Answer:
9.4 m
Explanation:
We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.
If the ballon is launched at 9.7 m/s at 39 degrees of elevation.
Vx0 = 9.7 * cos(39) = 7.5 m/s
Vy0 = 9.7 * sin(39) = 6.1 m/s
If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
a = -9.81 m/s^2
It will fall when Y(t) = 0
0 = 6.1 * t - 4.9 * t^2
0 = t * (6.1 - 4.9 * t)
t1 = 0 (this is when the balloon was launched)
0 = 6.1 - 4.9 * t2
4.9 * t2 = 6.1
t2 = 6.1 / 4.9 = 1.25 s
The distance from the car will be the horizonta distance it travelled in that time
X(t) = X0 + Vx0 * t
X(1.25) = 7.5 * 1.25 = 9.4 m
You are standing 1.2 m from a mirror, and you want to use a classic camera to take a photo of yourself. This camera requires you to select the distance of whatever you focus on. What distance do you choose?
Answer:
2.4 m
Explanation:
Plane Mirror: It is a type of mirror whose reflecting surface is plane. it produces an image of an object behind it at a distance equal to the object lying in front of the mirror.
According to the question, the person has to take the photo of himself who is looking at the mirror.
Let us assume that the person is standing in front of a plane mirror at a distance of 1.2 m.
Due to the reflection of light, the image of the person will be formed at a distance 1.2 m behind the mirror. This means the image of the person is 2.4 m apart from him.
As the person could take the picture of his own image from the classic camera in his hand. This means the image of the person will act as an object for the classic camera to which he is going to focus on which is at a distance of 2.4 m.
Hence, the person will choose a distance of 2.4 m.
Focus length is the distance of optics from the point of convergence of light rays to create a sharp image of the object.
The distance of the focus must be 2.4 meters.
How to select the distance for ideal focus length?
Focus length is the distance of optics from the point of convergence of light rays to create a sharp image of the object. Focus length represents in the millimetres.
In the given problem, the person is standing in front of the plane mirror. A plane mirror is the mirror with a flat reflective surface.
The image formed by the plane mirror is virtual, erect and the equal to the size of the person or thing.
Given information-
The distance of person from the mirror is 1.2 m.
As the image in the plane mirror appears at the equivalent distance to the object, thus the distance of the image of the person is,
[tex]d_i=1.2 \rm m[/tex]
Thus the distance of the image of the person appears is 1.2 meters behind the mirror.
To take the photo of this image, the distance of the focus should be equal to the sum of the distance of the image inside the mirror to the distance of person front of the mirror. Thus,
[tex]d_f=d_p+d_i\\d_f=1.2+1.2\\d_f=2.4 \rm m[/tex]
Hence the distance of the focus must be 2.4 meters.
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A parallel-plate capacitor is formed from two 2.0 cm x 2.0 cm electrodes spaced 2.2 mm apart. The electric field strength inside the capacitor is 1.0×10^6 N/C. Part A What is the charge (in nC) on the positive electrode? Express your answer in nanocoulombs.
Answer:
3.54 nC.
Explanation:
Given:
Dimensions of the electrodes of parallel-plate capacitor = [tex]2.0\ cm \times 2.0\ cm.[/tex]Separation between the electrodes, [tex]d=2.2\ mm = 2.2\times 10^{-3}\ m.[/tex]Electric field strength inside the capacitor, [tex]E = 1.0\times 10^6\ N/C.[/tex]The surface area of each of the electrode is given by
[tex]A=2.0\ cm \times 2.0\ cm = 4.0\ cm^2 = 4.0\times 10^{-4}\ m^2.[/tex]
The strength of the electric field inside a capacitor is given by
[tex]E = \dfrac{\sigma}{\epsilon_o}[/tex]
where,
[tex]\epsilon_o[/tex] = electrical permittivity of free space = [tex]8.85\times 10^{-12}\ C^2N^{-1}m^{-2}[/tex][tex]\sigma[/tex] = surface charge density of the electrode = [tex]\dfrac qA[/tex][tex]q[/tex] = charge on the electrode.Therefore,
[tex]E = \dfrac{q}{A\epsilon_o}\\q=EA\epsilon_o\\=1.0\times 10^6\times 4.0\times 10^{-4}\times 8.85\times 10^{-12}\\=3.54\times 10^{-9}\ C\\=3.54\ nC.[/tex]
It is the charge on the positive electrode.
The value of an electric field at a particular point is known as the strength of the electric field. the value of the charge will be 3.54 NC.
What is the strength of the electric field?
The electric field is the area around an electric charge where its impact may be felt. The force encountered by a unit positive charge put at a spot is the electric field strength at that point.
