Answer:
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Explanation:
An electron starts from rest and falls through a potential rise of 100V. What is its final
speed?
Answer:
The final speed of electron=[tex]5.93\times 10^6m/s[/tex]
Explanation:
We are given that
Initial velocity, u=0
Potential, V=100 V
We have to find the final speed.
Mass of electron, [tex]m=9.1\times 10^{-31} kg[/tex]
Charge on electron, q=[tex]1.6\times 10^{-19}C[/tex]
We know that
[tex]qV=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
Using the formula
[tex]1.6\times 10^{-19}\times 100=\frac{1}{2}\times 9.1\times 10^{-31} v^2-0[/tex]
[tex]v^2=\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}[/tex]
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}}[/tex]
[tex]v=5.93\times 10^6m/s[/tex]
Hence, the final speed of electron=[tex]5.93\times 10^6m/s[/tex]
Ann and Bob are carrying a 18.5 kg table that is 2.25 m long. A 8.33 kg box sits on the table 0.750 m from Ann. How much lift force does Ann exert? Use 9.80 for gravity and answer in Newtons
Answer:
F = 118 N
Explanation:
Assume Ann and Bob lift at their respective ends of the table
Sum moments about Bob's position to zero.
Let F be Ann's upward force
F[2.25] - 18.5(9.80)[2.25 / 2] - 8.33(9.80)[0.750] = 0
F = 117.86133333...
Given:
Mass of table, [tex]m_t = 18.5 \ kg[/tex]Mass of box, [tex]m_b = 8.33 \ kg[/tex]Length of table, [tex]2.25 \ m[/tex]Length of box, [tex]0.75 \ m[/tex]The weight of table will be:
→ [tex]W_t = m_t g[/tex]
[tex]= 18.5\times 9.8[/tex]
[tex]= 181.3 \ N[/tex]
Now,
→ [tex]\sum M_A_{nn} = -81.634\times 0.750-181.3\times 1.125+R_{bob}\times 1.125[/tex]
or,
→ [tex]R_{bob} = \frac{61.2255+203.9625}{2.25}[/tex]
[tex]= \frac{265.188}{2.25}[/tex]
[tex]= 117.9 \ N[/tex]
Thus the above answer is right.
Learn more about force here:
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The mass of a hot-air balloon and its occupants is 381 kg (excluding the hot air inside the balloon). The air outside the balloon has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 480 m3. The pressure of the heated air remains the same as that of the outside air. To what temperature in kelvins must the air be heated so that the balloon just lifts off
Answer:
In order to lift off the ground, the air in the balloon must be heated to 710.26 K
Explanation:
Given the data in the question;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
let F represent the force acting upward.
Now in a condition where the hot air balloon is just about to take off;
F - Mg - m[tex]_g[/tex]g = 0
where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.
the force acting upward F = Vρg
so
Vρg - Mg - m[tex]_g[/tex]g = 0
solve for m[tex]_g[/tex]
m[tex]_g[/tex] = ( Vρg - Mg ) / g
m[tex]_g[/tex] = Vρg/g - Mg/g
m[tex]_g[/tex] = ρV - M ------- let this be equation 1
Now, from the ideal gas law, PV = nRT
we know that number of moles n = m[tex]_g[/tex] / μ
where μ is the molecular mass of air
so
PV = (m[tex]_g[/tex]/μ)RT
solve for T
μPV = m[tex]_g[/tex]RT
T = μPV / m[tex]_g[/tex]R -------- let this be equation 2
from equation 1 and 2
T = μPV / (ρV - M)R
so we substitute in our values;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]
T = 1405920 / 1979.442
T = 710.26 K
Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K
The temperature required for the air to be heated is 710.26 K.
Given data:
The mass of a hot air-balloon is, m = 381 kg.
The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].
The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].
The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].
The condition where the hot air balloon is just about to take off is as follows:
[tex]F-mg - m'g =0[/tex]
Here,
m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,
[tex]F = V \times \rho \times g[/tex]
Solving as,
[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]
Now, apply the ideal gas law as,
PV = nRT
here, R is the universal gas constant and n is the number of moles and its value is,
[tex]n=\dfrac{m'}{M}[/tex]
M is the molecular mass of gas. Solving as,
[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]
Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,
[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]
Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.
