Complete question is;
One end of a uniform well-lagged metal bar of length 0.80 m and cross-sectional area of 4.0 x 10-3m2 is kept in steam at 100 ˚C while the other end is in melting ice in a well-lagged container. The ice melts at a steady rate of 5.5 x 10- 4 kgs-1 and the thermal conductivity of the material of the bar is 401 Wm-1K-1. Calculate the specific latent heat of fusion of ice
Answer:
Specific Latent heat of fusion;
L_f = 13.6 × 10^(5) J/kg
Explanation:
We are given;
Length of bar; L = 0.8 m
Area;A = 4 × 10^(-3) m²
Temperature;ΔT= 100°C = 100 + 273 = 373 K
Rate of melting;m/t = 5.5 × 10^(-4) kg/s
Thermal conductivity;k = 401 W/m·K
Latent heat of fusion has a formula;
ΔQ/Δt = (m/t)•L_f
So, L_f = (ΔQ/Δt)/(m/t) - - - (1)
We also know that ;
ΔQ/Δt = (ΔT × k × A)/L
Plugging in the relevant values, we have;
ΔQ/Δt = (373 × 401 × 4 × 10^(-3))/0.8
ΔQ/Δt = 747.865 J/S
Plugging this value for ΔQ/Δt in equation 1 gives;
L_f = 747.865/(5.5 × 10^(-4))
L_f = 13.6 × 10^(5) J/kg
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 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)
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 steel wire with mass 25.3 g and length 1.62 m is strung on a bass so that the distance from the nut to the bridge is 1.10 m. (a) Compute the linear density of the string. kg/m (b) What velocity wave on the string will produce the desired fundamental frequency of the E1 string, 41.2 Hz
Answer:
(a) μ = 0.015kg/m
(b) v = 90.64m/s
Explanation:
(a) The linear density of the string is given by the following relation:
[tex]\mu=\frac{m}{L}[/tex] (1)
m: mass of the string = 25.3g = 25.3*10-3 kg
L: length of the string = 1.62m
[tex]\mu=\frac{25.3*10^{-3}kg}{1.62m}=0.015\frac{kg}{m}[/tex]
The linear density of the string is 0.015kg/m
(b) The velocity of the string for the fundamental frequency is:
[tex]f_1=\frac{v}{2l}[/tex] (2)
f1: fundamental frequency = 41.2 Hz
vs: speed of the wave
l: distance between the fixed extremes of the string = 1.10m
You solve for v in the equation (2) and replace the values of the other parameters:
[tex]v=2lf_1=2(1.10m)(41.2Hz)=90.64\frac{m}{s}[/tex]
The speed of the wave for the fundamental frequency is 90.64m/s
A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration of 4 m/s2 by reducing the throttle. What is the velocity (in m/s) of the boat when it reaches the buoy
Answer:
7.5 m/s
Explanation:
We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:
[tex]v^2 = u^2 + 2as[/tex]
where v = final velocity
u = initial velocity
a = acceleration
s = distance traveled
From the question:
u = 28 m/s
a = -4 [tex]m/s^2[/tex]
s = 91 m
Therefore:
[tex]v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s[/tex]
The velocity of the boat when it reaches the buoy is 7.5 m/s.
Bonnie and Clyde are sliding a 325 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 377 N of force while Bonnie pulls forward on a rope with 353 N of force.
Required:
What is the safe's coefficient of kinetic friction on the bank floor?
Answer:
the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]
Explanation:
GIven that:
Bonnie and Clyde are sliding a 325 kg bank safe across the floor to their getaway car.
So ,let assume they are sliding the bank safe on an horizontal direction
Clyde → Δ(bank safe) → Bonnie
Also; from the above representation; let not forget that the friction force [tex]F_{friction}[/tex] is acting in the opposite direction ←
where;
[tex]F_{friction}[/tex] = [tex]\mu_k mg[/tex]
The safe slides with a constant speed
If Clyde pushes from behind with 377 N of force while Bonnie pulls forward on a rope with 353 N of force.
Thus; since the safe slides with a constant speed if the two conditions are met; then the net force acting on the slide will be equal to zero.
SO;
[tex]F_{net} = F_{Cylde} + F_{Bonnie} - F_{frition}[/tex]
[tex]F_{net} = F_{Cylde} + F_{Bonnie} - \mu_k \ mg[/tex]
Since the net force acting on the slide will be equal to zero.
Then; [tex]F_{net} =0[/tex]
Also; let [tex]F_{Cylde} = F_c[/tex] and [tex]F_{Bonnie} = F_B[/tex]
Then;
[tex]0 = F_c + F_B - \mu_k \ mg[/tex]
[tex]\mu_k \ mg= F_c + F_B[/tex]
[tex]\mu_k = \dfrac{F_c + F_B}{\ mg}[/tex]
where;
[tex]F_c = 377 \ N \\ \\ F_B = 353 \ N \\ \\ mass (m) = 325 \ kg[/tex]
Then;
[tex]\mu_k = \dfrac{377 + 353}{325*9.81}[/tex]
[tex]\mu_k = \dfrac{730}{3188.25}[/tex]
[tex]\mathbf{\mu_k =0.2290}[/tex]
Thus; the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]
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
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
Move the magnet at a relatively constant frequency back and forth through the coil. The voltage displayed is proportional to the current flowing in the coil. What happens as you move the magnet through the coils with different number of loops?