Electrode dimension = 2.0 cm×2.0 cm
Area of electrode = 4.0 cm²
seperated distance beween electrode = 2.2 mm = 2.2 × 10⁻³ m
strength of electric field = E = 1.0 × 10⁶ N/C
Electric field strength is given by
[tex]\rm E = \frac{\sigma}{\varepsilon_0}[/tex]
where,
[tex]\varepsilon_0[/tex]= electrical permittivity of free space
[tex]\sigma[/tex] = surface charge density of the electrode
q=Electrode charge
[tex]\rm \sigma =\RM{ \frac{q}{A} }[/tex]
[tex]\rm E = \frac{q}{A\varepsilon_0}[/tex]
[tex]\rm q = \sigma{A\varepsilon_0}[/tex]
[tex]\rm q =1.0\times 10^-6\times 4.0\times10^{-4}\times 8.85 \times 10^-12[/tex]
[tex]\rm q = 3.54\times 10^-9 \; C[/tex]
[tex]\rm q = 3.54\; nC[/tex]
Hence the value of the charge will be 3.54 NC.
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A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.5 m/s 2 . A green car arrives at the position of the stop-light 5 s after the light had turned green. If t = 0 when the light turns green, at what time does the green car catch the blue car if the green car maintains the slowest constant speed necessary to catch up to t
Answer:
The green car will catch the blue one after 10 s of the stop-light turning green.
Explanation:
The equation for the position of objects moving in a straight line is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the object at time t
x0 = initial position
t = time
a = acceleration
v0 = initial velocity
For the blue car, this will be its position at time t
x = 0 m + 0 m/s · t + 1/2 · 0.5 m/s² · t²
x = 0.25 m/s² · t²
For the green car, that moves at constant speed, its position will be:
x = 0 m + v · t (where v = velocity)
x = v · t
Now let´s find how much distance the blue car has traveled until the green car arrives at the stop-light:
x = 0.25 m/s² · (5s)² = 6.25 m
When the persecution begins (5 s after the stop-light turns green), the blue car is 6.25 ahead of the green car.
When the green car catches the blue one, its position is the same as the position of the blue car.
Then:
Position of the blue car = position of the green car
6.25 m + 0.25 m/s² · t² = v · t
0.25 m/s² · t² - v · t + 6.25 m = 0
Let´s use the formula for solving quadratic equations:
a = 0.25
b = -v
c = 6.25
(-b ± √(b²-4·a·c))/2·a
Let´s replace with the data:
(v ± √(v² - 4 · 0.25 m/s² · 6.25 m))/2· 6.25 m
(v ± √(v² - 6.25 m²/s²))/12.5 m
The minimum value of v that solves this equation is the value that makes the content inside the root to be null. A lower value than that and the content inside the root will be negative and have no solution (real solution).
Then:
v² - 6.25 m²/s² = 0
v² = 6.25 m²/s²
v = 2.5 m/s
Now the quadratic equation will be as follows:
0.25 m/s² · t² - 2.5 m/s · t + 6.25 m = 0
Solving this equation:
t = 5 s
Remember that this is the time that takes the green car to catch the blue one after the green car arrived at the stop-light (5 s after it turned green). If we count from when the light turns green, the elapsed time will be 10 s.
A 1.0-μm-diameter oil droplet (density 900 kg/m^3) is negatively charged with the addition of 27 extra electrons. It is released from rest 2.0 mm from a very wide plane of positive charge, after which it accelerates toward the plane and collides with a speed of 3.5 m/s . What is the surface charge density of the plane?
Answer:
surface charge density = 5.91 µC/m²
Explanation:
given data
diameter = 1 μm
radius = 0.5 × [tex]10^{-6}[/tex] m
speed = 3.5 m/s
density 900 kg/m³
distance = 2 mm = 2 × [tex]10^{-3}[/tex] m
to find out
surface charge density of the plane
solution
volume is express as
volume = [tex]\frac{4}{3} \pi r^3[/tex]
volume = [tex]\frac{4}{3} \pi (0.5*10^{-6})^3[/tex]
volume = 5.23 × [tex]10^{-19}[/tex] m³
and
mass = density × volume
mass = 900 ×5.23× [tex]10^{-19}[/tex]
mass = 4.712 × [tex]10^{-16}[/tex] kg
and
from motion of equation
v² -u² = 2×a×s
here v is speed 3.5 and u initial is 0 and a is acceleration and s is distance
3.5² = 2×a×2 × [tex]10^{-3}[/tex]
acceleration = 3062.5 m/s²
so from newton second law
Force = mass × acceleration
force = 4.712 × [tex]10^{-16}[/tex] ×3062.5
force = 1.44305 × [tex]10^{-12}[/tex] N
so
surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]
here q is charge with addition 27 extra electron and f is force and ∈ = 8.85× [tex]10^{-12}[/tex]
surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]
surface charge density = [tex]\frac{2*8.85*10^{-12}*1.443*10^{-12}}{27*1.60*10^{-19}}[/tex]
surface charge density = 5.91 × [tex]10^{-6}[/tex] C/m²
so surface charge density = 5.91 µC/m²
Change a speed of 72.4 miles per hour to its equivalent in meters per second.
Final answer:
To convert 72.4 miles per hour to meters per second, multiply by 1609.34 to convert miles to meters, then divide by 3600 to convert hours to seconds. The result is approximately 32.34 meters per second.