Learn more about the ideal gas equation here:
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A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 13 years have elapsed on earth, and 7.9 years have elapsed on board the ship. How far away (in meters) is the planet, according to observers on earth
Answer:
[tex]L=9.76*10^{16}m[/tex]
Explanation:
From the question we are told that:
Time on earth [tex]T_e= 13yrs[/tex]
Time on ship [tex]T_s= 7.9yrs[/tex]
Therefore
[tex]r=\frac{t_e}{t_s}[/tex]
[tex]r=\frac{13}{7.9}[/tex]
[tex]r=1.65[/tex]
Generally the equation for Constant Velocity is mathematically given by
[tex]V=C\sqrt{1-\frac{1}{r^2}}[/tex]
[tex]V=3*10^8\sqrt{1-\frac{1}{1.64^2}}[/tex]
[tex]V=2.38*10^8m/s[/tex]
Therefore
[tex]L=V*t[/tex]
Where
[tex]t=(13*365.25*24*3600)s[/tex]
[tex]t=4.1*10^8[/tex]
[tex]L=2.38*10^8m/s*4.1*10^8[/tex]
[tex]L=9.76*10^{16}m[/tex]
The driver provides a constant force on the engine through the foot pedal. Eventually the van stops accelerating and reaches a constant speed.
c Explain why the van reaches a constant speed if the driver provides a constant driving force to the van.
It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is
∑ F = (force from engine) - (resistive forces) = 0
abrief history of hand writing
A child weighing 200 N is being held back in a swing by a horizontal force of 125 N, as shown in the image. What is the tension T in the rope that supports the swing in units of Newtons? Note: Please enter only the numerical answer. If you include any units in your answer, your answer will be counted as incorrect. T F= 125 N Weight = 200 N
Answer:
75
Explanation:
i am not sure but if 200N boy is being held back then the force that's holding him back must be equal to or greater than his weight. if 125N is already exerted then the tension will be:
T=200-125
= 75
A 0.20 kg mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to be 0.30 m/s. If, instead, a 0.40 kg mass were used in this same experiment, choose the correct value for the maximum speed.
a. 0.40 m/s.
b. 0.20 m/s.
c. 0.28 m/s.
d. 0.14 m/s.
e. 0.10 m/s.
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]
Therefore, the maximum speed is 0.20 m/s
When a car's starter is in use, it draws a large current. The car's lights draw much less current. As a certain car is starting, the current through the battery is 54.0 A and the potential difference across the battery terminals is 9.18 V. When only the car's lights are used, the current through the battery is 2.10 A and the terminal potential difference is 12.6 V. Find the battery's emf.
Answer:
12.74 V
Explanation:
We are given that
Current, I1=54 A
Potential difference, V1=9.18V
I2=2.10 A
V2=12.6 V
We have to find the battery's emf.
[tex]E=V+Ir[/tex]
Using the formula
[tex]E=9.18+54r[/tex] ....(1)
[tex]E=12.6+2.10r[/tex] .....(2)
Subtract equation (1) from (2)
[tex]0=3.42-51.9r[/tex]
[tex]3.42=51.9r[/tex]
[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]
Using the value of r in equation (1)
[tex]E=9.18+54(0.0659)[/tex]
[tex]E=12.74 V[/tex]
A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of refraction of the film is 1.35, what is the minimum thickness the soap film can be if it is surrounded by air
Answer:
the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Explanation:
Given the data in the question;
wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m
Index of refraction; n = 1.35
Now, the thinnest thickness of the soap film can be determined from the following expression;
[tex]t_{min[/tex] = ( λ / 4n )
so we simply substitute in our given values;
[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)
[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 5.4
[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)
[tex]t_{min[/tex] = 8.574 × 10⁻⁸ m
[tex]t_{min[/tex] = 85.74 × 10⁻⁹ m
[tex]t_{min[/tex] = 85.74 nm
Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloon and wall. Does the balloon move, if so which way
Answer:
Move towards the wall.
Explanation:
When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.
As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.
What is total resistor formula
Answer:
If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.
Explanation:
How far from a 10 µC point charge will the potential be 1000 V?
a.
10 m
b.
90 m
c.
80 m
d.
