Answer:
540C.
Explanation:
A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;
Q = CV ----------(i)
From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.
Where;
C = 6F
V = 90V
Substitute these values into equation (i) as follows;
Q = 6 x 90
Q = 540 C
Therefore, the initial charge on the capacitor is 540C.
Two parallel plates 0.800 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate.
How far from the negative plate is the point at which the electron and proton pass each other?
Express your answer with the appropriate units.
Answer:
0.79 cm
Explanation:
The computation is shown below:-
Particle acceleration is
[tex]a = \frac{qE}{m}[/tex]
We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time
[tex]d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}[/tex]
After solving the equation we will get 0.79 cm from the negative plate.
Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other
The point at which the electron and proton pass each other will be 0.79 cm.
What is the charge?When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.
The given data in the problem is;
d' is the distance between the two parallel plates= 0.800 cm
The acceleration is given as;
[tex]\rm a= \frac{qE}{m} \\\\[/tex]
The distance from Newton's law is found as;
[tex]d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm[/tex]
Hence the point at which the electron and proton pass each other will be 0.79 cm.
To learn more about the charge refer to the link;
https://brainly.com/question/24391667
A 25-kilogram object is placed on a compression spring, and it creates a displacement of 0.15 meters. What is the weight of an object that creates a displacement of 0.23 m on the same spring? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
I hope it is correct ✌️
A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the bottom of hill if it rolled without slipping all the way down
Answer: The wheel is moving 29.26 m/s fast
Explanation: Please see the attachments below
A student in her physics lab measures the standing-wave modes of a tube. The lowest frequency that makes a resonance is 30 Hz. As the frequency is increased, the next resonance is at 90 Hz.
What will be the next resonance after this?
Answer:
The next resonance will be 150 Hz.
Explanation:
The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.
The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.
Frecuency of the standing sound wave modes in a open-closed tube is:
fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)
The next resonance is at 90 Hz. This means that it occurs when n = 3:
f₃=3*30 Hz= 90 Hz
This means that the next resonance occurs when n = 5:
f₅=5*30 Hz= 150 Hz
The next resonance will be 150 Hz.
Zuckerman’s test for sensation seeking measures which of the following characteristics?
dangerousness, antisocial traits, “letting loose,’ and intolerance for boredom
thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
adventurousness, physical prowess, creative morality, and charisma
dangerousness, adventurousness, creativity, and thrill and adventure seeking
The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
Explanation:
Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.
Use the slider to apply a force of about 400 N. After 2 s have elapsed in the simulation, decrease the Applied Force (force exerted) slowly back to zero. Try to do this adjustment in roughly 2 s . While the Applied Force (force exerted) is decreasing, the velocity is:______.
a. constant.
b. increasing.
c. decreasing.
Answer:
c. decreasing.
Explanation:
Force produces acceleration or deceleration. Force is the product of a body's mass and its acceleration. When a force is applied to an object, the force tends to cause the body to move if the body was originally stagnant, cause the body to accelerate if applied in the direction of the body's velocity, or decelerate the body if applied in opposite direction to the velocity of the body. When the force that is exerted on a moving body is slowly reduced to zero, frictional forces between the body and the floor surface gradually decelerates the body. When this deceleration occurs, the velocity of the body gradually decreases t a stop.
The only factor connecting horizontal and vertical components of projectile motion is _____.
Answer:
VelocityExplanation:
When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.
Along the horizontal Ux = Ucos(theta)
Along the vertical Uy = Ucos(theta)
Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.
This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.
A hungry 177 kg lion running northward at 81.8 km/hr attacks and holds onto a 32.0 kg Thomson's gazelle running eastward at 59.0 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.
Answer:
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
Explanation:
Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:
[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]
Where:
[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.
[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.
[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.
After using the definition of momentum, the system is expanded:
[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]
Vectorially speaking, the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]
Where:
[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.
[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.
