Answer: Option d.
Explanation:
The force between charges can be expressed as:
F = k*q1*q2/r^2
where k is a constant, q1 and q2 are the charges and r is the distance between them.
We hare in a equilateral triangle, so al the distances are equal.
and the charges are:
qa = q
qb = q
qc = 2q
Now, for example, the force that experiments charge A is (in the y axis)
F = (k*qa*qb/r^2 + k*qa*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2
for charge B, the force is again, in the y axis.
F = (k*qb*qa/r^2 + k*qb*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2
for particle C, we have:
F = (k*qc*qa/r^2 + k*qc*qb/r^2)*cos(30°) = cos(30°)*(q + q)*k*2q/r^2 = cos(30°)*4*k*q^2/r^2
Wher the cosine of 30° comes because we have a equilatiral triangle, where all the internal angles are 60°, so if we draw a line that cuts the angle by half (our y-axis) the angles to each side are 30°.
We can do a similar process for the forces in the x-axis, and we will reach the same conclusion:
Now, this means that the force that experiences the charge C is the biggest force, so the correct option is c.
A coil is connected to a galvanometer, which can measure the current flowing through the coil. You are not allowed to connect a battery to this coil. Given a magnet, a battery and a long piece of wire, can you induce a steady current in that coil?
Answer:
Yes we can induce current in the coil by moving the magnet in and out of the coil steadily.
Explanation:
A current can be induced there using the magnetic field and the coil of wire. Moving the bar magnet around the coil can induce a current and this is called electromagnetic induction.
What is electromagnetic induction ?The generation of an electromotive force across an electrical conductor in a fluctuating magnetic field is known as electromagnetic or magnetic induction.
Induction was first observed in 1831 by Michael Faraday, and James Clerk Maxwell mathematically named it Faraday's law of induction. The induced field's direction is described by Lenz's law.
Electrical equipment like electric motors and generators as well as parts like inductors and transformers have all found uses for electromagnetic induction.
Here, moving the bar magnet around the coil generates the electronic movement followed by a generation of electric current.
Find more on electromagnetic induction :
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An aluminum rod is designed to break when it is under a tension of 600 N. One end of the rod is connected to a motor and a 12-kg spherical object is attached to the other end. When the motor is turned on, the object moves in a horizontal circle with a radius of 5.78 m. If the speed of the motor is continuously increased, at what speed will the rod break
Answer:
17 m/s
Explanation:
Given:
Tension = 600 N
Mass of object, M= 12 kg
Radius, r = 5.78 m
Required:
Find the speed the rod will break
Here, the motor is continuously increased. To find the speed the rod will break (speed of centripetal force), we have:
Tension = Centripetal force
Where centripetal force = [tex] \frac{mv^2}{r} [/tex]
Therefore,
[tex] T = \frac{mv^2}{r} [/tex]
Make v subject of the formula:
[tex] v = \sqrt{\frac{T*r}{m}} [/tex]
[tex] = \sqrt{\frac{600*5.78}{12}} [/tex]
[tex] = \sqrt{\frac{3468}{12} [/tex]
[tex] = \sqrt{289} [/tex]
[tex] = 17 m/s [/tex]
Speed the rod will break is 17 m/s.
The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 0.700 m and an outer radius of 1.20 m. The toroid has 900 turns of large diameter wire, each of which carries a current of 13.0 kA. Find the difference in magnitudes of the magnetic fields of the toroid along the inner and outer radii. (Enter your answer in T.)
Answer:
The difference is [tex]\Delta B = 1.39 \ T[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r_i = 0.700 \ m[/tex]
The outer radius is [tex]r_o = 1.20 \ m[/tex]
The number of turns is [tex]N = 900 \ turns[/tex]
The current on each wire is [tex]I = 13.0 kA = 13*10^{3} \ A[/tex]
Generally magnetic field of a toroid along the outer radius is mathematically evaluated as
[tex]B_o = \frac{\mu_o * N * I}{2 \pi r_o}[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o= 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]B_o = \frac{ 4\pi * 10^{-7} * 13*10^{3} * 900}{ 2 * 3.142 * 1.20}[/tex]
[tex]B_o = 1.95 \ T[/tex]
Generally magnetic field of a toroid along the inner radius is mathematically evaluated as
[tex]B_i = \frac{\mu_o * N * I}{2 \pi r_i}[/tex]
substituting values
[tex]B_i = \frac{ 4\pi * 10^{-7} * 900 * 13*10^{3}}{2 *3.142 *0.700}[/tex]
[tex]B_i = 3.34 \ T[/tex]
The difference in magnitudes of the magnetic fields of the toroid along the inner and outer radii is mathematically evaluated as
[tex]\Delta B = B_i - B_o[/tex]
[tex]\Delta B = 3.34 -1.95[/tex]
[tex]\Delta B = 1.39 \ T[/tex]
As particle motion decreases, thermal energy does what?
