Answer:
The maximum speed of the mass is 4.437 m/s.
Explanation:
Given;
length of pendulum, L = 1.62 m
attached mass, m = 117 g
angle of inclination, θ = 38°
This mass was raised to a height of
h = 1.62 - cos38° = 1.0043 m
Apply the principle of conservation of mechanical energy
PE = KE
mgh = ¹/₂mv²
v = √(2gh)
v = √(2 * 9.8 * 1.0043)
v = 4.437 m/s.
Therefore, the maximum speed of the mass is 4.437 m/s.
As you drive down the road at 13 m/s , you press on the gas pedal and speed up with a uniform acceleration of 1.02 m/s2 for 0.70 s. If the tires on your car have a radius of 33 cm, what is their angular displacement during this period of acceleration?
Answer:
The angular displacement is [tex]\theta = 28.33 \ rad[/tex]
Explanation:
From the question we are told that
The speed of the driver is [tex]v =13 \ m/ s[/tex]
The acceleration of the driver is [tex]a = 1.02 \ m/s^2[/tex]
The time taken is [tex]t = 0.70 \ s[/tex]
The radius of the tire is [tex]r = 33 cm = 0.33 \ m[/tex]
The distance covered by the car during this acceleration can be calculated using the equation of motion as follows
[tex]s = v*t +\frac{1}{2} * a * t^2[/tex]
Now substituting values
[tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]
[tex]s = 9.35 \ m[/tex]
Now the angular displacement of the car with respect to the tire movement can be represented mathematically as
[tex]\theta = \frac{s}{r}[/tex]
substituting values
[tex]\theta = \frac{9.35}{0.33}[/tex]
[tex]\theta = 28.33 \ rad[/tex]
a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?
Answer:
6.44 s
Explanation:
Given:
v₀ = 119 m/s
v = 233 m/s
a = 17.7 m/s²
Find: t
v = at + v₀
(233 m/s) = (17.7 m/s²) t + (119 m/s)
t = 6.44 s
Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?
Answer:
94.248 g/sec
Explanation:
For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:
[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]
And, the velocity of blood pumping is 30 cm^2
Now apply the following formula to solve the total current
[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]
Q = 94.248 g/sec
Basically we applied the above formula So, that the total current could come
Suppose that the separation between two speakers A and B is 4.30 m and the speakers are vibrating in-phase. They are playing identical 103-Hz tones and the speed of sound is 343 m/s. An observer is seated at a position directly facing speaker B in such a way that his line of sight extending to B is perpendicular to the imaginary line between A and B. What is the largest possible distance between speaker B and the observer, such that he observes destructive interference
Answer:
The largest possible distance is [tex]x = 4.720 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 4.30 \ m[/tex]
The frequency of the tone played by both speakers is [tex]f = 103 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s[/tex]
The wavelength of the tone played by the speaker is mathematically evaluated as
[tex]\lambda = \frac{v}{f}[/tex]
substituting values
[tex]\lambda = \frac{343}{103}[/tex]
[tex]\lambda = 3.33 \ m[/tex]
Let the the position of the observer be O
Given that the line of sight between observer and speaker B is perpendicular to the distance between A and B then
The distance between A and the observer is mathematically evaluated using Pythagoras theorem as follows
[tex]L = \sqrt{d^2 + x^2}[/tex]
Where x is the distance between the observer and B
For the observer to observe destructive interference
[tex]L - x = \frac{\lambda}{2}[/tex]
So
[tex]\sqrt{d^2 + x^2} - x = \frac{\lambda}{2}[/tex]
[tex]\sqrt{d^2 + x^2} = \frac{\lambda}{2} +x[/tex]
[tex]d^2 + x^2 = [\frac{\lambda}{2} +x]^2[/tex]
[tex]d^2 + x^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} + x^2][/tex]
[tex]d^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} ][/tex]
substituting values
[tex]4.30^2 = [\frac{3.33^2}{4} +2 * x * \frac{3.33}{2} ][/tex]
[tex]x = 4.720 \ m[/tex]
A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule with an energy of about 0.8 × 10-19 J. If the velocity of the kinesin is 800 nm/s, can you determine the force the kinesin is exerting, if you assume that all the ATP energy is used (100% efficiency)? If you can, find it and give your answer in newtons. If not, answer with 0.
