Answer:
weight
*the objects weight*
A person is standing on a spring bathroom scale on the floor of an elevator which is moving up and slowing down at the rate of 3 m/s 2 . The acceleration of gravity is 9.8 m/s 2 . If the person’s mass is 93.3 kg, what does the scale read? Answer in units of n.
Answer:
[tex]F_N=1194.24\ N[/tex]
Explanation:
Given that,
The elevator is moving up and slowing down at the rate of, [tex]a=3\ m/s^2[/tex]
The acceleration due to gravity, [tex]g=9.8\ m/s^2[/tex]
Mass of the person, m = 93.3 kg
To find,
The reading of the scale.
Solution,
As the elevator is moving down with some acceleration. The net force acting on it is given by :
[tex]F_N=ma+mg[/tex]
[tex]F_N=m(a+g)[/tex]
[tex]F_N=93.3(3+9.8)[/tex]
[tex]F_N=1194.24\ N[/tex]
So, the scale will read 1194.24 N.
Friction a. is slow but steady movement along a fault. b. happens when rock is weak and can slip smoothly, without creating shock waves. c. is the process by which faults release energy. d. is the force that resists sliding along a surface
Answer:
option D
Explanation:
The correct answer is option D
The friction force is the force that holds back the surface from sliding. It acts in the opposite direction of the motion of the body.
It can be explained with an example,
when you throw a ball on the ground it stops automatically this is because of friction which is acting opposite to the ball.
Friction also has benefits like due to friction we can walk, run, etc.
So, from the given option
The force that resists sliding is friction force.
A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astronaut turns away to look at Earth and several seconds later is 38.1 m behind the ship, at rest relative to the spaceship. The only way to return to the ship without a thruster is to throw a wrench directly away from the ship. The wrench has a mass of 0.570 kg, and the astronaut throws the wrench with a speed of 22.4 m/s. How long does it take the astronaut to reach the ship?
Answer:
259.62521 seconds
Explanation:
[tex]m_1[/tex] = Mass of astronaut = 87 kg
[tex]m_2[/tex] = Mass of wrench = 0.57 kg
[tex]v_1[/tex] = Velocity of astronaut
[tex]v_2[/tex] = Velocity of wrench = 22.4 m/s
Here, the linear momentum is conserved
[tex]m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s[/tex]
Time = Distance / Speed
[tex]Time=\frac{38.1}{0.14675}=259.62521\ s[/tex]
The time taken to reach the ship is 259.62521 seconds
(a) What is the length of a simple pendulum that oscillates with a period of 1.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
LE = m LM = m
(b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 1.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
mE = kg mM = kg
Explanation:
a) Period of simple pendulum
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
l₁ = length of pendulum in Earth = ?
l₂ = length of pendulum in Mars = ?
T = Period of pendulum = 1.2 s
g₁ = Acceleration due to gravity in Earth = 9.80 m/s²
g₂ = Acceleration due to gravity in Mars = 3.70 m/s²
For Earth :-
[tex]T=2\pi \sqrt{\frac{l_1}{g_1}}\\\\1.2=2\pi \sqrt{\frac{l_1}{9.8}}\\\\l_1=0.36m[/tex]
Length of pendulum in Earth = 0.36 m
For Mars :-
[tex]T=2\pi \sqrt{\frac{l_2}{g_2}}\\\\1.2=2\pi \sqrt{\frac{l_2}{3.7}}\\\\l_2=0.13m[/tex]
Length of pendulum in Mars = 0.13 m
b) Period of spring
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
Here period is independent of acceleration due to gravity, so mass value is same in Earth and Mars.
m = Mass = ?
T = Period = 1.2 s
k = Spring constant = 20 N/m
[tex]T=2\pi \sqrt{\frac{m}{k}}\\\\1.2=2\pi \sqrt{\frac{m}{20}}\\\\m=0.73kg[/tex]
Mass in Earth = Mass in Mars = 0.73 kg
The length of a simple pendulum and the mass to suspend the spring are mathematically given as
a L1 = 0.36m for the earth and L2 = 0.13m for mars
b m=0.73kg
What is the length of a simple pendulum?Question Parameter(s):
period of 1.2 s on Earth
where the acceleration due to gravity is 9.80 m/s2
and on Mars, where the acceleration due to gravity is 3.70 m/s2?
a force constant of 20 N/m
Generally, the equation for the Period of a simple pendulum is mathematically given as
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
therefore the length of Earth is
[tex]\\\\1.2=2\pi \sqrt{\frac{l_1}{9.8}}\\\\l_1=0.36m[/tex]
and for mars
[tex]\\\\1.2=2\pi \sqrt{\frac{l_2}{3.7}}\\\\l_2=0.13m[/tex]
In conclusion length of a simple pendulum is
L1 = 0.36m for earth and L2 = 0.13m for mars
What mass would you need to suspend from a spring?Period of spring
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]}\\\\1.2=2\pi \sqrt{\frac{m}{20}}\\\\m=0.73kg[/tex]
In conclusion, the mass to suspend the spring
m = 0.73kg
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What is the phase ϕ(x,t) of the wave? Express the phase in terms of one or more given variables (A, k, x, t, and ω) and any needed constants like π. Throughout this problem, be sure to use 'k' instead of the greek letter 'κ' (kappa), and the greek letter 'ω' instead of 'w'.