Answer:
Changes of state. The kinetic theory of matter can be used to explain how solids, liquids and gases are interchangeable as a result of increase or decrease in heat energy. ... If it is cooled the motion of the particles decreases as they lose energy.13 Nov 2000
Explanation:
A soccer ball is released from rest at the top of a grassy incline. After 2.2 seconds, the ball travels 22 meters. One second later, the ball reaches the bottom of the incline. (Assume that the acceleration was constant.) How long was the incline
Answer:
x = 46.54m
Explanation:
In order to find the length of the incline you use the following formula:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
vo: initial speed of the soccer ball = 0 m/s
t: time
a: acceleration
You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):
[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex] (2)
Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:
[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex] (3)
The length of the incline is 46.54 m
If a bar magnet is falling through a loop of wire, the induced current in the loop of wire sets up a field which exerts a force on the magnet. This force between the magnet and the loop will be attractive when:
Answer:
When the magnet is leaving the loop
Explanation:
According to Lenz's law the direction of an induced current in a conductor will oppose the effect which produces it. As the current is induced in the wire loop and force is exerted on the magnet, the force between the magnet and the loop will be attractive when the magnet is leaving the loop because it's is the one that produces the effect which create the current.
Which best describes friction?
Answer:
It is the force that opposes motion between two surfaces touching each other. ( OR ) The force between two surfaces that are sliding or trying to slide across each other.
Explanation:
Answer:
a constant force that acts on objects that rub together
Explanation:
a constant force that acts on objects that rub together
A long horizontal hose of diameter 3.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.8 cm. Water squirts from the nozzle at velocity 14 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
A) What is the velocity of the water in the hose ?
B) What is the pressure differential between the water in the hose and water in the nozzle ?
C) How long will it take to fill a tub of volume 120 liters with the hose ?
Answer:
a) v₁ = 3.92 m / s , b) ΔP = = 9.0 10⁴ Pa, c) t = 0.0297 s
Explanation:
This is a fluid mechanics exercise
a) let's use the continuity equation
let's use index 1 for the hose and index 2 for the nozzle
A₁ v₁ = A₂v₂
in area of a circle is
A = π r² = π d² / 4
we substitute in the continuity equation
π d₁² / 4 v₁ = π d₂² / 4 v₂
d₁² v₁ = d₂² v₂
the speed of the water in the hose is v1
v₁ = v₂ d₂² / d₁²
v₁ = 14 (1.8 / 3.4)²
v₁ = 3.92 m / s
b) they ask us for the pressure difference, for this we use Bernoulli's equation
P₁ + ½ ρ v₁² + m g y₁ = P₂ + ½ ρ v₂² + mg y2
as the hose is horizontal y₁ = y₂
P₁ - P₂ = ½ ρ (v₂² - v₁²)
ΔP = ½ 1000 (14² - 3.92²)
ΔP = 90316.8 Pa = 9.0 10⁴ Pa
c) how long does a tub take to flat
the continuity equation is equal to the system flow
Q = A₁v₁
Q = V t
where V is the volume, let's equalize the equations
V t = A₁ v₁
t = A₁ v₁ / V
A₁ = π d₁² / 4
let's reduce it to SI units
V = 120 l (1 m³ / 1000 l) = 0.120 m³
d1 = 3.4 cm (1 m / 100cm) = 3.4 10⁻² m
let's substitute and calculate
t = π d₁²/4 v1 / V
t = π (3.4 10⁻²)²/4 3.92 / 0.120
t = 0.0297 s
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag is negligible. If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when the ball is at its peak height?
