Answer:
C
Explanation:
the correct answer is c) Pre Schematic.
It is in the Pre-schematic stage of development that the children learn to draw human figures by sketching a circle with dangly lines for legs. It remains till the age of six. After pre schematic stage comes the schematic stage, where in children can draw complex structures.
Children drawing human figures with a circle and lines for legs are in the Pre-schematic stage of art development, indicating an early attempt at representational drawing before achieving greater accuracy and detail in later stages.
Explanation:When children draw human figures by sketching a circle with dangly lines for legs, they are demonstrating the Pre-schematic developmental stage in art. This stage is distinct from the earlier scribbling phase, signaling that children have begun to recognize and attempt to represent the human form in a more organized albeit simplified manner. However, this representation lacks the complexity and accuracy found in later stages, such as the Schematic stage, where more detailed and proportionate figures are drawn.
The Pre-schematic stage typically occurs in children aged between 4 and 7 years old. During this time, their cognitive and motor skills are developing rapidly, but they are still learning how to translate what they see and imagine onto paper with greater fidelity. Thus, the circle and lines approach is a developmentally appropriate method for them to begin exploring human figures in their art.
A hiker treks 30 degrees south of east at a speed of 15 m/s for 30 min and then turns due west and hikes at a speed of 8m/s for another 20 min. What is the displacement of this explorer (magnitude and direction)?
Answer:19.3 km,[tex]\theta =44.40^{\circ}[/tex] south of east
Explanation:
Given
Hiker treks [tex]30 ^{\circ}[/tex] south of east at a speed of 15 m/s for 30 min and then turns due to west and hikes at speed of 8 m/s for another 20 min
Let position vector of Hiker at the end of 30 min
[tex]r=27000cos30\hat{i}-27000sin30\hat{j}[/tex]
after he turns west so new position vector of hiker is
[tex]r'=27000cos30\hat{i}-27000sin30\hat{j}-9600\hat{i}[/tex]
[tex]r'=13782.68\hat{i}-13500\hat{j}[/tex]
Therefore Displacement is given by |r'|
[tex]|r'|=\sqrt{13782.68^2+13500^2}[/tex]
[tex]|r'|=19,292.8035 m\ or 19.3 km[/tex]
for direction
[tex]tan\theta =\frac{13500}{13782.68}[/tex]
[tex]\theta =44.40^{\circ}[/tex] south of east
An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at the top 16.0°C, what is the volume of the bubble just before it reaches the surface?
Final answer:
The volume of the air bubble just before it reaches the surface is 3.32 cm^3.
Explanation:
To find the volume of the air bubble just before it reaches the surface, we can use the ideal gas law equation: PV = nRT. Since the pressure and amount of gas remain constant, we can rewrite the equation as V/T = k, where V is the volume, T is the temperature, and k is a constant.
Using the given temperatures at the bottom and top of the lake (5.9°C and 16.0°C) and the initial volume of the bubble (1.22 cm^3), we can set up the following equation:
(1.22 cm^3) / (5.9°C) = V / (16.0°C).
Solving for V, the volume of the bubble just before it reaches the surface, we get:
V = (1.22 cm^3)(16.0°C) / (5.9°C) = 3.32 cm^3.
An engineer is undertaking some reconnaissance by pacing on a construction site. The engineer takes 26 steps per 20m, the ground he is walking on has a vertical slope angle of 6. The engineer takes 152 steps. How far has he travelled horizontally (in metres)?(only enter numeric answer to 2 decimal places, no alpha characters)
Answer:
Thus the distance traveled horizontally is 116.28 m
Solution:
As per the question:
Distance covered in 26 steps = 20 m
Thus
Distance covered in one step, step size = [tex]\frac{20}{26}[/tex]
Total steps taken = 152
Now, the distance covered in 152 steps = [tex]one step size\times 152[/tex]
The distance covered in 152 steps, D = [tex]\frac{20}{26}\times 152 = 116.92 m[/tex]
The above distances are on a slope of [tex]6^{\circ}[/tex] above horizontal.
Thus the horizontal component of this distance is given by:
[tex]d_{H} = Dcos6^{\circ}[/tex]
where
[tex]d_{H}[/tex] = horizontal component of distance, D
[tex]d_{H} = 116.92cos6^{\circ} = 116.28 m[/tex]
A Car travel at a speed of 200 km/hr. How far it will go in 15 mins? 27.7 km 66.1 km7.70 km50.0 km8.33 km"
Answer:
So car will go 50 km in 15 minutes
Explanation:
We have given speed of the car = 200 km/hour
Time t = 15 minutes
We know that 1 hour = 60 minute
So 15 minute = [tex]\frac{15}{60}=0.25hour[/tex]
We have to find the distance
We know that distance = speed ×time = 200×0.25=50 km
So car will go 50 km in 15 minutes
So option (d) will be the correct option
What is the longest wavelength of light that will
emitelectrons from a metal whose work function is 3.10 eV?