Answer:
The force is [tex]F = 8*10^{-12} \ N[/tex]
Explanation:
From the question we are told that
The rate at which ATP molecules are used is [tex]R = 80 ATP/ s[/tex]
The energy provided by a single ATP is [tex]E_{ATP} = 0.8 * 10^{-19} J[/tex]
The velocity of the kinesin is [tex]v = 800 nm/s = 800*10^{-9} m/s[/tex]
The power provided by the ATP in one second is mathematically represented as
[tex]P = E_{ATP} * R[/tex]
substituting values
[tex]P = 80 * 0.8*10^{-19 }[/tex]
[tex]P = 6.4 *10^{-18}J/s[/tex]
Now this power is mathematically represented as
[tex]P = F * v[/tex]
Where F is the force the kinesin is exerting
Thus
[tex]F = \frac{P}{v}[/tex]
substituting values
[tex]F = \frac{6.4*0^{-18}}{800 *10^{-9}}[/tex]
[tex]F = 8*10^{-12} \ N[/tex]
The force exerted by the kinesin is 8 × 10-12 N.
Let us recall that power is defined as the rate of doing work. Hence, power = Energy/Time.
Since;
Energy = 0.8 × 10-19 J/molecule
Number ATP molecules transported per second = 80 ATP molecules/s
Power = 0.8 × 10-19 J/molecule × 80 ATP molecules/s
Power = 6.4 × 10-18 J
Again, we know that;
Power = Force × Velocity
Velocity of the ATP molecules = 800 nm/s or 8 × 10-7 m/s
Force = Power/velocity
Force = 6.4 × 10-18 J/ 8 × 10-7 m/s
Force = 8 × 10-12 N
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If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of
Answer:
The object has an excess of [tex]10^{13}[/tex] electrons.
Explanation:
When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:
[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]
The object has an excess of [tex]10^{13}[/tex] electrons.
What if a solid cylinder of mass M = 2.50 kg, radius R = 2.18 cm, and length L = 2.7 cm, is rolling down from rest instead? With h = 79.60 m and x = 4.64 m, what is the center of mass velocity when the cylinder reaches the bottom?
Answer:
The center of mass velocity is [tex]v = 32.25 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the cylinder is [tex]m = 2.50 \ kg[/tex]
The radius is [tex]r = 2.18 \ cm = 0.0218 \ m[/tex]
The length is [tex]l = 2.7 \ cm = 0.027 \ m[/tex]
The height of the plane is h = 79.60 m
and the distance covered is [tex]d = 4.64 \ m[/tex]
The center of mass velocity o the cylinder when it reaches the bottom is mathematically represented as
[tex]v = \sqrt{\frac{4gh}{3} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{4 * 9.8 * 79.60}{3} }[/tex]
[tex]v = 32.25 \ m/s[/tex]
A crane lifts a 425 kg steel beam vertically a distance of 64 m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s2
Answer:
work done= 48.96 kJExplanation:
Given data
mass of load m= 425 kg
height/distance h=64 m
acceleration a= 1.8 m/s^2
The work done can be calculated using the expression
work done= force* distance
but force= mass *acceleration
hence work done= 425*1.8*64= 48,960 J
work done= 48.96 kJ
Find the change in the force of gravity between two planets when the distance between them becomes 10 times smaller. g
Answer:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
Explanation:
The force of gravity between two planets, is given by the following formula:
F = Gm₁m₂/r² ----------- equation 1
where,
F = Force of gravity between two planets
G = Gravitational Constant
m₁ = Mass of one planet
m₂ = Mass of other plant
r = Distance between two planets
Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:
F' = Gm₁m₂/(4r)²
F' = (1/16) (Gm₁m₂/r²)
using equation 1:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.
Answer:
D
Explanation:
The power equation is P= V^2/R
Please let me know if this helped! Please rate it the brainlist if possible!
As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.
What is a resistors?A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.
Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.
A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.
The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.
Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.
Thus, the correct option is D.
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Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Answer:
C is the correct answer
Explanation:
A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back toward your friend. What should you do in order to maximize your speed right after your interaction with the ball?