60 m
• Point charge (Q) = 10 μC = 10 × 10⁻⁶ C
• Potential (V) = 1000 V
• Distance (r) = ?
[tex]\implies V = \dfrac{KQ}{r} \\ [/tex]
[tex]\implies r= \dfrac{KQ}{V}[/tex]
[tex]\implies r= \dfrac{9 \times {10}^{9} \times 10 \times {10}^{ - 6} }{1000} \\ [/tex]
[tex]\implies r= \dfrac{90 \times {10}^{3} }{1000} \\ [/tex]
[tex]\implies r= \dfrac{90 \times {10}^{3} }{ {10}^{3} } \\ [/tex]
[tex]\implies r= 90 \times {10}^{3} \times {10}^{ - 3} \\ [/tex]
[tex]\implies\bf r= 90\:m \\ [/tex]
Hence,the option B) 90 m is the correct answer.
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A stone is dropped from a cliff 64 feet above the ground. Answer the following, using -32 ft/sec2 as the acceleration due to gravity. Show all work and submit to D2L.
a. Find functions that represent the acceleration, velocity, and position of the stone above the ground at time t.
b. How long does it take the stone to reach the ground?
c. With what velocity does the stone hit the ground?
Answer:
(a) v = 32 t
h = 16 t^2
g = 32 ft/s^2
(b) 64 ft/s
Explanation:
height, h = 64 feet
g = - 32 ft/s^2
(a) Let the time is t .
Let the velocity after time t is v.
Use first equation of motion
v = u + at
- v = 0 - 32 t
v = 32 t
Let the distance is h from the top.
Use second equation of motion
[tex]h = u t + 0.5 at^2 \\\\- h = 0 - 0.5\times 32 \times t^2\\\\h = 16 t^2[/tex]
The acceleration is constant for entire motion.
(b) Let the velocity is v as it hits the ground. Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 32\times 64\\\\v = 64 feet/s[/tex]
Hi,A body changes its velocity from 60 km/hr to 72 km/hr in 2 sec.Find the acceleration and distance travelled.
Answer:
Initial velocity, u = 60 km/h = 16.7 m/s
Final velocity, v = 72 km/h = 20 m/s
time, t = 2 sec
From first equation of motion:
[tex]{ \bf{v = u + at}}[/tex]
Substitute the variables:
[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]
Consider an airplane with a total wing surface of 50 m^2. At a certain speed the difference in air pressure below and above the wings is 4.0 % of atmospheric pressure.
Required:
Find the lift on the airplane.
Answer:
[tex]F=202650N[/tex]
Explanation:
From the question we are told that:
Area [tex]a=50m^2[/tex]
Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]
Generally the equation for Force is mathematically given by
[tex]F=dP*A[/tex]
[tex]F=4053*50[/tex]
[tex]F=202650N[/tex]
A cylinder within a piston expands from a volume of 1.00 L to a volume of 2.00 L against an external pressure of 1.00 atm. How much work (in Joule) was done by the expansion?
Answer:
1.671L
Explanation:
In the given case the pressure is constant therefore it is an isobaric process, the process in which the pressure remains constant is called as isobaric process.
Work done= external pressure× change in volume.
288= 2×( final volume-intial volume)×101.32
144÷101.32=(finalvolume-0.250)
1.421=final volume-0.250
final volume=1.671L
In which type of mixture do the physically distinct component parts each have distinct properties?
Answer:
In heterogeneous mixture do the physically distinct component parts each have distinct properties.
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt
This question is incomplete, the complete question is;
Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.
1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?
Fwith belt =
2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.
Fwithout belt =
Answer:
1) The Net force on the driver with seat belt is 10.3 KN
2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Explanation:
Given the data in the question;
from the equation of motion, v² = u² + 2as
we solve for a
a = (v² - u²)/2s ----- let this be equation 1
we know that, F = ma ------- let this be equation 2
so from equation 1 and 2
F = m( (v² - u²)/2s )
where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.