Answer:
The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]
Explanation:
According to Einstein's photoelectric equation we have
[tex]E_{incident}\geq \phi [/tex]
where
[tex]E_{incident}[/tex] is the energy of the incident light
[tex]\phi [/tex] is the work function of the metal
The incident energy of the light with wavelength [tex]\lambda [/tex] is given by
[tex]E_{incident}=h\cdot \frac{c}{\lambda}[/tex]
Thus the photoelectric equation reduces to
[tex]h\cdot \frac{c}{\lambda}\geq \phi\\\\h\cdot c\geq \lambda \times \phi\\\\\therefore \lambda\leq \frac{h\cdot c}{\phi}[/tex]
Thus applying values we get
[tex]\lambda\leq \frac{6.62\times 10^{-34}\times 3\times 10^{8}}{3.10\times 1.602\times 10^{-19}}\\\\\therefore \lambda\leq 0.4\times 10^{-6}m[/tex]
Hence The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]
The acceleration of a particle which is moving along a straight line is given by a = –kv0.5, where a is in meters per second squared, k is a constant, and v is the velocity in meters per second. Determine the velocity as a function of both time t and position s. Evaluate your expressions for t = 2.7 sec and at s = 6 m, if k = 0.3 m0.5 sec–1.5 and the initial conditions at time t = 0 are s0 = 2.6 m and v0 = 2.7 m/sec.
Answer:
[tex]v(t)=2.7*e^{0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7-1.3k\\\\[/tex]
For t=2.7s and k=0.3 m/s:
[tex]v(2.7s)=1.80m/s[/tex]
For s=6m and k=0.3 m/s:
[tex]v(6m)=6.69m/s\\\\[/tex]
Explanation:
Definition of acceleration:
[tex]a=\frac{dv}{dt} =0.5kv[/tex]
we integrate in order to find v(t):
[tex]\frac{dv}{v} =-0.5kdt[/tex]
[tex]\int\limits^v_0 { \frac{dv}{v}} \, =-0.5k\int\limits^t_0 {dt} \,[/tex]
[tex]ln(v)=-0.5kt+C\\\\ v=A*e^{-0.5kt}[/tex] A=constant
Definition of velocity:
[tex]v=\frac{ds}{dt} =A*e^{-0.5kt}[/tex]
We integrate:
[tex]v=\frac{ds}{dt} \\s=- (2/k)*A*e^{-0.5kt}+B[/tex] B=constant
But:
[tex]v=A*e^{-0.5kt}[/tex]⇒[tex]s= -(2/k)*v+B[/tex]
[tex]v(s)=-(\frac{k}{2} )(s-B)=D-\frac{k}{2}*s[/tex] D=other constant
Initial conditions:t = 0 are s0 = 2.6 m and v0 = 2.7 m/sec:
[tex]v(t)=A*e^{-0.5kt}\\ 2.7=Ae^{-0.5k*0}\\ 2.7=A\\[/tex]
[tex]v(s)=D-\frac{k}{2}*s\\2.7=D-\frac{k}{2}*2.6\\D=2.7+1.3k[/tex]
So:
[tex]v(t)=2.7*e^{-0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7+1.3k\\\\[/tex]
For t=2.7s and k=0.3 m/s:
[tex]v(2.7s)=2.7*e^{-0.5*0.3*2.7}=1.80m/s[/tex]
For s=6m and k=0.3 m/s:
[tex]v(6m)=\frac{0.3}{2}*6+2.7+1.3*0.3=6.69m/s\\\\[/tex]
A 75 kg astronaut has become detached from their her space ship. To get back to the ship she throws a tool in the opposite direction to the spaceship with a force of 16 N. What is her acceleration during the throw assuming that distances going away from the spaceship are positive?
Answer:
0.21 m/s/s.
Explanation:
Whenever there is an action force acting on a body, there will be a reaction force.
Here the force with which the astronaut throws the tool is given as 16 N.
Force is measured in newtons and is equal to the rate of change of momentum.
Since the astronaut has a mass, she experience a reaction force. It is given by F = ma, according to Newton's 2nd law.
