Answer:
Explanation:
Magnetic field B = 6 X 10⁻⁵ T.
Width of car = L (Let )
Velocity of car v = 100 km/h
= 27.78 m /s
induced emf across the body ( width ) of the car
= BLv
= 6 X 10⁻⁵ L X 27.78
166.68 X 10⁻⁵ L
Induced electric field across the width
= emf induced / L
E = 166.68 X 10⁻⁵ N/C
We suppose breadth of a typical car = 1.5 m
potential difference induced
= 166.68 x 1.5 x 10⁻⁵
250 x 10⁻⁵ V
= 2.5 milli volt.
The side of the car which is positively charged depends on the direction in which car is moving , whether it is moving towards the north or south.
Starting from rest, a runner reaches a speed of 2.8 m/s in 2.1 s. In the same time 2.1 s time, a motorcycles increases speed from 37. 0 to 43.0 m/s. In both cases, assume the acceleration is constant: (a). What is the acceleration (magnitude only) of the runner?(b). What is the acceleration (magnitude only) of the motorcycle?(c). Does the motorcycle travel farther than the runner during the 2.1 s? (yes or no)If som how much father? (if not, enter zero)?
Answer:
a) Acceleration of runner is 1.33 m/s²
b) Acceleration of motorcycle is 2.85 m/s²
c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2[/tex]
Acceleration of runner is 1.33 m/s²
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2[/tex]
Acceleration of motorcycle is 2.85 m/s²
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m[/tex]
The runner moves 2.94 m
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m[/tex]
The motorcycle moves 84.21 m
The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree. At some instant, it moves with a velocity of −1.29 m/s−1.29 m/s . Then, 2.91 s2.91 s later, it moves with a velocity of 1.77 m/s1.77 m/s . What is the chipmunk's average acceleration during the 2.91 s2.91 s time interval?
Answer:
1.05 ms⁻²
Explanation:
Acceleration = change in velocity / Time
Change in velocity = Final velocity - initial velocity
= 1.77 - (-1.29)
= 1.77 + 1.29
= 3.06 m/s
Time = 2.91
Acceleration = 3.06 / 2.91
= 1.05 ms⁻² .
A woman on a bridge 90.0 m high sees a raft floating at
aconstant speed on the river below. She drops a stone fromrest in
an attempt to hit the raft. The stone is releasedwehn the raft has
6.00 m more to travel before passing under thebridge. The stone
hits the water 2.00 m in front of theraft. Find the speed of the
raft.
Answer:
0.93 m/s
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement = 90 m
a = Acceleration = 9.81 m/s²
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 90=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{90\times 2}{9.81}}\\\Rightarrow t=4.3\ s[/tex]
So, the raft covered 6-2 = 4 m in 4.3 seconds
Speed = Distance / Time
[tex]\text{Speed}=\frac{4}{4.3}=0.93\ m/s[/tex]
Speed of the raft is 0.93 m/s
To find the speed of the raft, we can use the principle of conservation of energy. When the woman drops the stone, it starts with potential energy due to its height and then converts to kinetic energy as it falls.
Explanation:To find the speed of the raft, we can use the principle of conservation of energy. When the woman drops the stone, it starts with potential energy due to its height and then converts to kinetic energy as it falls. The kinetic energy of the stone when it hits the water is equal to the potential energy it had initially. We can use the equation:
mgh = 0.5mv^2
Where m is the mass of the stone, g is the acceleration due to gravity, h is the height of the bridge, and v is the speed of the stone when it hits the water. Rearranging the equation, we can solve for v:
v = √(2gh)
Substituting the given values h = 90.0 m and g = 9.8 m/s^2, we can calculate the speed of the stone when it hits the water. This speed is equal to the speed of the raft.
A motorboat is moving at 4.0 m/s when it begins to accelerate at 1.0 m/s^2. To the nearest tenth of a second, how long does it take for the boat to reach a speed of 17.0 m/s? Please show work.
Answer:
Time taken by motorboat to reach [tex]17.0m/s[/tex] equals 13 seconds.
Explanation:
From the first equation of kinematics we have
[tex]v=u+at[/tex]
where,
'v' is the final speed of the accelerating object
'u' is the initial speed of the object
'a' is the accleration of the object
't' is the time for which the object accelerates
Applying the given values in the equation above we get
[tex]17=4+1.0\times t\\\\\\\therefore t=17-4=13seconds[/tex]
A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North of East. a) Sketch a plot of the vector sum of this motion. b) Use vector math to find his total displacement in component form. c) Convert to magnitude and direction form. d) How far is the hippo from his starting point? Note: this is distance, a scalar. What total distance has the hippo traveled?