The phase φ(x,t) of a wave represents its phase shift. It can be obtained from the wave function y (x, t) = A sin (kx - wt + p) used to model the wave's behavior. The expression for the phase is φ(x,t) = kx - wt + p.
Explanation:The phase φ(x,t) of a wave can be expressed from the given sinusoidal wave function y (x, t) = A sin (kx - wt + φ). Here, φ represents the phase of the wave, which is a shift that allows for initial conditions other than x = +A and v = 0. The aforementioned variables stand for: A - amplitude of the wave, k - wave number, ω - angular frequency, x - position, and t - time.
Now, looking at the wave equation, the phase shift, φ, is the amount by which the wave function is shifted. This shift can be to the right or the left and is usually represented by the Greek letter phi (p). In the wave function, this phase shift may result because the motion being modeled by the function isn't starting from the rest position. For example, if considering the motion of a pendulum, if it was released from an elevated position, there would be a phase shift in the resulting wave function, unlike if it had been released from a resting position.
So for the waveform y (x, t) = A sin (kx - wt + p), the phase φ(x,t) = kx - wt + p.
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In physics, the phase ϕ of a wave accounts for the initial conditions and describes the state of oscillation relative to a specific reference point. It is part of the wave function formula: y (x, t) = A sin (kx - ωt + ϕ). Here, ϕ = kx - ωt + ϕ, with k and x related to position and ωt to elapsed time.
Explanation:In physics, often when discussing waves, the term 'phase' is used. The phase of a wave, represented by ϕ in wave equations, is a measure of the relative position in a wave cycle of a particle, taking into account the initial point of the cycle (which can involve a phase shift) and the time elapsed since this initial point.
It allows us to describe the current state of oscillation of any point in the wave, relative to a specific reference point.
For a sinusoidal wave moving in the x direction, its wave function, which models the wave's displacement over time, can be represented as y (x, t) = A sin (kx - ωt + ϕ). Here, y (x, t) is the amplitude of the wave at any point x and at any time t. The variable A signifies the peak amplitude of the wave, k is the wave number, ω is the angular frequency, and ϕ is the phase.
The purpose of ϕ in this equation is to account for the initial state of the system. For example, if our wave is not starting at its equilibrium position, we can incorporate these initial conditions into the wave equation using the phase constant ϕ.
So, in conclusion, the phase ϕ of a wave in terms of the given variables would simply be ϕ = kx - ωt + ϕ, with k and x related to spatial position and ωt related to the elapsed time.
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When 0.1375 g of solid magnesium is burned in a constant-volume bomb calorimeter, the temperature increases by 1.126°C. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 3024 J/°C. Calculate the heat given off by the burning magnesium in kJ/mol.
Answer:
-24.76 kJ/mol
Explanation:
given,
mass of solid magnesium burned = 0.1375 g
the temperature increases by(ΔT) 1.126°C
heat capacity of of bomb calorimeter (C_{cal})= 3024 J/°C
heat absorbed by the calorimeter
[tex]q_{cal} = C_{cal}\DeltaT[/tex]
[tex]q_{cal} = 3024 \times 1.126[/tex]
[tex]q_{cal} =3405.24\ J[/tex]
[tex]q_{cal} =3.405\ kJ[/tex]
heat released by the reaction
[tex]q_{rxn} = -q_{cal}[/tex]
[tex]q_{rxn} = -3.405\ kJ[/tex]
energy density will be equal to heat released by the reaction divided by the mass of magnesium
Energy density = [tex]\dfrac{-3.405}{0.1375}[/tex]
Energy density = -24.76 kJ/mol
heat given off by burning magnesium is equal to -24.76 kJ/mol
How do i calculate acceleration? pls help! thanks :)
As I drive down South Hill toward downtown Spokane, my car goes from a velocity of 25 mph north to a velocity of 35 miles an hour north in 5 minutes. What is my acceleration during this time?
5 mph / minute
2 mph / min squared
10 mph
The acceleration is 2 mph/min
Explanation:
The acceleration of an object is the rate of change of velocity. It is defined as follows:
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken for the velocity to change from u to v
For the car in this problem, we have (taking North as positive direction):
u = 25 mph
v = 35 mph
t = 5 min
Substituting into the equation, we find the acceleration:
[tex]a=\frac{35 mph-25mph}{5min}=2 mph/min[/tex]
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Four of the wavelengths of the Balmer series occur in the visible spectrum (656 nm, 486 nm, 434 nm, and 410 nm). In which region of the electromagnetic spectrum does the fifth line, with a wavelength of 397 nm, occur?
a) X rays
b) gamma rays
c) radio waves
d) ultraviolet
e) visible light
f) infrared
g) microwaves
Final answer:
The fifth line of the Balmer series, with a wavelength of 397 nm, occurs in the ultraviolet region of the electromagnetic spectrum. Option D is correct.
Explanation:
The fifth line of the Balmer series with a wavelength of 397 nm falls into the ultraviolet (UV) region of the electromagnetic spectrum. The visible part of the spectrum ranges approximately from 380 nm to 740 nm. Since the wavelength of the fifth line is just below 400 nm, it is just outside the range for visible light and therefore must be in the ultraviolet range.
In the context of the Balmer series, there are four lines within the visible spectrum, specifically at wavelengths of 656 nm, 486 nm, 434 nm, and 410 nm. The line at 397 nm would be the first one in the UV part of the spectrum, making the answer to the student's question d) ultraviolet.