Explanation:
To convert a speed of 72.4 miles per hour to its equivalent in meters per second, you can use the conversion factors that 1 mile is equal to 1609.34 meters and 1 hour is equal to 3600 seconds. Since the conversion from kilometers per hour to meters per second involves dividing by 3.6, and knowing from the reference material that 60 miles per hour is approximately equal to 27.8 meters per second, we can calculate the speed in meters per second as follows:
72.4 mi/h * (1609.34 m / 1 mi) * (1 h / 3600 s) = 72.4 * 1609.34 / 3600 m/s.Doing the math:
72.4 * 1609.34 / 3600 = 32.34 m/s (rounded to two decimal places).Check if the answer is reasonable: Since 72.4 mi/h is a little more than 60 mi/h, and we know that 60 mi/h is approximately 27.8 m/s, it seems reasonable that 72.4 mi/h would be slightly over 27.8 m/s, so our answer of 32.34 m/s seems plausible within the context.
Preston pushes a wheelbarrow weighing 500 N to the top of a 50.0-m ramp, inclined at 20.0° with the horizontal, and leaves it. Tamara accidentally bumps the wheelbarrow. It slides back down the ramp, during which an 80.0-N frictional force acts on it over the 50.0 m. What is the wheelbarrow's kinetic energy at the bottom at of the ramp
The wheelbarrow's final kinetic energy at the bottom of the ramp is 4500 J, found by calculating the work done on the wheelbarrow by gravity and friction and applying the work-energy theorem.
Explanation:To determine the wheelbarrow's kinetic energy at the bottom of the ramp, we need to calculate the work done on the wheelbarrow by both gravity and friction as it slides down the ramp, and then use this to find the change in kinetic energy.
The work done by gravity is simply the component of the wheelbarrow's weight (500 N) that acts parallel to the ramp (i.e., 500 N * sin(20) * 50 m = 8500 J).
The work done by friction is the frictional force (80 N) times the distance (50 m), but since friction opposes the motion it’s negative, so -4000 J.
By the work-energy theorem, the total work done on the wheelbarrow equals its change in kinetic energy. The wheelbarrow starts from rest, so its initial kinetic energy is zero. Hence, its final kinetic energy (at the bottom of the ramp) is the total work done on it: 8500 J - 4000 J = 4500 J.
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Part A Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant (Kf) for water is 1.86 °C/m. Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86 °C/m. +0.23 °C -0.23 °C 1.23 °C -0.45 °C +0.45 °C
Answer: [tex]-0.45^0C[/tex]
Explanation:-
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(0-T_f)^0C[/tex] = Depression in freezing point
i= vant hoff factor = 2 (for electrolyte undergoing complete dissociation, i is equal to the number of ions produced)
[tex]KCl\rightarrow K^++Cl^-[/tex]
[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (water)= 550 g = 0.55 kg (1kg=1000g)
Molar mass of solute (KCl) = 74.5 g/mol
Mass of solute (KCl) = 5.0 g
[tex](0-T_f)^0C=2\times 1.86\times \frac{5g}{74.5g/mol\times 0.55kg}[/tex]
[tex](0-T_f)^0C=0.45[/tex]
[tex]T_f=-0.45^0C[/tex]
Thus the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water is [tex]-0.45^0C[/tex]
The freezing point of a solution : -0.45 °C
Further explanationSolution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.
Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles and break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes
The term is used in the Solution properties
• 1. molal
that is, the number of moles of solute in 1 kg of solvent
[tex]\large {\boxed {\bold {m = mole. \frac {1000} {mass \: of \: solvent (in \: grams)}}}[/tex]
• 2. mole fraction
the ratio of the number of moles of solute to the mole of solution
[tex]\large {\boxed {\bold {Xa = \frac {na} {na + nb}}}[/tex]
a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86
• Step 1: determine molal
molar mass KCl = 39 + 35.5 = 74.5
mole KCl = mass: molar mass
mole KCl = 5 gr: 74.5
mole KCl = 0.067
molal = m = 0.067 x (1000: 550 gr water)
molal = 0.122
• Step 2: determine the freezing point of the solution
i = 1 + (n-1) a
i = 1 + (2-1) 1
i = 2
[tex]\displaystyle \Delta T_f=K_f.m.i[/tex]
---> KCl is an electrolyte solution
[tex]\Delta T_f=1.86.0.122.2[/tex]
[tex]\displaystyle \Delta T_f=0.454[/tex]
freezing point of water = 0 °C
[tex]0.454=0-T~solution[/tex]
[tex]T_f~solution=- 0.454 ^oC[/tex]
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what is the necessary condition on a force the result the conservation of angular momentum for a particle affected by that force?
Answer:
The necessary condition is that the force that is applied should not be able to produce any torque about the axis of rotation which is achieved only if the force passes through the axis of rotation.
Explanation:
Mathematically the angular momentum is related to force as
[tex]\overrightarrow{F}\times \overrightarrow{r}=\frac{d\overrightarrow{l}}{dt}[/tex]
where
'r' is arm of the force 'F'
For angular momentum to be conserved the right hand side of the above equation should be zero which can happen only if:
1) F = 0
2) r = 0
3) F and r are parallel.
Thus fulfilling any of the above condition shall conserve the angular momentum of a particle.
If a 1kg English swallow is carrying a 1kg coconut and flies northward from rest to a final velocity of 10m/s in 7.5 seconds, what is its acceleration? A. What force was exerted by the bird during that acceleration? B. How far did the bird travel during that time?
