Answer:
initial: 1654.6 J, final: 0 J, change: -1654.6 J
Explanation:
The length of the slide is
L = 8.80 ft = 2.68 m
So the height of the child when he is at the top of the slide is (with respect to the ground)
[tex]h = L sin \theta = (2.68 m)sin 25.0^{\circ}=1.13 m[/tex]
The potential energy of the child at the top is given by:
[tex]U = mgh[/tex]
where
m = 63.0 kg is the mass of the child
g = 9.8 m/s^2 is the acceleration due to gravity
h = 1.13 m
Substituting,
[tex]U=(63.0 kg)(9.8 m/s^2)(2.68 m)=1654.6 J[/tex]
At the bottom instead, the height is zero:
h = 0
So the potential energy is also zero: U = 0 J.
This means that the change in potential energy as the child slides down is
[tex]\Delta U = 0 J - (1654.6 J) = -1654.6 J[/tex]
The potential energy at the top is 700.9 J. At the bottom is 0 J. Change is -700.9 J (loss).
The length of the slide [tex]\( L = 8.80 \, \text{ft} \)[/tex] and the angle with the horizontal [tex]\( \theta = 25.0^\circ \)[/tex], we can use the sine function to find the vertical height h from the top to the bottom of the slide:
[tex]\[ h = L \sin \theta \][/tex]
[tex]\[ h = 8.80 \, \text{ft} \times \sin 25.0^\circ \][/tex]
First, calculate [tex]\( \sin 25.0^\circ \)[/tex]:
[tex]\[ \sin 25.0^\circ \approx 0.4226 \][/tex]
Now, calculate h:
[tex]\[ h = 8.80 \times 0.4226 \approx 3.719 \, \text{ft} \][/tex]
Potential energy PE is given by:
[tex]\[ PE = mgh \][/tex]
where [tex]\( m = 63.0 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \)[/tex] , and h is the height
Convert h from feet to meters (1 ft = 0.3048 m):
[tex]\[ h = 3.719 \, \text{ft} \times 0.3048 \, \text{m/ft} = 1.133 \, \text{m} \][/tex]
The potential energy at the top:
[tex]\[ PE_{\text{top}} = 63.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.133 \, \text{m} \][/tex]
[tex]\[ PE_{\text{top}} \approx 63.0 \times 9.81 \times 1.133 \approx 700.9 \, \text{J} \][/tex]
Since the child is at the bottom of the slide, where h = 0, the potential energy at the bottom [tex](\( PE_{\text{bottom}} \))[/tex] is 0 Joules.
The change in potential energy
The change in potential energy [tex](\( \Delta PE \))[/tex] is:
[tex]\[ \Delta PE = PE_{\text{bottom}} - PE_{\text{top}} \][/tex]
[tex]\[ \Delta PE = 0 - 700.9 \, \text{J} \][/tex]
[tex]\[ \Delta PE = -700.9 \, \text{J} \][/tex]
Suppose you wanted to change the fundamental frequency of an oscillating string. What are three parameters that you could alter and how would these alter the oscillation frequency?
Answer:
Linear density, Length, and tension in the string
Explanation:
The fundamental frequency of an oscillating string is given by:
[tex]f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]
where
L is the length of the string
T is the tension in the string
[tex]\mu[/tex] is the linear density of the string, which can also be rewritten as
[tex]\mu = \frac{m}{L}[/tex]
where m is the mass of the string.
Therefore, we can say that in order to change the fundamental frequency of the string, we can change either its lenght, or its tension or its linear density.
Carlos attempts to get a refrigerator magnet to stick on the S end of a bar magnet. He is unable to do so and states that the bar magnet is broken. Evaluate his argument: is this correct or incorrect? Explain.
Answer:
It is incorrect.
Explanation:
Refrigerator magnets are actually multiple layers of magnets, and they are stacked using opposite polarities, so the south end of the magnet might be under a north, so if he puts the magnet on its side it will probably attract.
The argument that Carlos attempts to get a refrigerator magnet to stick on the S end of a bar magnet. He is unable to do so and states that the bar magnet is broken is incorrect.
What is magnet?A substance which is negative charged at one end and positive charged at another end creating the magnetic field around it.
A bar is a magnet which can attract the particle which are charged. It has two poles North and South.
Refrigerator magnets are having multiple layers of magnets. They are stacked using opposite polarities. So, south pole of the magnet will definitely attract a north even if it is broken.
Thus, the argument is incorrect.
Learn more about magnet.
https://brainly.com/question/2841288
#SPJ2
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1, the rocket engine is shut down and the sled moves with constant velocity v for a time t2. If the total distance traveled by the sled is 18,350 ft and the total time is 90 s, find the following. (a) the times t1 and t2 t1 = s t2 = s (b) the velocity v
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
A 211-Ω and a 221-Ω resistor are connected in series across an unspecified power supply. If the current through the 211-Ω resistor is 0.22 A, what is the exact current (in A) through the 221-Ω resistor? Do not include units with your answer.
Answer:
0.22 (A)
Explanation:
Two resistors are connected in series when they are connected in the same branch of the circuit. When this occurs, we have the following:
- The current flowing through each resistor is the same (1)
- The total voltage of the circuit is the sum of the voltage drops across each resistor
- The total resistance of the circuit is equal to the sum of the individual resistances:
[tex]R=R_1 + R_2[/tex]
So, given statement (1), in this case the current flowing through each resistor is the same, so the current flowing throught the 221-Ω resistor is also 0.22 A.