Answer:
P.E = 0.068 J = 68 mJ
Explanation:
First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = -9.8 m/s² (negative sign due to upward motion)
h = height attained by the ball toy = ?
Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)
Vi = Initial Velocity = 3 m/s
Therefore,
2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²
h = (9 m²/s²)/(19.6 m/s²)
h = 0.46 m
Now, the gravitational potential energy of ball at its peak is given by the following formula:
P.E = mgh
P.E = (0.015 kg)(9.8 m/s²)(0.46 m)
P.E = 0.068 J = 68 mJ
a car travels 12 miles due north and then 12 miles due west going from town A to town B. The magnitude of the car's displacement is --- miles
Answer:
The magnitude of the displacement of the car = 16.97 miles (North-West of A)
Explanation:
Attached to this answer is a diagram to give you a visual on what is going on i the question
Let the magnitude of the car's displacement be 'd'
The triangle formed is a right angled triangle, using the Pythagoras theorem:
d² = 12² + 12² (Hyp² = Opp² + Adj²)
d² = 144 +144 = 288
d =√ 288 = 16.97 miles
Therefore the magnitude of the displacement of the car = 16.97 miles (North-West of A)
You throw a ball straight up into the air from the top of a building. The building has a height of 15.0 m. The ball reaches a height (measured from the ground) of 25.0 m and then it starts to fall back down.
a) Determine the initial velocity of the ball.
b) What is the velocity of the ball when it comes back down and is at the same height from which it was thrown?
c) How long will it take the ball to come back down to this height from the time at which it was first thrown?
d) Let’s say that you missed catching the ball on the way back down and it fell to the ground. How long did it take to hit the ground from the moment you threw it up?
e) What was the ball’s final velocity the moment before it hit the ground?
Answer:
a) vo = 14m/s
b) v = 14m/s
c) t = 2.85s
d) t = 0.829s
e) v = 22.12 m/s
Explanation:
a) To find the initial velocity of the ball yo use the following formula:
[tex]h_{max}=\frac{v_o^2}{2g}[/tex] (1)
hmax: maximum height reached by the ball but measured from the point at which the ball is thrown = 25.0m - 15.0m = 10.0m
vo: initial velocity of the ball = ?
g: gravitational acceleration = 9.8m/s^2
You solve the equation (1) for vo and replace the values of the other parameters:
[tex]v_o=\sqrt{2gh_{max}}}=\sqrt{2(9.8m/s^2)(10.0m)}=14\frac{m}{s}[/tex]
The initial velocity of the ball is 14m/s
b) To find the velocity of the ball when it is at the same position as the initial point where it was thrown, you can use the following formula:
[tex]v^2=2gh_{max}\\\\v=\sqrt{2gh_{max}}[/tex]
as you can notice, v = vo = 14m/s
The velocity of the ball is 14 m/s
c) The flight time of the ball is given by twice the time the ball takes to reach the maximum height. You use the following formula:
[tex]t=2\frac{v_o}{g}=2\frac{14m/s}{9.8m/s^2}=2.85s[/tex] (3)
The time is 2.85s
d) To find the time the ball takes to arrive to the ground after the ball passes the same height at which is was thrown, you can use the following formula:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (4)
y: 0 m (ball just after it impact the ground)
yo: initial position = 15.0 m
vo: in)itial velocity of the ball = 14m/s
t: time
You replace the values of the parameters in the equation (4) and obtain a quadratic formula:
[tex]0=15.0-14t-\frac{1}{2}(9.8)t^2\\\\[/tex]
You use the quadratic formula to find the roots t:
[tex]t_{1,2}=\frac{-(-14)\pm\sqrt{(-14)^2-4(4.9)(15)}}{2(-4.9)}\\\\t_{1,2}=\frac{14\pm22.13}{-9.8}\\\\t_1=0.829s\\\\t_2=-2.19s[/tex]
you choose the positive values because is has physical meaning
The time the ball takes to arrive to the ground is 0.829s
e) The final velocity is:
[tex]v=v_o+gt[/tex]