A. You should catch the ball.
B. You should let the ball go past you without touching it.
C. You should deflect the ball back toward your friend.
D. More information is required to determine how to maximize your speed.
E. It doesn't matter. Your speed is the same regardless of what you do.
Answer:
C You should deflect the ball back toward your friend.
Explanation:
This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,
with m= mass, V=velocity, i=initial, f=final:
mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)
So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision
Answer:
A. You should catch the ball.
Explanation:
Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.
Two vehicles approach an intersection, a 2500kg pickup travels from E to W at 14.0m/s and a 1500kg car from S to N at 23.0m/s. Find P net of this system (direction and magnitude)
Answer:
The magnitude of the momentum is 49145.19 kg.m/s
The direction of the two vehicles is 44.6° North West
Explanation:
Given;
speed of first vehicle, v₁ = 14 m/s (East to west)
mass of first vehicle, m₁ = 1500 kg
speed of second vehicle, v₂ = 23 m/s (South to North)
momentum of the first vehicle in x-direction (E to W is in negative x-direction)
[tex]P_x = mv_x\\\\P_x = 2500kg(-14 \ m/s)\\\\P_x = -35000 \ kg.m/s[/tex]
momentum of the second vehicle in y-direction (S to N is in positive y-direction)
[tex]P_y = m_2v_y\\\\P_y = 1500kg(23 \ m/s)\\\\P_y = 34500 \ kg.m/s[/tex]
Magnitude of the momentum of the system;
[tex]P= \sqrt{P_x^2 + P_y^2} \\\\P = \sqrt{(-35000)^2+(34500)^2} \\\\P = 49145.19 \ kg.m/s[/tex]
Direction of the two vehicles;
[tex]tan \ \theta = \frac{P_y}{|P_x|} \\\\tan \ \theta = \frac{34500}{35000} \\\\tan \ \theta = 0.9857\\\\\theta = tan^{-1} (0.9857)\\\\\theta = 44.6^0[/tex]North West
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?
Answer:
(a) 62.3 N
(b) 1.89 N
(c) 0.430 kg m²
Explanation:
(a) Find the acceleration of block B.
Δy = v₀ t + ½ at²
1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²
a = 0.90 m/s²
Draw a free body diagram of block B. There are two forces:
Weight force mg pulling down,
and tension force Tb pulling up.
Sum of forces in the -y direction:
∑F = ma
mg − Tb = ma
Tb = m (g − a)
Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)
Tb = 62.3 N
(b) Draw a free body diagram of block A. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and tension force Ta pulling right.
Sum of forces in the +x direction:
∑F = ma
Ta = ma
Ta = (2.10 kg) (0.90 m/s²)
Ta = 1.89 N
(c) Draw a free body diagram of the pulley. There are two forces:
Tension force Tb pulling down,
and tension force Ta pulling left.
Sum of torques in the clockwise direction:
∑τ = Iα
Tb r − Ta r = Iα
(Tb − Ta) r = I (a/r)
I = (Tb − Ta) r² / a
I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)
I = 0.430 kg m²
The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].
Given :
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.a) First, determine the acceleration of the B block.
[tex]\rm s = ut + \dfrac{1}{2}at^2[/tex]
[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]
[tex]\rm a = 0.9\; m/sec^2[/tex]
Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.
[tex]\rm \sum F=ma[/tex]
[tex]\rm mg-T_b=ma[/tex]
[tex]\rm T_b = m(g-a)[/tex]
[tex]\rm T_b = 7\times (9.8-0.9)[/tex]
[tex]\rm T_b = 62.3\;N[/tex]
b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.
[tex]\rm \sum F=ma[/tex]
[tex]\rm T_a=ma[/tex]
[tex]\rm T_a = 2.1\times 0.9[/tex]
[tex]\rm T_a = 1.89\;N[/tex]
c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.
[tex]\rm \sum \tau = I\alpha[/tex]
[tex]\rm T_br-T_ar = I\alpha[/tex]
[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]
Now, substitute the values of the known terms in the above expression.
[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]
[tex]\rm I = 0.430 \; kg\;m^2[/tex]
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symbol of science hhshsjsiwtwwisjzhJava
Answer:
is this a company name.? java is a computer software right..
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c
Answer:
0.85c
Explanation:
Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²
Rest mass of proton [tex]M_{0P}[/tex] = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV
for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV
Recall that the rest energy, and the total energy are related by..