Suppose a force of 60 N is required to stretch and hold a spring 0.1 m from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant k. b. How much work is required to compress the spring 0.5 m from its equilibrium position? c. How much work is required to stretch the spring 0.6 m from its equilibrium position? d. How much additional work is required to stretch the spring 0.1 m if it has already been stretched 0.1 m from its equilibrium? a. kequals 600
a. 600 N/m
Hooke's law states that:
F = kx
where
F is the force applied
k is the spring constant
x is the stretching/compression of the spring relative to the equilibrium position
In this problem we have
F = 60 N
x = 0.1 m
So the spring constant is
[tex]x=\frac{F}{x}=\frac{60 N}{0.1 m}=600 N/m[/tex]
b. 75 J
The work required to stretch a spring is equal to the elastic potential energy stored in the spring:
[tex]W=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the stretching/compression of the spring
Here we have
k = 600 N/m
x = 0.5 m
So the work done is
[tex]W=\frac{1}{2}(600 N/m)(0.5 m)^2=75 J[/tex]
c. 108 J
We can use the same formula used in the previous part:
[tex]W=\frac{1}{2}kx^2[/tex]
where here we have
k = 600 N/m
x = 0.6 m
So the work done is
[tex]W=\frac{1}{2}(600 N/m)(0.6 m)^2=108 J[/tex]
d. 9 J
In this case, the additional work required is the difference between the elastic potential energy in the two situations
[tex]W=\frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2[/tex]
where
k = 600 N/m
[tex]x_i = 0.1 m[/tex] is the initial stretching
[tex]x_f = 0.1 m + 0.1 m = 0.2 m[/tex] is the final stretching
Solving the equation,
[tex]W=\frac{1}{2}(600 N/m)(0.2 m)^2 - \frac{1}{2}(600 N/m)(0.1 m)^2=9 J[/tex]
Howuch work does it taje to move a 5kg box up/a 10m tall flight of stairs
Answer:
490 J
Explanation:
The work done to lift an object is equal to its increase in gravitational potential energy, therefore
[tex]W=mgh[/tex]
where
m is the mass of the object
g = 9.8 m/s^2 is the acceleration of gravity
h is the increase in height of the object
In this problem,
m= 5 kg
h = 10 m
Therefore, the work done is
[tex]W=(5 kg)(9.8 m/s^2)(10 m)=490 J[/tex]
|| Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down the runway. A smaller plane with the same acceleration has a takeoff speed of 40 m/s. Starting from rest, after what distance will this smaller plane reach its takeoff speed?
The distance at which the smaller plane will reach its takeoff speed, we can use the equation of motion:
[tex]\[ v^2 = u^2 + 2as \][/tex]
v is the final velocity (takeoff speed) of the smaller plane (40 m/s)
u is the initial velocity (rest) of the smaller plane (0 m/s)
a is the acceleration of both planes (which is the same)
s is the distance covered by the smaller plane before reaching its takeoff speed (what we want to find)
[tex]\[ s = \frac{{v^2 - u^2}}{{2a}} \][/tex]
[tex]\[ s = \frac{{(40 \, \text{m/s})^2 - (0 \, \text{m/s})^2}}{{2a}} \][/tex]
[tex]\[ a = \frac{{v - u}}{{t}} = \frac{{(80 \, \text{m/s}) - (0 \, \text{m/s})}}{{30 \, \text{s}}} \][/tex]
[tex]\[ s = \frac{{(40 \, \text{m/s})^2}}{{2 \times \left(\frac{{80 \, \text{m/s}}}{30 \, \text{s}}\right)}} \]\[ s = \frac{{1600 \, \text{m}^2/\text{s}^2}}{{\frac{{160 \, \text{m/s}}}{30 \, \text{s}}}} \]\[ s = \frac{{1600 \, \text{m}^2/\text{s}^2 \times 30 \, \text{s}}}{{160 \, \text{m/s}}} \]\[ s = \frac{{48,000 \, \text{m}^2/\text{s}}}{{160 \, \text{m/s}}} \]\[ s = 300 \, \text{m} \][/tex]
Therefore, the smaller plane will reach its takeoff speed after traveling a distance of 300 meters.
Learn more about Distance, refer to the link:
https://brainly.com/question/34824016
#SPJ12
By using the equation of motion v² = u² + 2as, we can calculate the acceleration of the first plane based on its initial velocity, final velocity, and distance covered. Then, using the same acceleration and applying it to the second plane's parameters, we can find the distance it covers before reaching its takeoff speed.
Explanation:To solve this problem, we'll make use of the equation of motion usually represented as v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered.
For the first plane, we have v = 80 m/s, u = 0 m/s (because the plane starts from rest), and s = 1200 m. Plugging these into our equation, we can solve for a.
Once we have the acceleration, we can then apply it to the second plane to find the distance it covers (s) before reaching its takeoff speed of 40 m/s. Given a = acceleration calculated above and v = 40 m/s, u = 0 m/s (since the second plane also starts from rest), we can rearrange our equation of motion to solve for s.
Learn more about Physics Motion Problems here:https://brainly.com/question/30706637
#SPJ3
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance of each bulb does not vary with current. (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.) A. Find the current through the bulbs. B. Find the power dissipated in the 60 W bulb. C. Find the power dissipated in the 200 W bulb. D. One bulb burns out very quickly. Which one? 60-W bulb, 200-W bulb.
A. 0.77 A
Using the relationship:
[tex]P=\frac{V^2}{R}[/tex]
where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.
