3.0cm
Explanation:
For lenses in an achromatic combination, the following condition holds, assuming the two lenses are of the same materials;
d = [tex]\frac{f_1 + f_2}{2}[/tex] ---------(i)
Where;
d= distance between lenses
f₁ = focal length of the first lens
f₂ = focal length of the second lens
From the question;
f₁ = 4.5cm
f₂ = 1.5cm
Substitute these values into equation (i) as follows;
d = [tex]\frac{4.5+1.5}{2}[/tex]
d = [tex]\frac{6.0}{2}[/tex]
d = 3.0cm
Therefore, the distance between the two lenses is 3.0cm
If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
14.79 kgm/s
Explanation:
Data provided in the question
Let us assume the mass of baseball = m = 0.145 kg
The Initial velocity of pitched ball = [tex]v_i[/tex] = 47 m/s
Final velocity of batted ball in the opposite direction = [tex]v_f[/tex]= -55m/s
Based on the above information, the change in momentum is
[tex]\Delta P = m(v_f -v_i)[/tex]
[tex]= 0.145 kg(-55m/s - 47m/s)[/tex]
= 14.79 kgm/s
Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evaporate
For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.
When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.
For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.
Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.
Learn more: https://brainly.com/question/5019199
The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.
Answer:
(c) always equal to the weight of the liquid displaced.
Explanation:
Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.
Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.
A certain dam generates 120 MJ of mechanical (hydroelectric) energy each minute. If the conversion from mechanical to electrical energy is then 15% efficient, what is the dam's electrical power output in W?
Answer:
electric energy ( power ) = 300000 W
Explanation:
given data
mechanical (hydroelectric) energy = 120 MJ/min = 2000000 J/s
efficiency = 15 % = 0.15
solution
we know that Efficiency of electric engine is expression as
Efficiency = Mechanical energy ÷ electric energy ......................1
and here dam electrical power output is
put here value in equation 1
electric energy ( power ) = Efficiency × Mechanical energy ( power )
electric energy ( power ) = 0.15 × 2000000 J/s
electric energy ( power ) = 300000 W
hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac slow twitch intermediate fast twitch
Answer:
Dead lifting uses tho muscle fundamentals
Explanation:
Answer:
Fast twitch
Explanation:
Edmentum
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
describe the relation among density, temperature, and volume when the pressure is constant, and explain the blackbody radiation curve
Answer:
in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body,
Explanation:
Let's start for ya dream gas
PV = nRT
Since it indicates that the pressure is constant, we see that the volume is directly proportional to the temperature.
The density of is defined by
ρ = m / V
As we saw that volume increases with temperature, this is also true for solid materials, using linear expansion. Therefore in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body, since all the radiation that falls on it is absorbed or emitted.
This type of construction has a characteristic curve where the maximum of the curve is dependent on the tempera, but independent of the material with which it is built, to explain the behavior of this curve Planck proposed that the diaconate in the cavity was not continuous but discrete whose energy is given by the relationship
E = h f
Which of the following is not a benefit of improved cardiorespiratory fitness
Answer:
C - Arteries grow smaller
Explanation:
The option choices are:
A. Faster post-exercise recovery time
B. Lungs expand more easily
C. Arteries grow smaller
D. Diaphragm grows stronger
Explanation:
There are many advantages of cardiorespiratory fitness. It can decrease the risk of heart disease, lung cancer, type 2 diabetes, stroke, and other diseases. Cardiorespiratory health helps develop lung and heart conditions and enhances feelings of well-being.
To understand the standard formula for a sinusoidal traveling wave.
One formula for a wave with a y displacement (e.g., of a string) traveling in the x direction is
y(x,t)=Asin(kxâÏt).
All the questions in this problem refer to this formula and to the wave it describes.
1) What is the phase Ï(x,t) of the wave?
Express the phase in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï
Ï(x,t)=
2) What is the wavelength λ of the wave?
Express the wavelength in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï.
λ=
3) What is the period T of this wave?
Express the period in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï.
T=
4) What is the speed of propagation v of this wave?
Express the speed of propagation in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï.
v=
Answer:
1) Φ=zero
2) λ = 2π / k
3) T = 2π / w
4) v = w / k
Explanation:
The equation of a traveling wave is
y = A sin (ka - wt + Ф)
Let's answer using this equation the different questions
1) we see that the equation given in the problem the phase is zero
2) wavelength
k = 2π /λ
λ = 2π / k
3) The perido
angular velocity is related to frequency
w = 2π f
frequency and period are related
f = 1 / T
w = 2 π / T
T = 2π / w
4) the wave speed is
v = λ f
λ = 2π / k
f = w / 2π
v = 2π /k w /2π
v = w / k
Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?