Answer:
a) Please, see the attched figure
b) Total displacement R = (78.3 km; -4.8 km)
c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))
d) The hippo is 78.4 km from his starting point.
The total distance traveled is 102 km
Explanation:
a)Please, see the attached figure.
b) The vector A can be expressed as:
A = (magnitude * cos α; magnitude * sin α)
Where
magnitude = 42 km
α= 0
Then,
A = (42 km ; 0) or 42 km i
In the same way, we can proceed with the other vectors:
B = ( Bx ; By)
where
(apply trigonometry of right triangles: sen α = opposite / hypotenuse and
cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)
Bx = 28 km * sin 25 = 11.8 km
By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)
B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j
C = (Cx; Cy)
where
Cx = 32 km * cos 40° = 24.5 km
Cy = 32 km * sin 40 = 20.6 km
C = (24.5 km; 20.6 km)
Then:
R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)
= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j
c) R = (78.3 km; -4.8 km)
The magnitude of R is:
[tex]magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km[/tex]
Using trigonometry, we can calculate the angle:
Knowing that
tan α = opposite / adjacent
and that
opposite = Ry = -4.8 km
adjacent = Rx = 78.3 km
Then:
tan α = -4.8 km / 78.4 km
α = -3.5°
We can now write the vector R in magnitude and direction form:
R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))
d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):
magnitude R = 78. 4 km
The total distance traveled is the sum of the magnitudes of each vector:
Total distance = 42 km + 28 km + 32 km = 102 km
Tom Sawyer runs 10 m/s down the dock and leaps on to his floating raft already moving 1.5 m/s away from shore. If Tom weighs 70 kg and the raft weighs 130 kg, what speed will they both be moving? A. 895 kg m/sB. 11.5 m/sC. 6.4 m/sD. 5.75 m/s
Answer:
Not the right answer in the options, speed is 4.47 m/s, and the procedure is coherent with option A
Explanation:
Answer A uses mass and velocity units, which are momentum units. By using the conservation of momentum:
.[tex]p_{initial} =p_{final} \\m_{Tom}*v_{Tom}+m_{raft}*v_{raft}=(m_{Tom}+m_{raft})*v_{both} \\70*10+130*1.5 kg*m/s=895kg*m/s\\v_{both}=\frac{895 kg*m/s}{200 kg} =4.47 m/s[/tex]
Since Tom stays in the raft, then both are moving with the same speed. From the options, the momentum is in agreement with option A, however, the question asks for speed.
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?
b)How long are they in the air?
Answer:
a) 4.45 m/s
b) 0.9 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s[/tex]
a) The vertical speed when the player leaves the ground is 4.45 m/s
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s[/tex]
Time taken to reach the maximum height is 0.45 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s[/tex]
Time taken to reach the ground from the maximum height is 0.45 seconds
b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds
Suppose your hair grows at the rate of 1/26 inches per day. Find the rate at which it rows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, you answer suggests how rapidly layers of atoms are assembled in this protein synthesis. Your units should be "atomic layers/sec" Hint : Use dimensional analysis
Answer:
11.306 nm/s
or
113.06 atomic layers/sec
Explanation:
Hello!
First we need to know how much an inch equals in nanometers and a day in seconds:
Since 1inch = 2.54cm and 1cm=10^7nm
1 inch = 2.54 * 10^7 nm
Also 1day = 24hours = 24*60minutes = 24*60*60seconds
1 day = 86.4 * 10^3 s
Therefore the rate at which the hair grows in nanometers per seconds is:
1/26 in/day = (1/26) * (2.54*10^7)/(86.4*10^3) = 11.306 nm/s
Now, if 1 atomic layer = 0.1 nm this means that 1 nm = 10 atomic layers.
Therefore:
The rate in atomic layers is
11.306 nm/s = 11.306 (10 atomic layers)/s = 113.06 atomic layers/sec
13. You throw a ball vertically upward, and as it leaves your hand, its speed is 37.0 m/s. How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point? (A) 1.38 seconds (B) 2.28 seconds (C) 3.78 seconds (D) 4.38 seconds (E) 5.18 seconds
Answer:
(C) 3.78 seconds
Explanation:
At the highest point, the velocity is equal to 0m/s
[tex]v_{f}=v_{o}-gt[/tex]
[tex]t=\frac{v_{o}}{g}[/tex] ; t is the time to reach the highest point
The time the ball takes to return to its starting point after the ball reach its maximum height is the same:
[tex]T_{descent}=t=\frac{v_{o}}{g}=\frac{37}{9.81}=3.78s[/tex]
If the length of a wire is increased by 20% keeping its volume constant. what will be the % change in heat produced when connected across same potential difference. please explain properly!!