The Balmer series is a series of spectral lines in the hydrogen emission spectrum. Four of the wavelengths of the Balmer series occur in the visible spectrum: 656 nm, 486 nm, 434 nm, and 410 nm. Since the fifth line has a wavelength of 397 nm, it falls in the ultraviolet region of the electromagnetic spectrum.
A system has three macrostates. Macrostates 1 and 3 are least likely and have one basic state each. Macrostate 2 is the equilibrium state and is four times more likely to occur than either of the other two macrostates.
Part A
What is the probability that the equilibrium state 2 occurs?
Express your answer using three significant digits.
p2=
Part B
What is the probability that macrostate 1 occurs?
Express your answer using three significant digits.
p1=
Part C
What is the probability that macrostate 3 occurs?
Express your answer using three significant digits.
p3=
Part D
What is the sum of the probabilities for all macrostates?
Express your answer using three significant digits.
p=
Answer:
[tex]p2=0.667\\p1=0.167\\p3=0.167\\p=1[/tex]
Explanation:
Let's start writing the sample space for this exercise :
Let be ''M'' an abbreviation for Macrostate
Ω = { M1 , M2 , M3 }
Let be P(M1) the probability of Macrostate 1.
Reading the exercise, we know that ⇒
[tex]P(M1)=P(M3)[/tex]
Let's note this probability as ''p''.
[tex]P(M1)=P(M3)=p[/tex]
Macrostate 2 is four times more likely to occur than either of the other two macrostates ⇒
[tex]P(M2)=4p[/tex]
The sum of all probabilities must be equal to 1 for this sample space.Therefore,
[tex]p+p+4p=1[/tex]
[tex]6p=1[/tex]
[tex]p=\frac{1}{6}[/tex]
Finally :
[tex]P(M1)=\frac{1}{6}[/tex]
[tex]P(M2)=\frac{4}{6}[/tex]
[tex]P(M3)=\frac{1}{6}[/tex]
For Part A :
[tex]p2=\frac{4}{6}=0.667[/tex]
For Part B and C :
[tex]p1=p3=0.167[/tex]
For Part D :
The sum of the probabilities for all macrostates is equal to 1 :
[tex]p=p1+p2+p3=\frac{1}{6}+\frac{4}{6}+\frac{1}{6}=\frac{6}{6}=1[/tex]
Final answer:
The probability that equilibrium state 2 occurs is approximately 0.667. The probability that macro state 1 or macro state 3 occurs is each approximately 0.167. The sum of the probabilities for all macro states is exactly 1, as expected for a complete set of possible outcomes.
Explanation:
To find the probability of each macro state, we first recognize that the probabilities must sum to 1 for all macro states. Given that the equilibrium state, Macro state 2, is four times more likely than either of the other two Macro states 1 and 3, we can assign a probability 'p' to Macro states 1 and 3 and '4p' to Macro state 2.
Let's assume:
Probability of Macro state 1, p1 = p
Probability of Macro state 2, p2 = 4p
Probability of Macro state 3, p3 = p
Since the total probability must equal 1, we write the equation:
p1 + p2 + p3 = 1
Substituting the assumed probabilities gives:
p + 4p + p = 1
6p = 1
Therefore, p = 1/6
Now, we can calculate each probability:
p1 = 1/6 ≈ 0.167
p2 = 4/6 ≈ 0.667
p3 = 1/6 ≈ 0.167
The sum of all probabilities, p, is as expected:
p = p1 + p2 + p3 = 0.167 + 0.667 + 0.167 = 1
A 3.00 kg pool ball is moving to the left with a speed of 4.30 m/s without friction. If it experiences an impulse of -4.00 Ns, what is the object's speed and direction after the impulse occurs?
Explanation:
Given that,
Mass of thee pool ball, m = 3 kg
Initial speed of the ball, u = -4.3 m/s
Impulse experienced by thee ball, J = -4 N-s
To find,
Speed of the object after impulse occurs and its direction.
Solution,
Let left side is negative and right side is positive. So, the change in momentum or the impulse is given by the following expression as :
[tex]J=m(v-u)[/tex]
[tex]v=\dfrac{J}{m}+u[/tex]
[tex]v=\dfrac{-4}{3}+(-4.33)[/tex]
v = -5.663 m/s
So, the speed of the object is 5.663 m/s and it is towards left.
The pool ball experiences an impulse in the opposite direction of its initial motion. After the impulse, the speed of the ball reduces to 2.97 m/s and it is still moving to the left.
Explanation:The subject of this question is Physics, specifically impulse and momentum. Firstly, we need to calculate the initial momentum which is the mass of the object multiplied by its initial velocity, so that will be 3.00 kg * 4.30 m/s = 12.90 kg*m/s to the left since the direction is towards left.
An impulse is the change in momentum, and in this case, the impulse is -4.00 Ns which means it acts in the opposite direction of the initial motion, i.e to the right.
After the impulse, the final momentum will be the sum of initial momentum and impulse: 12.90 kg*m/s (to the left) + -4.00 kg*m/s (to the right) = 8.90 kg*m/s (to the left).
Finally, to calculate the final speed, divide the final momentum by the mass: 8.90 kg*m/s ÷ 3.00 kg = 2.97 m/s. So, after the impulse, the object's speed is 2.97 m/s and it is still moving to the left.