For the first light bulb, P = 60 W and V = 120 V, so the resistance is
[tex]R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega[/tex]
For the second light bulb, P = 200 W and V = 120 V, so the resistance is
[tex]R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega[/tex]
The two light bulbs are connected in series, so their equivalent resistance is
[tex]R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega[/tex]
The two light bulbs are connected to a voltage of
V = 240 V
So we can find the current through the two bulbs by using Ohm's law:
[tex]I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A[/tex]
B. 142.3 W
The power dissipated in the first bulb is given by:
[tex]P_1=I^2 R_1[/tex]
where
I = 0.77 A is the current
[tex]R_1 = 240 \Omega[/tex] is the resistance of the bulb
Substituting numbers, we get
[tex]P_1 = (0.77 A)^2 (240 \Omega)=142.3 W[/tex]
C. 42.7 W
The power dissipated in the second bulb is given by:
[tex]P_2=I^2 R_2[/tex]
where
I = 0.77 A is the current
[tex]R_2 = 72 \Omega[/tex] is the resistance of the bulb
Substituting numbers, we get
[tex]P_2 = (0.77 A)^2 (72 \Omega)=42.7 W[/tex]
D. The 60-W bulb burns out very quickly
The power dissipated by the resistance of each light bulb is equal to:
[tex]P=\frac{E}{t}[/tex]
where
E is the amount of energy dissipated
t is the time interval
From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.
Final answer:
To solve for the current and power dissipated in each bulb, use the power rating to find their resistances, then apply Ohm's Law to calculate current, and subsequently, determine power dissipated in each. Finally, assess which bulb is likely to burn out first by the dissipated power relative to each bulb's rating.
Explanation:
The question involves finding the current through, and the power dissipated in, a series circuit with two light bulbs with different power ratings. It also inquires which bulb will burn out quickly.
A. Current through the bulbs
Using the power rating (P) and voltage (V) for each bulb, we can find their resistances (R) using the formula P = V^2/R. From there, we can find the total resistance in the series circuit and calculate the current (I) using Ohm's Law, I = V/R.
B. & C. Power dissipated in each bulb
Once the current is known, we can determine the power dissipated (P) in each bulb with the formula P = I^2 * R.
D. Which bulb burns out quickly?
We can infer which bulb burns out based on the power dissipated in each bulb compared to their rated power.
A 28-kg particle exerts a gravitational force of 8.3 x 10^-9 N on a particle of mass m, which is 3.2 m away. What is m? A) 140 kg B) 8.5 x 10^-10 kg C) None of the choices are correct D) 46 kg E)1300 kg
Answer:
Mass of another particle, m = 46 kg
Explanation:
it is given that,
Mass of first particle, m₁ = 28 kg
Gravitational force, [tex]F=8.3\times 10^{-9}\ N[/tex]
Distance between the particles, d = 3.2 m
We need to find the mass m of another particle. It is given by the formula as follows :
[tex]F=G\dfrac{m_1m}{d^2}[/tex]
[tex]m=\dfrac{Fd^2}{Gm_1}[/tex]
[tex]m=\dfrac{8.3\times 10^{-9}\ N\times (3.2\ m)^2}{6.67\times 10^{-11}\times 28\ kg}[/tex]
m = 45.5 kg
or
m = 46 kg
So, the correct option is (d) "46 kg". Hence, this is the required solution.
The correct answer is B) 8.5 x 10⁻¹⁰ kg is the mass.
[tex]\[ F = G \frac{M m}{r^2} \][/tex]
Where G is the gravitational constant, approximately equal to [tex]\( 6.674 \times 10^{-11} \) N(m/kg)^2[/tex].
Given that the force F is [tex]\( 8.3 \times 10^{-9} \)[/tex] N, the mass M is 28kg, and the distance ris 3.2 m, we can rearrange the equation to solve for m:
[tex]\[ m = \frac{F r^2}{G M} \][/tex]
Plugging in the given values:
[tex]\[ m = \frac{(8.3 \times 10^{-9} \text{ N}) \times (3.2 \text{ m})^2}{6.674 \times 10^{-11} \text{ N(m/kg)}^2 \times 28 \text{ kg}} \] \[ m = \frac{(8.3 \times 10^{-9}) \times (3.2)^2}{6.674 \times 10^{-11} \times 28} \] \[ m = \frac{(8.3 \times 10^{-9}) \times 10.24}{1.86872 \times 10^{-9}} \][/tex]
[tex]\[ m = \frac{8.5056 \times 10^{-8}}{1.86872 \times 10^{-9}} \] \[ m \approx 4.55 \times 10^{-1} \] \[ m \approx 8.5 \times 10^{-10} \text{ kg} \][/tex]
Therefore, the mass m of the second particle is approximately [tex]\( 8.5 \times 10^{-10} \)[/tex] kg, which corresponds to option B.
If Vx = 7.00 units and Vy = -7.60 units, determine the magnitude of V⃗ .
Determine the direction of V⃗ .
|V| = 10.33 units and the direction θ = -47.35° or 312.65°.
Given the x and y components of a vector, we can calculate the magnitude and direction from these components.
Applying the Pythagorean theorem we have that the magnitude of the vector is:
|V| = [tex]\sqrt{Vx^{2}+Vy^{2} }[/tex]
|V| = [tex]\sqrt{(7.00units)^{2}+(-7.60units)^{2}} = \sqrt{106units^{2}} = 10.33units[/tex]
The expression for the direction of a vector comes from the definition of the tangent of an angle:
tan θ = [tex]\frac{Vy}{Vx}[/tex] ------> θ = arc tan [tex]\frac{Vy}{Vx}[/tex]
θ = arc tan [tex]\frac{-7.60units}{7.00units}[/tex]
θ = -47.35° or 312.65°
Final answer:
The magnitude of vector V is found using the Pythagorean theorem, and the direction of V is determined using the arctan function with the known x and y components of V.
Explanation:
To find the magnitude of V, we use the Pythagorean theorem to combine the orthogonal components Vx and Vy. This is given by the equation V = √(Vx² + Vy²). Substituting the given values, we get V = √(7.00² + (-7.60)²) units.
After calculating the magnitude, we determine the direction of V by finding the angle θ relative to the x-axis, using the inverse tangent function (arctan). The formula is θ = tan⁻¹ (Vy/Vx). Since Vy is negative, the vector is in the 4th quadrant, with an angle measured clockwise from the positive x-axis.