Answer:decreases by 30.55%
Explanation:
Given
length of wire is increased by 20 % keeping volume constant
Let the length of wire be L and its area of cross section be A
Thus new length=1.2 L
Volume is constant
[tex]AL=1.2 L\times A'[/tex]
A'=0.833 A
and resistance is given by
[tex]R=\frac{\rho L}{A}[/tex]
where [tex]\rho [/tex]=resistivity
New resistance [tex]R'=\frac{\rho\times 1.2L}{0.833A}[/tex]
R'=1.44 R
heat produced for same potential
[tex]H_1=\frac{V^2t}{R}[/tex]
[tex]H_2=\frac{V^2t}{1.44R}=0.694H_1[/tex]
% change in heat
[tex]\frac{H_2-H_1}{H_1}\times 100[/tex]
[tex]=\frac{0.694-1}{1}[/tex]
=30.55 decreases
Answer:
30.55 %
Explanation:
Assumptions:
l = initial length of the wireL = final length of the wirev = initial volume of the wireV = final volume of the wirea = initial cross sectional area of the wireA = final cross sectional area of the wireh = initial heat of generated by the wireH = final heat generated by the wireP = potential difference across the wiret = time for which the potential difference is created across the wirer = initial resistance of the wireR = final resistance of the wire[tex]\Delta H[/tex] = change in heat producedAccording to the question, we have
[tex]L = l + 20\ \% l = \dfrac{120l}{100}\\V=v\\\Rightarrow LA=la\\\Rightarrow A= \dfrac{la}{L}\\\Rightarrow A= \dfrac{la}{\dfrac{120l}{100}}\\\Rightarrow A= \dfrac{100a}{120}[/tex]
Using the formula of resistance of a wire in terms of its length, cross sectional area and the resistivity of the material, we have
[tex]r = \dfrac{\rho l}{a}\\R=\dfrac{\rho L}{A}=\dfrac{\rho\times \dfrac{120l}{100} }{\dfrac{100a}{120}}=(\dfrac{120}{100})^2\dfrac{\rho l}{a}= 1.44r\\[/tex]
Using the formula of heat generated by the wire for potential diofference created across its end for time t, we have
[tex]h = \dfrac{P^2}{r}t\\H = \dfrac{P^2}{R}t= \dfrac{P^2}{1.44r}t\\\therefore \Delta H = h-H\\\Rightarrow \Delta H = \dfrac{P^2}{r}t-\dfrac{P^2}{1.44r}t\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(-\dfrac{1}{1.44})\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(\dfrac{0.44}{1.44})\\\therefore \textrm{Percentage change in the heat produced}= \dfrac{\Delta H}{h}\times 100\ \%= \left (\dfrac{\dfrac{P^2t}{r}(\dfrac{0.44}{1.44})}{\dfrac{P^2}{r}t} \right )\times 100\ \% = 30.55\ \%[/tex]
Hence, the percentage change in the heat produced in the wire is 30.55 %.
If a lens has a power of -14.50, what is the focal length in mm?
Answer:
Focal length of the lens, f = - 68 mm
Explanation:
Given that,
Power of a lens, P = -14.50 D
We need to find the focal length of the lens. We know that the focal length and the power of lens has inverse relationship. Mathematically, it is given by :
[tex]f=\dfrac{1}{P}[/tex]
f is the focal length of the lens
[tex]f=\dfrac{1}{-14.50}[/tex]
f = -0.068 m
or
f = -68 mm
So, the focal length of the lens is (-68 mm). Hence, this is the required solution.
An apple falls (from rest) from a tree. It hits the ground at a speed of about 4.9 m/s. What is the approximate height (in meters) of the tree above the ground? The magnitude of the gravitational acceleration g = 9.8 m/s2 Enter your answer in meters. Keep 2 decimal places.
Answer:
The inicial height of the apple is 1.22 meters
Explanation:
Using the equation for conservarion of mechanical energy:
[tex]E=V+K=constant[/tex]
[tex]K_i=\frac{1}{2}mv_i^2[/tex] where v is the velocity
[tex]V=mgh[/tex]where h is the height
We equate the initial mechanical energy to the final:
Since [tex]v_0=0\ and h_f=0 [/tex]:
[tex]\frac{1}{2}mv_0^2+mgh_0= \frac{1}{2}mv_f^2+mgh_f\\gh_0= \frac{1}{2}v_f^2[/tex]
Solving for h:
[tex]h_0=\frac{4.9^2}{2g}= 1.22 m[/tex]
For a positive point charge, the electric field vectors point in what direction? a) Point charges cannot create an electric field.
b) Along a circle around it.
c) Toward it.
d) Away from it.
e) None of the above.
Answer:d- Away from it
Explanation:
For a positive point charge, the electric field vectors point away from the charge. Electric field line radiates out of positive charge and could terminate to a negative charge if it is placed in its vicinity.