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A museum curator moves artifacts into place on various different display surfaces. If the curator moves a 145 kg aluminum sculpture across a horizontal steel platform what is fs, fmax, and fk
Answer:
Static friction is, [tex]0\ N\leq f_s\leq 866.81\ N[/tex]
Maximum static friction is, [tex]f_{max}= 866.81\ N[/tex]
Kinetic friction is, [tex]f_k=667.87\ N[/tex]
Explanation:
The value of coefficient of static friction between Aluminium and Steel is [tex]\mu_s=0.61[/tex] and the value of coefficient of kinetic friction is [tex]\mu_k=0.47[/tex]
Given:
Mass of the sculpture is, [tex]m=145\ kg[/tex].
As the sculpture is moved along the horizontal direction only, there will no net force acting along the vertical direction. Therefore,
Normal force by the steel platform will be equal to the weight of the aluminium sculpture. That is,
[tex]N=mg[/tex]
Now, the static friction is given as the product of normal force and coefficient of static friction. The static friction varies from a value of 0 to limiting friction.
The limiting friction is the maximum static friction that acts on a body to resist the motion.
The value of the limiting friction or the maximum static friction is given as:
[tex]f_{max}=\mu_sN=\mu_smg=0.61\times 145\times 9.8=866.81\ N[/tex]
Thus, static friction is given as:
[tex]0\leq f_s\leq f_{max}\\0\leq f_s\leq \mu_smg\\0\ N\leq f_s\leq 866.81\ N[/tex]
Now, kinetic friction is the frictional force acting on a body when the body is moving under the action of forces. Thus kinetic friction is the product of normal force and coefficient of kinetic friction. Kinetic friction is a constant opposing force.
Thus, kinetic friction is given as:
[tex]f_k=\mu_kN\\f_k=\mu_kmg\\f_k=0.47\times 145\times 9.8=667.87\ N[/tex]
What is an extrasolar planet?
a. A planet that is considered an "extra," in that it was not needed for the formation of its solar system.
b. A planet that is extra large compared to what we'd expect.
c. A planet that is larger than the Sun.
d. A planet that orbits a star that is not our own Sun
Answer:
option (d)
Explanation:
The extra solar planet is also called as exo planet.
the planet which orbits around the star other than sun is called exo planet.
So, a planet that orbits a star that is not our own sun.
Thus, option (d) is correct.
An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first started, it applies a constant torque of 1930 Nm to the propeller, which starts from rest.
a. What is the angular acceleration of the propeller? Treat the propeller as a slender rod.
b. What is the propeller's angular speed after making 5.00 rev?
c. How much work is done by the engine during the first 5.00 rev?
d. What is the average power output of the engine during the first 5.00 rev?
e. What is the instantaneous power output of the motor at the instant that the propeller has turned through 5.00 rev?
Answer:
a)30.14 rad/s2
b)43.5 rad/s
c)60633 J
d)42 kW
e)84 kW
Explanation:
If we treat the propeller is a slender rod, then its moments of inertia is
[tex] I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2[/tex]
a. The angular acceleration is Torque divided by moments of inertia:
[tex]\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2[/tex]
b. 5 revolution would be equals to [tex]10\pi[/tex] rad, or 31.4 rad. Since the engine just got started
[tex]\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5[/tex]
[tex]\omega = \sqrt{1893.5} = 43.5 rad/s[/tex]
c. Work done during the first 5 revolution would be torque times angular displacement:
[tex]W = T*\theta = 1930 * 31.4 = 60633 J[/tex]
d. The time it takes to spin the first 5 revolutions is
[tex]t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s[/tex]
The average power output is work per unit time
[tex]P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W[/tex] or 42 kW
e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:
[tex]P_i = T*\omega = 1930*43.5=83983 W[/tex] or 84 kW
A room has a wall with an R value of 13 F sq.ft. hr/BTU. The room is 18 feet long and 8 feet wide with walls that are 8 ft high. On a particular day the average outside temperature is 31 degrees F. How much heat is transferred through the walls on that day (in BTUs)?
Answer:
Carbon electrons are arranged in a way that makes it possible for them to form long chains. That also helps carbon maintain its properties.
Explanation:
Carbon electrons are arranged in a way that makes it possible for them to form long chains. That also helps carbon maintain its properties.
The ball's gravitational potential energy as a function of its height after release. The ball's kinetic energy as a function of its height after release. The ball's total mechanical energy as a function of its height after release.
Answer:
a) U = - G m₁m₂ / r , b) K = ½ m (v₀² + 2gy)² or K = 2 mg² y² c) Em = m g y (2 g y + 1)
Explanation:
Let's write the functions that are requested
a) Gravitational power energy
U = - dF / dr
F = G m₁ m₂ / r²
U = - G m₁m₂ / r
r is the height
b) The scientific enrgia
K = ½ m v²
Cinematic
v² = v₀² + 2 g y
K = ½ m (v₀² + 2gy)²
If the initial velocity is zero, the ball is released
K = ½ m 4 g² y²
K = 2 mg² y²
c) Mechanical energy
Em = K + U
Em = 2 m g² y² + m g y
Em = m g y (2 g y + 1)
A girl attaches a rock to a string, which she then swings counter-clockwise in a horizontal circle. The string breaks at point P in the figure, which shows a bird's-eye view (as seen from above). Which path (A-E) will the rock follow?