A medium-sized pear provides about 102 Calories of energy. HINT (a) Convert 102 Cal to joules. J (b) Suppose that amount of energy is transformed into kinetic energy of a 2.03 kg object initially at rest. Calculate the final speed of the object (in m/s). m/s (c) If that same amount of energy is added to 3.79 kg (about 1 gal) of water at 23.7°C, what is the water's final temperature (in °C)? The specific heat of water is c = 4186 J (kg · °C) . °C
Answer:
(a) 426.8 J
(b) 20.5 m/s
(c) 23.727 °C
Explanation:
(a)
E = Energy of medium-sized pear = 102 cal
we know that , 1 cal = 4.184 J
hence
E = 102 (4.184 J)
E = 426.8 J
(b)
KE = kinetic energy of the object = E = 426.8 J
m = mass of the object = 2.03 kg
v = speed of the object = ?
Kinetic energy of the object is given as
KE = (0.5) m v²
inserting the values
426.8 = (0.5) (2.03) v²
v = 20.5 m/s
(c)
Q = Amount of heat added to water = E = 426.8 J
m = mass of water = 3.79 kg
c = specific heat of water = 4186 J/(Kg °C)
T₀ = initial temperature = 23.7 °C
T = Final temperature = ?
Using the equation
Q = m c (T - T₀)
426.8 = (3.79) (4186) (T - 23.7)
T = 23.727 °C
We first convert Calories to Joules and get 426,768 J. Then, using the kinetic energy equation and the given mass, we find the final speed of the object to be approximately 460 m/s. Using the heat formula to calculate the temperature change, we find that the final temperature of water would be 49.4°C.
Explanation:Let's tackle these parts one by one:
(a) Converting Calories to Joules: The conversion factor is given as 1 Calorie = 4184 Joules (noting that a Calorie in food is actually a kilocalorie). When we multiply 102 Cal by this conversion factor, we get 102 Cal x 4184 J/Cal = 426,768 J.
(b) Kinetic Energy: The formula for kinetic energy is KE = 1/2 mv², where m is mass and v is speed. If all the energy from the pear is transformed to kinetic energy, we can set this equal to 426,768 J, then solve for v. We get: √((2*426768 J) / 2.03 kg) ≈ 460 m/s as the final speed of the object.
(c) Temperature Change in Water: The formula to calculate temperature change is Q = mcΔT, where Q is the heat added (in Joules), m is the mass of the water, c is the specific heat capacity, and ΔT is the temperature change. We know that Q = 426,768 J, m = 3.79 kg, and c = 4186 J/kg°C. We can rearrange the formula to solve for ΔT: ΔT = Q / (m*c). So, ΔT = 426768 J / (3.79 kg * 4186 J/Kg°C) ≈ 25.7°C. So, the final temperature would be the initial temperature plus this change, or 23.7°C + 25.7°C = 49.4°C.
Learn more about Energy Conversion here:https://brainly.com/question/20458806
#SPJ11
A small object has a mass of 3.0 × 10-3 kg and a charge of -32C. It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of 2.4 × 103 m/s2 in the direction of the +x axis. Determine the electric field, including sign, relative to the +x axis.
The electric field is [tex]-0.225N/C[/tex] towards the negative x-axis.
The electric force (F) acting on a charged object in an electric field (E) is given by the following formula:
[tex]F=qE[/tex]
Here, q is the charge of the object and E is the electric field.
Given:
Acceleration, [tex]a=2.4 \times 10^3 m/s^2[/tex]
Mass, [tex]m=3.0 \times 10^{-3} kg[/tex]
Charge, [tex]q=-32C[/tex]
The force on the charge is computed as:
[tex]qE=ma\\E=\frac{3.0 \times 10^{-3} \times 2.4 \times 10^3}{-32}\\E=-0.225N/C[/tex]
The electric field is towards the negative x-axis.
Therefore, the electric field is [tex]-0.225N/C[/tex] towards the negative x-axis.
To know more about the electric force, click here:
https://brainly.com/question/20935307
#SPJ12
The electric field that causes the object with a mass of 3.0 × 10⁻³kg and a charge of -32C to accelerate at 2.4 × 10³m/s^2 is -7.5 × 10² N/C, and it is in the +x axis direction.
Explanation:The student's question involves determining the electric field that causes an object to experience a specific acceleration. The object has a mass of 3.0 × 10⁻³ kg and a charge of -32 C. To find the electric field E, we can use Newton's second law of motion and the definition of the electric force: F = ma = qE, where F is the force, m is the mass, a is the acceleration, q is the charge, and E is the electric field.
To calculate the electric field, we set F equal to the product of mass and acceleration and then solve for E:
E = F / q
= (m × a) / q
= (3.0 × 10^⁻³ kg × 2.4 × 10³ m/s²) / (-32 C)
= -7.5 × 10²N/C
The negative sign indicates that the electric field's direction is opposite to the charge's motion. Since the charge is negative and it accelerates in the direction of the +x axis, the electric field must be in the +x axis direction.
A cord, 6 m long and fixed at both ends, supports a standing wave with a total of 4 nodes. What is the wavelength of the standing wave?
Answer:
The wavelength of the standing wave is 3 m.
Explanation:
Given that,
Length = 6 m
Nodes = 4
We need to calculate the wave length
Using formula of nodes
[tex]L=\dfrac{n}{2}\lamda[/tex]
[tex]\lambda=\dfrac{2L}{n}[/tex]
Where, l = length
n = number of nodes
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]\lambda=\dfrac{2\times6}{4}[/tex]
[tex]\lambda=3\ m[/tex]
Hence, The wavelength of the standing wave is 3 m.
Answer:
Should be....