Similarly for negative charge electric field lines seems to come inside of negative charge. It is basically opposite of positive charge.
The density of a rock will be measured by placing it into a graduated cylinder partially filled with water, and then measuring the volume of water displaced. The density D is given by D = m/(V1 − V0), where m is the mass of the rock, V0 is the initial volume of water, and V1 is the volume of water plus rock. Assume the mass of the rock is 750 g, with negligible uncertainty, and that V0 = 500.0 ± 0.1 mL and V1 = 813.2 ± 0.1 mL. Estimate the density of the rock, and find the uncertainty in the estimate.
Answer:
[tex]\rho = 2.39 g/mL[/tex]
[tex]\Delta \rho = 1.53 \times 10^{-3} mL[/tex]
Explanation:
As we know that density is the ratio of mass and volume of the object
here we know that
mass of the rock is
[tex]m = 750 g[/tex]
volume of the rock is given as
[tex]V = V_1 - V_o[/tex]
here we know that
[tex]V_1 = 813.2 \pm 0.1 mL[/tex]
[tex]V_2 = 500.0 \pm 0.1 mL[/tex]
now we have
[tex]V = 313.2 \pm 0.2 mL[/tex]
now density is given as
[tex]\rho = \frac{750}{313.2}[/tex]
[tex]\rho = 2.39 g/mL[/tex]
now uncertainty of density is given as
[tex]\Delta \rho = \frac{\Delta V}{V} \rho[/tex]
[tex]\Delta \rho = \frac{0.2}{313.2}(2.39)[/tex]
[tex]\Delta \rho = 1.53 \times 10^{-3} mL[/tex]
The density of the rock, given its mass is 750 g and the volume of water displaced is 313.2 mL, is 2.394 g/mL. The uncertainty in this measurement is ± 0.0015 g/mL, considering an uncertainty of ± 0.1 mL for both the initial and final volume measurements.
Given that the mass (m) of the rock is 750 g, the initial volume of water (V0) is 500.0 mL, and the volume of water plus the rock (V1) is 813.2 mL, we can determine the density (D) and the uncertainty in the density.
Using the formula:
D = m / (V1 - V0)
Therefore, D = 750 g / (813.2 mL - 500.0 mL)
= 750 g / 313.2 mL
= 2.394 g/mL.
Using the uncertainties in V0 and V1, which are both ± 0.1 mL. Since we subtract these volumes, the total volume uncertainty is ± (0.1 mL + 0.1 mL) = ± 0.2 mL. Thus, the uncertainty in the density (ΔD) can be approximated by the formula:
ΔD = D × (ΔV / (V1 - V0))
where ΔV is the total volume uncertainty. Substituting the values, we get ΔD = 2.394 g/mL × (0.2 mL / 313.2 mL) = ± 0.00153 g/mL (rounded to four significant figures).
Therefore, the estimated density of the rock is 2.394 ± 0.0015 g/mL.
A car is driven east for a distance of 47 km, then north for 23 km, and then in a direction 32° east of north for 27 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
Answer:
(a). The car's total displacement from its starting point is 76.58 m.
(b). The angle of the car's total displacement measured from its starting direction is 36.81°.
Explanation:
Given that,
Distance = 47 km in east
Distance = 23 km in north
Angle = 32° east of north
Distance = 27 km
According to figure,
Angle = 90-32 = 58°
(a). We need to calculate the magnitude of the car's total displacement from its starting point
Using Pythagorean theorem
[tex]AC=\sqrt{AB^2+BC^2}[/tex]
[tex]AC=\sqrt{(47+27\cos58)^2+(23+27\sin58)^2}[/tex]
[tex]AC=76.58\ m[/tex]
The magnitude of the car's total displacement from its starting point is 76.58 m.
(b). We need to calculate the angle (from east) of the car's total displacement measured from its starting direction
Using formula of angle
[tex]\tan\theta=\dfrac{y}{x}[/tex]
put the value into the formula
[tex]\theta=tan^{-1}\dfrac{23+27\sin58}{47+27\cos58}[/tex]
[tex]\theta=tan^{-1}0.7486[/tex]
[tex]\theta=36.81^{\circ}[/tex]
Hence, (a). The car's total displacement from its starting point is 76.58 m.
(b). The angle of the car's total displacement measured from its starting direction is 36.81°.
What is the acceleration of a 20 kg cart if the net force on it is 40 N?