[tex]E[/tex] = γ[tex]E_{0}[/tex]
which can be written in this case as
[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1
where [tex]E[/tex] = total energy of the kaon, and
[tex]E_{0}[/tex] = rest energy of the kaon
γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]
where [tex]\beta = \frac{v}{c}[/tex]
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2
where [tex]E_{K}[/tex] is the total energy of the kaon, and
[tex]E_{0P}[/tex] is the rest energy of the proton.
From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89
1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1
squaring both sides, we get
3.57( 1 - [tex]\beta^{2}[/tex]) = 1
3.57 - 3.57[tex]\beta^{2}[/tex] = 1
2.57 = 3.57[tex]\beta^{2}[/tex]
[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72
[tex]\beta = \sqrt{0.72}[/tex] = 0.85
but, [tex]\beta = \frac{v}{c}[/tex]
v/c = 0.85
v = 0.85c
A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon's radius increasing when the radius is 7ft
Answer:
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Explanation:
Given;
Rate of inflation dV = 140pift^3/min
Radius r = 7 ft
Change in radius = dr
Volume of a spherical balloon is;
V = (4/3)πr^3
The change in volume can be derived by differentiating both sides;
dV = (4πr^2)dr
Making dr the subject of formula;
dr = dV/(4πr^2)
Substituting the given values;
dr = 140π/(4π×7^2)
dr = 0.714285714285 ft/min
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Defination coulomb's law.
Answer:
a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
Find the magnitude of the resultant of forces 6N and 8N acting at 240° to each other
Answer:
magnitude of the resultant of forces is 11.45 N
Explanation:
given data
force F1 = 6N
force F2 = 8N
angle = 240°
solution
we get here resultant force that is express as
F(r) = [tex]\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}[/tex] ..............1
put here value and we get
F(r) = [tex]\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}[/tex]
F(r) = 11.45 N
so magnitude of the resultant of forces is 11.45 N
a point charge q is located at the center of a cube with edge length d. whatis the value of the flux over one face of the cube
Answer:
q/6Eo
Explanation:
See attached file pls
The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
I hope it helps you!
Can an object travel at the speed of
light? Why or why nbt?
Answer:
no the only things that can travel at the speed of light are waves in the electromagnetic spectrum
What length (in m) should an oboe have to produce a fundamental frequency of 294 Hz on a day when the speed of sound is 350 m/s? It is open at both ends.
Answer:
L = 0.60 m
The length in metres should be 0.60 m
Explanation:
A pipe open at both ends can have a standing wave pattern with resonant frequency;
f = nv/2L ........1
Where;
v = velocity of sound
L = length of pipe
n = 1 for the fundamental frequency f1
Given;
Fundamental frequency f1 = 294 Hz
Velocity v = 350 m/s
n = 1
From equation 1;
Making L the subject of formula;
L = nv/2f1
Substituting the given values;
L = 1×350/(2×294)
L = 0.595238095238 m
L = 0.60 m
The length in metres should be 0.60 m
Near the surface of Earth, an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (magnitude and sign) would have to be placed on a penny that has a mass of 3 g to cause it to rise into the air with an upward acceleration of 0.19 m/s2?
Answer:
q = 2.997*10^-4C
Explanation:
In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.
You use the second Newton law to sum both electrical and gravitational forces:
[tex]F_e-W=ma\\\\qE-mg=ma[/tex] (1)
Fe: electric force
W: weight of the penny
q: required charge = ?