λ = 2(6)/3 = 4 m
Explanation:
λ = 2L/n where n is the antinode
n = 3 instead of 4 (4 is where the wave crosses zero)
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s
Answer:
4.42 x 10⁷ W/m²
Explanation:
A = energy absorbed = 500 J
η = efficiency = 0.90
E = Total energy
Total energy is given as
E = A/η
E = 500/0.90
E = 555.55 J
t = time = 4.00 s
Power of the beam is given as
P = E /t
P = 555.55/4.00
P = 138.88 Watt
d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m
Area of the circular spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Intensity of the beam is given as
I = P /A
I = 138.88 / (3.14 x 10⁻⁶)
I = 4.42 x 10⁷ W/m²
The intensity of a laser beam : 4.42.10⁷ W/m²
Further explanationThe energy transferred by waves per unit area per unit time is called wave intensity
Because energy per unit time is Power, the intensity of the wave is equal to Power divided by area
[tex]\rm P=\dfrac{W}{t}[/tex]
P = power, watt
W = energy, J
t = time, s
For waves that spread in all directions, the intensity at the distance R from the source can be formulated
[tex]\rm I=\dfrac{Power}{Area}=\dfrac{P}{\pi .R^2}[/tex]
From the equation above shows the intensity of the wave is inversely proportional to the square of the distance from the source.
[tex]\rm I\approx \dfrac{1}{R^2}[/tex]
The farther the wave spreads, the smaller the intensity
Cancerous tissue area:
[tex]\rm A=\dfrac{1}{4}\pi d^2\\\\A=\dfrac{1}{4}\pi(2.10^{-3})^2\Rightarrow d=2~mm=2.10^{-3}\:m\\\\A=\frac{1}{4}\pi 4.10^{-6}\\\\A=\pi .10^{-6}\\\\A=3.14.10^{-6}[/tex]
So that the intensity
[tex]\rm 0.9I(only~90\%~absorbed)=\dfrac{W}{A.t}\\\\0.9I=\dfrac{500}{3.14.10^{-6}.4}\\\\I=4.42.10^7~\dfrac{W}{m^2}[/tex]
Learn moreelectric field
brainly.com/question/2080732
magnetism
brainly.com/question/10809295
An electric device delivers a current of 5 a to a device.
brainly.com/question/4438943
A race car goes around a level, circular track with a diameter of 1.00 km at a constant speed of 89 km/h. What is the car's centripetal acceleration in m/s^2?
Answer:
The car's centripetal acceleration is 1.22 m/s².
Explanation:
Given that,
Diameter of circular track [tex]d= 1.00\ km=1000\ m[/tex]
Radius r = 500 m
Constant speed [tex]v = 89\ km/h = 89\times\dfrac{5}{18}=24.722\ m/s[/tex]
The centripetal acceleration is defined as,
[tex]a_{c} = \dfrac{v^2}{r}[/tex]
Where, v = tangential velocity
r = radius
Put the value into the formula
[tex]a_{c}=\dfrac{(24.722)^2}{500}[/tex]
[tex]a_{c}=1.22\ m/s^2[/tex]
Hence, The car's centripetal acceleration is 1.22 m/s².
To find the car's centripetal acceleration, convert the speed to meters per second, then use the formula a = v^2 / r. With a speed of 24.71 m/s and a track radius of 500 m, the acceleration is approximately 1.22069 m/s².
Explanation:The student is asking to determine the centripetal acceleration of a race car moving at a constant speed along a circular track. To calculate this, we can use the formula for centripetal acceleration: a = v^2 / r. First, we need to convert the speed from km/h to m/s, which can be done by multiplying by ⅟ (≅ 0.27778). Thus, 89 km/h equates to 89 × 0.27778 m/s, which is about 24.71 m/s. The diameter of the track is 1.00 km (1000 m), hence the radius r is half of that, 500 m.
Using the formula:
a = (24.71 m/s)² / 500 m
a = 610.3441 m²/s² / 500 m
a = 1.22069 m/s² (approximately)
The centripetal acceleration of the race car is approximately 1.22069 m/s².
Water flowing through a garden hose of diameter 2.71 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose? Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle? m/s
Answer:
i) [tex]v_1 = 0.40 m/s[/tex]
ii) [tex]v_2 = 3.60 m/s[/tex]
Explanation:
Part A)
As we know that diameter of the hose pipe is 2.71 cm
Now the area of crossection of the pipe will be
[tex]A = \pi (\frac{D}{2})^2[/tex]
[tex]A = \pi (\frac{0.0271}{2})^2 = 5.77 \times 10^{-4} m^2[/tex]
Now the flow rate is defined as the rate of volume
It is given as
[tex]Q = \frac{Volume}{time} = Area \times speed[/tex]
[tex]\frac{20 L\times \frac{10^{-3} m^3}{1L}}{1.45 \times 60 seconds} = 5.77 \times 10^{-4} \times v[/tex]
[tex]v = 0.40 m/s[/tex]
Part b)
As per equation of continuity we know
[tex]A_1 v_1 = A_2 v_2[/tex]
now we have
[tex]\pi (\frac{d_1}{2})^2 v_1 = \pi (\frac{d_2}{2})^2v_2[/tex]
[tex](2.71)^2 (0.40) = (\frac{2.71}{3})^2 v_2[/tex]
[tex]v_2 = 3.60 m/s[/tex]
An astronaut of 66.0kg mass is in preparation for launch to a distant planet. a) knowing that the gravitational acceleration on the surface of that planet is 0.33 of the gravitational acceleration g, what would be the weight of the astronaut on the planet? b) what is the mass of the astronaut on the planet? c) What would be the weight of the astronaut on the earth?