A. Path A
B. Path B
C. Path C
D. Path D
E. Path E
Answer:
B. Path B
Explanation:
Since the velocity v is tangent to the circular path, the rock will follow the path B.
You can see the paths in the pic.
A 50.9 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resistance of 1492 N to the diver’s fall. If the diver comes to rest 6 m below the water’s surface, what is the total distance between the diving board and the diver’s stopping point underwater? The acceleration due to gravity is 9.81 m/s 2 .
Answer:
T = 23.92 m
Explanation:
given,
mass of the diver = 50.9 Kg
Resistant force from the water = f = 1492 N
diver come top rest under water at a distance = 6 m
acceleration due to gravity = 9.81 m/s²
final velocity = v = 0 m/s
initial velocity = u = ?
total distance = ?
Now acceleration of body under water
f = m a
[tex]a = \dfrac{1492}{50.9}[/tex]
[tex]a = 29.31\ m/s^2[/tex]
using equation of motion
v² = u² + 2 a s
0 = u² - 2 x 29.31 x 6
[tex]u = \sqrt{2\times 29.31\times 6}[/tex]
u = 18.75 m/s
now.
calculating distance of the diver in air
v² = u² + 2 a s
0 = 18.75² - 2 x 9.81 x s
s = 17.92 m
total distance
T = 17.92 + 6
T = 23.92 m
the total distance between the diving board and the diver’s stopping point underwater T = 23.92 m
A person exerts a horizontal force of 190 n in the test apparatus shown in the figure. Find the horizontal force that his flexor muscle exerts on his forearm.
Answer:
Check explanation.
Explanation:
From the question, we know that The person exerted 190N, force on the flexor is unknown. Since, we don't have access to our diagram, we have to make one or two assumption; (1) that ba= 0.3 m, ac= 0.05.
Therefore, the horizontal force that his flexor muscle exerts on his forearm,F(flexor) = (190N) × (0.3m) / 0.05m.
The horizontal force that his flexor muscle exerts on his forearm,F(flexor) = 57 Nm/ 0.05 m.
The horizontal force that his flexor muscle exerts on his forearm,F(flexor) = 1140N.
Using principles of physics, specifically equilibrium, we know the force exerted by muscles can be greater than the load they support, especially when the load is distant from the joint. In the case of a 190 N horizontal force, the exact force of the flexor muscle can't be calculated without more information, but it's likely larger than 190 N.
Explanation:To answer the student's question, we must refer to concepts in physics, specifically the principle of equilibrium and the analysis of free-body diagrams. Given that the student is trying to find the force that a person's flexor muscle exerts horizontally when the person exerts a horizontal force of 190 N.
As mentioned in the provided information, forces and tensions in muscles and joints can be quite substantial. In this scenario, the flexor muscle in the forearm is acting against the applied force to maintain balance or equilibrium. This can be seen in the free-body diagram where forces are broken down into their x- and y-components.
It's important to note that the force exerted by muscles can be far greater than the load they support, especially when the load is a considerable distance from the joint, as stated in the provided text snippet. This may be relevant when analyzing the muscle force in this scenario.
Unfortunately, without more specific details about the angles involved and the exact configuration of the arm and forearm in the test apparatus, we can't calculate the exact force exerted by the flexor muscle. However, we can say that it's likely to be substantially larger than the 190 N applied force, considering the principles discussed above.
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You are presently taking a weather observation. The sky is full of wispy cirrus clouds estimated to be about 10 km overhead. If a warm front is approaching from the south, about how far is it (assuming a slope of 1:200)? Enter only the numeric value of the distance.
Answer:
x = 2000 Km
Explanation:
Given
y = 10 km
Slope: 1 : 200
x = ?
We can apply the formula
y / x = 1 / 200 ⇒ x = 200*y = 200*10 Km
⇒ x = 2000 Km
Given the slope ratio of 1:200 for a warm front and an observed cloud height of 10 km, the warm front is around 2000 km away.
Explanation:In meteorological terms, a warm front slope can be visualized as a ramp, approximating a 1:200 ratio in this case, where the height (vertical change) is comparable to the 'rise' and the horizontal distance (or reach) is comparable to the 'run'. Here, you mentioned cirrus clouds at an approximate altitude of 10 kilometers - this is the vertical elevation or 'rise'. The slope ratio 1:200 is a simplification indicating that for every 1 km rise, the front extends 200 km in the horizontal direction or the 'run'.
So, if the 'rise' is 10 km (the altitude at which you observe the cirrus clouds), we can calculate the 'run' by multiplying the 'rise' by the slope ratio (200). Thus, the approaching warm front is approximately 10 km * 200 = 2000 km away.
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Star-forming clouds appear dark in visible-light photos because the light of stars behind them is absorbed by __________.
Answer:
Interstellar Dust
Explanation:
In the solar system Interstellar dust is a type of cosmic dust and unlike Interstellar gas they have large dust particles enabling them to block visible light. Unlike smoke and fog, they contain large number of closely packed clumps of atoms and molecules
Answer:
interstellar dust
Explanation:
Dust particles interact with light through scattering and absorption. Reducing the amount of starlight received
The colour of the sky during the day is blue, because the blue light is scattered by atmosphere particles. The increased path length causes the change of colour of sun light . This is due to dust grains in the atmosphere, a similar process happens with interstellar dust
A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R 0.500 cm. (a) If each wire carries 2.00 A, what are the magnitude and direction of the magnetic force per unit length acting on a wire located 0.200 cm from the center of the bundle? (b) What If? Would a wire on the outer edge of the bundle experi- ence a force greater or smaller than the value calculated in part (a)? Give a qualitative argument for your answer.