[tex]v=14m/s+(9.8m/s^2)(0.829s)=22.12\frac{m}{s}[/tex]
The final velocity is 22.14 m/s
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at 6.00 m/s due south. The second car has a mass of 900 kg and was approaching at 25.0 m/s due west. (a) Calculate the final velocity (magnitude in m/s and direction in degrees counterclockwise from the west) of the cars. (Note that since both cars have an initial velocity, you cannot use the equations for conservation of momentum along the x-axis and y-axis; instead, you must look for other simplifying aspects.) magnitude m/s direction ° counterclockwise from west (b) How much kinetic energy (in J) is lost in the collision? (This energy goes into deformation of the cars.) J
Answer:
a) v = 11.24 m / s , θ = 17.76º b) Kf / K₀ = 0.4380
Explanation:
a) This is an exercise in collisions, therefore the conservation of the moment must be used
Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved
Recall that moment is a vector quantity so it must be kept on each axis
X axis
initial moment. Before the crash
p₀ₓ = m₁ v₁
where v₁ = -25.00 me / s
the negative sign is because it is moving west and m₁ = 900 kg
final moment. After the crash
[tex]p_{x f}[/tex]= (m₁ + m₂) vx
p₀ₓ = p_{x f}
m₁ v₁ = (m₁ + m₂) vₓ
vₓ = m1 / (m₁ + m₂) v₁
let's calculate
vₓ = - 900 / (900 + 1200) 25
vₓ = - 10.7 m / s
Axis y
initial moment
[tex]p_{oy}[/tex]= m₂ v₂
where v₂ = - 6.00 m / s
the sign indicates that it is moving to the South
final moment
p_{fy}= (m₁ + m₂) [tex]v_{y}[/tex]
p_{oy} = p_{fy}
m₂ v₂ = (m₁ + m₂) v_{y}
v_{y} = m₂ / (m₁ + m₂) v₂
we calculate
[tex]v_{y}[/tex] = 1200 / (900+ 1200) 6
[tex]v_{y}[/tex] = - 3,428 m / s
for the velocity module we use the Pythagorean theorem
v = √ (vₓ² + v_{y}²)
v = RA (10.7²2 + 3,428²2)
v = 11.24 m / s
now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)
tan θ = [tex]v_{y}[/tex] / Vₓ
θ = tan-1 v_{y} / vₓ)
θ = tan -1 (3,428 / 10.7)
θ = 17.76º
This angle is from the west to the south, that is, in the third quadrant.
b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship
Kf = 1/2 (m1 + m2) v2
K₀ = ½ m₁ v₁² + ½ m₂ v₂²
Kf = ½ (900 + 1200) 11.24 2
Kf = 1.3265 105 J
K₀ = ½ 900 25² + ½ 1200 6²
K₀ = 2,8125 10⁵ + 2,16 10₅4
K₀ = 3.0285 105J
the wasted energy is
Kf / K₀ = 1.3265 105 / 3.0285 105
Kf / K₀ = 0.4380
this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10−12C/(V⋅m) for the permittivity of space and c=3.00×108m/s for the speed of light.
Complete Question
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
Answer:
The electric field vector of the satellite broadcast as measured at the surface of the earth is [tex]E_o = 6.995 *10^{-6} \ V/m[/tex]
Explanation:
From the question we are told that
The height of the satellite is [tex]r = 35000 \ km = 3.5*10^{7} \ m[/tex]
The power output of the satellite is [tex]P = 1 \ KW = 1000 \ W[/tex]
Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is mathematically represented as
[tex]I = \frac{P}{4 \pi r^2}[/tex]
substituting values
[tex]I = \frac{1000}{4 * 3.142 (3.5*10^{7})^2}[/tex]
[tex]I = 6.495*10^{-14} \ W/m^2[/tex]
This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be mathematically represented as
[tex]I = c * \epsilon_o * E_o^2[/tex]
Where [tex]E_o[/tex] is the amplitude of the electric field vector of the satellite broadcast so
[tex]E_o = \sqrt{\frac{2 * I}{c * \epsilon _o} }[/tex]
substituting values
[tex]E_o = \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }[/tex]
[tex]E_o = 6.995 *10^{-6} \ V/m[/tex]
Two wheels initially at rest roll the same distance without slipping down identical planes. Wheel B has twice the radius, but the same mass as wheel A. All the mass is concentrated in their rims so that the rotational inertias are I = mR2. Which has more translational kinetic energy when it gets to the bottom?
