Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:vₓ₀ = v * cos θ (1)where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:[tex]x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)[/tex]
Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:[tex]cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)[/tex]
⇒θ = cos⁻¹ (0.526) = 58.3º (4)b)
At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)c)
At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)d)
Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)[/tex]
Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)Replacing (7) in (6), we get:[tex]\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)[/tex]
e)
When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.The horizontal component, since it keeps constant, is just v₀x:v₀ₓ = 13.7 m/sThe vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:[tex]v_{fy} = v_{oy} - g*t (9)[/tex]
Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:[tex]v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)[/tex]
Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:[tex]v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)[/tex]
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
In which part of a lab report would be the following sentence most likely occur? “Since the data showed that the
Answer:
most likely be included in the analysis section of a lab report
Explanation:
Sultan walks for 15 km at 35° south of east.
Which of the following journeys would result in the same displacement?
Answer:
☝
Explanation:
Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire?
Answer:
Explanation:
Expression for fundamental frequency of tone produced in a wire under tension of T and length L is given as follows
[tex]f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
m is mass per unit length .
We shall apply this formula for given wires .
For shorter wire
[tex]60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
For longer wire for second harmonic
length of wire is 2L , tension is 4T ,
[tex]f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }[/tex]
[tex]f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }[/tex]
f = 2 x 60 = 120 Hz .
Various amplifier and load combinations are measured as listed below using rms values. For each, find the voltage, current, and power gains ( A v , Ai , and Ap, respectively) both as ratios and in dB:
(a) vI= 100 mV, iI = 100 μA, vO = 10 V, RL = 100Ω
(b) vI = 10 μV, iI = 100 nA, vO =1 V, RL= 10 kΩ
(c) vI =1 V, iI = 1 mA, vO =5 V, RL = 10Ω
Answer:
The solution to this question can be defined as follows:
Explanation:
In point (a):
[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]
In point (b):
[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]
In point (C):
[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in N ). b. Find the work done by the student (in J). c. Find the power exerted by the student (in W)
Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
How long ago did most Middle Eastern countries gain their independence?
A.
10-20 years ago
B.
50-100 years ago
C.
200-300 years ago
D.
400-500 years ago
Please select the best answer from the choices provided
A
B
C
D
Answer:
the correct answer is B
Explanation:
Answer:
.B
Explanation:
This is the answer on edge 2021
Have a good day!
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
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help me help me help me
In the figure, if Q = 52 µC q =10 µC and d = 55 cm, what is the magnitude of the electrostatic force on q?
Answer:
F = 15.47 N
Explanation:
Given that,
Q = 52 µC
q = 10 µC
d = 55 cm = 0.55 m
We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N[/tex]
So, the magnitude of the electrostatic force is 15.47 N.
The magnitude of electrostatic force will be "15.47 N".
Electrostatic forceAccording to the question,
Charges, Q = 52 μC
q = 10 μC
Distance, d = 55 cm, or
= 0.55 m
Constant, k = 9 × 10⁹
We know the relation,
→ Electrostatic force, F = k [tex]\frac{q_1 q_2}{d^2}[/tex]
By substituting the values, we get
= 9 × 10⁹ × [tex]\frac{10\times 10^{-6}}{(0.55)^2}[/tex]
= 15.47 N
Thus the above answer is correct.
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Solve each of the following problems to 3 sig figs and correct Sl units, showing all work.
1. A cart with a mass of 45.0 kg is being pulled to the right with a force of 250 N giving it an
acceleration of 1.30 m/s2. The wheels of the cart are locked and the cart must be dragged.
a) Draw a free body diagram of the cart.
b) Calculate the net force acting on the cart.
c) Create a force table and fill it in.
d) Find the coefficient of kinetic friction.
Answer:
every number to 3 sf = 1) 45.0 2) 250 3) 1.30
Explanation:
your welcome :)
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
Particle A with charge q and mass ma and particle B with charge 2q and mass
mb, are accelerated from rest by a potential difference AV and subsequently
deflected by a uniform magnetic field into semicircular paths. The radii of the
trajectories by particle A and B are R and 3R, respectively. The direction of
the magnetic field is perpendicular to the velocity of the particle. Determine
their mass ratio?
the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your wheels is goingnto turn by an angle of
1. An engine absorbs 600 J of heat while doing 650 J of work. What is the change
in internal energy of the enginge? *
1250 J
-50 J
-1250 J
50 J
The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The absorb energy: Q = 600 Joule
Work done: W = 650 Joule.
Let, the change in internal energy of the engine= dU.
According to conservation of energy:
The absorb energy = change in internal energy + Work done
Q = dU + W
dU = Q - W
= 600 joule - 650 joule
= - 50 joule.
Hence, the change in internal energy of the engine is -50 joule.
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Why is damage from sound waves is an issue on the launchpad but not in the air
(I would have done more points for answering but I'm almost out sry. just pls answer and help.)