16 = 75 a
⇒ Acceleration = a = F/m = 16/75 = 0.21 m/s/s
The force exerted on a bridge pier in a river is to be tested in a 1:10 scale model using water as the working fluid. In the prototype the depth of water is 2.0 m, the velocity of flow is 1.5 m/s and the width of the river is 20 m. If the hydrodynamic force on the model bridge pier is 5 N, what would it be on the prototype? (hint: pressure ratio is equal to the length-scale ratio) (5 points; Ans: 5000 N)
Answer:
[tex]f_p = 5000 N[/tex]
Explanation:
GIVEN DATA:
pressure ratio = length ratio
force = 5 N
scale = 1:10
velocity = 1.5 m/s
[tex]B_p = 20 m[/tex]
[tex]h_p = 2m[/tex]
As pressure ratio = length ratio so we have
[tex]\frac{p_m}{p_p} =\frac{l_m}{l_p} =\frac{1}{10}[/tex]
[tex]\frac{f_m *A_m}{f_p *A_p} ==\frac{1}{10}[/tex]
[tex]\frac{f_m}{f_p} * \frac{A_p}{A_m} = \frac{1}{10}[/tex]
[tex] \frac{f_m}{f_p} * \frac{b_p*h_p}{b_m*h_m} =\frac{1}{10}[/tex]
[tex]5 * \frac{1}{\frac{1}{10}} *\frac{1}{\frac{1}{10}} = \frac{f_p}{10}[/tex]
solving F_p
[tex]f_p = 5000 N[/tex]
Consider a charged particle at a pointS whose coordinates are (8 m, 4 m, 10 m). We would like to find the electric field vector at a point P whose coordinates are (7 m, 2 m, 6 m). The "unit vector" r ^ is a vector that points from S to P that has length of 1 (or "unity"). What is its y component, in meters
Answer:
E=[tex]k*\frac{q}{21}*u[/tex]
[tex]u=\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]
[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]
Explanation:
q: particle's charge
k: coulomb constant
E=E*u
r=r*u
r=distancia vectorial entre P y S
r=distancia escalar entre P y S
E: Electric field vector
E: magnitud of magnetic field vector
u: unit vector radial
then:
[tex]E=k*q/r^{2}[/tex]
r=r*u
r=P-S=(-1,-2,-4)m
[tex]r^{2}=(Magnitude(P-S))^2=(-1)^2+(-2)^2+(-4)^2=21[/tex]
[tex]r=\sqrt{21}[/tex]
E=[tex]k*\frac{q}{21}*u[/tex]
u=r/r=[tex]\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]
[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]
A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS coordinates are used to determine his displacement as a function of time, x = 0.5t3 + t2 + 2t, where x and t are expressed in feet and seconds, respectively. Determine the position, velocity, and acceleration of the boarder expressed when t = 12 seconds. (Round the final answer to one decimal place.)
To find the position, velocity, and acceleration of the snowboarder at t = 12 seconds, evaluate the given displacement equation and its first and second derivatives at t = 12 seconds, then round the final answers to one decimal place.
Explanation:To determine the position, velocity, and acceleration of the snowboarder at t = 12 seconds, we need to evaluate the given displacement function x(t) = 0.5t³ + t² + 2t and its derivatives at that specific time value.
Position at t = 12 seconds
By substituting t = 12 into the displacement function, we get:
x(12) = 0.5(12)³ + (12)² + 2(12), which can be calculated to give the position x.
Velocity at t = 12 seconds
The velocity v(t) is the first derivative of the displacement function, v(t) = 1.5t² + 2t + 2. Evaluate v(12) to find the velocity at t = 12 seconds.
Acceleration at t = 12 seconds
Acceleration a(t) is the second derivative of the displacement function, a(t) = 3t + 2. Evaluate a(12) to find the acceleration at t = 12 seconds.
Using these steps, you can calculate the exact values and round them to one decimal place as asked in the question.
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic, (b) fcc, (c) bcc, and (d) diamond lattice. As- suming that nearest atoms are touching each other, what is the lattice constant of each lattice?
Answer:
(a) A = [tex]3.90 \AA[/tex]
(b) [tex]A = 4.50 \AA[/tex]
(c) [tex]A = 5.51 \AA[/tex]
(d) [tex]A = 9.02 \AA[/tex]
Solution:
As per the question:
Radius of atom, r = 1.95 [tex]\AA = 1.95\times 10^{- 10} m[/tex]
Now,
(a) For a simple cubic lattice, lattice constant A:
A = 2r
A = [tex]2\times 1.95 = 3.90 \AA[/tex]
(b) For body centered cubic lattice:
[tex]A = \frac{4}{\sqrt{3}}r[/tex]
[tex]A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA[/tex]
(c) For face centered cubic lattice:
[tex]A = 2{\sqrt{2}}r[/tex]
[tex]A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA[/tex]
(d) For diamond lattice:
[tex]A = 2\times \frac{4}{\sqrt{3}}r[/tex]
[tex]A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA[/tex]
What is the density of water vapor in g/m^3 on a hot dry day in the desert when the temperature is 50.0°C and the relative humidity is 6.50%?
Answer:
[tex]Density = 538 \frac{g}{m3} [/tex]
Explanation:
To get the density you need the vapor pressure for the moisture, to get this first you need to find in tables (found in internet, books, apps) the saturation vapor pressure for water at 50°C:
[tex]P_{sat} = 12350 Pa @ 50°C[/tex]
Now, a relative humidity of 6,5% means that the actual vapor pressure is 6,5% that of the saturated air so:
[tex]P_{vapor} = 12350 Pa * 0,065 = 802,75 Pa = 0,792 atm [/tex]
According to the ideal gases formula density can be calculated as:
[tex]d = \frac{P*M}{R*T}[/tex]
Where:
Ideal gases constant [tex]R = 0,082 \frac{atm L}{mol k}[/tex]
Pressure P = 0,792 atm
Temperature T = 50°C = 323 K
molar mass M = 18 g/mol
[tex]d = 0,538 \frac{g}{L} = 538 \frac{g}{m3} [/tex]
Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart. What is the magnitude of the electric force between two protons separated by 2.00 ✕ 10-15 m
Answer:
The magnitude of the electric force between the to protons will be 57.536 N.