Answer:
a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Explanation:
a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:
First arrow
[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
Now, the system is expanded and simplified:
[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]
[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]
Where:
[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.
[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
The initial vertical speed of the arrow is:
[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]
Where:
[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.
[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:
[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]
[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{2} - y_{1} = 56.712\,m[/tex]
Second arrow
[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]
[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{3} - y_{1} = 342.816\,m[/tex]
The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.
b) The total energy of each system is determined hereafter:
First arrow
The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:
[tex]E = U + K_{x}[/tex]
The expression is now expanded:
[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]
Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.
[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:
[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]
[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]
If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 201.720\,J[/tex]
Second arrow:
The total mechanical energy is equal to the potential gravitational energy. That is:
[tex]E = m\cdot g \cdot y_{max}[/tex]
[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]
[tex]E = 201.720\,J[/tex]
Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
A uniform 2.0-kg rod that is 0.92 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 29 N/m and 66 N/m. Find the angle that the rod makes with the horizontal.
Answer:
11.7°
Explanation:
See attached file
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
(a) Find the magnitude of the angular acceleration of the wheel.
(b) Find the angle in radians through which it rotates in this time interval.
Explanation:
(a) Find the magnitude of the angular acceleration of the wheel.
angular acceleration = angular speed /timeangular acceleration = 12.9/2.98 = 4.329rad/s²(b) Find the angle in radians through which it rotates in this time interval.
angular speed = 2x3.14xf12.9rad = 2 x3.14rad = 6.28/12.9rad = 0.487Now we convert rad to angle
1 rad = 57.296°0.487 = unknown angleunknown angle =57.296 x 0.487 = 27.9°The angle in radians = 27.9°
As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94 rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Answer:
I₂/I₁ = 0.53
Explanation:
During the motion the angular momentum of the skater remains conserved. Therefore:
Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms
L₁ = L₂
but, the formula for angular momentum is:
L = Iω
Therefore,
I₁ω₁ = I₂ω₂
I₂/I₁ = ω₁/ω₂
where,
I₁ = Initial Moment of Inertia
I₂ = Final Moment of Inertia
ω₁ = Initial Angular Velocity = 3.14 rad/s
ω₂ = Final Angular velocity = 5.94 rad/s
Therefore,
I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)
I₂/I₁ = 0.53
When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly
Answer:
Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force, air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.
A ball bouncing against the ground and rebounding is an example of an elastic collision. Describe two different methods of evaluating this interaction, one for which momentum is conserved, and one for which momentum is not conserved. Explain your answer.
Answer:
Momentum is conserved when there are no outside forced present and it has an equal and opposite reaction, also momentum is conserved the ball's momentum is transferred to the ground. This first instance is the case of a Closed system.
The second case where momentum is not conserved is when there is a variation or difference in the moment of the ball because of influence of external forces
The cost of energy delivered to residences by electrical transmission varies from $0.070/kWh to $0.258/kWh throughout the United States; $0.110/kWh is the average value.
Required:
At this average price, calculate the cost of:
a. leaving a 40-W porch light on for two weeks while you are on vacation?
b. making a piece of dark toast in 3.00 min with a 970-W toaster
c. drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer.
Answer:
Cost = $ 1.48
Cost = $ 0.005
Cost = $ 0.38
Explanation:
given data
electrical transmission varies = $0.070/kWh to $0.258/kWh
average value = $0.110/kWh
solution
when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be
first we get here energy consumed that is express as
E = Pt .................1
here E is Energy Consumed and Power Delivered is P and t is time
so power is here 0.04 KW and t = 2 week = 336 hour
so
put value in 1 we get
E = 0.04 × 336
E = 13.44 KWh
so cost will be as
Cost = E × Unit Price .............2
put here value and we get
Cost = 13.44 × 0.11
Cost = $ 1.48
and
when you making a piece of dark toast in 3.00 min with a 970-W toaster
so energy consumed will be by equation 1 we get
E = Pt
power is = 0.97 KW and time = 3 min = 0.05 hour
put value in equation 1 for energy consume
E = 0.97 × 0.05 h
E = 0.0485 KWh
and we get cost by w\put value in equation 2 that will be
cost = E × Unit Price
cost = 0.0485 × 0.11
Cost = $ 0.005
and
when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer
from equation 1 we get energy consume
E = Pt
Power Delivered = 5.203 KW and time = 40 min = 0.67 hour
E = 5.203 × 0.67
E = 3.47 KWh
and
cost will by put value in equation 2
Cost = E × Unit Price
Cost = 3.47 × 0.11
Cost = $ 0.38
what is thermodynamic?
Answer:
Thermodynamics is a branch of physics which deals with the energy and work of a system. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiment.
Answer:
thermodynamics is the branch of physics which deals with the study of heat and other forms energy and their mutual relationship(relation ship between them)
Explanation:
i hope this will help you :)
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch is of width 62.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?