Answer:
Part a)
Weight on surface of other planet = 213.4 N
Part b)
Mass of the Astronaut = 66.0 kg
Part c)
Weight of the Astronaut on Earth = 646.8 N
Explanation:
Part A)
Weight of the Astronaut on the surface of the planet is given as
[tex]F_g = mg[/tex]
here we will have
[tex]m = 66.0 kg[/tex]
also we have
[tex]g = 0.33(9.8) m/s^2[/tex]
[tex]g = 3.23 m/s^2[/tex]
now we have
[tex]F = 66 \times 3.23 = 213.4 N[/tex]
Part B)
Mass of the Astronaut will always remains the same
So it will be same at all positions and all planets
So its mass will be
m = 66.0 kg
Part C)
Weight of the Astronaut on Earth is given as
[tex]F_g = mg[/tex]
[tex]F_g = (66.0 kg)(9.8 m/s^2)[/tex]
[tex]F_g = 646.8 N[/tex]
Suppose that your mass is 59.1 kg, and you are standing on a scale fastened to the floor of an elevator. The scale measures force and is calibrated in newtons. What does the scale read when the elevator is rising and its speed is decreasing at a rate of 6.87 m/s2? Use 9.80 m/s2 for acceleration due to gravity.
Answer:
173.2 Newtons
Explanation:
Since the scale is a reading of the normal force, the force of gravity and the normal force are in opposite directions, which means the forces can be subtracted from each other. In this situation, we can use Newtons 2nd Law: Net Force =(mass)( acceleration). The net force in this situation would be the Force of Gravity minus the Normal Force. So the equation you will begin with is Fg(which is mg) - Normal Force = ma. We want the normal Force, so our new equation derived for Fnormal will be mg - ma = Fnormal.
The force that a scale in a rising and decelerating elevator reads can be calculated with the equation F= m*(g-a). Given that the mass, acceleration due to gravity, and elevator deceleration are 59.1 kg, 9.80 m/s2, and 6.87 m/s2 respectively, the force would be 172.8 Newtons.
Explanation:The subject of this question is Physics, and it seeks to investigate the reading on a scale in an elevator when it is rising and decelerating. The force that the scale reads can be represented by the equation F= m*(g-a), where 'F' is the force, 'm' is the mass, 'g' is acceleration due to gravity, and 'a' is the rate at which the elevator is decelerating. In your case:
mass (m) = 59.1 kggravitational acceleration (g) = 9.80 m/s2deceleration of the elevator (a) = 6.87 m/s2So, the force the scale reads is F = 59.1 kg * (9.80 m/s2 - 6.87 m/s2). This is equal to 172.8 Newtons.
Learn more about Physics here:https://brainly.com/question/32123193
#SPJ12
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
We use the principles of Kinetic and Potential Energy to establish an energy conservation equation, balancing the initial kinetic energy of the box against the potential energy stored in the spring at rest. Solving this equation gives the maximum additional deformation of the spring.
Explanation:From the problem, we know that the mass of the package (m) is 50 kg, the initial speed (v) is 2 m/s, the spring's constant (k) is 30 kN/m or 30000 N/m, and the initial compression of the spring (xi) is 50 mm or 0.05 m. We are tasked to determine the maximum additional deformation (xf) of the spring in bringing the package to rest.
Your problem involves Kinetic Energy and Potential Energy concepts. If we consider the instant when the package begins to hit the spring and the moment when it comes to rest, the energy conservation law can be applied, stating that the initial kinetic energy of the box equals the work done by the spring (which is the potential energy stored in it).
So, the kinetic energy of the package when it begins to hit the spring equals 1/2 * m * v^2, and the potential energy of the spring when the package comes to rest equals 1/2 * k * (xi+xf)^2.
Therefore, equating both expressions and resolving the quadratic equation for xf, provides us with the maximum additional deformation of the spring.
Learn more about Energy Conservation here:https://brainly.com/question/32490774
#SPJ2
A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm is applied to the merry-go-round with a moment of inertia of 50.0 kg m2, what is the angular acceleration in rad/s2?
Answer:
[tex]-0.25 rad/s^2[/tex]
Explanation:
The equivalent of Newton's second law for rotational motions is:
[tex]\tau = I \alpha[/tex]
where
[tex]\tau[/tex] is the net torque applied to the object
I is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
In this problem we have:
[tex]\tau = -12.5 Nm[/tex] (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)
[tex]I=50.0 kg m^2[/tex] is the moment of inertia
Solving for [tex]\alpha[/tex], we find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2[/tex]
The average flow rate in the Niagara River is 6.0 × 106 10. kg/s, and the water drops 50 m over Niagara Falls. If all this energy could be harnessed to generate hydroelec- tric power at 90% efficiency, what would be the electric power output?
Answer:
[tex]2.65\cdot 10^9 W[/tex]
Explanation:
The power produced by the water falling down the Falls is
[tex]P=\frac{E}{t}=\frac{mgh}{t}[/tex]
where
E = mgh is the potential energy of the water, with m being the mass, g the gravitational acceleration, h the height
t is the time
In this problem we have
[tex]\frac{m}{t}=6.0\cdot 10^6 kg/s[/tex] us the mass flow rate
h = 50 m is the height
g = 9.8 m/s^2 is the acceleration of gravity
Substituting,
[tex]P=(6.0\cdot 10^6 kg/s)(9.8 m/s^2)(50 m)=2.94\cdot 10^9 W[/tex]
And since the efficiency is only 90%, the power output is
[tex]P_{out} = (0.90) (2.94\cdot 10^9 W)=2.65\cdot 10^9 W[/tex]
Heavy water usually refers to water that : (A) has been frozen, and so is more dense (B) has had its hydrogen removed (C) is radioactive (D) contains deuterium
Answer:
Option (D)
Explanation:
The chemical formula for normal water is H2O and the chemical formula for heavy water is D2O.
Where D is deuterium which is the isotope of hydrogen.
There are three isotopes of hydrogen.
1H1 it is called protium.