If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]
[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]
The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:
[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]
[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
4. 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
Answer:
Explanation:
Pls see diagram in attached file
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12 + 35 t + 1
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1
Answer:
Velocity = 131 m/s
Speed = 131 m/s
Explanation:
Equation of motion, s = f(t) = 12t² + 35 t + 1
To get velocity of the particle, let us find the first derivative of s
v (t) = ds/dt = 24t + 35
At t = 4
v(4) = 24(4) + 35
v(4) = 131 m/s
Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s
A 75kg passenger at the bottom of a roller coaster loop that has a radius of 20m. If the roller coaster car is moving 10m/s, what is the apparent weight of the passenger? g
Answer:
The apparent weight of the passenger is 360 N
Explanation:
Given;
The mass of the passenger, m = 75 kg
radius of the loop, r = 20 m
velocity of the roller coaster, v = 10 m/s
Centripetal force acting on this passenger is given as;
[tex]F = \frac{mv^2}{r}[/tex]
where;
F is the centripetal force acting on the passenger
m is the mass of the passenger
v is the velocity of the passenger
r is the radius of the track
[tex]F = \frac{mv^2}{r} \\\\F = \frac{75*10^2}{20} \\\\F = 375 \ N[/tex]
Real weight of the passenger,
W = mg
where;
g is acceleration due to gravity
W = 75 x 9.8
W = 735 N
Apparent weight of the passenger = Real weight - Centripetal force
Apparent weight of the passenger = 735 N - 375 N
Apparent weight of the passenger = 360 N
Therefore, the apparent weight of the passenger is 360 N
Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water level in the pan drops by 1.45 cm in 18.6 min, determine the rate of heat transfer to the pan in watts. (Give your answer in 3 significant digits.)
Answer:
Q = 20.22 x 10³ W = 20.22 KW
Explanation:
First we need to find the volume of water dropped.
Volume = V = πr²h
where,
r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m
h = height drop = 1.45 cm = 0.0145 m
Therefore,
V = π(0.151 m)²(0.0145 m)
V = 1.038 x 10⁻³ m³
Now, we find the mass of the water that is vaporized.
m = ρV
where,
m = mass = ?
ρ = density of water = 1000 kg/m³
Therefore,
m = (1000 kg/m³)(1.038 x 10⁻³ m³)
m = 1.038 kg
Now, we calculate the heat required to vaporize this amount of water.
q = mH
where,
H = Heat of vaporization of water = 22.6 x 10⁵ J/kg
Therefore,
q = (1.038 kg)(22.6 x 10⁵ J/kg)
q = 23.46 x 10⁵ J
Now, for the rate of heat transfer:
Rate of Heat Transfer = Q = q/t
where,
t = time = (18.6 min)(60 s/1 min) = 1116 s
Therefore,
Q = (23.46 x 10⁵ J)/1116 s
Q = 20.22 x 10³ W = 20.22 KW
Air bags greatly reduces the chance of injury in a car accident.explain how they do so in terms of energy transfer
Answer:
Airbags reduce chances of injury by absorbing most of the impact force from the body during a car crash
Explanation:
In a car collision, the speed of the vehicle is suddenly bought to rest. All the kinetic energy is suddenly converted into other forms of energy.
The body of the driver keeps travelling forward under his inertia force due to his mass until he is slammed against the steering wheel. The steering wheel is a very rigid component, and so when the body slams against it, the body takes the deformation, absorbing some of the energy of the moving car. This sudden impact of energy can be fatal enough to gravely injure the driver because the body does not undergo much deformation. When an airbag is used, the crash automatically triggers the release of the airbag. Instead of the body colliding against the rigid steering wheel, it is now collided against the soft air bag. The airbag is very collapsible, and some of the kinetic energy of the car on the driver is converted into the deformation energy used to deform the airbag when they collide. In the process of deformation, the time of impact is extended, reducing the force impacted on the driver, reducing the fatality of the impact.
Two small charged spheres are 7.59 cmcm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now?