Answer:
[tex]a=1.33 m/s^{2}[/tex]
A. [tex]F=2.67N[/tex]
B. [tex]x=37.50m[/tex]
Explanation:
From the exercise we know that the swallow:
[tex]m_{total} =2kg[/tex]
[tex]v_{o}=0\\v_{f}=10m/s\\t=7.5s[/tex]
To find its acceleration we must calculate:
[tex]a=\frac{v_{f}-v_{o} }{t_{f}-t_{o} }=\frac{(10-0)m/s}{(7.5-0)s}=1.33m/s^{2}[/tex]
A. Now, from Newton's second law we know that force is:
[tex]F=m*a[/tex]
[tex]F=(2kg)*(1.33m/s^{2})=2.67N[/tex]
B. To find the distance which the bird travels we need to find how long does it take
[tex]v_{f}=v_{o}+at[/tex]
Solving for t
[tex]t=\frac{v_{f} }{a}=\frac{10m/s}{1.33m/s^{2} }=7.51s[/tex]
Now, from the equation of position we know that
[tex]x=x_{o}+v_{o}t+\frac{1}{2}at^{2}[/tex]
[tex]x=\frac{1}{2}(1.33m/s^2)(7.51s)^2=37.50m[/tex]
A lead ball is dropped into a lake from a diving board 16.0 ft above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. If it reaches the bottom 5.30 s after it is dropped, how deep is the lake (in feet)?
Answer:
138.46 ft
Explanation:
When the ball is dropped until the moment it hits the water, the ball moves in a uniform acceleration motion. Therefore, the equation that describes the movement of the ball is:
[tex]X = \frac{1}{2}*g*t^{2} + V_{0} *t + x_0[/tex]
Where X is the distance that the ball has fallen at a time t. [tex]V_0[/tex] is the initial velocity, which is 0 ft/s as the ball was simply dropped. [tex]x_0[/tex] is the initial position, we will say that this value is 0 in the position where the ball was dropped for simplicity, and it increases as the ball is falling. Now, we replace x with 16 feets and solves for t:
[tex]16 ft = \frac{1}{2} * 32.2 \frac{ft}{s^{2}} *t^{2} +0 \frac{ft}{s} * t + 0 ft[/tex]
[tex]t = \sqrt{2* \frac{16 ft}{32.2 \frac{ft}{s^2}}} = 1 s[/tex]
The velocity that the ball will have at the moment the ball that the ball hits the water will be:
[tex]V = V_o+g*t=0\frac{ft}{s}+32.2\frac{ft}{s^2}*1s =32.2\frac{ft}{s}[/tex]
The time that will take the ball to reach the bottom from the top of the lake will be t = 5.3s - 1s = 4.3s. And as the ball will travel with constant velocity equal to 32.2 ft/s^2, the depth of the lake will be:
[tex]d = v*t = 32.2\frac{ft}{s} * 4.3s = 138.46 ft[/tex]
The depth of the lake is approximately 153.6 feet.
Step 1: Understanding the Problem
We have a lead ball that is dropped from a height of 16.0 ft. After being dropped, it hits the water and sinks to the bottom with a constant velocity. We are told that it reaches the bottom of the lake 5.30 seconds after it is dropped. Our goal is to calculate the depth of the lake.
Step 2: Analyzing the Time
The total time from the moment the ball is dropped until it hits the bottom of the lake is 5.30 seconds. This time includes:
The time it takes to fall to the water's surface.The time it takes to sink to the bottom of the lake once it hits the water.Step 3: Calculate the Fall Time
We can first determine the time it takes for the ball to drop 16.0 ft. The formula for the distance fallen under gravity is:
[tex]d = \frac{1}{2} g t^2[/tex]
where:
[tex]d[/tex] is the distance (16.0 ft),[tex]g[/tex] is the acceleration due to gravity (approximately 32 ft/s²), and [tex]t[/tex] is the time in seconds.Rearranging the formula for time:
[tex]t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \times 16.0\text{ ft}}{32 \text{ ft/s}^2}} = \sqrt{1} = 1 \text{ second}[/tex]
Step 4: Calculate Time to Sink
If the total time from the drop to reaching the bottom is 5.30 seconds, the time taken to sink in the water is:
[tex]\text{Time to sink} = 5.30 \text{ s} - 1 \text{ s} = 4.30 \text{ s}[/tex]
Step 5: Calculate Depth of the Lake
Since the ball sinks at a constant velocity, we need to find the velocity of the ball once it is in the water. The velocity of the ball just before entering the water can be found using the formula:
[tex]v = g t = 32 \text{ ft/s}^2 \times 1 \text{ s} = 32 \text{ ft/s}[/tex]
This means the ball will sink at a velocity of 32 ft/s. Now, we can calculate the depth of the lake (d):
[tex]d = \text{velocity} \times \text{time} = 32 \text{ ft/s} \times 4.30 \text{ s} = 137.6 \text{ ft}[/tex]
Step 6: Add the Initial Drop
Finally, to find the total depth of the lake, we need to add the initial drop of 16.0 ft:
[tex]\text{Total depth} = 137.6 \text{ ft} + 16.0 \text{ ft} = 153.6 \text{ ft}[/tex]
A plane wall of thickness 0.1 mm and thermal conductivity 25 W/m K having uniform volumetric heat generation of 0.3 MW/m3 is insulated on one side, while the other side is exposed to a fluid at 92 oC. The convection heat transfer coefficient between the wall and the fluid is 500 W/m2 K. Determine the maximum temperature in the wall.
