kPa or 1.88 psi
A 10-cm-long thin glass rod uniformly charged to 11.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 11.0 nC are placed side by side, 4.10 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
The electric field strengths at various points around a charged object can be calculated using Coulomb's Law. They depend on the charge of the object and the distance from it. Fields produced by multiple charges need to be added vectorially.
Explanation:You're asking about the electric field strengths at varying distances from a charged glass rod, located next to a plastic rod with a charge of equal magnitude, but of opposite sign. The electric field strength E at a point in the vicinity of a charged object can be derived using Coulomb's Law. This law states that the force F between two charges q1 and q2 is proportional to the product of the charges and inversely proportional to the square of the distance r between them.
Given this, the electric field E created by a charge q at a distance r from the charge is given by E = k|q| / r², where k is Coulomb's constant, approximately 9 × 10⁹ N•m²/C². The direction of the electric field is determined by the sign of the charge. When it's positive, the field lines are going outward, and when it's negative, they are directed inward.
To calculate the net electric field at a point on the line connecting the midpoints of the rods, we consider the electric fields produced by both rods. Since they're opposite in sign, the fields will have opposite directions, and they should be added vectorially. The magnitudes and directions of the fields will also depend on the specific distances (1.0 cm, 2.0 cm, and 3.0 cm) from the glass rod.
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A startled armadillo leaps upward, rising 0.540 m in the first 0.216 s. (a) What is its initial speed as it leaves the ground? (b)What is its speed at the height of 0.540 m? (c) How much higher does it go? Use g=9.81 m/s^2.
Answer:
a) 3.6 m/s
b) 1.53 m/s
c) 0.12 m
Explanation:
t = Time taken = 0.216 s
u = Initial velocity
v = Final velocity
s = Displacement = 0.54 m
a = Acceleration due to gravity = 9.81 m/s² (negative upward)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0.54=u\times 0.216+\frac{1}{2}\times -9.81\times 0.216^2\\\Rightarrow u=\frac{0.54+\frac{1}{2}\times 9.81\times 0.216^2}{0.216}\\\Rightarrow u=3.6\ m/s[/tex]
Initial speed as it leaves the ground is 3.6 m/s
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 0.54+3.6^2}\\\Rightarrow v=1.53\ m/s[/tex]
Speed at the height of 0.540 m is 1.53 m/s
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-3.6^2}{2\times -9.81}\\\Rightarrow s=0.66\ m[/tex]
The total height the armadillo leaps is 0.66 m
So, the additional height is 0.66-0.54 = 0.12 m
To make a secure fit, rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.871 cm in diameter is to be placed in a hole 1.869 cm in diameter in a metal at 24 ∘C.To what temperature must the rivet be cooled if it is to fit in the hole?
Answer:
-65 degC
Explanation:
ΔD=[tex]D_{0}[/tex]αΔT
1.869-1.871=1.871(1.2E-5)ΔT
ΔT≈-89 degC
24-89=-65 degC
A particle moves along the x axis according to the equation x = 2.08 + 3.06t − 1.00t^2, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 3.30 s. (b) Find its velocity at t = 3.30 s. (c) Find its acceleration at t = 3.30 s.
Explanation:
The position of a particle along x - axis is given by :
[tex]x=2.08+3.06t-1t^2[/tex]
(a) Position at t = 3.3 s
[tex]x=2.08+3.06(3.3)-1(3.3)^2[/tex]
x = 1.288 m
(b) Velocity at t = 3.3 s
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(2.08+3.06t-1t^2)}{dt}[/tex]
[tex]v=3.06-2t[/tex]
at t = 3.3 s
[tex]v=3.06-2(3.3)[/tex]
v = -3.54 m/s
(c) Acceleration,
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]a=\dfrac{d(3.06-2t)}{dt}[/tex]
[tex]a=-2\ m/s^2[/tex]
Hence, this is the required solution.
Final answer:
At t = 3.30 s, the particle is at the position of 1.288 meters, moves with a velocity of -3.54 m/s, and is subjected to a constant acceleration of -2 m/s^2.
Explanation:
The student's question about a particle's motion along the x-axis can be answered using kinematic equations from physics, which relate to the position, velocity, and acceleration of the particle.
Position at t = 3.30 s
To find the position of the particle at t = 3.30 s, we substitute the time into the position function x = 2.08 + 3.06t - 1.00t2.
2)>
Calculating this gives x(3.30 s) = 2.08 + 10.098 - 10.89 = 1.288 meters. The particle is at 1.288 meters from the origin at 3.30 seconds.
Velocity at t = 3.30 s
To find the velocity, we differentiate the position function with respect to time: v(t) = 3.06 - 2.00t.
The particle's velocity at 3.30 seconds is -3.54 meters per second.
Acceleration at t = 3.30 s
The acceleration is the second derivative of the position, which in this case is a constant because the t2 term has a constant coefficient: a(t) = -2.00 m/s2.
Therefore, the acceleration of the particle at any time, including at t = 3.30 s, is -2 meters per second squared.