Answer:
The two small charged spheres are now 4.382 cm apart
Explanation:
Given;
distance between the two small charged sphere, r = 7.59 cm
The force on each of the charged sphere can be calculated by applying Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
F is the force on each sphere
q₁ and q₂ are the charges of the spheres
r is the distance between the spheres
[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]
Therefore, the two small charged spheres are now 4.382 cm apart.
A car has a mass of 1200 kg and an acceleration of 4 m/s^2. If the friction on the car is 200 N, how much force is the thrust providing?
Answer:
5000N
Explanation:
According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e
∑F = m x a -------------(i)
From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e
∑F = F + Fₓ -------------(ii)
Where;
Fₓ = -200N [negative sign because the friction force opposes motion]
Combine equations(i) and (ii) together to get;
F + Fₓ = m x a
F = ma - Fₓ -------------(iii)
Where;
m = mass of car = 1200kg
a = acceleration of the car = 4m/s²
Now substitute the values of m, a and Fₓ into equation (iii) as follows;
F = (1200 x 4) - (-200)
F = 4800 + 200
F = 5000N
Therefore, the force the thrust is providing is 5000N
Earth orbiting the Sun The Earth is 1.5 ⋅ 10 8 km from the Sun (on average). How fast is the Earth orbiting the Sun in kilometers per second (on average)? You can assume the orbit of the Earth is a circle and that the circumference of a circle is equal to C = 2 π R where R is the radius of a circle (the distance between the center and the edge. Note that for our purposes, it is perfectly fine to assume π = 3 which allows for a pretty good approximation C = 6 R . Your answer does not need to be put into scientific notation, but if you choose to do so it will be marked correct! kilometers per second
Answer:
1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec
C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km circumference of orbit
v = C / t = 9 * 10E8 km / 3 * 10E7 sec = 30 km / sec = 18 mi/sec
Parallel rays of monochromatic light with wavelength 583 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 5.00×10^−4W/m^2. What is the intensity at a point on the screen that is 0.900 mm from the center of the central maximum?
Answer:
I = 2.18 10⁻⁴ W / m²
Explanation:
The two-slit interference pattern is described by the expression for constructive interference.
d sin θ = m λ
If we also want to know the distribution of intensities we must perform the su of the electric field of the two waves, and find the intensity as the square of the velvet field, obtaining the expression
I = I_max cos² ((π d /λ L) y)
where d is the separation of the slits, λ the wavelength, L the distance to the screen e and the separation of the interference line with respect to the central maximum
let's reduce the magnitudes to the SI system
λ = 583 nm = 583 10⁻⁹ m
L = 75.0 cm = 75.0 10⁻² m
d = 0.640 mm = 0.640 10⁻³ m
y = 0.900 mm = 0.900 10⁻³ m
let's calculate the intensity of this line
I = 5 10⁻⁴ cos² ((π 0.640 10⁻³ /583 10⁻⁹ 0.75 10⁻²) 0.900 10⁻³)
I = 5 10⁻⁴ cos2 (413.84)
I = 5 10⁻⁴ 0.435
I = 2.18 10⁻⁴ W / m²
A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is:
Answer:
Total charge = 1800C
Explanation:
Q= IT
I = currentt = timeQ = chargeA street light is at the top of a pole that has a height of 17 ft . A woman 5 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?
Answer:
8 ft/s
Explanation:
This is a straight forward question without much ado.
It is given from the question that she walks with a speed of 8 ft/s
A woman who weighs 500 N stands on an 8 m long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times bigger than the support force at the left end. How far from the right end is the woman standing
Answer:
[tex]1.6\; \rm m[/tex].
Explanation:
Let [tex]x[/tex] denote the distance (in meters) between the person and the right end of the board.
To keep the calculations simple, consider another unknown: let [tex]y[/tex] denote the support force (in Newtons) on the left end. The support force on the right end of this board would be [tex]3 \, y[/tex] (also in Newtons.)