Answer:
The maximum temperature is 90.06° C
Explanation:
Given that
t= 0.1 mm
Heat generation
[tex]q_g=0.3\ MW/m^3[/tex]
Heat transfer coefficient
[tex]h=500\ W/m^2K[/tex]
Here one side(left side) of the wall is insulated so the all heat will goes in to right side .
The maximum temperature will at the left side.
Lets take maximum temperature is T
Total heat flux ,q
[tex]q=q_g\times t[/tex]
[tex]q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2[/tex]
[tex]q=30\ W/m^2[/tex]
So the total thermal resistance per unit area
[tex]R=\dfrac{t}{K}+\dfrac{1}{h}[/tex]
[tex]R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}[/tex]
R=0.002 K/W
We know that
q=ΔT/R
30=(T-90)/0.002
T=90.06° C
The maximum temperature is 90.06° C
A european sports car dealer claims that his car
willaccelerates ar a constant rate from rest to 100 km/hr in8.00 s.
If so, what is the acceleration?
Answer:
Acceleration of the car, [tex]a=3.47\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 0 (at rest)
Final speed of the car, v = 100 km/h = 27.77 m/s
Time, t = 8 s
We need to find the acceleration of the car. Mathematically, it is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{27.77\ m/s}{8\ s}[/tex]
[tex]a=3.47\ m/s^2[/tex]
So, the acceleration of the car is [tex]3.47\ m/s^2[/tex]. Hence, this is the required solution.
Answer:
acceleration is 3.47 m/s²
Explanation:
given data
speed = 100 km/hr = 27.78 m/s
time = 8 s
to find out
acceleration?
solution
we will apply here first equation of motion
v = u +at ..............1
here v is final speed and u is initial speed that is 0 and t is time that is 8 sec
so put all value in equation 1
v = u +at
27.78 = 0 +a(8)
a = [tex]\frac{27.78}{8}[/tex]
a = 3.47 m/s²
so acceleration is 3.47 m/s²
Consider a gray squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel's velocity (in m/s) just before hitting the ground, assuming it fell from a height of 3.0 m
Answer: 7.66 8 m/s
Explanation:
Since the squirrel fell out and it is not thrown, we assume the initial velocity is zero ([tex]V_{o}=0[/tex]). On the other hand, the squirrel only experiences the acceleration due gravity, which is constant and in the downward direction [tex]g=-9.8 m/s^{2}[/tex]. So, the following equation will be useful to find the squirrel's final velocity [tex]V_{f}[/tex]:
[tex]{(V_{f})}^{2}={(V_{o})}^{2}-2gd[/tex]
Where [tex]d=3 m[/tex] is the height from which the squirrel fell
As [tex]V_{o}=0[/tex]:
[tex]{(V_{f})}^{2}=-2gd[/tex]
Then:
[tex]{(V_{f})}^{2}=-2(-9.8 m/s^{2})(3 m)[/tex]
[tex]V_{f}=\sqrt{58.8 m^{2}/s^{2}}[/tex]
Finally:
[tex]V_{f}=7.668 m/s[/tex]
A car is traveling 20 m/s and slows down at a uniform rate. It stops in 6 seconds. How far has it traveled in this interval?
Answer:
Car travel a distance of 60.06 m in 6 sec
Explanation:
We have given initial velocity v = 20 m/sec
Time = 6 sec
As the car stops finally so final velocity v = 0
From the first equation of motion
v = u+at (as the car velocity is slows down means it is a case of deceleration)
So v = u-at
[tex]0=20-a\times 6[/tex]
[tex]a=3.33m/sec^2[/tex]
Now from second equation of motion [tex]s=ut-\frac{1}{2}at^2=20\times 6-\frac{1}{2}\times 3.33\times 6^2=60.06m[/tex]
A car traveling 85 km/h slows down at a constant 0.47 m/s^2 just by "letting up on the gas." Calculate the distance it travels during the first second. Express your answer to two significant figures and include the appropriate units. Calculate the distance it travels during the fifth second.
Answer:
Explanation:
u = 85 km / h = 23.61 m /s
Acceleration a = - 0.47 m /s²
Distance travelled in n th sec
S_n = u + ( 2t -1 )a/2
S_n is distance travelled in n th seond
Distance traveled in 1 st second
= 23.61 - .5 x .47
= 23.37 m
=23 m
Distance travelled when t = 5 th
= 23.61 - (2x5 -1)/2 x .47
= 23.61 - 4.5 x .47
23.61 -2.115
=21.495
21 m is distance travelled in 5 th second .
Solids in which electrons are able to move easily from one atom to another are: a) conductors b) electrically charged c) capacitors d) insulators
Answer:
Option a) conductors
Explanation:
Conductors are the solids which allows the movement of electrons from atom to atom.