The magnetic force on a wire in a bundled configuration can be calculated using the formula F = I × B. The force per unit length on a wire located 0.200 cm from the center of the bundle can be calculated using this formula. A wire on the outer edge of the bundle experiences a smaller force compared to a wire closer to the center.
Explanation:a. The magnetic force per unit length on a wire located 0.200 cm from the center of the bundle can be calculated using the formula F = I × B. Since the current and magnetic field are perpendicular, we can simplify the formula to F = I × B × sin(90°) = I × B. Plug in the values to get F = (2.00 A) × (μ × I/(2πR)) = (2 × 10^(-7) T·m/A) × (2.00 A) / (2π × 0.005 m). Calculate this to get the magnitude of the force per unit length.
b. A wire on the outer edge of the bundle will experience a smaller force than the wire 0.200 cm from the center. This is because the magnetic field strength decreases as the distance from the wire increases. Therefore, the force on a wire decreases as its distance from the center of the bundle increases.
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The magnetic force per unit length on a wire [tex]0.200 cm[/tex] from the center is [tex]1.28 \times 10^{-2} \, \text{N/m}[/tex]. A wire at the outer edge experiences a greater force of [tex]1.6 \times 10^{-2} \, \text{N/m}\\ \\[/tex] due to the higher magnetic field there.
This problem involves calculating the magnetic force per unit length acting on a current-carrying wire located within a bundle of wires, as well as comparing it to the force on a wire at the bundle's outer edge.
(a) Force on a Wire [tex]0.200 cm[/tex] from the CenterFirst, we calculate the magnetic field at a distance of [tex]0.200 cm[/tex] from the center. The magnetic field inside the bundle can be found using Ampère's Law:
The current enclosed by a circular path of radius [tex]r[/tex] inside the bundle is:
[tex]I_{\text{enclosed}} = (I_{\text{total}}) \times \left( \frac{\pi r^2}{\pi R^2} \right) \\= 200 \, \text{A} \times \left( \frac{0.200^2}{0.500^2} \right) \\= 32 \, \text{A}[/tex]
The magnetic field B at this radius is given by:
[tex]B = \frac{\mu_0 I_{\text{enclosed}}}{2\pi r} \\= \frac{4\pi \times 10^{-7} \times 32}{2\pi \times 0.002} \\ = 6.4 \times 10^{-3} \, \text{T}[/tex]
The force per unit length acting on a wire carrying a current I₁ is:
[tex]\frac{F}{L} = I_1 \times B \\= 2 \, \text{A} \times 6.4 \times 10^{-3} \, \text{T} \\= 1.28 \times 10^{-2} \, \text{N/m}[/tex]
The direction of the force follows the right-hand rule and is perpendicular to both the wire's current and the magnetic field.
(b) Force on a Wire at the Outer EdgeA wire on the outer edge will experience a magnetic field generated by all the currents inside the circle of radius R. Therefore:
[tex]I_{\text{total}} = 100 \, \text{wires} \times 2.00 \, \text{A} \\= 200 \, \text{A}[/tex]
The magnetic field at the edge [tex]B_{edge}[/tex] is:
[tex]B_{\text{edge}} = \frac{\mu_0 I_{\text{total}}}{2\pi R} \\= \frac{4\pi \times 10^{-7} \times 200}{2\pi \times 0.005} \\= 8 \times 10^{-3} \, \text{T}[/tex]
The force per unit length [tex]F_{new}[/tex] on a wire at the edge is:
[tex]\frac{F}{L}_{\text{new}} = I_1 \times B_{\text{edge}} \\= 2 \, \text{A} \times 8 \times 10^{-3} \, \text{T} \\= 1.6 \times 10^{-2} \,[/tex]
Thus, the force experienced by a wire at the edge is greater than the force calculated in part (a).
In an old-fashioned amusement park ride, passengers stand inside a 4.9-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.63 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed." What is the minimum angular speed, in rpm, for which the ride is safe?