Answer:
14.1 m/s
Explanation:
From the question,
μk = a/g...................... Equation 1
Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.
make a the subject of the equation
a = μk(g).................. Equation 2
Given: μk = 0.160, g = 9.8 m/s²
Substitute into equation 2
a = 0.16(9.8)
a = 1.568 m/s²
Using,
F = ma
Where F = force, m = mass.
Make m the subject of the equation
m = F/a................... Equation 3
m = 160/1.568
m = 102.04 kg.
Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.
Assuming the initial velocity of the skier to be zero
Fd+mgμ = 1/2mv²........................Equation 4
Where v = speed of the skier after crossing the patch, d = distance/width of the patch.
v = √2(Fd+mgμ)/m)................ Equation 5
Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16
Substitute these values into equation 5
v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]
v = √197.57
v = 14.1 m/s
v = 9.86 m/s
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
The speed of the arrow after passing through the target is 30.1 meters per second.
Explanation:
The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:
[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]
Where:
[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.
[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.
[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.
The final speed of the arrow is now cleared:
[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]
[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]
If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:
[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]
[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]
The speed of the arrow after passing through the target is 30.1 meters per second.
5) what is the weight of a
if its weight
is 5N in moon?
body in the earth,
Answer:
Weight of object on moon is 5N ,as we know. Weight of object on moon is 1/3 the of object on earth,so
let weight of object on earth = X
5N= X/3
X = 3×5 = 15N
Hence the weight of the object on earth will
be 15N
What would be the Roche limit (in units of Earth radii) if the Earth had the same mass, but its radius was increased to 1.5 Earth radii?
First calculate the density of this new, larger, Earth. Now use this new density and the new radius in the calculator above to determine the Roche limit for this new larger 'Earth.
Answer:
Roche limit = 1.89 of earth radius
Explanation:
We know that,
Mass of earth = 5.972 × 10²⁷ g
New radius = 1.5(old radius) = 1.5(6.371 × 10⁸) = 9.5565 × 10⁸
Density of earth = 5.5132 g/cm³
New density of earth = Mass of earth / (4/3)πr³
New density of earth = 5.972 × 10²⁷ kg / (4/3)(22/7)( 9.5565 × 10⁸)³
New density of earth = 1.634 g/cm³
Roche limit = [2(Density of earth)/(New density of earth)]¹/³r
Roche limit = 1.89 of earth radius
An electric heater is constructed by applying a potential different of 120V across a nichrome wire that has a total resistant of 8 ohm .the current by the wire is
Answer:
15amps
Explanation:
V=IR
I=V/R
I = 120/8
I = 15 amps
When a nucleus at rest spontaneously splits into fragments of mass m1 and m2, the ratio of the momentum of m1 to the momentum of m2 is
Answer:
p₁ = - p₂
the moment value of the two particles is the same, but its direction is opposite
Explanation:
When a nucleus divides spontaneously, the moment of the nucleic must be conserved, for this we form a system formed by the initial nucleus and the two fragments of the fission, in this case the forces during the division are internal and the moment is conserved
initial instant. Before fission
p₀ = 0
since they indicate that the nucleus is at rest
final moment. After fission
[tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂
p₀ = p_{f}
0 = m₁ v₁ + m₂v₂
m₁ v₁ = -m₂ v₂
p₁ = - p₂
this indicates that the moment value of the two particles is the same, but its direction is opposite
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to
Complete Question
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)
Answer:
The velocity is [tex]v = 3.79 *10^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 144 \ kV /m = 144*10^{3} \ V/m[/tex]
The magnetic field is [tex]B = 0.38 \ T[/tex]
The force due to the electric field is mathematically represented as
[tex]F_e = E * q[/tex]
and
The force due to the magnetic field is mathematically represented as
[tex]F_b = q * v * B * sin(\theta )[/tex]
Now given that it is perpendicular , [tex]\theta = 90[/tex]
=> [tex]F_b = q * v * B * sin(90)[/tex]
=> [tex]F_b = q * v * B[/tex]
Now given that it is not deflected it means that
[tex]F_ e = F_b[/tex]
=> [tex]q * E = q * v * B[/tex]
=> [tex]v = \frac{E}{B }[/tex]
substituting values
[tex]v = \frac{ 144 *10^{3}}{0.38 }[/tex]
[tex]v = 3.79 *10^{5} \ m/s[/tex]
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
Answer:
Explanation:
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other? What would the general shape of the field lines look like? What would the field lines look like in between the two pieces?
Answer:
Explanation:
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If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?
Answer:
The sun will be located near the Gemini constellation at sunset
A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.
By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
What is experiment ?An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.
What is hydrochloric acid?Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.
Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.
By the experiment "By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
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