Answer:
Acceleration of the cart will be [tex]a=2m/sec^2[/tex]
Explanation:
We have given force F = 40 N
Mass of the cart = 20 kg
From newton's second law we know that force, mass and acceleration are related to each other
From second law of motion force on any object moving with acceleration a is given by
F = ma, here m is mass and a is acceleration
So [tex]40=20\times a[/tex]
[tex]a=2m/sec^2[/tex]
A ball is thrown vertically into the air with a initial velocity of 20 m/s. Find the maximum height of the ball and find the amount of time needed to reach the maximum height.
Answer:
The maximum height of the ball is 20 m. The ball needs 2 s to reach that height.
Explanation:
The equation that describes the height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the ball at time t
y0 = initial height
v0 = initial velocity
t = time
g = acceleration
v = velocity at time t
When the ball is at its maximum height, its velocity is 0, then, using the equation of the velocity, we can calculate the time at which the ball is at its max-height.
v = v0 + g · t
0 = 20 m/s - 9.8 m/s² · t
-20 m/s / -9.8 m/s² = t
t = 2.0 s
Then, the ball reaches its maximum height in 2 s.
Now, we can calculate the max-height obtaining the position at time t = 2.0 s:
y = y0 + v0 · t + 1/2 · g · t²
y = 0 m + 20 m/s · 2 s - 1/2 · 9,8 m/s² · (2 s)²
y = 20 m
The maximum height reached by the ball is 20.4 meters, and it takes approximately 2.04 seconds to reach this height.
When a ball is thrown vertically into the air with an initial velocity of 20 m/s, we can calculate the maximum height using the kinematic equation:
[tex]v^2 = u^2 + 2gh,[/tex]
where v is the final velocity (0 m/s at the highest point), u is the initial velocity (20 m/s), g is the acceleration due to gravity (9.81 m/s2), and h is the maximum height. Solving for h gives us:
[tex]h = u^2 / (2g).[/tex]
By substituting the values we get:
[tex]h = (20 m/s)^2 / (2 * 9.81 m/s^2) = 20.4 m.[/tex]
To find the time needed to reach the maximum height, we use the equation:
v = u + gt,
Solving for t when v is 0 m/s, we get:
t = u / g = 20 m/s / 9.81 m/s2 = approx. 2.04 seconds.
Thus, the maximum height of the ball is 20.4 meters and the time needed to reach the maximum height is approximately 2.04 seconds.
Which of the following combinations of position (x) and direction of motion would give a velocity in the x-direction that has a negative value? a. Positive x, moving towards the origin.b. Negative x, moving away from the origin.c. Both a and b.d. Positive x, moving away from the origin.e. Negative x, moving towards the origin.
Answer:
c) Both a) and b) are the combinations that have a negative velocity.
Explanation:
The velocity is given by this equation:
v = Δx / Δt
Where
Δx = final position - initial position
Δt = elapsed time
Now let´s evaluate the options. We have to find those combinations in which Δx < 0 since Δt is always positive.
a) If the initial position, x, is positive and you move towards the origin, the final position will be a smaller value than x. Then:
final position < initial position
final position - initial position < 0 The velocity will be negative.
b) If x is negative and you move away from the origin, the final position will be a more negative number than x. Again:
final position < initial position
final position - initial position < 0 The velocity will be negative.
Let´s do an example to show it:
initial position = -5
final position = -10 (since you moved away from the origin)
final position - initial position = -10 -(-5) = -5
d) If x is positive and you move away from the origin, the final position will be a greater value than the initial position. Then:
final position > initial position
final position - initial position > 0 The velocity will be positive.
e) If x is negative and you move towards the origin, the final position will be a greater value than the initial position. Then:
final position > initial position
final position - initial position > 0 The velocity will be positive.
Let´s do an example:
initial position = -10
final position = -5
final postion - initial position = -5 - (-10) = 5
or with final position = 0
final postion - initial position = 0 -(-10) = 10
And so on.
The right answer is c) Both a) and b) are the combinations that have a negative velocity.
A thin metal bar, insulated along its sides, is composed of five different metal connected together. The left end bar is immersed in a heat bath at 100°C and right end in a heat bath at 0°C. Starting at the left end, the pieces and lenghts are steel(2cm), brass(3cm), copper(1cm), aluminum(5cm) and silver(1cm). What is the temperature of the steel/brass interface?