Answer:
Their translational kinetic energies are the same
Explanation:
The translational kinetic energy of an object is given by the formula:
[tex]KE = 0.5 mv^2[/tex]
Where m = the mass of the object and
v = the linear speed of the object
From the question, it is stated that wheel A has the same mass as wheel B, that is [tex]m_A = m_B[/tex]
Linear speed is also a function of the distance covered. Since both wheels cover the same distance within the same interval, we can conclude that [tex]v_A = v_B[/tex]
Both wheels A and B have equal speed and mass, this means that their translational kinetic energy is the same.
Note that translational kinetic energy is not a function of the radius
QUESTION ONE
(a) Zindhile and Phindile are rowing a boat across a river which is 40m wide. They row in a direction
perpendicular to the bank. However, the river is flowing downstream and by the time they reach the other
side, they end up 30m downstream from their starting point. Over what distance did the boat travel?
Water molecules are made of slightly positively charged hydrogen atoms and slightly negatively charged oxygen atoms. Which force keeps water molecules stuck to one another? strong nuclear gravitational weak nuclear electromagnetic
Answer:
The answer is electromagnetic
Answer:
electromagnetic
Explanation:
edge 2021
airground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 11 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration 2.3 times that due to gravity
Answer:
N = 13.65 rpm
Explanation:
given data
radius = 11 m
centripetal acceleration = 2.3 times that due to gravity
to find out
how many revolutions per minute
solution
we know here centripetal accel = 2.3 × g
ω²r = 2.3 × 9.8
ω² × 11 = 2.3 × 9.8
solve it we get
ω² = 2.0490
ω = 1.43 rad/s
and
ω = [tex]\frac{2\pi N}{60}[/tex]
1.43 = [tex]\frac{2\pi N}{60}[/tex]
solve it we get
N = 13.65 rpm
A cylindrical pulley with a mass of 8 kg, radius of 0.561 m and moment of inertia 1 2 M r2 is used to lower a bucket with a mass of 1.9 kg into a well. The bucket starts from rest and falls for 2.6 s. r M m How far does it drop
Answer:
s = 15.84m
Explanation:
In order to calculate the distance traveled by the bucket, you first use the formula for the torque exerted on the pulley by the weight of the bucket:
[tex]\tau=I\alpha[/tex] (1)
I: moment of inertia of the pulley
α: angular acceleration of the pulley
You can calculate the angular acceleration by taking into account that the torque is also:
[tex]\tau=Wr[/tex] (2)
W: weight of the bucket = Mg = (1.9kg)(9,8m/s^2) = 18.62N
r: radius of the pulley = 0.561m
[tex]\tau=(18.62N)(0.561m)=10.44Nm[/tex]
The moment of inertia is given by:
[tex]I=\frac{1}{2}M_pr^2[/tex] (3)
Mp: mass of the pulley = 8kg
[tex]I=\frac{1}{2}(8kg)(0.561m)^2=1.25kg.m^2[/tex]
You solve the equation (1) for α and replace the values of the moment of inertia and the torque to obtain the angular acceleration:
[tex]\alpha=\frac{\tau}{I}=\frac{10.44Nm}{1.25kgm^2}=8.35\frac{rad}{s^2}[/tex]
Next, you use the following formula to find the angular displacement:
[tex]\theta=\frac{1}{2}\alpha t^2[/tex]
[tex]\theta=\frac{1}{2}(8.35rad/s^2)(2.6s)^2=28.24rad[/tex]
Finally, you calculate the arc length traveled by the pulley, this arc length is equal to the vertical distance traveled by the bucket:
[tex]s=r\theta =(0.561m)(28.24rad)=15.84m[/tex]
The distance traveled by the bucket is 15.84m
Oh football player kicks a football from the height of 4 feet with an initial vertical velocity of 64 ft./s use the vertical motion model H equals -16 tea to the power of 2+ VT plus S where V is initial velocity and feet per second and S is the height and feet to calculate the amount of time the football is in the air before it hits the ground round your answer to the nearest 10th if necessary.