1H2 it is called deuterium.
1H3 it is called tritium.
Heavy water usually refers to water that contains deuterium.
When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that R remains constant when the power consumption changes.
Answer:
4.41 W
Explanation:
P = IV, V = IR
P = V² / R
Given that P = 0.0625 when V = 1.50:
0.0625 = (1.50)² / R
R = 36
So the resistor is 36Ω.
When the voltage is 12.6, the power consumption is:
P = (12.6)² / 36
P = 4.41
So the power consumption is 4.41 W.
To find the power consumption of the resistor when connected to a 12.6-V car battery, we can use the formula P = IV. By substituting the given voltage into the formula and using the resistance value from the initial scenario, we can calculate the power consumed.
Explanation:If a resistor with resistance R consumes 0.0625 W of electrical power when connected to a 1.50-V flashlight battery, we can determine the power consumption when the resistor is connected to a 12.6-V car battery by using the formula P = IV. Since R remains constant, we can use the same resistance value. By substituting the given voltage into the formula, we can find the current flowing through the resistor when connected to the car battery, and then calculate the power consumed.
Let's calculate the current flowing through the resistor when connected to the flashlight battery:
I = sqrt(P/R) = sqrt(0.0625/R) A
Now, let's calculate the power consumed when connected to the car battery:
P' = IV' = (sqrt(0.0625/R))(12.6)
https://brainly.com/question/22103646
#SPJ2
(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is 0.301 T. Each has a speed of 7.92 × 10^5 m/s, but one ion is singly charged and the other is doubly charged. Find the radius of the circular path followed by the singly charged ion in the field. Answer in units of cm. (b) Find the radius of the circular path followed by the doubly charged ion in the field. (c) Find the distance of separation when they have moved through one-half of their circular path and strike a piece of photographic paper.
Answer:
Part a)
[tex]R_1 = 0.072 m[/tex]
Part b)
[tex]R_2 = 0.036 m[/tex]
Part c)
d = 0.072 m
Explanation:
Part a)
As we know that the radius of the charge particle in constant magnetic field is given by
[tex]R = \frac{mv}{qB}[/tex]
now for single ionized we have
[tex]R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}[/tex]
[tex]R_1 = 0.072 m[/tex]
Part b)
Similarly for doubly ionized ion we will have the same equation
[tex]R = \frac{mv}{qB}[/tex]
[tex]R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}[/tex]
[tex]R_2 = 0.036 m[/tex]
Part c)
The distance between the two particles are half of the loop will be given as
[tex]d = 2(R_1 - R_2)[/tex]
[tex]d = 2(0.072 - 0.036)[/tex]
[tex]d = 0.072 m[/tex]
A 0.16kg stone attached to a string of length I = 0.22m, is whirled in a horizontal circle with a constant velocity of 4.0m/s. a) Calculate the radial i.e the centripetal acceleration of the stone. b) The tension in the string while the stone is rotating. c) The horizontal force on the stone.
Answer:
PART A)
[tex]a_c = 72.7 m/s^2[/tex]
PART B)
T = 11.6 N
PART C)
[tex]F_{horizontal} = 11.6 N[/tex]
Explanation:
PART A)
Centripetal acceleration is given by
[tex]a_c = \frac{v^2}{R}[/tex]
now we have
[tex]a_c = \frac{4^2}{0.22}[/tex]
[tex]a_c = 72.7 m/s^2[/tex]
PART B)
Here Tension force of string is providing centripetal force
so we can say
[tex]T = ma_c[/tex]
[tex]T = 0.16(72.7) = 11.64 N[/tex]
PART C)
Force on the stone in horizontal direction is only tension force
so here we have
[tex]F_{horizontal} = T = 11.64 N[/tex]
A rock is thrown vertically upward from ground level at time t = 0. At t = 1.6 s it passes the top of a tall tower, and 1.0 s later it reaches its maximum height. What is the height of the tower?
The rock has height [tex]y[/tex] at time [tex]t[/tex] according to
[tex]y=v_0t-\dfrac g2t^2[/tex]
where [tex]v_0[/tex] is the velocity with which it was thrown, and g = 9.8 m/s^2 is the acceleration due to gravity.
Complete the square to get
[tex]y=\dfrac{{v_0}^2}{2g}-\dfrac g2\left(t-\dfrac{v_0}g\right)^2[/tex]
which indicates a maximum height of [tex]\dfrac{{v_0}^2}{2g}[/tex] occurs when [tex]t=\dfrac{v_0}g[/tex]. We're told this time is 2.6 s after the rock is thrown:
[tex]2.6\,\mathrm s=\dfrac{v_0}{9.8\frac{\rm m}{\mathrm s^2}}\implies v_0=25.48\dfrac{\rm m}{\rm s}[/tex]
So when t = 1.6 s, the rock reaches the tower's height of
[tex]y=v_0(1.6\,\mathrm s)-\dfrac g2(1.6\,\mathrm s)^2\approx\boxed{28\,\mathrm m}[/tex]
The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock.
Explanation:The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock. Since the rock reaches its maximum height 1.0 s after passing the top of the tower, we can use the formula:
Max height = y1 - y0 - (1/2) * g * t^2
where y1 is the maximum height, y0 is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes to reach the maximum height.
Since the rock is 8.10 m above its starting point at t = 1.00 s, we can plug in the values:
Max height = 8.10 m - 0 m - (1/2) * (9.8 m/s^2) * (1.0 s)^2
Simplifying this equation, we find that the maximum height reached by the rock is 3.15 m. Therefore, the height of the tower is 8.10 m + 3.15 m = 11.25 m.
A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output power of the charger in unit of watt?
Answer:
Output power of the charger is 9.85 watts.