Answer:
T = 61.06 °C
Explanation:
given data:
a thin metal bar consist of 5 different material.
thermal conductivity of ---
K {steel} = 16 Wm^{-1} k^{-1}
K brass = 125 Wm^{-1} k^{-1}
K copper = 401 Wm^{-1} k^{-1}
K aluminium =30Wm^{-1} k^{-1}
K silver = 427 Wm^{-1} k^{-1}
[tex]\frac{d\theta}{dt} = \frac{KA (T_2 -T_1)}{L}[/tex]
WE KNOW THAT
[tex]\frac{l}{KA} = thermal\ resistance[/tex]
total resistance of bar = R steel + R brass + R copper + R aluminium + R silver
[tex]R_{total} =\frac{1}[A} [\frac{0.02}{16} +\frac{0.03}{125} +\frac{0.01}{401} +\frac{0.05}{30} +\frac{0.01}{427}][/tex]
[tex]R_{total} =\frac{1}[A} * 0.00321[/tex]
let T is the temperature at steel/brass interference
[tex]\frac{d\theta}{dt}[/tex] will be constant throughtout the bar
therefore we have
[tex]\frac{100-0}{R_{total}} = \frac{100-T}{R_{steel}}[/tex]
[tex]\frac{100-0}{0.00321} *A = \frac{100-T}{0.00125} *A[/tex]
solving for T we get
T = 61.06 °C
A vacuum gage attached to a power plant condenser gives a reading of 27.86 in. of mercury. The surrounding atmospheric pressure is 14.66 lbf/in. Determine the absolute pressure inside the condenser, in lbf/in. The density of mercury is 848 lb/ft and the acceleration of gravity is g = 32.0 ft/s?.
Answer:
absolute pressure = 1.07 lbft/in^2
Explanation:
given data:
vaccum gauge reading h = 27.86 inch = 2.32 ft
we know that
gauge pressure p is given as
[tex]p = \rho gh[/tex]
[tex]p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft = 62955.52 \ lbft/s^2 * 1/ft^2[/tex]
we know that [tex]1\ lb ft\s^2 = \frac{1}{32.174}\ lbft[/tex]
1 ft = 12 inch
therefore [tex]p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2[/tex]
[tex]p = 13.59\ lbft/in^2[/tex]
[tex]P_{atm} = 14.66\ lbf/in^2[/tex]
so absolute pressure = [tex]P_{atm} - p[/tex]
= 14.66 - 13.59 = 1.07 lbft/in^2
Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L = 2 cm apart. Assume that the surfaces are black (emissivity ε = 1). Determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is filled with atmospheric air.
Answer:
[tex]Q=81.56\ W/m^2[/tex]
Explanation:
Given that
[tex]T_1= 210 K[/tex]
[tex]T_2= 150 K[/tex]
Emissivity of surfaces(∈) = 1
We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies
[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]
So now by putting the values
[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]
[tex]Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2[/tex]
[tex]Q=81.56\ W/m^2[/tex]
So rate of heat transfer per unit area
[tex]Q=81.56\ W/m^2[/tex]
An object is dropped from rest and falls through height h. It travels 0.5h in the last 1 second of fall. Find the total time & height of the fall. (Hint: use two triangles!)
Answer:
3.41 s
114 m
Explanation:
The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.
It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.
0.5*h = 1/2 * a * t1^2
h = a * t1^2
It travels 0.5*h in 1 second.
h = X(t1 + 1) = 1/2 * a * (t1+1)^2
Equating both equations:
a * t1^2 = 1/2 * a * (t1+1)^2
We simplify a and expand the square
t1^2 = 1/2 * (t1^2 + 2*t1 + 1)
t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0
1/2 * t1^2 - t1 - 1/2 = 0
Solving electronically:
t1 = 2.41 s
total time = t1 + 1 = 3.41.
Now
h = a * t1^2
h = 9.81 * 3.41^2 = 114 m
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of v0 = 25.0 m/s and at an angle of θ0 = 41.0°. What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?
(a) 18.9 m/s
The motion of the stone consists of two independent motions:
- A horizontal motion at constant speed
- A vertical motion with constant acceleration ([tex]g=9.8 m/s^2[/tex]) downward
We can calculate the components of the initial velocity of the stone as it is launched from the ground:
[tex]u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s[/tex]
The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.
When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore
[tex]v=18.9 m/s[/tex]
(b) 22.2 m/s
We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by
[tex]v_y^2 - u_y^2 = 2ah[/tex] (1)
where
[tex]u_y = 16.4 m/s[/tex] is the initial vertical velocity
[tex]v_y[/tex] is the vertical velocity at height h
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration due to gravity (negative because it is downward)
At the top of the parabolic path, [tex]v_y = 0[/tex], so we can use the equation to find the maximum height
[tex]h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m[/tex]
So, at half of the maximum height,
[tex]h = \frac{13.7}{2}=6.9 m[/tex]
And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:
[tex]v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s[/tex]
And so, the speed of the stone at half of the maximum height is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s[/tex]
(c) 17.4% faster
We said that the speed at the top of the trajectory (part a) is
[tex]v_1 = 18.9 m/s[/tex]
while the speed at half of the maximum height (part b) is
[tex]v_2 = 22.2 m/s[/tex]
So the difference is
[tex]\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s[/tex]
And so, in percentage,
[tex]\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%[/tex]
So, the stone in part (b) is moving 17.4% faster than in part (a).