1)
Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.
i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m
so we substitute the given values into the equation;
F = 70( ((0)² - (18)²) / 2 × 1.1 )
F = 70 × ( -324 / 2.4 )
F = 70 × -147.2727
F = -10309.09 N
F = -10.3 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwith belt = 10.3 KN
Therefore, Net force of the driver is 10.3 KN
2)
No sit belt,
m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m
we substitute
F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )
F = 70 × ( -324 / 0.022 )
F = 70 × -14727.2727
F = -1030909.08 N
F = -1030.9 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwithout belt = 1030.9 KN
Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Two objects moving with a speed v travel in opposite directions in a straight line. The objects stick together when they collide, and move with a speed of v/2 after the collision.
Required:
a. What is the ratio of the final kinetic energy of the system to the initial kinetic energy?
b. What is the ratio of the mass of the more massive object to the mass of the less massive object?
Answer:
Explanation:
Let the mass of objects be m₁ and m₂ .
Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²
Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4 = 1/2 ( m₁ + m₂ ) v² x .25
final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²
= 0.25
b )
Applying law of conservation of momentum to the system . Let m₁ > m₂
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ - m₂ = ( m₁ + m₂ ) / 2
2m₁ - 2 m₂ = m₁ + m₂
m₁ = 3m₂
m₁ / m₂ = 3 / 1
Answer:
(a) The ratio is 1 : 4.
(b) The ratio is 1 : 3.
Explanation:
Let the mass of each object is m and m'.
They initially move with velocity v opposite to each other.
Use conservation of momentum
m v - m' v = (m + m') v/2
2 (m - m') = (m + m')
2 m - 2 m' = m + m'
m = 3 m' .... (1)
(a) Let the initial kinetic energy is K and the final kinetic energy is K'.
[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]
[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]
The ratio is
K' : K = 1 : 4
(b) m = 3 m'
So, m : m' = 3 : 1
An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10 m/s. (Assume the speed of sound is 343 m/s.)
(a) What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?
Hz
(b) What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?
Hz
Answer:
a) [tex]F=475.7Hz[/tex]
b) [tex]F'=410.899Hz[/tex]
Explanation:
From the question we are told that:
Velocity of eagle [tex]V_1=35m/s[/tex]
Frequency of eagle [tex]F_1=440Hz[/tex]
Velocity of Black bird [tex]V_2=10m/s[/tex]
Speed of sound [tex]s=343m/s[/tex]
a)
Generally the equation for Frequency is mathematically given by
[tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]
[tex]F=440(\frac{343-10}{343-35})[/tex]
[tex]F=475.7Hz[/tex]
b)
Generally the equation for Frequency is mathematically given by
[tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]
[tex]F'=440(\frac{343+10}{343+35})[/tex]
[tex]F'=410.899Hz[/tex]
A 2600 kg truck travelling at 72 km/h slams on the brakes and skids to a stop. The frictional force from the road is 8200 N. Use the relationship between kinetic energy and mechanical work to determine the distance it takes for the truck to stop.
Answer:
Approximately [tex]63\; \rm m[/tex].
Explanation:
Convert the initial speed of this truck to standard units:
[tex]\begin{aligned} v &= 72\; \rm km \cdot h^{-1} \times \frac{1\; \rm h}{3600\; \rm s} \times \frac{1000\; \rm m}{1\; \rm km} \\ &= 20\; \rm m \cdot s^{-1}\end{aligned}[/tex].
Calculate the initial kinetic energy of this truck:
[tex]\begin{aligned}\text{KE} &= \frac{1}{2} \, m \cdot v^{2} \\ &= \frac{1}{2} \times 2600\; \rm kg \times (20\; \rm m \cdot s^{-1})^{2} \\ &= 5.2 \times 10^{5} \; \rm J\end{aligned}[/tex].
The kinetic energy of the truck would be [tex]0[/tex] when the truck is not moving. Hence, the friction on the truck would need to do [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work on the truck to bring the truck to a stop.
Calculate the displacement required to achieve [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work at [tex]8200\; \rm N[/tex]:
[tex]\begin{aligned}x &= \frac{W}{F} \\ &= \frac{-5.2 \times 10^{5}\; \rm J}{8200\; \rm N} \approx -63\; \rm m\end{aligned}[/tex].
This result is negative because the direction of friction is opposite to the direction of the motion of the truck.
Hence, the truck would come to a stop after skidding for approximately [tex]63\; \rm m[/tex].