The electric output of a power plant is 669 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows at a rate of 1.17x 108 L/Hr. The water enters the plant at 23.6°C and exits at 29.8°C. (a) What is the power plant's total thermal power? (b) What is the efficiency of the power plant? (a) MWT (Megawatt thermal) O Answer part (a) (b) Answer part (b)
Answer:
a) 1511 MW
b) 44%
Explanation:
The thermal power will be the electric power plus the heat taken away by the cooling water.
Qt = P + Qc
The heat taken away by the water will be:
Qc = G * Cp * (t1 - t0)
The Cp of water is 4180 J/(kg K)
The density of water is 1 kg/L
Then
G = 1.17 * 10^8 L/h * 1 kg/L * 1/3600 h/s = 32500 kg/s
Now we calculate Qc
Qc = 32500 * 4180 * (29.8 - 23.6) = 842*10^6 W = 842 MW
The total thermal power then is
Qt = 669 + 842 = 1511 MW
The efficiency is
η = P / Qt
η = 669 / 1511 = 44%
When 19.3 J was added as heat to a particular ideal gas, the volume of the gas changed from 56.7 cm^3 to 104 cm^3 while the pressure remained constant at 0.947 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 1.97 x 10^-3 mol, find the molar specific heat of the gas at (b) constant pressure and (c) constant volume.
Answer:
a)[tex] C_p=35.42\ \rm J/mol.K[/tex]
b)[tex] C_v=27.1\ \rm J/mol.K[/tex]
Explanation:
Given:
Heat given to the gas, [tex]Q=19.3\ \rm J[/tex]Initial volume of the gas,[tex]V_i=56.7\ \rm cm^3[/tex]Final volume of the gas, [tex]V_f=104\ \rm cm^3[/tex]Constant pressure of the gas,[tex]P=0.947\ \rm atm[/tex]Number of moles of the gas,[tex]n=1.97\times10^{-3}[/tex]Let R be the gas constant which has value [tex]R=8.31432\times10^3\ \rm N\ m\ kmol^{-1}K^{-1}[/tex]
Work done in the process
[tex]W=PdV\\W=0.947\times(104-56.7)\times10^{-3}\ \rm L\ atm\\W=44.8\times10^{-3}\times101.33\ \rm J\\W=4.43\ \rm J[/tex]
Now Using First Law of thermodynamics
[tex]Q=W+\Delta U\\19.3=4.43+\Delta U\\\Delta U=14.77\ \rm J[/tex]
Let[tex]C_v[/tex] be the molar specific heat of the gas at constant volume given by
[tex]\Delta U=nC_v\Delta T\\14.77=\dfrac{C_v}{R}{PdV}\\14.77=\dfrac{C_v}{R}(0.987\times10^5(104-56.7)\times10^-6})\\C_v=3.26R\\C_v=27.1\ \rm J/mol.K[/tex]
Also we know that
Let[tex]C_p[/tex] be the molar specific heat of the gas at constant pressure given by
[tex]C_p-C_v=R\\C_p=R+C_v\\C_p=4.26R\\C_p=35.42\ \rm J/mol.K[/tex]
Electron kinetic energies are often measured in units of electron-volts (1 eV 1.6 x 10-19 J), which is the kinetic energy of an electron that is accelerated through a 1 volt potential. When an aluminum plate is irradiated with UV light of 253.5 nm wavelength the ejected electrons are observed to have an average kinetic energy of about 0.8 eV. Use these results to determine the electron binding energy (or "work function") o of aluminum (in eV units).
Answer:
4.1 eV
Explanation:
Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J
wavelength, λ = 253.5 nm = 253.5 x 10^-9 m
According to the Einstein energy equation
[tex]E = W_{o}+K[/tex]
Where, E be the energy incident, Wo is the work function and K is the kinetic energy.
h = 6.634 x 10^-34 Js
c = 3 x 10^8 m/s
[tex]E=\frac{hc}{\lambda }=\frac{6.634 \times 10^{-34} \times 3 \times 10^{8}}{253.5\times 10^{-9}}=7.85 \times 10^{-19} J[/tex]
So, the work function, Wo = E - K
Wo = 7.85 x 10^-19 - 1.28 x 10^-19
Wo = 6.57 x 10^-19 J
Wo = 4.1 eV
Thus, the work function of the metal is 4.1 eV.
A highway curve forms a section of a circle. A car goes around the curve. Its dashboard compass shows that the car is initially heading due east. After it travels 830. m, it is heading 15.0° south of east. Find the radius of curvature of its path. (Use the correct number of significant figures.)
Answer:
R = 3170.36m or R = 186.5m
Explanation:
For this problem, we have either trajectory (a), assuming that the car was going south-east, or trajectory (b), assuming the car was going north-east.
In both cases, we know that S = 830m = θ * R. Finding θ, will lead us to the value of R.