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.1 × 10^-7 C/m^2, and the plates are separated by a distance of 1.2 × 10^-2 m. How fast is the electron moving just before it reaches the positive plate?
Explanation:
An electron is released from rest, u = 0
We know that charge per unit area is called the surface charge density i.e. [tex]\sigma=\dfrac{q}{A}=2.1\times 10^{-7}\ C/m^2[/tex]
Distance between the plates, [tex]d=1.2\times 10^{-2}\ m[/tex]
Let E is the electric field,
[tex]E=\dfrac{\sigma}{\epsilon_o}[/tex]
[tex]E=\dfrac{2.1\times 10^{-7}}{8.85\times 10^{-12}}[/tex]
E = 23728.81 N/C
Now, [tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 23728.81}{9.1\times 10^{-31}}[/tex]
[tex]a=4.17\times 10^{15}\ m/s^2[/tex]
Let v is the speed of the electron just before it reaches the positive plate. So, third equation of motion becomes :
[tex]v^2=2ad[/tex]
[tex]v^2=2\times 4.17\times 10^{15}\times 1.2\times 10^{-2}[/tex]
[tex]v=10.003\times 10^6\ m/s[/tex]
Hence, this is the required solution.
A 3.0 mg bead with a charge of 2.9 nC rests on a table. A second bead, with a charge of -5.3 nC is directly above the first bead and is slowly lowered toward it. What is the closest the centers of the two beads can be brought together before the lower bead is lifted off the table?
Answer:
6.86 cm
Explanation:
Given:
q = charge on the first bead on the table= [tex]2.9\ nC = 2.9\times 10^{-9}\ C[/tex]m = mass of bead on the table = [tex]3.0\ mg = 3.0\times 10^{-6}\ kg[/tex]Q = charge on the second bead = [tex]-5.3\ nC = -5.3\times10^{-9}\ C[/tex]Assume:
r = the closest distance between the centers of the beadsF = electrostatic force of attraction between the two beadsW = weight of the first beadg = acceleration due to gravity = 9.8\ m/s^2N = normal force on the first beadWhen the first bead rests on the table, then electrostatic force due to the second bead acts on it in the upward direction, Normal force acts in the upward direction and its weight in the downward direction.
So, using Newton's second law on the first bead resting on the table, we have
[tex]F+N-W=0\\[/tex]
At the closest distance of the second bead to the first bead, it just lifts off the table and the normal force becomes zero.
[tex]\therefore F-W=0\\\Rightarrow F=W\\\Rightarrow \dfrac{kqQ}{r^2}=mg\\\Rightarrow r^2=\dfrac{kqQ}{mg}\\\Rightarrow r^2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}\times 5.3\times 10^{-9}}{3\times 10^{-6}\times 9.8}\\\Rightarrow r^2=4.70\times 10^{-3}\\\textrm{Taking square root on both the sides}\\r = \pm 0.0686\ m\\\textrm{Since the distance is never negative}\\\therefore r = 0.0686\ m\\\Rightarrow r = 6.86\ cm[/tex]
Hence, the centers of the two beads must be brought closest to 6.86 cm before the lower bead is lifted off the table.
Force is a vector, while mass is a scalar. Why can we use mass as an indicator of the magnitude of the force vector?
Answer:
Explanation:
Force = mass x acceleration
[tex]\overrightarrow{F}=m\overrightarrow{a}[/tex]
Force is always vector and acceleration also vector but the mass is a saclar quanity.
here, the direction of force vector is same as the direction of acceleration vector but the magnitude of force depends on the magnitude of mass of the body.
Is mass is more, force is also more.
Thus, the mass is like an indicator of the magnitude of force.
If the speed of an object in uniform circular motion is tripled, the magnitude of the centripetal acceleration increases by a factor of: (A) 2 (B) 3 (C) 9 (D) 6 (E) 8
Answer:
The correct option is 'D': 9
Explanation:
We know that the magnitude of the centripetal acceleration of a body moving in circular orbit of radius 'r' with speed 'v' is given by
[tex]a_{c}=\frac{v^{2}}{r}[/tex]
Now when the speed of the body is tripled the speed becomes [tex]3v[/tex]
Hence the new centripetal acceleration is obtained as
[tex]a'_{c}=\frac{(3v)^{2}}{r}\\\\a'_{c}=\frac{9v^{2}}{r}=9a_{c}[/tex]
Thus we can see that the new centripetal acceleration becomes 9 times the oroginal value.
The acceleration of a body traveling in a circular route is known as centripetal acceleration. The magnitude of the centripetal acceleration increases by a factor of 9.