Explanation:
We can use Coulomb's law to find out the force, in scalar form, will be:
[tex]F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}[/tex].
Now, making the substitutions
[tex]d \ = \ 2.00 * 10 ^{-15} \ m[/tex],
[tex]q_1 = q_2 = 1.60 * 10 ^ {-19} \ C[/tex],
[tex]\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}[/tex],
we can find:
[tex]F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}[/tex].
[tex]F \ = 57.536 N[/tex].
Not so big for everyday life, but enormous for subatomic particles.
The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton
[tex]\texttt{ }[/tex]
Further explanationElectric charge consists of two types i.e. positively electric charge and negatively electric charge.
[tex]\texttt{ }[/tex]
There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
distance between protons = d = 2 × 10⁻¹⁵ m
charge of proton = q = 1.6 × 10⁻¹⁹ C
Unknown:
electric force = F = ?
Solution:
[tex]F = k \frac{q_1 q_2}{(d)^2}[/tex]
[tex]F = k \frac{q^2}{(d)^2}[/tex]
[tex]F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(2 \times 10^{-15})^2}[/tex]
[tex]F = 57.6 \texttt{ Newton}[/tex]
[tex]\texttt{ }[/tex]
Learn moreThe three resistors : https://brainly.com/question/9503202A series circuit : https://brainly.com/question/1518810Compare and contrast a series and parallel circuit : https://brainly.com/question/539204[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Static Electricity
Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another reference frame that is moving with a speed of 0.85 relative to the original frame, in the positive x-direction
Answer:
The components of the moving frame is (8.07c, -2, 3, 9.493)
Solution:
As per the question:
Velocity of moving frame w.r.t original frame [tex]v_{m}[/tex] 0.85c
Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane
a = (0, - 2, 3, 5)
Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):
New coordinates are given by:
X = [tex]\frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}[/tex]
X = [tex]\frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}[/tex]
X = [tex]8.07 c[/tex]
Now,
Y = y = - 2
Z = z = 3
Now,
[tex]t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}[/tex]
[tex]t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s[/tex]
A ball is thrown horizontally with a speed of 15 m/s, from the top of a 6.0 m tall hill. How far from the point on the ground directly below the launch point does the ball strike the ground?
Answer:16.59 m
Explanation:
Given
initial horizontal speed of ball(u)=15 m/s
Height of building =6 m
Consider vertical motion first
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
here initial vertical velocity is zero
[tex]6=0+\frac{1}{2}\times 9.81\times t^2[/tex]
[tex]t=\sqrt{\frac{12}{9.81}}=1.106 s[/tex]
Thus time taken will also be 1.106 s in horizontal motion
[tex]R_x=u_xt+\frac{1}{2}at^2[/tex]
here a=0
[tex]R_x=15\times 1.106=16.59 m[/tex]
The total time that the ball is in the air can be calculated using the formula for free fall, which is 1.10 seconds. Given that time and the constant horizontal speed, we can then calculate the total horizontal distance traveled by the ball, which is 16.5 meters.
Explanation:To calculate the horizontal distance the ball travels before striking the ground, we must first determine how long the ball is in the air. This time is only affected by the vertical motion, thus we can treat the ball as if it were dropped from the hill with no initial horizontal velocity. The formula used to calculate the time is based on free fall, where t = sqrt((2*h)/g). In this case, h = 6.0 m (the height of the hill) and g = 9.8 m/s² (the acceleration due to gravity). This gives us t = sqrt((2*6)/9.8) = 1.10 s.
Once we have the time of flight, we can calculate the horizontal distance traveled using the horizontal speed of the ball. As the horizontal motion occurs at a constant velocity we can use the formula d = v*t, where v = 15 m/s (the speed the ball is thrown) and t = 1.10 s. This gives us the horizontal distance d = 15*1.10 = 16.5 m.
Learn more about Projectile Motion here:https://brainly.com/question/29545516
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What is the electrostatic force between and electron and a proton separated by 0.1 mm? 2.3 × 10^-20N, attractive.
2.3 × 10^-20N, repulsive.
2.3 × 10^-18N, attractive.
2.3 × 10^-18N, repulsive.
2.3 × 10^-16N, attractive.
The electrostatic force between an electron and a proton separated by 0.1mm, calculated using Coulomb's law, is approximately 2.3 × 10^-20N, and it is attractive due to the opposite charges.