m: mass of the penny = 3g = 0.003kg
E: magnitude of the electric field = 100N/C
g: gravitational acceleration = 9.8m/s^2
a: acceleration of the penny = 0.19m/s^2
You solve the equation (1) for q, and replace the values of the other parameters:
[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]
It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magnitude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x
Answer:
The change in potential energy is [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]
Explanation:
From the question we are told that
The magnitude of the uniform electric field is [tex]E = 950 \ N/C[/tex]
The distance traveled by the electron is [tex]x = 2.50 \ m[/tex]
Generally the force on this electron is mathematically represented as
[tex]F = qE[/tex]
Where F is the force and q is the charge on the electron which is a constant value of [tex]q = 1.60*10^{-19} \ C[/tex]
Thus
[tex]F = 950 * 1.60 **10^{-19}[/tex]
[tex]F = 1.52 *10^{-16} \ N[/tex]
Generally the work energy theorem can be mathematically represented as
[tex]W = \Delta KE[/tex]
Where W is the workdone on the electron by the Electric field and [tex]\Delta KE[/tex] is the change in kinetic energy
Also workdone on the electron can also be represented as
[tex]W = F* x *cos( \theta )[/tex]
Where [tex]\theta = 0 ^o[/tex] considering that the movement of the electron is along the x-axis
So
[tex]\Delta KE = F * x cos (0)[/tex]
substituting values
[tex]\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)[/tex]
[tex]\Delta KE = 3.8*10^{-16} J[/tex]
Now From the law of energy conservation
[tex]\Delta PE = - \Delta KE[/tex]
Where [tex]\Delta PE[/tex] is the change in potential energy
Thus
[tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]
A medieval city has the shape of a square and is protected by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round your answers to one decimal place. Use g ≈ 9.8 m/s2. Enter your answer using interval notation. Enter your answer in terms of degrees without using a degree symbol.)
Answer:
θ₁ = 85.5º θ₂ = 12.98º
Explanation:
Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.
Let's write the equations for motion for this point
X axis
x = v₀ₓ t
x = v₀ cos θ t
Y axis
y = [tex]v_{oy}[/tex] t - ½ g t2
y = v_{o} sin θ t - ½ g t²
let's substitute the values
100 = 80 cos θ t
15 = 80 sin θ t - ½ 9.8 t²
we have two equations with two unknowns, so the system can be solved
let's clear the time in the first equation
t = 100/80 cos θ
15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²
15 = 100 tan θ - 7.656 sec² θ
we can use the trigonometric relationship
sec² θ = 1- tan² θ
we substitute
15 = 100 tan θ - 7,656 (1- tan² θ)
15 = 100 tan θ - 7,656 + 7,656 tan² θ
7,656 tan² θ + 100 tan θ -22,656=0
let's change variables
tan θ = u
u² + 13.06 u + 2,959 = 0
let's solve the quadratic equation
u = [-13.06 ±√(13.06² - 4 2,959)] / 2
u = [13.06 ± 12.599] / 2
u₁ = 12.8295
u₂ = 0.2305
now we can find the angles
u = tan θ
θ = tan⁻¹ u
θ₁ = 85.5º
θ₂ = 12.98º
A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between the two ends of the wire.
Answer:
V = 0.45 Volts
Explanation:
First we need to find the total current passing through the wire. That can be given by:
Total Current = I = (Current Density)(Surface Area of Wire)
I = (Current Density)(2πrL)
where,
r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m
L = Length of Wire = 6.5 m
Therefore,
I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]
I = 1.25 x 10⁻⁴ A
Now, we need to find resistance of wire:
R = ρL/A
where,
ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm
A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²
Therefore,
R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)
R = 0.36 Ω
From Ohm's Law:
Voltage = V = IR
V = (1.25 x 10⁻⁴ A)(0.36 Ω)
V = 0.45 Volts
An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?
Answer:
Towards the west.
Explanation:
The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.
Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.
According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward
A uniform disk of 10 kg and radius 4.0 m can rotate in a horizontal plane about a vertical axis through its center. The disk is rotating at an angular velocity of 15 rad/s when a 5-kg package is dropped vertically on a point that is 2.0 m from the center of the disk. What is the angular velocity of the disk/package system
Answer:
18.75 rad/s
Explanation:
Moment of inertia of the disk;
I_d = ½ × m_disk × r²
I_d = ½ × 10 × 4²
I_d = 80 kg.m²
I_package = m_pack × r²
Now,it's at 2m from the centre, thus;
I_package = 5 × 2²
I_package = 20 Kg.m²
From conservation of momentum;
(I_disk + I_package)ω1 = I_disk × ω2
Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.
Thus;
Plugging in the relevant values, we obtain;
(80 + 20)15 = 80 × ω2
1500 = 80ω2
ω2 = 1500/80
ω2 = 18.75 rad/s
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistance is 207.4 What is the temerature (in degree Celsius of this solution
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
T₂ = 95.56°C