Answer:
The induced EMF, and hence the induced current produced will increase or decrease with the number of loops on the coil.
Explanation:
According to Faraday' law of electromagnetic induction, the induced EMF increases with the speed with which we turn the coil, the surface area of the coil, the number of loops, and the strength of the magnetic field. From this, we can see that increasing the number of loops also increases the surface area involved. This means that if we move the magnet in this experiment through the coils with different number of loops, the induced EMF, and hence the induced current, will increase or decrease with an increase or decrease in the number of loops respectively.
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
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
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.
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
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"
A capacitor of 2mF is charged with a DC Voltage source of 100 V . There is a resistor of 1 kilo ohms in series with the capacitor. What will be the time taken by the capacitor so that the voltage across the capacitor is 70 V
Answer:
t = 0.731s
Explanation:
In order to calculate the time that the capacitor takes to have a voltage of 70V, you use the following formula:
[tex]V=V_oe^{-\frac{t}{RC}}[/tex] (1)
V: final voltage across the capacitor = 70V
Vo: initial voltage across the capacitor = 100V
R: resistance of the resistor in the circuit = 1kΩ = 1*10^3Ω
C: capacitance of the capacitor = 2mF = 2*10^-3F
t: time
You use properties of logarithms to solve the equation (1) for t:
[tex]\frac{V}{V_o}=e^{-\frac{t}{RC}}\\\\ln(\frac{V}{V_o})=ln(e^{-\frac{t}{RC}})\\\\ln(\frac{V}{V_o})=-\frac{t}{RC}\\\\t=-RCln(\frac{V}{V_o})[/tex]
Next, you replace the values of the parameters:
[tex]t=-(1*10^3\Omega)(2*10^{-3})ln(\frac{70V}{100V})\\\\t=0.713s[/tex]
The capacitor takes 0.731s to reache a voltage of 70V
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
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
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a sled. The first child exerts a force of 79 N, the second a force of 92 N, kinetic friction is 5.5 N, and the mass of the third child plus sled is 24 kg.
1. Using a coordinate system where the second child is pushing in the positive direction, calculate the acceleration in m/s2.
2. What is the system of interest if the accelaration of the child in the wagon is to be calculated?
3. Draw a free body diagram including all bodies acting on the system
4. What would be the acceleration if friction were 150 N?
Answer:
Please, read the anser below
Explanation:
1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:
[tex]F_2-F_1-F_f=Ma[/tex] (1)
Where is has been used that the motion is in the direction of the applied force by the second child
F2: force of the second child = 92N
F1: force of the first child = 79N
Ff: friction force = 5.5N
M: mass of the third child = 24kg
a: acceleration of the third child = ?
You solve the equation (1) for a, and you replace the values of the other parameters:
[tex]a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}[/tex]
The acceleration is 0.48m/s^2
2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.
3. An image with the diagram forces is attached below.
4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.
Then, the acceleration is zero
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 :
https://brainly.com/question/13369951
#SPJ6
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
Which of the following statements are true?
1. Liquid water expands with increasing temperature above 4°C.
2. Liquid water expands with increasing temperature between 0°C and 4°C.
3. Water contracts as it freezes at 0°C.
4. Solid ice is less dense than liquid water.
Answer:
water contracts as it freezes at 0°C
Answer:
weeve
Explanation:
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
which of the following best describes a stable atom?
An ideal (non-viscous, incompressible) fluid flows through a horizontal pipe. The fluid density is 900 kg/m3. Initially, the pipe has a diameter of 0.7 cm and the fluid flows at a speed of 9 m/s at a pressure of 13,000 N/m2. Then, the pipe widens to a diameter of 2.1 cm. What is the speed of the fluid in the wider section of the pipe, in units of m/s
Answer:
the speed of the fluid in the wider section of the pipe is 1m/s.
Explanation:
By equation of continuity we can write (for ideal (non-viscous, in-compressible).
[tex]A_1v_1 =A_2v_2[/tex]
A_1,A_2 are areas of the pipe at inlet and outlet of the pipe.
[tex]\Rightarrow \pi d_1^2v_1=\pi d_2^2v_2[/tex]_1
Here, d_1 , d_2 are diameters of inlet and outlet, also v_1, v_2 are velocities at inlet and outlet.
putting values we get
[tex]\Rightarrow \p 0.7^2\times9=\pi 2.1^2\timesv_2[/tex]
solving we get
[tex]v_2= 1m/s[/tex]
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
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.
A charged Adam or particle is called a
Answer:
A charged atom or particle is called an ion :)
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
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