Now there are two unknowns. At least two equations will be required for finding the exact solutions. For that, consider this board as a lever, but with two possible fulcrums. Refer to the two diagrams attached. (Not to scale.)
In the first diagram, the support at the left end of the board is considered as the fulcrum. In the second diagram, the support at the right end of the board is considered as the fulcrum.Calculate the torque in each situation. Note that are four external forces acting on this board at the same time. (Two support forces and two weights.) Why does each of the two diagrams show only three? In particular, why is the support force at each "fulcrum" missing? The reason is that any force acting on the lever at the fulcrum will have no direct impact on the balance between torques elsewhere on the lever. Keep in mind that the torque of each force on a lever is proportional to [tex]r[/tex], the distance between the starting point and the fulcrum. Since that missing support force starts right at the fulcrum, its [tex]r[/tex] will be zero, and it will have no torque in this context.
Hence, there are three (non-zero) torques acting on the "lever" in each diagram. For example, in the first diagram:
The weight of the board acts at the center of the board, [tex](1/2) \times 8\; \rm m = 4\; \rm m[/tex] from the fulcrum. This force will exert a torque of [tex]\tau(\text{weight of board}) = 4\; \rm m \times (-100\; \rm N) = (-400\; \rm N \cdot m)[/tex] on this "lever". The negative sign indicates that this torque points downwards.The weight of the person acts at [tex]x\;\rm m[/tex] from the right end of the board, which is [tex](8 - x)\; \rm m[/tex] from the fulcrum at the other end of this board. This force will exert a torque of [tex]\tau(\text{weight of person}) = (8 - x)\; {\rm m \times (-500\; \rm N)} = (-500\, \mathnormal{(8 - x)})\; \rm N \cdot m[/tex] on this "lever". This torque also points downwards.The support on the right end of the board acts at [tex]8\; \rm m[/tex] from the fulcrum (i.e., the left end of this board.) This force will exert a torque of [tex]\tau(\text{support, right}) = 8\; {\rm m} \times (3\, \mathnormal{y})\; {\rm N} = (24\, y)\; \rm N \cdot m[/tex] on the "lever". This torque points upwards.If the value of [tex]x[/tex] and [tex]y[/tex] are correct, these three torques should add up to zero. That is:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\, (8 - x))}_{\text{person}} + \underbrace{24\, y}_{\text{support}} = 0[/tex].
That gives the first equation of this system. Similarly, a different equation can be obtained using the second diagram:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\,x)}_{\text{person}} + \underbrace{8\, y}_{\text{support}} = 0[/tex].
Combine these two equations into a two-by-two system. Solve the system for [tex]x[/tex] and [tex]y[/tex]:
[tex]\left\lbrace\begin{aligned}&x = 1.6\\ &y = 150\end{aligned}\right.[/tex].
In other words, the person is standing at about [tex]1.6\; \rm m[/tex] from the right end of the board. The support force at the left end of the board is [tex]150\; \rm N[/tex].
A noisy channel needs to transfer 87 kbps, but has a SNR of 11 dB (decibels). Calculate the minimum Bandwidth required , in kHz, according to Shannon.
Answer:
24KHz
Explanation:
See attached file
A spherical balloon contains a charge +Q uniformly distributed over its surface. When it has a diameter D, the electric field at its surface has magnitude E. If the balloon is not blown up to thrice this diameter without changing the charge, the electric field at its surface is?
Answer:
E = 1/9 E₀
Explanation:
In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression
we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge
For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere
Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q
A = 4π r²
E 4π r² = Q /ε₀
E = 1 /4πε₀ Q / r²
the value of the indicated distance is 3 times the initial diamete
r = 3 D / 2
we substitute
E = 1/4 πε₀ Q (2/ 3D)²
for the initial conditions
E₀ = 1 / 4πε₀ Q (2/D)²
subtitled in the equation above
E = 1/9 E₀
If two radio telescope dishes are wired together in the right way, the "D" used in determining the angular resolution is determined by
Answer:
D is determined by distance between the telescopes.