What is centripetal acceleration?The acceleration of a body traveling in a circular route is known as centripetal acceleration. Because velocity is a vector quantity. It has both a magnitude and a direction.
When a body moves on a circular route, its direction changes constantly, causing its velocity to vary, resulting in acceleration.
Mathematically it is given as,
[tex]\rma_c=\frac{v^2}{r} \\\\ a_c'=\frac{(3v)^2}{r} \\\\ \rm v=9\frac{v^2}{r}\\\\ a_c'=9a_c[/tex]
Hence the magnitude of the centripetal acceleration increases by a factor of 9. Option c is correct.
To learn more about centripetal acceleration refer to the link;
https://brainly.com/question/17689540
Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball at 8 m/s from 50 m above the ground at the same time Gary throws a ball up. At what speed must Gary throw a ball up in order for the two balls to cross paths 25 m above the ground? The starting height of the ball thrown upward is 1.5 m above the ground. Ignore the effects of air resistance. whats the answer in m/s?
Answer:22.62 m/s
Explanation:
Given
two balls are separated by a distance of 50 m
Alex throws the ball from a height of 50 m with a velocity of 8 m/s and Gary launches a ball with some velocity exactly at the same time.
ball from ground travels a distance of 25 m in t sec
For Person on tree
[tex]25=ut+\frac{1}{2}gt^2[/tex]
[tex]25=8t+\frac{1}{2}\times 9.81\times t^2--------1[/tex]
For person at ground
[tex]23.5=ut-\frac{1}{2}gt^2---------2[/tex]
Solve equation (1)
[tex]50=16t+9.81t^2[/tex]
[tex]9.81t^2+16t-50=0[/tex]
[tex]t=\frac{-16\pm\sqrt{256+4\times 50\times 9.81}}{2\times 9.81}=\frac{47.1-16}{19.62}=1.58 s[/tex]
put the value of t in equation 2
[tex]23.5=u\times 1.58-\frac{9.81\times 1.58^2}{2}[/tex]
[tex]u=\frac{35.744}{1.58}=22.62 m/s[/tex]
A man pushes a lawn mower on a level lawn with a force of 195 N. If 37% of this force is directed downward, how much work is done by the man in pushing the mower 5.7 m?
Final answer:
The work done by the man in pushing the lawn mower is 699.245 J, calculated by determining the horizontal force component and multiplying by the distance pushed.
Explanation:
To calculate how much work is done by the man in pushing the lawn mower, we need to consider only the component of the force that acts in the direction of the movement. Since 37% of the 195 N force is directed downward, only the remaining 63% is contributing to the horizontal movement. Therefore, the horizontal component of the force is 0.63 × 195 N = 122.85 N.
The formula to calculate work (W) is W = force (F) × distance (d) × cosine(θ), where θ is the angle between the force and the direction of movement. In this case, the force and movement are in the same direction, so θ = 0 and cosine(θ) = 1. Thus, the work done is:
W = 122.85 N × 5.7 m × 1 = 699.245 J
In terms of energy expended while pushing a lawn mower, this work is a relatively small amount when compared to a person's daily intake of food energy.
A woman is sitting at a bus stop when an ambulance with a siren wailing at 317 Hz approaches at 69 miles per hour (mph). Assume the speed of sound to be 343 m/s. a) How fast is the ambulance moving in meters per second? (perform the necessary unit conversion) Vs= 69 mph = m/s b) What frequency does the woman hear? fa = Hz c) What speed (vs) would the ambulance be traveling in order for the woman to hear the siren at an approaching frequency of 350 Hz? Vs= m/s d) What frequency would she hear as the siren moves away from her at the same speed (as in part c)? fa = Hz
Answer:
a) 30.84m/s
b) 348.32Hz
c) 32.34m/s
d) 289.69Hz
Explanation:
a) If 1 mile=1609,34m, and 1 hour=3600 seconds, then 69mph=69*1609.34m/3600s=30.84m/s
b) Based on Doppler effect:
/*I will take as positive direction the vector [tex]\vec r_{observer}-\vec r_{emiter}[/tex] */
[tex]f_{observed}=(\frac{v_{sound}-v_{observed}}{v_{sound}-v_{emited}})f_{emited}[/tex]
[tex]f_{observed}=(\frac{343m/s-0m/s}{343m/s-30.84m/s})317Hz=348.32Hz[/tex]
c) [tex]350Hz=(\frac{343m/s-0m/s}{343m/s-v_{ambulance}})317Hz, V_{ambulance}=343m/s-\frac{317Hz}{350Hz}.343m/s=32.34m/s[/tex]
d) [tex]f_{observed}=(\frac{343m/s-0m/s}{343m/s+32.34m/s})317Hz=289.69Hz[/tex]