The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.
When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.
Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.
Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
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describe measurement in our daily life
Question 1 of 10
What might happen to personal information when it is transferred using
digital signals?
A. Some information might be changed when the data are copied.
B. It might be accessed by someone who was not the intended
recipient.
C. The information might change while being transmitted because of
noise.
D. The information might change to analog, making it less reliable.
Answer:
its b for sure
Explanation:
Answer:
B. It might be accessed by someone who was not the intended
recipient
An object is released from rest at a height H near and above the surface of Earth. As the object falls toward the surface, Earth’s atmosphere exerts a resistive force on the object such that it reaches a terminal velocity before it reaches the ground. Which of the following claims is true? Select two answers.
The system consisting of only the object is an open system.
The system consisting of only the object is an open system.
A
Earth’s atmosphere does negative work on the object as it falls toward the surface.
Earth’s atmosphere does negative work on the object as it falls toward the surface.
B
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
C
The total mechanical energy of the object-Earth system remains constant at all times in which the object is in motion.
Answer:
Second and Last Option Are Correct
Explanation:
What is the function
of second plate in
parallel plate capacitor?
This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.
(i) Including the water in its internal cavity, the squid has a total mass of 6.50 kg.
(ii) The mass of the water in its cavity is 1.75 kg.
(iii) In order to escape its predators, the squid needs to achieve an escape speed of 2.5 m/s.
Answer:
6.79 m/s
Explanation:
By applying the principle of conservation of momentum.
The total momentum = MV - mv = 0 (since the squid is beginning at rest)
the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg
= 4.75 kg
speed of the squid (V) = 2.5 m/s
mass of the water expelled (m) = 1.75 kg
speed of the water (v) = ???
∴
4.75 × 2.5 = 1.75 × v
[tex]v = \dfrac{4.75 \times 2.5}{1.75 }[/tex]
v = 6.79 m/s
What is the instantaneous velocity of a freely falling object 11 s after it is released from a position of rest
Answer:
v= -107.8 m/s
Explanation:
Since the object is in free fall, this means that is moving at an accceleration equal to the one due to gravity.Since it starts at rest, we can apply the definition of acceleration, rearranging terms as follows:[tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]
(Assuming as positive the upward direction)I need this done by tonight!! Can anyone help me please? Answer these 4 questions
Answer:
1. 14 g of chocolate mixture.
2. 24 fl oz of chocolate milk
3. 10 cups of chocolate milk.
4. 12½ cups.
Explanation:
From the question given above, the following data were obtained:
1 TBSP = 7 g
1 Cup = 8 fl oz
2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.
1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.
From the question given above,
1 Cup required 2 Table spoons (TBSP)
But
1 TBSP = 7 g
Therefore,
2 TBSP = 2 × 7 = 14 g
Thus, 1 Cup required 14 g of chocolate mixture.
2. Determination of the number fl oz of chocolate milk in 3 cups
1 Cup = 8 fl oz
Therefore,
3 Cups = 3 × 8
3 Cups = 24 fl oz
Thus, 24 fl oz of chocolate milk are in 3 cups.
3. Determination of the number of cups of chocolate milk produce from 20 TBSP.
2 TBSP is required to produce 1 cup.
Therefore,
20 TBSP will produce = 20/2 = 10 Cups.
Thus, 10 cups of chocolate milk produce from 20 TBSP.
4. Determination of the number of cups obtained from 100 fl oz chocolate milk.
8 fl oz is required to produce 1 cup.
Therefore,
100 fl oz will produce = 100 / 8 = 12½ cups.
Thus, 12½ cups is obtained from 100 fl oz chocolate milk.
PLZ HELP WILL MARK BRAINLIEST!!
Amy has a mass of 50 kg, and she is riding a skateboard traveling 10 meters per second. What is her momentum?
5 kg·m/s
10 kg·m/s
50 kg·m/s
500 kg·m/s
Answer:
[tex]500 \: \mathrm{kg} \cdot \mathrm{m/s}[/tex]
Explanation:
The momentum of an object is given as [tex]p=mv[/tex]. Since Amy has a mass of 50 kg and is travelling 10 m/s, her momentum is [tex]p=mv=50\cdot 10 =\fbox{$500\: \mathrm{kg\cdot m/s}$}[/tex].
Answer:
500
Explanation:
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q>0.5 will the electrostatic force between the two parts have 1/3 of the maximum possible value?
Answer:
Explanation:
Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q
F( max ) = k .5Q x .5Q / R²
=.25kQ² /R²
For the second case
F = k q ( Q-q)/ R²
F = .25kQ² /3R²
.25kQ² /3R² = k q ( Q-q)/ R²
.25 Q² = 3qQ - 3q²
3q² - 3qQ + .25 Q² = 0
q =