Explanation:The electrostatic force between charged particles can be calculated using Coulomb's law: F = k(Q1*Q2)/r², where F is the force, k is Coulomb's constant (~8.99 * 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the separation in meters. An electron and a proton have charges of -1.6*10^-19 C and +1.6*10^-19 C respectively. Plugging these values into Coulomb's law with a separation of 0.1 mm or 0.1*10^-3 meters, we find that the force is approximately 2.3 × 10^-20N, and it is attractive because the charges are opposite.
Learn more about Electrostatic Force here:https://brainly.com/question/31042490
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"The correct answer is c. [tex]\(2.3 \times 10^{-18}\)[/tex] N, attractive.
To find the electrostatic force between an electron and a proton separated by a distance, we use Coulomb's law, which is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
The charge of an electron is approximately [tex]\(-1.602 \times 10^{-19}\) C[/tex]and the charge of a proton is approximately [tex]\(+1.602 \times 10^{-19}\) C[/tex]. The separation distance is given as 0.1 mm, which we need to convert to meters: [tex]\(0.1 \times 10^{-3}\)[/tex] m or [tex]\(1 \times 10^{-4}\)[/tex] m.
Plugging in the values, we get:
[tex]\[ F = (8.9875 \times 10^9) \frac{(1.602 \times 10^{-19})(1.602 \times 10^{-19})}{(1 \times 10^{-4})^2} \] \[ F = (8.9875 \times 10^9) \frac{2.5664 \times 10^{-38}}{1 \times 10^{-8}} \] \[ F = (8.9875 \times 10^9) (2.5664 \times 10^{-30}) \] \[ F = 2.3017 \times 10^{-19} \][/tex]
Since the force is attractive (opposite charges attract), the magnitude of the force is [tex]\(2.3 \times 10^{-19}\) N[/tex], but we need to express it in the format given in the options, which is [tex]\(2.3 \times 10^{-18}\) N[/tex].
a car is travelling at 25m/s when it begins to slow down and eventually stops in a 98.5m A. What is the car’s acceleration?
B. At what time after the brakes were applied was the car traveling at 12.5m/s?
C. At what time did the car travel only half the stopping distance?
Answer:
(a) -3.173 m/s^2
(b) 3.94 s
(c) 2.47 s
Explanation:
initial velocity, u = 25 m/s
final velocity, v = 0
distance, s = 98.5 m
(a) Let a be the acceleration of the car
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
0 = 625 + 2 x a x 98.5
a = -3.173 m/s^2
(b) v = 12.5 m/s
u = 25 m/s
a = - 3.173 m/s^2
Let the time is t.
Use first equation of motion
v = u + a t
12.5 = 25 - 3.173 t
t = 3.94 s
(c) s = 98.5 / 2 = 49.25 m
u = 25 m/s
a = - 3.173 m/s^2
Let the time be t.
Let v be the velocity at this distance.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=25^{2}+2\times {-3.173}\ times 49.25tex]
v = 17.17 m/s
Use first equation of motion
v = u + at
17.17 = 25 - 3.173 x t
t = 2.47 s
What is the magnitude of the electric field at a point midway between a −5.6μC and a +5.8μC charge 9.0cm apart? Assume no other charges are nearby. Express your answer using two significant figures.
Answer:
Magnitude of electric field at mid way is [tex]5.07\times 10^{7} N/C[/tex]
Solution:
As per the question:
Q = [tex]- 5.6\mu m = - 5.6\times 10^{- 6} C[/tex]
Q' = [tex]5.8\mu m = 5.8\times 10^{- 6} C[/tex]
Separation distance, d = 9.0 cm = 0.09 m
The distance between charges at mid-way, O is [tex]\farc{d}{2} = 0.045 m[/tex]
Now, the electric at point O due to charge Q is:
[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]
[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E = - 2.49\times 10^{7} N/C[/tex]
Here, negative sign is indicative of the direction of electric field which is towards the point O
Now, the electric at point O due to charge Q' is:
[tex]E' = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]
[tex]E' = (9\times 10^{9})\frac{5.8\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E' = 2.58\times 10^{7} N/C[/tex]
Refer to Fig 1.
Since, both the fields are in the same direction:
[tex]E_{net} = E + E' = 2.49\times 10^{7} + 2.58\times 10^{7} = 5.07\times 10^{7} N/C[/tex]
The magnitude of the electric field at a point midway between two charges can be calculated using the formula E = k * (Q1-Q2) / r^2. Plugging in the values -5.6μC, +5.8μC, and 9.0cm, the magnitude of the electric field is -2.03 x 10^4 N/C.
Explanation:The magnitude of the electric field at a point midway between two charges can be calculated using the formula:
E = k * (Q1-Q2) / r^2
where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
In this case, the charges are -5.6μC and +5.8μC, and the distance is 9.0cm (0.09m).