Explanation:
While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 52-kg person develops a power of 80 W during a race, how fast is the person running
Answer:
The person is running at a speed of 2.564 m/s
Explanation:
Given;
mechanical energy dissipated per kilogram per step, E/kg/S = 0.6 J/kg/S
mass of the person, m = 52 kg
power developed by the person, P = 80 W
mechanical energy of the person per step, E = 0.6 J/kg x 52 kg
[tex]E_{step}[/tex] = 31.2 J
mechanical energy for the total step, [tex]E_{total}[/tex] = 31.2 J x S
P = E / t
[tex]P_{avg} = \frac{E_{total}}{t} \\\\P_{avg} = \frac{E_{step}*S}{t}\\\\\frac{P_{avg}}{E_{step}} = \frac{S}{t} \\\\\frac{S}{t} = \frac{80}{31.2} \\\\\frac{S}{t} = 2.564 \ m/s[/tex]
Therefore, the person is running at a speed of 2.564 m/s
A diver wants to jump from a board, the initial height is 10 meters and he wants to reach a horizontal distance of 2 meters. What minimum speed must he have when jumping from the board to achieve his goal?
Answer:
1.4 m/s
Explanation:
The minimum speed will be when the diver's initial velocity is horizontal.
First, find the time it takes for the diver to fall 10 meters.
Given:
Δy = 10 m
v₀ᵧ = 0 m/s
aᵧ = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
10 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.43 s
Now find the initial horizontal velocity.
v = (2 m) / (1.43 s)
v = 1.4 m/s
How many components do vectors have, and what are they?
Answer:
There are two components for a two-dimensional coordinate system/vector.
Explanation:
For two-dimensional vectors, such as velocity, acceleraton, etc, there are two components, the x- and y-components.
These components could be rotated or translated, depending on the coordinate system.
Instead of rectangular cartesian system, the components could also be in the form of polar coordinates, such as radius and theta (angle).
For three-dimensional vectors, such as velocity in space, there are three components, in various coordinate systems.
g If the interaction of a particle with its environment restricts the particle to a finite region of space, the result is the quantization of ____ of the particle.
Answer:
the result is the quantization of __Energy__ of the particle
Explanation:
PLEASE! PLEASE! PLEASE! HELP! I need this by Monday! It's a project DUE on Monday! I'm giving 75 points for the Brainiest :-)
Answer:
Please see below for all the numbers to be entered in the table:
Explanation:
Coaster World: F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J; V = 40/10 = 4 m/s
Wally: F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s
Elijah: F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s
George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s
Answer:
Down below
Explanation:
Coaster World: F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J; V = 40/10 = 4 m/s
Wally: F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s
Elijah: F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s
George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s
A car is traveling down a highway. It was moving with a velocity of 50m/s when the driver reads the speed limit and has to decelerate with an acceleration of -5m/s for 2 seconds. What is the momentum of this 500kg car after it decelerates?
Answer:
20,000 kg m/s
Explanation:
Given:
v₀ = 50 m/s
a = -5 m/s²
t = 2 s
Find: v
v = at + v₀
v = (-5 m/s²) (2 s) + (50 m/s)
v = 40 m/s
p = mv
p = (500 kg) (40 m/s)
p = 20,000 kg m/s
A motorcyclist changes his speed from 20 km / h to 100 km / h in 3 seconds, maintaining a constant acceleration in that time interval. If the mass of the motorcycle is 200 kg and that of its rider is 80 kg, what is the value of the net force to accelerate the motorcycle? Help!
Answer:
2000 N
Explanation:
20 km/h = 5.56 m/s
100 km/h = 27.78 m/s
F = ma
F = m Δv/Δt
F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)
F = 2074 N
Rounded to one significant figure, the force is 2000 N.