Answer:
24.07415 rpm
Explanation:
[tex]\mu[/tex] = Coefficient of friction = 0.63
v = Velocity
d = Diameter = 4.9 m
r = Radius = [tex]\frac{d}{2}=\frac{4.9}{2}=2.45\ m[/tex]
m = Mass
g = Acceleration due to gravity = 9.81 m/s²
Here the frictional force balances the rider's weight
[tex]f=\mu F_n[/tex]
The centripetal force balances the weight of the person
[tex]\mu m\frac{v^2}{r}=mg\\\Rightarrow \mu \frac{v^2}{r}=g\\\Rightarrow v=\sqrt{\frac{gr}{\mu}}\\\Rightarrow v=\sqrt{\frac{9.81\times 2.45}{0.63}}\\\Rightarrow v=6.17656\ m/s[/tex]
Velocity is given by
[tex]v=\omega r\\\Rightarrow \omega=\frac{v}{r}\\\Rightarrow \omega=\frac{6.17656}{2.45}\\\Rightarrow \omega=2.52104\ rad/s[/tex]
Converting to rpm
[tex]2.52104\times \frac{60}{2\pi}=24.07415\ rpm[/tex]
The minimum angular speed for which the ride is safe is 24.07415 rpm
The minimum angular speed for which the ride will be safe is ≈ 24.07 rpm
Given data :
Diameter of hollow steel cylinder = 4.9 m. Radius ( r ) = 4.9 / 2 = 2.45 m
coefficient of friction of clothing ( [tex]\alpha[/tex] ) = 0.63
g = 9.81 m/s²
First step : Determine the velocity using the centripetal forces relation
v = [tex]\sqrt{\frac{g*r}{\alpha } }[/tex] ----- ( 1 )
where ; g = 9.81 m/s, r = 2.45 m , [tex]\alpha = 0.63[/tex]
Insert values into equation 1
V = [√( 9.81 * 2.45 )/0.63 ]
= 6.177 m/s
Next : convert velocity to rad/sec ( angular velocity )
V = ω*r
∴ ω = V / r
= 6.177 / 2.45 = 2.52 rad/sec
Final step: The minimum angular speed expressed in rpm
angular velocity ( ω ) * [tex]\frac{60}{2\pi }[/tex]
= 2.52 * [tex]\frac{60}{2\pi }[/tex] ≈ 24.07 rpm
Hence the minimum angular speed in rpm = 24.07 rpm.
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A 6.0 × 103 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 19.3 ◦ with the horizontal. An average frictional force of 4.4×103 N impedes the car’s motion so that the car’s speed at the bottom of the driveway is 4.3 m/s. The acceleration of gravity is 9.81 m/s 2 . What is the length of the driveway? Answer in units of m.
Answer:
s = 3.68 m
Explanation:
given,
mass of the car = 6 x 10³ Kg
angle of slope with horizontal = 19.3°
Average frictional force = 4.4×10³ N
Speed of the car at the bottom = 4.3 m/s
acceleration due to gravity = 9.81 m/s²
length of driveway = ?
first calculating the net force acting on the car
F = mg sin θ - f
(mg sin θ) is the weight component of the body
(f) is the frictional force which is acting opposite to the car
F = 6000 x 9.81 x sin 19.3° - 4400
F = 19454 - 4400
F = 15054 N
we know,
Force = mass x acceleration
15054 = 6000 x a
a = 2.51 m/s²
using formula
v² = u² + 2 as
4.3² = 0² + 2 x 2.51 x s
s = 3.68 m
the length of the driveway = s = 3.68 m
Final answer:
To find the length of the driveway, we can use the equations of motion. Calculate the acceleration using trigonometry, find the net force, and use the equation v^2 = u^2 + 2as to determine the distance.
Explanation:
To find the length of the driveway, we can use the equations of motion. The acceleration of the car can be found by considering the forces acting on it. The force of gravity can be resolved into two components: one along the slope and one perpendicular to the slope. The force along the slope, which is responsible for the acceleration of the car, can be found using trigonometry. The net force acting on the car is the force along the slope minus the frictional force. We can then use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, to find the length of the driveway.
Let's break the problem down step by step:
Calculate the acceleration of the car along the slope using trigonometry: acceleration = g * sin(angle).Find the net force acting on the car: net force = m * acceleration - frictional force.Use the equation v^2 = u^2 + 2as to find the length of the driveway: distance = (v^2 - u^2) / (2 * acceleration).Substituting the given values into the equations will give you the length of the driveway.
A manufacturer claims to have built a home stereo speaker that is 4 feet tall but uses an average of only 30 watts of power. A sample of 40 speakers from the manufacturer found that they used an average of 32 watts of power. The appropriate hypotheses to test the manufacturer’s claim are?
Answer:
Original claim is H_o: M = 30
Alternate claim is H_a: M > 30.
Explanation:
Given data:
4 ft tall speaker consume 30 watt power
a group of 40 speaker consume average power of 32 watt
calculation:
A home installed 4 ft tall speaker that use 30 watt Power
40 speakers used 32 watt Power
Original claim is H_o: M = 30
Alternate claim is H_a: M > 30.
A gymnast is in a tucked position to complete her somersaults. While tucked her moment of inertia about an axis through the center of her body is 16.0 kg-m² and she rotates at 2.5 rev/s. When she kicks out of her tuck into a straight position, her moment of inertia becomes 19.5 kg-m². What is her rate of rotation after she straightens out?
Answer:
[tex]\omega_s=12.8886\ rad.s^{-1}[/tex]
Explanation:
Given that:
moment of inertia of tucked body, [tex]I_t=16\ kg.m^2[/tex]rotational speed of the body, [tex]N_t=2.5\ rev.s^{-1}[/tex]i.e. [tex]\omega_t=2\pi\times 2.5=15.708\ rad.s^{-1}[/tex]moment of inertia of the straightened body, [tex]I_s=19.5\ kg.m^2[/tex]Now using the law of conservation of angular momentum:
angular momentum of tucked body=angular momentum of straight body
[tex]I_t.\omega_t=I_s.\omega_s[/tex]
[tex]16\times 15.708=19.5\times \omega_s[/tex]
[tex]\omega_s=12.8886\ rad.s^{-1}[/tex]
How much work is done by a force vector F = (4x N)i hat +(5 N)j, with x in meters, that moves a particle from a position vector r i = (5 m)i hat +(6 m)j, to a position vector r f = -(2 m)i hat -(5 m)j?