Plugging these values into the formula:
E = (9 x 10^9 Nm^2/C^2) * ((-5.6μC) - (+5.8μC)) / (0.09m)^2
Simplifying the equation:
E = -2.03 x 10^4 N/C
To practice Problem-Solving Strategy 16.2 Doppler effect. The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.
Answer:
Δf=73Hz
Explanation:
From the question we know that:
C = 1482 m/s
Vs = 0 m/s
Vr = -4.95 m/s (it's negative because it is in the opposite direction to the waves)
f0 = 22000 Hz
Applying the formula for the doppler effect:
[tex]f=(\frac{C-Vr}{C-Vs} )*fo[/tex]
f = 22073 Hz. So the difference is only 73Hz
The driver of a pick up truck going 100 km/h applies the brakes, giving the truck a form deceleration of 6.50 m/s^2 while it travels 20.0 m. What is the speed of the truck in km/h at the end of this distance?
Answer:
81.42 km/h
Explanation:
t = Time taken for the car to slow down
u = Initial velocity = 100 km/h
v = Final velocity
s = Displacement = 20 m = 0.02 km
a = Acceleration = -6.5 m/s² = -0.0065×60×60×60×60 = -84240 km/h²
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -84240\times 0.02+100^2}\\\Rightarrow v=81.42\ km/h[/tex]
Speed of the truck at the end of this distance is 81.42 km/h
Using the equation of motion v^2 = u^2 + 2as, and converting the initial speed from km/hr to m/s, the final velocity is approximately 15.22 m/s when deceleration and distance are considered. Converted back to km/hr, this is approximately 54.8 km/hr.
Explanation:The final speed of the truck can be calculated using one of the equations of motion, specifically v^2 = u^2 + 2as, where 'v' is final velocity, 'u' is initial velocity, 'a' is acceleration (which in this case is -6.50 m/s^2 due to deceleration), and 's' is distance. Convert the initial speed from km/hr to m/s: 100 km/hr = 27.78 m/s.
Then, we can plug the values into the formula: v^2 = (27.78 m/s)^2 - 2(6.50 m/s^2)(20.0 m). Solving the equation gives a final velocity of approximately 15.22 m/s. This speed in km/h is then obtained by converting m/s back, 15.22 m/s * (3600 / 1000) = 54.8 km/h. So, the speed of the truck at the end of the distance is approximately 54.8 km/h.
Learn more about Final Speed Calculation here:https://brainly.com/question/38113096
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In a compression test, a steel test specimen (modulus of elasticity = 30 x 106lb/in2) has a starting height= 2.0 in and diameter = 1.5 in. The metal yields (0.2% offset) at a load = 140,000 lb. At a load of 260,000 lb, the height has been reduced to 1.6 in. Determine (a)yield strength and (b) flow curve parameters (strength coefficient and strain-hardening exponent). Assume that the cross-sectional area increases uniformly during the test.
Answer:
Explanation:
A) we know that volume is given as V
[tex]V =\frac{\pi}{4} D^2 h[/tex]
where D = 1.5 in , h = 2.0 in
so [tex]V = \frac{\pi}{4} 1.5^2\times 2 = 3.53in^3[/tex]
[tex]Area =\frac{\pi}{4} D^2 = \frac{\pi}{4} \times 1.5^2 = 1.76 in^4[/tex]
yield strenth is given as[tex] \sigma_y = \frac{force}{area} = \frac{140,000}{1.76}[/tex]
[tex]\sigma_y = 79.224 ksi[/tex]
b)
elastic strain[tex] \epsilon = \frac{\sima_y}{E} = \frac{79.224}{30\times 10^3} = 0.00264[/tex]
strain offsets = 0.00264 + 0.002 = 0.00464 [where 0.002 is offset given]
[tex]\frac{\delta}{h} = 0.00464[/tex]
[tex]\frac{h_i -h_o}{h_o} = 0.00464[/tex]
[tex]h_i = 2\times(1-0.00464) = 1.99 inch[/tex]
area [tex]A = \frac{volume}{height} = \frac{3.534}{1.9907} = 1.775 in^2[/tex]
True strain[tex] \sigma = \frac{force}{area} = \frac{140,000}{1.775 in^2} = 78,862 psi[/tex]
At P= 260,000 lb ,[tex] A = \frac{3.534}{1.6} = 2.209 inc^2[/tex]
true stress [tex]\sigma = \frac{260,000}{2.209} = 117,714 psi[/tex]
true strain [tex]\epsilon = ln\frac{2}{1.6} = 0.223[/tex]
flow curve is given as \sigma = k\epsilon^n
[tex]\sigma_1 = 78,862 psi[/tex]
[tex]\epsilon_1 = 0.00464[/tex]
[tex]\sigma_2 = 117,714 psi[/tex]
[tex]\epsilon_2 = 0.223[/tex]
so flow curve is
[tex]78,868 = K 0.00464^n[/tex] .........1
[tex]117,714 = K 0.223^n[/tex] .........2
Solving 1 and 2
we get
n = 0.103
and K =137,389 psi
Strength coffecient = K = 137.389ksi
strain hardening exponent = n = 0.103
A point of charge 4.9 μC is placed at the origin (x1 = 0) of a coordinate system, and another charge of -1.1 μC is placed on the x-axis at x2 = 0.28 m. a) Where on the x-axis can a third charge be placed in meters so that the net force on it is zero?
b) What if both charges are positive; that is, what if the second charge is 1.9 μC?