Answer:
4.1 seconds
Explanation:
The height of the football is given by the equation:
[tex]H = -16t^2 + V*t + S[/tex]
Using the inicial position S = 4 and the inicial velocity V = 64, we can find the time when the football hits the ground (H = 0):
[tex]0 = -16t^2 + 64*t + 4[/tex]
[tex]4t^2 - 16t - 1 = 0[/tex]
Using Bhaskara's formula, we have:
[tex]\Delta = b^2 - 4ac = (-16)^2 - 4*4*(-1) = 272[/tex]
[tex]t_1 = (-b + \sqrt{\Delta})/2a[/tex]
[tex]t_1 = (16 + 16.49)/8 = 4.06\ seconds[/tex]
[tex]t_2 = (-b - \sqrt{\Delta})/2a[/tex]
[tex]t_2 = (16 - 16.49)/8 = -0.06\ seconds[/tex]
A negative time is not a valid result for this problem, so the amount of time the football is in the air before hitting the ground is 4.1 seconds.
The amount of time the football spent in air before it hits the ground is 4.1 s.
The given parameters;
initial velocity of the ball, V = 64 ft/sthe height, S = 4 ftTo find:
the amount of time the football spent in air before it hits the groundUsing the vertical model equation given as;
[tex]H = -16t^2 + Vt + S\\\\[/tex]
the final height when the ball hits the ground, H = 0
[tex]0 = -16t^2 + 64t + 4\\\\16t^2 - 64t - 4 = 0\\\\divide \ through \ by\ 4\\\\4t^2 - 16t - 1= 0\\\\solve \ the \ quadratic \ equation \ using \ the \ formula \ method;\\\\\\a = 4, \ b = -16, \ c = - 1\\\\t = \frac{-b \ \ + /- \ \ \ \sqrt{b^2 - 4ac} }{2a} \\\\[/tex]
[tex]t = \frac{-(-16) \ \ + /- \ \ \ \sqrt{(-16^2 )- 4(4\times -1)} }{2\times 4}\\\\t = \frac{16 \ \ + /- \ \ \sqrt{272} }{8} \\\\t = \frac{16 \ \ +/- \ \ 16.49}{8} \\\\t = \frac{16 - 16.49}{8} \ \ \ \ or \ \ \ \frac{16 + 16.49}{8} \\\\t = -0.61 \ s \ \ or \ \ \ 4.06 \ s\\\\t\approx 4.1 \ s[/tex]
Thus, the amount of time the football spent in air before it hits the ground is 4.1 s.
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As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it can be excluded from the calculations. For such a system, the potential energy is stored in the spring and is given by
U = 12k x 2
where k is the force constant of the spring and x is the distance from the equilibrium position. The kinetic energy of the system is, as always,
K = 12mv2
where m is the mass of the block and v is the speed of the block.
A) Find the total energy of the object at any point in its motion.
B) Find the amplitude of the motion.
C) Find the maximum speed attained by the object during its motion.
Answer:
a) [tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex], b) Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex], c) The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].
Explanation:
a) The total energy of the object is equal to the sum of potential and kinetic energies. That is:
[tex]E = K + U[/tex]
Where:
[tex]K[/tex] - Kinetic energy, dimensionless.
[tex]U[/tex] - Potential energy, dimensionless.
After replacing each term, the total energy of the object at any point in its motion is:
[tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex]
b) The amplitude of the motion occurs when total energy is equal to potential energy, that is, when objects reaches maximum or minimum position with respect to position of equilibrium. That is:
[tex]E = U[/tex]
[tex]E = \frac{1}{2} \cdot k \cdot A^{2}[/tex]
Amplitude is finally cleared:
[tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex]
Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex].
c) The maximum speed of the motion when total energy is equal to kinetic energy. That is to say:
[tex]E = K[/tex]
[tex]E = \frac{1}{2}\cdot m \cdot v_{max}^{2}[/tex]
Maximum speed is now cleared:
[tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex]
The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].