For option a:
θ = 15° = 0.2618 rad
[tex]R = \frac{S}{\theta} = 3170.36m[/tex]
For option b:
θ = 270° - 15° = 4.45 rad
[tex]R = \frac{S}{\theta} = 186.5m[/tex]
A red train traveling at 72km/h and a green train traveling at 144km/h are headed towards each other along a straight, level track. When they are 950m apart, each engineer sees the other's train and applies the breaks. The breaks slow each train at the rate of 1.0m/s^2. is there a collision? if so, give the speed of the red train and the speed of the green train at impact, respectively. If not, give the separation between the trains when they stop.
Answer:
given,
speed of red train = 72 km/h = 72× 0.278 = 20 m/s
speed of green train = 144 km/h = 144 × 0.278 = 40 m/s
deceleration of both the train = 1 m/s²
distance between the train when they start decelerating = 950 m
using equation of motion
v² = u² + 2 a s
distance taken by the red train to stop
v² = u² + 2 a s
0 = 20² - 2×1×s
s = 200 m
distance taken by the blue train to stop
v² = u² + 2 a s
0 = 40² - 2×1×s
s = 800 m
so, both train will be cover 200 + 800 = 1000 m
hence, there will be collision between both the trains.
distance traveled by the green train will be 750 m
distance traveled by the red train will be 200 m
so, velocity of the red train will be zero
velocity of the green train will be
v² = u² + 2 a s
v² = 40² - 2 × 1× 750
v = 10 m/s
hence, velocity of the green train will be 10 m/s.
A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energy of the group of charges ?
Answer:
- 2.7 x 10^-6 J
Explanation:
q1 = 1 nC at x = 0 cm
q2 = - 1 nC at x = 1 cm
q3 = 4 nC at x = 2 cm
The formula for the potential energy between the two charges is given by
[tex]U=\frac{Kq_{1}q_{2}}{r}[/tex]
where r be the distance between the two charges
By use of superposition principle, the total energy of the system is given by
[tex]U = U_{1,2}+U_{2,3}+U_{3,1}[/tex]
[tex]U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}[/tex]
[tex]U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}[/tex]
U = - 2.7 x 10^-6 J
You and a friend are driving to the beach during spring break. You travel 14.5 km east and then 66.5 km south in a total time of 50 minutes. (a) What is the average speed of the trip? (b) What is the magnitude of the average velocity? -In this question for part B, why isnt it necessay to find the unit vectors for displacement?
Answer:
(a). The average speed is 97.23 km/hr.
(b). The average velocity is 81.70 km/min.
Explanation:
Given that,
Distance in east = 14.5 km
Distance in south = 66.5 km
Time = 50 min = 0.833 hr
(a). We need to calculate the average speed
Using formula of average speed
[tex]v_{avg}=\dfrac{D}{T}[/tex]
Where, D = total distance
T = total time
Put the value into the formula
[tex]v_{avg}=\dfrac{14.5+66.5}{0.833}[/tex]
[tex]v_{avg}=97.23\ Km/hr[/tex]
(b). We need to calculate the displacement
Using Pythagorean theorem
[tex]d=\sqrt{(d_{e})^2+(d_{s})^2}[/tex]
Put the value into the formula
[tex]d=\sqrt{(14.5)^2+(66.5)^2}[/tex]
[tex]d=68.06\ km[/tex]
(b). We need to calculate the average velocity
Using formula of average velocity
[tex]v_{avg}=\dfrac{d}{t}[/tex]
Where, d = displacement
t = time
Put the value into the formula
[tex]v_{avg}=\dfrac{68.06}{0.833}[/tex]
[tex]v_{avg}=81.70\ km/min[/tex]
Here, We can not necessary to find the unit vectors for displacement because we need to the displacement for find the average velocity.
Hence, (a). The average speed is 97.23 km/hr.
(b). The average velocity is 81.70 km/hr.
An archer standing on a 15 degree slope shoots an arrow at an angle of 26 degrees above the horizontal. How far below its original point of release does the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground?
Answer:
The arrow will hit 112.07 m from the point of release.
Explanation:
The equation for the position of an object in a parabolic movement is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position
v0 = initial velocity
α = launching angle
y0 = initial vertical position
t = time
g = acceleration due to gravity
We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).
Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)
-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t² (g is downward)
0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m
Solving the quadratic equation:
t = 4.5 s ( the negative value is discarded)
Now, with this time we can calculate the horizontal distance:
x = x0 + v0 · t · cos α (x0 = 0, the same as y0)
x = 33 m/s · 4.5 s · cos 41° = 112.07 m
The object comes to 112.07 m below its original point of release the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground.
The equation for the position of an object in a parabolic movement is as follows:
r = (x₀ + v₀ t cos α, y₀ + v₀ t sin α + 1/2 g t²)
y = y₀ + v₀ t sin α + 1/2 g t²
Since the origin of the frame of reference is located at the point where the arrow is released, y₀ = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)
-1.88 m = 33 m/s sin 41° t - 1/2 9.8 m/s² t²
0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m
Solving the quadratic equation:
t = 4.5 s
Now, with this time we can calculate the horizontal distance:
x = x₀ + v₀ t cos α
x = 33 m/s · 4.5 s · cos 41° = 112.07 m
The object comes to 112.07 m below its original point of release the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground.
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