Explanation:
It is given that,
Output current of the USB charger, I = 1.97 A
The standard USB output voltage, V = 5 V
We need to find the output power of the charger. It can be determined using the following formula as :
P = V × I
[tex]P=5\ V\times 1.97\ A[/tex]
P = 9.85 watts
The output power of the charger is 9.85 watts. Hence, this is the required solution.
The USB charger's output power, in watts, is calculated using the formula P = IV. Substituting the given values of current (1.97 A) and voltage (5 V) gives a power output of 9.85 W.
Explanation:The power output of any device is calculated using the formula P = IV where I is the current in amperes and V is the voltage in volts. Given that the quick USB charger has an output current I of 1.97 amps and a standard USB output voltage V of 5 volts, we can substitute these values into our formula.
Thus, the power is P = IV = (1.97 A)(5 V) = 9.85 W. This means the output power of the quick USB charger is 9.85 watts. It's crucial to remember that power is measured in watts, where 1 watt is equivalent to 1 Ampere Volt (1 A.V = 1W).
https://brainly.com/question/33277947
#SPJ12
The weights of bags filled by a machine are normally distributed with a standard deviation of 0.05 kilograms and a mean that can be set by the operator. At what level should the mean weight be set if it required that only 1% of the bags weigh less than 9.5 kilograms? Round the answer to 2 decimal places.
Answer:
9.62 kg
Explanation:
From a z-score table, P(z<-2.33) ≈ 0.01. So 9.5 should be 2.33 standard deviations below the mean.
z = (x − μ) / σ
-2.33 = (9.5 − μ) / 0.05
-0.1165 = 9.5 − μ
μ = 9.6165
Rounding to 2 decimal places, the mean should be set to 9.62 kg.
An object with initial temperature 130 ∘ F is submerged in large tank of water whose temperature is 50 ∘ F . Find a formula for F ( t ) , the temperature of the object after t minutes, if the cooling constant is k = − 0.2 . Remember Newton's Law of Cooling (the rate of change of temperature with respect to time is equal to k times the difference between the temperature of the object and the surrounding temperature) ! :)
Answer:
T = 50 + 80e^(-0.2t)
Explanation:
Newton's law of cooling says the rate of change of temperature with respect to time is proportional to the temperature difference:
dT/dt = k (T − Tₐ)
Separating the variables and integrating:
dT / (T − Tₐ) = k dt
ln (T − Tₐ) = kt + C
T − Tₐ = Ce^(kt)
T = Tₐ + Ce^(kt)
Given that Tₐ = 50 and k = -0.2:
T = 50 + Ce^(-0.2t)
At t = 0, T = 130.
130 = 50 + Ce^(0)
130 = 50 + C
C = 80
Therefore:
T = 50 + 80e^(-0.2t)
The formula for the object cooling is [tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex], where [tex]t[/tex] is in minutes.
The object is cooled by heat mechanism of Convection, Convection is a Heat Transfer mechanism in which is a solid object is cooled due to a fluid in motion and is described by the Newton's Law of Cooling, whose Differential Equation is:
[tex]\frac{dT}{dt} = -r\cdot (T-T_{m})[/tex] (1)
Where:
[tex]T[/tex] - Temperature of the solid, in degrees Fahrenheit.
[tex]r[/tex] - Cooling rate, in [tex]\frac{1}{min}[/tex].
[tex]T_{m}[/tex] - Water temperature, in degrees Fahrenheit.
The solution of this Differential Equation is:
[tex]T(t) = T_{m} + (T_{o}-T_{m})\cdot e^{-r\cdot t}[/tex] (2)
Where [tex]T_{o}[/tex] is the initial temperature of the solid, in degrees Fahrenheit.
If we know that [tex]T_{m} = 50\,^{\circ}F[/tex], [tex]T_{o} = 130\,^{\circ}F[/tex] and [tex]r = 0.2[/tex], then the formula for the object cooling is:
[tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex]
The formula for the object cooling is [tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex], where [tex]t[/tex] is in minutes.
Here is a question related to the Newton's Law of Cooling: https://brainly.com/question/13724658
A 1400kg automobile moving at a maximum speed of 23m/s on a level circular track of readius of 95m. What is the coefficient of friction?
Answer:
The coefficient of friction is 0.56
Explanation:
It is given that,
Mass of the automobile, m = 1400 kg
Speed of the automobile, v = 23 m/s
Radius of the track, r = 95 m
The automobile is moving in a circular track. The centripetal force is given by :
[tex]F_c=\dfrac{mv^2}{r}[/tex]............(1)
Frictional force is given by :
[tex]F_f=\mu mg[/tex]...................(2)
[tex]\mu[/tex] = coefficient of friction
g = acceleration due to gravity
From equation (1) and (2), we get :
[tex]\dfrac{mv^2}{r}=\mu mg[/tex]
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(23\ m/s)^2}{95\ m\times 9.8\ m/s^2}[/tex]
[tex]\mu=0.56[/tex]
So, the coefficient of friction is 0.56. Hence, this is the required solution.
To find the coefficient of friction, calculate the maximum speed at which the car can negotiate the curve without slipping using the formula v = √(μrg). Plug in the values and solve for μ to get the coefficient of friction. The approximate value is 0.24.
Explanation:To find the coefficient of friction, we will first calculate the maximum speed at which the car can negotiate the curve without slipping. The maximum speed is given by the formula:
v = √(μrg)
Where v is the maximum speed, μ is the coefficient of friction, r is the radius of the curve, and g is the acceleration due to gravity.
Plugging in the values, we get:
23 = √(μ * 1400 * 9.8 * 95)
Squaring both sides of the equation, we have:
529 = μ * 1400 * 9.8 * 95
Simplifying the equation, we find:
μ = 529 / (1400 * 9.8 * 95)
Calculating the value, we get:
μ ≈ 0.24
Therefore, the coefficient of friction is approximately 0.24.