If you tethered a space station to the earth by a long cable, you could get to space in an elevator that rides up the cable much simpler and cheaper than riding to space on a rocket. There's one big problem, however: There is no way to create a cable that is long enough. The cable would need to reach 36,000 km upward, to the height where a satellite orbits at the same speed as the earth rotates; a cable this long made of ordinary materials couldn't even support its own weight. Consider a steel cable suspended from a point high above the earth. The stress in the cable is highest at the top; it must support the weight of cable below it.
What is the greatest length the cable could have without failing?
Answer:
[tex]l=12916.5m[/tex]
Explanation:
Distance [tex]d=3600km[/tex]
Since
Density of steel [tex]\rho=7900kg/m^3[/tex]
Stress of steel [tex]\mu= 1*10^9[/tex]
Generally the equation for Stress on Cable is mathematically given by
[tex]S=\frac{F}{A}[/tex]
[tex]S=\frac{\rho Alg}{A}[/tex]
Therefore
[tex]l=\frac{s}{\rhog}[/tex]
[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]
[tex]l=12916.5m[/tex]
Two positive charges ( 8.0 mC and 2.0 mC) are separated by 300 m. A third charge is placed at distance r from the 8.0 mC charge in such a way that the resultant electric force on the third charge due to the other two charges is zero. The distance r is
Answer:
[tex]r=200m[/tex]
Explanation:
From the question we are told that:
Charges:
[tex]Q_1=8.0mC[/tex]
[tex]Q_2=2.0mC[/tex]
[tex]Q_3=8.mC[/tex]
Distance [tex]d=300m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Therefore
[tex]F_{32}=F_{31}[/tex]
[tex]\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}[/tex]
[tex]\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}[/tex]
[tex]r=2(300-r)[/tex]
[tex]r=200m[/tex]
A 50.0 kg person is walking horizontally with constant acceleration of 0.25 m/s² inside an elevator. The elevator is also accelerating downward at a rate of 1.0 m/s². Sketch the path of the man as it is observed from someone on the ground. Explain your choice.
Answer:
The acceleration is in 2 D as in between east and south.
Explanation:
mass, m = 50 kg
acceleration, a = 0.25 m/s^2 horizontal
acceleration of elevator, a' = 1 m/s^2 downwards
When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:
Question 4 of 5
How can the Fitness Logs help you in this class?
O A. They can't; the Fitness Logs are only useful to your teacher.
B. They show your parents how much you're learning.
C. They let you keep track of your thoughts, feelings, and progress.
D. They help you evaluate yourself for your final grade.
SUBMIT
Answer:
C is the right answer
Explanation:
fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success
hope it was useful for you
When an external magnetic flux through a conducting loop decreases in magnitude, a current is induced in the loop that creates its own magnetic flux through the loop. How does that induced magnetic flux affect the total magnetic flux through the loop
Answer:
Len's law
Explanation:
We can explain this exercise using Len's law
when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.
During the 5 minute period described in question 4, the water in the insulated vessel undergoes a temperature increase of 20 C. Assuming all of the electrical energy dissipated by the resistor was transferred to the water as heat, what is the mass of the water
Answer:
Please find the complete question in the attached file.
Explanation:
[tex]V=12\ V\\\\I=1.2\ A\\\\T=5\times 60=300\ second\\\\[/tex]
Calculating the electrical energy dissipated:
[tex]w=p\cdot t=V\cdot I \cdot t\\\\[/tex]
[tex]=12\times 1.2 \times 300 \ J\\\\=4320\ J[/tex]
[tex]\Delta T=20^{\circ}\ C\\\\W=m\cdot c\cdot \Delta T\\\\4320=m(4186 \times 20)\\\\m=\frac{4320}{4186 \times 20}=51.6 \ grams=0.516 \ kg\\\\[/tex]
1) Define Mechanical Advantage?
2) What factor affect the mechanical advantage of a machine?
3) Define ideal machine?
4) What are output work and input work?
5) What is moment?
Answer:
1) ans: The ratio of load to effort in a simple machine is called Mechanical Advantage.
2) ans: Frictuon produced in Simple machine affect the mechanical advantage of a machine.
3) ans: The machine whose efficiency is 100% is called ideal machine.
4) ans: The work done by the machine is called output work.
ans: The work done in the machine is called input work.
5) ans: The turning effect of force is called moment.