In conductors, electrons are free to move in the conductor's vicinity thus these free electrons contributes to the electrical conductivity of the conductor. A body is said to be electrically charged when the electrons in the atoms of the body are not evenly shared.Capacitors are the elements for the storage of electric charge. Insulators are the materials with no free electrons and have a high energy gap which do not allow the conduction of electricity.Answer:
a) Conductors.
Explanation:
Conductors are solids that have free electrons. Some of the examples of solids that are conductors are Steel, Iron, Silver, Copper etc. How best a metallic solid can conduct depends on the number of conduction electrons present in it.These electrons are called conduction electrons.
Not all solids are conductors. For example, wood is a solid, but it is not a conductor. It is an insulator.
Insulators are those materials that do not have free electrons.
Semiconductors are the materials that behave as conductors under certain conditions
A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wheel is immersed in 600 g of water at 15°C. What is the maximum temperature that the water could be increased to? Is this a very efficient way of heating water? The specific heat of water is 4.186 J/(g* ˚C).
Answer:[tex]16.234^{\circ}C[/tex]
Explanation:
Given
Mass of object (m)=6 kg
falling height(h)=10 m
mass of water([tex]m_w[/tex])=600 gm
temperature of water =15
specific heat of water [tex]=4.186 j/g-^{\circ}C[/tex]
Let T be the Final Temperature of water
Here Object Potential Energy is converted into Heat energy which will be absorbed by water
Potential Energy(P.E.)[tex]=mgh=6\times 9.81\times 10=588.6 J[/tex]
Heat supplied[tex]=m_wc(\Delta T)[/tex]
[tex]H.E.=600\times 4.186\times (T-16) [/tex]
[tex]588.6=2511.6\times (T-16)[/tex]
T-16=0.234
[tex]T=16.234^{\circ}C[/tex]
This is not an efficient way of heating water as there is only[tex] 0.234^{\circ}C [/tex]increase in temperature.
A cheetah is walking at 5.0m/s when it sees a zebra 35m away. What acceleration would be required for the cheetah to reach 25.0 m/s in that distance?
Answer:
Acceleration, [tex]a=8.57\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the Cheetah, u = 5 m/s
Distance covered, s = 35 m
Final speed of the cheetah, v = 25 m/s
We need to find the acceleration required for the cheetah to reach its final speed. The formula is as follows :
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(25)^2-(5)^2}{2\times 35}[/tex]
[tex]a=8.57\ m/s^2[/tex]
So, the acceleration of the cheetah is [tex]8.57\ m/s^2[/tex]. Hence, this is the required solution.
Three moles of ideal gas is inside a sealed rectangular container of size 0.2 m x 0.2 m x 0.4 m. If the pressure inside the container is 498,840 N/m^2, what is the temperature of the gas inside the container?
Answer:
320 K
Explanation:
Given:
Number of moles of gas , n = 3
Dimensions of the rectangular container = 0.2 m × 0.2 m × 0.4 m
Thus,
Volume of the container = 0.2 m × 0.2 m × 0.4 m = 0.016 m³
Pressure inside the container, P = 498,840 N/m²
Now,
From the ideal gas law, we have
PV = nRT
here,
T is the temperature
R is the ideal gas constant = 8.314 J / (mol·K)
on substituting the values, we get
498,840 × 0.016 = 3 × 8.314 × T
or
T = 320 K
Hence, the pressure inside the container is 320 K
If an amplitude modulation (AM) has an amplitude modulating signal of 0.5 V and carrier amplitude of 1.3 V, what is the modulation index?
Answer:
The modulation index in the amplitude modulation will be 0.384
Explanation:
We have given amplitude of adulating signal [tex]A_m=0.5volt[/tex]
Amplitude of carrier signal [tex]A_C=1.3volt[/tex]
We have to find modulation index
Modulation index is the ratio of amplitude of modulating signal and amplitude of carrier signal
So modulation index [tex]m=\frac{A_m}{A_c}=\frac{0.5}{1.3}=0.384[/tex]
So the modulation index in the amplitude modulation will be 0.384
You comb your hair and the comb becomes negatively charged. Strictly speaking, how will the mass of your hair change? A. it wil increase B. it will decrease C. it wil not change
Answer:
B. it will decrease
Explanation:
If the comb is negatively charged, it's because electrons went from the hair to the comb. These electrons have a certain mass, that is why the mass of the hair will decrease.
The mass of your hair will not change significantly when combing and the comb becomes negatively charged because the mass of electrons is extremely small. Small pieces of paper will be attracted to a negatively charged comb due to electrostatic forces.
Explanation:When you comb your hair and the comb becomes negatively charged, it does so by gaining excess electrons from your hair. However, the mass of electrons is extremely small, so when considering the mass change of your hair due to the transfer of electrons, it's accurate to say that the mass will not change significantly enough to be noticeable or measurable with standard equipment. Therefore the correct answer is C. the mass will not change.
In the contexts of the related information provided, when you bring a negatively charged comb close to small pieces of paper, they will be attracted to the comb because of electrostatic forces. This is because the paper, being neutral, becomes polarized in the presence of the charged comb, leading to an attraction. Thus, the observed effect would be b. Pieces of paper are attracted to the comb.