To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. If both charges are positive, the net force on a third charge placed on the x-axis will never be zero.
Explanation:a) To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. The net force will be zero when the two forces are equal in magnitude but opposite in direction. Using Coulomb's law, we can calculate the force between the first charge and the third charge at an unknown position x3. We can then set that force equal to the force between the second charge and the third charge at the same position x3, and solve for x3.
b) If both charges are positive, the net force on a third charge placed on the x-axis will never be zero. Positive charges repel each other, so the forces will always be in the same direction.
A commuting student leaves home and drives to school at an average speed of 39.0 km/h. After 23.0 min he realizes that he has forgotten his homework and returns home to get it at the same average speed. It takes 10.0 min to find the report, after which the trip to school 39.0 km away to the east is resumed at the same speed as before. What is the average speed for the entire trip ? What is the average velocity for the entire trip?
The average speed for the entire trip is 83.6 km/h. The average velocity for the entire trip is zero.
Explanation:The average speed for the entire trip can be calculated by finding the total distance traveled divided by the total time taken. In this case, the student traveled 39.0 km to school and then returned the same distance back home, resulting in a total distance of 78.0 km. The total time taken for the entire trip is the sum of the time taken to drive to school, the time taken to return home, and the time spent finding the report, which is 23.0 min + 23.0 min + 10.0 min = 56.0 min. To convert the total time to hours, divide by 60: 56.0 min / 60 = 0.933 hours. Therefore, the average speed for the entire trip is 78.0 km / 0.933 hours = 83.6 km/h.
The average velocity for the entire trip can be determined by considering both the magnitude and direction of the displacement. In this case, since the student drove in the same direction to school and back home, the displacement is zero, meaning there was no change in position. Thus, the average velocity for the entire trip is also zero.
A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?
Answer:
The no. of electrons is [tex]7.59\times 10^{21}[/tex]
Solution:
According to the question:
The rate at which the charge is delivered is given by:
[tex]\frac{dQ}{dt} = - 0.75[/tex]
Now,
[tex]\int_{0}^{Q}dQ = - 0.75\int_{0}^{27 min} dt[/tex]
[tex]Q = -0.75t|_{0}^{27 min}[/tex]
[tex]Q= -0.75\times 27\times 60 = - 1215 C[/tex]
No. of electrons, n can be calculated from the following relation:
Q = ne
where
e = electronic charge =[tex]1.6\times 10^{- 19} C[/tex]
Thus
[tex]n = \frac{Q}{e}[/tex]
[tex]n= \frac{1215}{1.6\times 10^{- 19}}[/tex]
[tex]n = 7.59\times 10^{21}[/tex]
A basketball player makes a bounce pass by throwing the ball with a speed of 5.3 m/s and at an angle 12° below the horizontal. It leaves her hands 1.0 m above the floor. Calculate how far omher feet the ball hits the ground.
Answer:
ball hit the ground from her feet is 1.83 m far away
Explanation:
given data
speed = 5.3 m/s
angle = 12°
height = 1 m
to find out
how far from her feet ball hit ground
solution
we consider here x is horizontal component and y is vertical component
so in vertical
velocity will be = v sin12
vertical speed u = 5.3 sin 12 = 1.1 m/s downward
and
in horizontal , velocity we know v = 5.3 m/s
so from motion of equation
s = ut + 0.5×a×t²
s is distance t is time a is 9.8
put all value
1 = 1.1 ( t) + 0.5×9.8×t²
solve it we get t
t = 0.353 s
and
horizontal distance is = vcos12 × t
so horizontal distance = 5.3×cos12 × ( 0.353)
horizontal distance = 1.83 m
so ball hit the ground from her feet is 1.83 m far away
What mass of steam at 100°C must be mixed with 499 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 33.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg.
Answer:
The mass of the steam is 91.2 g
Mass of the steam=91 grams
Explanation:
Given:
Mass of the ice=499 gFinal temperature of the liquid water[tex]=33^\circ \rm C[/tex]Latent heat of fusion=[tex]333\ \rm kJ\kg[/tex]Latent heat of vaporization =[tex]2256\ \rm kJ\kg[/tex]When steam is mixed with the ice then the heat loss by the steam will be gained by the ice so there will no overall heat gain or loss during the mixing
So According to question
Let M be the mass of the steam mixed with ice then we have
[tex]M\times2256\times10^3+M\times4186\times(100-33)=0.499\times222\times10^3+0.499\times4186\times(33-0)\\M\times2.58\times10^6=2.35\times10^5\\M=91.2\ \rm g[/tex]
A truck with a mass of 1650 kg and moving with a speed of 15.0 m/s rear-ends a 779 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
Truck's speed = 5.21 m/s
Car's speed = 20.2 m/s
Explanation:
Given:
Mass of truck = M = 1650 kg
Speed of the truck initially = U = 15 m/s
Mass of the car = m = 779 kg
Initial speed of the car =u = 0
From the momentum conservation, Total initial momentum = Total final momentum.