A 300-W computer (including the monitor) is turned on for 8.0 hours per day. If electricity costs 15¢ per kWh, how much does it cost to run the computer annually for a typical 365-day year? (Choose the closest answer)
Answer:
Cost per year = $131.4
Explanation:
We are given;
Power rating of computer with monitor;P = 300 W = 0.3 KW
Cost of power per KWh = 15 cents = $0.15
Time used per day by the computer with monitor = 8 hours
Thus; amount of power consumed per 8 hours each day = 0.3 × 8 = 2.4 KWh per day
Thus, for 365 days in a year, total amount amount of power = 2.4 × 365 = 876 KWh
Now, since cost of power per KWh is $0.15, then cost for 365 days would be;
876 × 0.15 = $131.4
A car travels north at 30 m/s for one half hour. It then travels south at 40 m/s for 15 minutes. The total distance the car has traveled and its displacement are: Group of answer choices 36 km; 36 km N. 90 km; 18 km N. 90 km; 36 km N. 36 km; 36 km S. 18 km; 18 km S.
Answer:
xtotal = 90km
displacement = 18km N
Explanation:
To find the total distance traveled by the car, you first calculate the distance traveled by the car when it travels to north. You use the following formula:
[tex]x=vt[/tex] (1)
x: distance
v: speed of the car = 30 m/s
t: time = one half hour
In order to calculate the distance you convert the time from hours to seconds:
[tex]t=0.5\ h*\frac{3600s}{1\ h}=1800s[/tex]
Then, you replace the values of t and v in the equation (1):
[tex]x=(30m/s)(1800s)=54000m[/tex] (2)
Next, you calculate the distance traveled by the car when it travels to south:
[tex]x'=v't'\\\\v'=40\frac{m}{s}\\\\t'=15\ min[/tex]
You convert the time from minutes to seconds:
[tex]t'=15\ min*\frac{60s}{1min}=900s[/tex]
[tex]x'=(40m/s)(900s)=36000m[/tex]
Finally, you sum both distances x and x':
[tex]x_{total}=x+x'=54000m+36000m=90000m=90km[/tex]
The total distance traveled by the car is 90km
The total displacement is the final distance of the car respect to the starting point of the motion. This is calculated by subtracting x' to x:
[tex]d=x-x'=54000m-36000m=18000m=18km[/tex]
The total displacement of the car is 18km to the north from its starting point of motion.
g The Trans-Alaskan pipeline is 1,300 km long, reaching from Prudhoe Bay to the port of Valdez, and is subject to temperatures ranging from -71°C to +35°C. How much does the steel pipeline expand due to the difference in temperature?
Answer:
ΔL = 1.653 km
Explanation:
The linear expansion of any object due to change in temperature is given by the following formula:
ΔL = αLΔT
where,
ΔL = Change in length or expansion of steel pipe line = ?
α = coefficient of linear expansion of steel = 12 x 10⁻⁶ /°C
L = Original Length of the steel pipe = 1300 km
ΔT = Change in temperature = 35°C - (- 71°C) = 35°C + 71°C = 106°C
Therefore,
ΔL = (12 x 10⁻⁶ /°C)(1300 km)(106°C)
ΔL = 1.653 km
The Bohr radius a0 is the most probable distance between the proton and the electron in the Hydrogen atom, when the Hydrogen atom is in the ground state. The value of the Bohr Radius is: 1 a0 = 0.529 angstrom. One angstrom is 10-10 m. What is the magnitude of the electric force between a proton and an electron when they are at a distance of 2.63 Bohr radius away from each other?
Answer:
The electric force is [tex]F = 11.9 *10^{-9} \ N[/tex]
Explanation:
From the question we are told that
The Bohr radius at ground state is [tex]a_o = 0.529 A = 0.529 ^10^{-10} \ m[/tex]
The values of the distance between the proton and an electron [tex]z = 2.63a_o[/tex]
The electric force is mathematically represented as
[tex]F = \frac{k * n * p }{r^2}[/tex]
Where n and p are charges on a single electron and on a single proton which is mathematically represented as
[tex]n = p = 1.60 * 10^{-19} \ C[/tex]
and k is the coulomb's constant with a value
[tex]k =9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.[/tex]
substituting values
[tex]F = \frac{9*10^{9} * [(1.60*10^{-19} ]^2)}{(2.63 * 0.529 * 10^{-10})^2}[/tex]
[tex]F = 11.9 *10^{-9} \ N[/tex]
If electrons are ejected from a given metal when irradiated with a 10-W red laser pointer, what will happen when the same metal is irradiated with a 5-W green laser pointer? (a) Electrons will be ejected, (b) electrons will not be ejected, (c) more information is needed to answer this question. Group of answer choices
Answer:
(b) electrons will not be ejected
Explanation:
Determine the number of photons ejected by 10 W red laser pointer.