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
Answer:
Time period, [tex]T=3.05\times 10^{-5}\ s[/tex]
Explanation:
Given that,
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz
We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :
[tex]T=\dfrac{1}{f}[/tex]
T is the time period of the crystal's motion.
Time period is given by :
[tex]T=\dfrac{1}{32768}[/tex]
[tex]T=3.05\times 10^{-5}\ s[/tex]
So, the time period of the crystal's motion is [tex]3.05\times 10^{-5}\ s[/tex]. Hence, this is the required solution.
A well hole having a diameter of 5 cm is to be cut into the earth to a depth of 75 m. Determine the total work (in joules) required to raise the earth material to the surface if the average mass of 1 m is 1830 kg. (Data: g = 9.81 m/s) (Hint: How much work is required to raise a volume of ad/4 x dx from a depth of x feet to the surface?
Answer:
total work is 99.138 kJ
Explanation:
given data
diameter = 5 cm
depth = 75 m
density = 1830 kg/m³
to find out
the total work
solution
we know mass of volume is
volume = [tex]\frac{\pi}{4} d^2 dx[/tex]
volume = [tex]\frac{\pi}{4} d^2 1830 dx[/tex]
so
work required to rise the mass to the height of x m
dw = [tex]\frac{\pi}{4} d^2 1830[/tex] gx dx
so total work is integrate it with 0 to 75
w = [tex]\int\limits^{75}_{0} {\frac{\pi}{4} d^2 1830 gx dx}[/tex]
w = [tex]\frac{\pi}{4}[/tex] × 0.05² × 1830× 9.81× [tex](\frac{x^2}{2})^{75}_0[/tex]
w = 99138.53 J
so total work is 99.138 kJ
The car traveling 56.0 km/h is 25.0 m towards a haystack when the driver brakes. The car hits the haystack 2.11 s later. How fast is the car traveling at impact? Answer in m/s
Answer:
36.13 m/s
Explanation:
initial speed, u = 56 km/h = 15.56 m/s
distance, s = 25 m
t = 2.11 s
Let the speed of the car at the impact is v and the acceleration is a.
Use second equation of motion
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
[tex]25=1.56 \times 2.11+\frac{1}{2}\times a \times 2.11^{2}[/tex]
a = 9.75 m/s^2
Use first equation of motion
v = u + at
v = 15.56 + 9.75 x 2.11
v = 36.13 m/s
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot can fall when the acceleration is more than 8g, with g being the gravitational constant. Find the radius of the smallest circle that the pilot can take. (HINT: Think about the centripetal acceleration).
Answer:
1472.98 m
Explanation:
Data provided:
Speed of circular looping, v = 340 m/s
Acceleration, a = 8g
here,
g is the acceleration due to the gravity = 9.81 m/s²
Now,
the centripetal acceleration is given as,
[tex]a=\frac{v^2}{r}[/tex]
r is the radius of the loop
on substituting the respective values, we get
[tex]8\times9.81=\frac{340^2}{r}[/tex]
or
r = 1472.98 m
In an experiment to measure the density of iron, the following data was obtained. Book value is 7.86 g/cm3. One of the two sets of data is more precise, while one is more accurate. Which of these two sets has the greater precision and which has the greater accuracy
(a) 7.72, 7.74. 7.73. 7.75, 7.74
(b) 7.86, 7.90, 7.78, 7.93, 7.83
Which is more precise?
Which is more accurate?
Answer:
a) is more precise
b) is more accurate
Explanation:
(a) 7.72, 7.74. 7.73. 7.75, 7.74
For this set the average is: (7.72+7.74+7.73+7.75+7.74)/5=7.736
(b) 7.86, 7.90, 7.78, 7.93, 7.83
For this set the average is: (7.86+7.90+7.78+7.93+7.83)/5=7.86
Accuracy refers to the closeness of a measured set of measures to a known value.In our case that the book value is 7.86g/cm3
We can see that the set (b) is more accurate.
Precision correspond to the closeness of the measurements to each other.
We can see that the set of measures a) is more precise, they are more close one of the each other than the set b)
Answer:
(a) is more precise
(b) is more accurate
Explanation:
When a given set of values are precise, this implies that all the values are close to each other but may not be accurate. But a set of values are accurate when they are compared and it can be observed that they are close to a given measured value. So we can infer that the values in option (a) are precise, while that in (b) are accurate when compared with the book value given.
Consider flipping N fair coins (assume N is even) (a) 1 point] What is the probability of getting half heads and half tails, in no particular order? 1, simplify your answer, and show the limiting behavior of the (b) [1.5 points] Assume that N probability as N -> co.