M V+m U = M V +m v
⇒ (1650)(15) + 779×0 = (1650)V + 779 v
⇒ 24750 = 1650 V+779 v →(1)
Since the collision is elastic, relative velocity of approach = relative velocity of separation. 15 = v - V
⇒ v =V + 15; This is now substituted in the equation(1) above.
24750 = 1650 V + (799) (V+15)
⇒ 24750 = 1650 V + 799 V + 11985
⇒ 2449 V = 12765
⇒ Final velocity of the truck = [tex]\frac{12765}{2449}[/tex] = 5.21 m/s
Final velocity of the car = v = V+15 = 5.21 + 15 = 20.2 m/s
A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if the image (formed) is real?
Answer:
Image distance is -52.5 cm
Image is virtual and forms on the same side of the lens and upright image is formed.
Explanation:
u = Object distance
v = Image distance
f = Focal length = 35
m = Magnification = 2.5
[tex]m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u[/tex]
Lens equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm[/tex]
[tex]v=-2.5\times 21=-52.5\ cm[/tex]
Image distance is -52.5 cm
Image is virtual and forms on the same side of the lens and upright image is formed.
According to the reading, which of the following is NOT an obstacle to using big data for analysis and decision making? O Poor reporting process b. Limited types of data Insufficient data d. Lack of talent to evaluate and analyze data All of the above
Answer:
B .Limited types of data Insufficient data
Explanation:
To analyse and making decision from big data we have following thing
- Sufficient data
- data analyst
- Accurate data
- data Privacy
- data storage
So, limited data or data that cannot be copied is not and obstacle in data handling. hence option B Limited types of data Insufficient data is correct.
A driver of a car going 60 km/h suddenly sees the lights of a barrier 40 m ahead. It takes the driver 0.80 seconds before he applies the brakes, and the average acceleration during braking is -9.5 m/s. (A) Does the car hit the barrier? Explain. (B) What is the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead?
Answer:
a) The car doesn't hit the barrier because after he sees the lights of a barrier he only travels for 27.95m, enough to miss the barrier. b) The maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s or 75.6km/hr.
Explanation:
a)
In order to solve this problem we must first do a drawing of the situation for us to visualize it better. (See picture attached).
As you may see on the drawing, the car still travels some distance when the driver notices the lights of the barrier. This distance is calculated for a constant velocity V:
x=Vt
the car has a velcity of 60km/h which is equivalent to:
[tex]\frac {60km}{hr} * \frac{1000m}{1km}*\frac{1hr}{3600s}=16.667m/s[/tex]
so on the first 0.80s the car travels a distance of:
x=(16.667m/s)(0.8s)=13.333m
Next, the car breaks, so we can say it moves with a constant acceleration of -9.5m/s, so the distance is found by using the following formula:
[tex]x=\frac {V_{f}^{2}-V_{0}^{2}}{2a}[/tex]
We know the final velocity will happen when the car stops, so the final velocity is zero, leaving us with:
[tex]x=\frac {-V_{0}^{2}}{2a}[/tex]
so we can substitute the provided values to find the distance traveled by the car during this time.
[tex]x=\frac{-(16.667m/s)^{2}}{2(-9.5m/s^{2})}[/tex]
which yields:
x=14.62m
so in total the car traveled:
[tex]x_{tot}=13.333m+14.62m = 27.95m[/tex]
Which is not enough for the car to hit the barrier.
b)
In order to solve this part of the problem, we must combine the two equations we got on the previous part to find a single equation that will represent the total displacement of the car:
[tex]x=vt-\frac{v^{2}}{2a}[/tex]
so we can now substitute the known values so we get:
[tex]40=0.8v-\frac{v^{2}}{2(-9.5)}[/tex]
which simplifies to:
[tex]40=0.8v+0.05263v^{2}[/tex]
this is a quadratic equation so it can be solved by using the quadratic formula, but first we must rewrite it in standard form, so we get:
[tex]0.05263v^{2}+0.8v-40=0[/tex]
now we can use the quadratic formula:
[tex]v=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
So we can substitute the given values:
[tex]v=\frac{-0.8\pm\sqrt{(0.8)^{2}-4(0.05263)(-40)}}{2(0.05263)}[/tex]
which returns two answers:
v=21m/s and v=-36.20m/s
We take the positive answer since is the one that represents a moving towards the right side of the drawing.
So the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s