The wavelength (λ) of red light is 700 nm = 700 x 10⁻⁹ m
Energy of a photon is given as;
[tex]E = \frac{hc}{\lambda}[/tex]
where;
h is Planck's constant, = 6.626 x 10⁻³⁴ J/s
c is speed of light, = 3 x 10⁸ m/s
[tex]E = \frac{6.626*10^{-34} *3*10^8}{700 X 10^{-9}} \\\\E = 2.8397 *10^{-19} \ J/photon[/tex]
The number of photons emitted by 10 W red laser pointer
10 W = 10 J/s
[tex]Number \ of \ photons = 10(\frac{ J}{s}) * \frac{1}{2.8397*10^{-19}} (\frac{photon}{J} ) = 3.522 *10^{19} \ photons/s[/tex]
Determine the number of photons ejected by 5 W red green pointer
The wavelength (λ) of green light is 500 nm = 500 x 10⁻⁹ m
[tex]E = \frac{hc}{\lambda} = \frac{6.626*10^{-34} *3*10^8}{500*10^{-9}} = 3.9756 *10^{-19} \ J/photon[/tex]
The number of photons emitted by 5 W green laser pointer
5 W = 5 J/s
[tex]Number \ of \ photons = \frac{5J}{s} *\frac{photon}{3.9756*10^{-19}J} = 1.258 *10^{19} \ Photons/s[/tex]
The number of photons emitted by 10 W red laser pointer is greater than the number of photons emitted by 5 W green laser pointer.
Thus, 5 W green laser pointer will not be able to eject electron from the same metal.
The correct option is "(b) electrons will not be ejected"
what is a push or a pull on an object known as
Answer:
Force
Explanation:
Force is simply known as pull or push of an object
A box weighing 180 newtons is hanging by rope as shown in the figure. Find the tension T2.
The question is incomplete, however, the correct question is attached
in the image format:
Answer:
B. 171 N
Explanation:
The equation of the forces along the
Horizontal direction:
[tex]T_{2} cos62^{0} = T_{1} cos20^{0}[/tex]...... 1
Verticalb direction:
[tex]T_{1} sin20^{0} = T_{2} sin62^{0}[/tex] = W . . . 2
Where W = 180 N is the weight of the box.
From equation (1),
[tex]= T_{1} =T_{2} \frac{cos62^{0}}{ cos20^{0}}[/tex]
Substituting into equation (2),
[tex](T_{2} \frac{cos62^{0}}{ cos20^{0}})[/tex][tex]sin20^{0} = T_{2} sin62^{0}[/tex]
= [tex]T = \frac{W}{cos62x^{0} tan20x^{0}+sin62x^{0} }[/tex]
=117 N
Thus, the correct answer is option B. 117 N
Suppose you are chatting with your friend, who lives on the moon. He tells you he has just won a Newton of gold in a contest. Excitedly, you tell him that you entered the Earth version of the same contest and also won a Newton of gold. Who is richer
Answer:
The friend on moon is richer.
Explanation:
The value of acceleration due to gravity changes from planet to planet. So the weight of 1 Newton of gold carries different mass on different places. So we need to calculate the mass of gold that both persons have.
FRIEND ON MOON:
W₁ = m₁g₁
where,
W₁ = Weight of Gold won by friend on moon = 1 N
m₁ = mass of gold won by friend on moon = ?
g₁ = acceleration due to gravity on moon = 1.625 m/s²
Therefore,
1 N = m₁(1.625 m/s²)
m₁ = 0.62 kg
ON EARTH:
W₂ = m₂g₂
where,
W₂ = Weight of Gold won by me on Earth = 1 N
m₂ = mass of gold won by me on Earth = ?
g₂ = acceleration due to gravity on Earth = 9.8 m/s²
Therefore,
1 N = m₁(9.8 m/s²)
m₁ = 0.1 kg
Since, the friend on moon has greater mass of gold than me.
Therefore, the friend on moon is richer.
which of the following best describes a stable atom?
A charged Adam or particle is called a
Answer:
A charged atom or particle is called an ion :)