Answer with Explanation:
The problem can be simplified as follows
The number of possible outcomes after tossing a coin N times is [tex]2^N[/tex] since for each toss 2 outcomes are possible
Since we need equal heads and equal tails the no of cases amont the [tex]2^n[/tex] cases are
[tex]\binom{N}{N/2}=\frac{N!}{(N-N/2)!(N/2)!}=\frac{n!^2}{(n/2)!^2}[/tex]
Thus the required probability is
[tex]P(E)=\frac{\frac{n!}{(n/2)!^2}}{2^n}=\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]
Part 2)
For the limit as N approaches infinity we have
[tex]P(E')=\lim_{n\rightarrow \infty }\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]
Using Stirling's approximation and solving we get
[tex]\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{n}{2\pi i(n-i)}}\times \frac{n^n}{i^i(n-i)^i}\\\\\lim_{n\rightarrow \infty }\binom{n}{n/2}\approx \sqrt{\frac{2n}{\pi n^2}}\times \frac{n^n}{(n/2)^{n/2}\cdot (n/2)^{n/2}}\\\\\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{2n}{\pi n^2}}\times 2^n\\\\P(E')=\frac{\sqrt{\frac{2n}{\pi n^2}}\times 2^n}{2^n}=\sqrt{\frac{2}{\pi N}}[/tex]
How does a capacitor store energy? Can a resistor store energy?
Answer:
A capacitor stores energy in the form of an electric field.A resistor can’t store energy.Explanation:
A capacitor stores energy in the form of an electric field formed between the plates of the capacitor. To establish this electric field ( or to move against it) the charges must do work, and this work its stored in the capacitor.
A resistor can’t store energy. There is a loss of energy in a current passing a resistor, but this is thanks to the heat dissipated by the resistor, and can’t be recovered.
Final answer:
A capacitor stores energy in its electric field between its plates, whereas a resistor dissipates energy as heat.
Explanation:
A capacitor stores energy in the electric field between its plates. When a voltage is applied across the capacitor, it charges up and stores electrical potential energy.
The energy stored in a capacitor can be calculated using the formula:
E = ½CV²
Where E is the energy stored in joules, C is the capacitance of the capacitor in farads, and V is the voltage across the capacitor in volts.
On the other hand, a resistor dissipates energy as heat when a current passes through it and cannot store energy like a capacitor.
Please help asap!!
Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?
Answer numerically in units of meters.
Answer:
Explanation:
Let's use the equation Δy = [tex]v(t)-\frac{1}{2}(g)(t)^{2}[/tex]
That would mean -h = [tex]v(\frac{9.8}{2})-\frac{1}{2}(9.80)(\frac{9.8}{2})^{2}.
Since the ball has stopped at t = 9.8 = g, then that would mean that the final velocity v = 0.
-h = [tex](0)(\frac{9.8}{2})-(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]
[tex]h = -(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]
[tex]h = (\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]
[tex]h = (\frac{9.8}{2})(\frac{9.8}{2})(\frac{9.8}{2})[/tex]
[tex]h = \frac{9.8^{3}}{2^{3}}[/tex]
[tex]h = \frac{941.192}{8}[/tex]
[tex]h = 117.649[/tex]
The height of the cannonball at [tex]t = \frac{9.8}{2}[/tex] should be 117.649 m
Hope this helps!
Answer:
[tex]h = 67.5\,m[/tex]
Explanation:
The position of the cannonball is given by the following expressions:
Half position
[tex]h = H -\frac{1}{2}\cdot g \cdot \left(\frac{t_{g}}{2} \right)^{2}[/tex]
Final position
[tex]0 = H -\frac{1}{2}\cdot g \cdot t_{g}^{2}[/tex]
The instant when the cannonball hits the ground is:
[tex]t_{g} = \sqrt{\frac{2\cdot H}{g} }[/tex]
Lastly, this result is applied in the other equation, which simplified afterwards:
[tex]h = H -\frac{1}{8}\cdot g \cdot \left(\frac{2\cdot H}{g} \right)[/tex]
[tex]h = H -\frac{1}{4}\cdot H[/tex]
[tex]h = \frac{3}{4}H[/tex]
[tex]h = 67.5\,m[/tex]
One cubic meter (1.00 m) of aluminum has a mass of 2.70 10 kg, and the same volume of iron has a mass of 7.8610'kg. Find the radus of a solid aluminum sphere that will balance a sod irn sphere o radu 14 squal-a balano rm
Answer:
Radius of aluminium sphere which has same mass as of sphere of iron with radius 14 m is 40.745 meters.
Explanation:
Let the radius of aluminium sphere be [tex]r_{a}[/tex]
From the relation between density, mass and volume we know that
[tex]Mass=Density\times volume...............(i)[/tex]
Applying equation 'i' separately to iron and aluminium sphere we get
[tex]M_{a}=\rho _{a}\times V_{a}[/tex]
[tex]M_{ir}=\rho _{ir}\times V_{ir}[/tex]
Equating the masses of iron and aluminium spheres we get
[tex]M_{a}=M_{ir}[/tex]
[tex]\rho _{a}\times \frac{4\pi r_a^3}{3}=\rho _{ir}\times \frac{4\pi r_{ir}^3}{3}[/tex]
[tex]\rho _{a}\times r_{a}^3=\rho _{ir}\times r_{ir}^{3}\\\\\therefore r_{a}=(\frac{\rho _{ir}}{\rho _{a}})^{1/3}\cdot r_{ir}\\\\r_{a}=\frac{7.8610}{2.701}\times 14\\\\\therefore r_a=40.745m[/tex]