Answer:
-83
Explanation:
A package of mass m is released from rest at a warehouse loading dock and slides down the h = 3.0 m - high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.
a. What is the speed of the package of mass m right before the collision?
b. Suppose the packages stick together. What is their common speed after the collision? Is the total mechanical energy for the system including both packages conserved in this case? If not, what is the difference of the mechanical energies before and after the collision?
c. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound? Is the total linear momentum for the system including both packages conserved before and after this perfectly elastic collision? Why or why not?
To find the speed of the package of mass m right before the collision, we can use the principle of conservation of energy. The common speed of the packages after the collision, if they stick together, can be found using the principle of conservation of momentum. If the collision is perfectly elastic, the rebounding height of the package of mass m can be determined using the principle of conservation of mechanical energy.
Explanation:To answer the given question, we can use the principles of conservation of energy, momentum, and Newton's laws of motion.
To find the speed of the package of mass m right before the collision, we can use the principle of conservation of energy, which states that the initial potential energy of the package is converted into kinetic energy as it slides down the chute. Therefore, we can use the formula for potential energy: mgh = 0.5mv^2, where m is the mass of the package, g is the acceleration due to gravity, h is the height of the chute, and v is the speed of the package. Rearranging the formula, we can solve for v:v = sqrt(2gh)To find the common speed of the packages after the collision, we can use the principles of conservation of momentum. The momentum before the collision is given by the sum of the momentum of the first package (mass m) and the momentum of the second package (mass 2m), both of which are moving in the same direction. After the collision, if the packages stick together, their total mass becomes 3m. Therefore, we can equate the initial momentum with the final momentum to find the common speed:(m)(v) + (2m)(0) = (3m)(v')v' = v/3If the collision between the packages is perfectly elastic, the total linear momentum for the system including both packages is conserved before and after the collision. However, the mechanical energy is not conserved in this case because some of the kinetic energy is converted into potential energy during the collision. To find the height to which the package of mass m rebounds, we can use the principle of conservation of mechanical energy. The initial mechanical energy is given by the sum of the kinetic energy and potential energy of the package before the collision. After the collision, the mechanical energy is given by the sum of the kinetic energy (0.5mv'^2) and potential energy (mgh') of the rebounding package. Equating the initial and final mechanical energies, we can solve for h':0.5mv^2 = 0.5mv'^2 + mgh'h' = (v^2 - v'^2)/(2g)Learn more about Conservation of energy, Conservation of momentum, Mechanical energy here:https://brainly.com/question/31825388
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The package will hit the ground with a speed of 7.67 m/s. If the packages stick together, they will move with a speed of 2.56 m/s. In the case of a perfectly elastic collision, the package of mass m will rebound to the original height.
Explanation:The subject of this question is Physics, specifically it applies the topics of kinetic energy, potential energy, momentum conservation and collisions in two scenarios: inelastic and elastic scenarios.
The speed of the package of mass m right before the collision can be found using energy conservation. The total mechanical energy of the system is conserved because there is no external force acting. This means the total potential energy at the beginning is equal to the total kinetic energy at the bottom of the chute. Mathematically, this can be represented as m*g*h = 0.5*m*v^2, solving for v (velocity), we get v=√(2*g*h) = √(2*9.8*3) ≈ 7.67 m/sIf the packages stick together, the common speed after the collision can be determined using momentum conservation. Before the collision, the momentum is m*v and after the collision, the momentum is (m+2m)*V, where V is the common velocity. By setting these two equal, we get that V=(m*v)/(m+2m) = (7.67)/(3) ≈2.56m/s. The total mechanical energy of the system is not conserved in this inelastic collision. To find the difference, we calculate the kinetic energy before and after the collision and subtract them, ∆E = 0.5*m*v^2 - 0.5*(3m)*V^2 ≈ 18J.If the collision is perfectly elastic, then both momentum and energy are conserved, which means that the package with mass m will rebound to the original height of 3.0 m. The linear momentum of the system is conserved before and after the collision because no external force is applied and according to the law of conservation of momentum, the total linear momentum remains constant.Learn more about Energy and Momentum Conservation here:https://brainly.com/question/35479656
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a. At what frequency would an inductor and a capacitor have the same reactance?
b. What would the reactance be?
c. Show that this frequency would be the natural frequency of an oscillating circuit with the same L and C.
Answer:
(A.) Resonance Frequency
(B.) Totally resistance
Explanation:
A. The resonance frequency is the frequency at which the net reactance of the inductor and capacitor is zero, this is the frequency at which the capacitive reactance of the capacitor and inductive reactance of the inductor cancels each other out. Resistance is low here and the frequency is high
B. Since the net reactance of the inductor and capacitor is zero as explained in (A) above, the total reactance of the circuit will be entirely from the resistor (resistance)
A spring with a force constant of 5.2 N/m has a relaxed length of 2.45 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.57 m. Calculate the elastic potential energy stored in the spring.
Answer:
Elastic potential energy, E = 3.26 J
Explanation:
It is given that,
Force constant of the spring, k = 5.2 N/m
Relaxed length of the spring, X = 2.45 m
When the mass is attached to the end of the spring, the vertical length of the spring is, x' = 3.57 m
To find,
The elastic potential energy stored in the spring.
Solution,
The extension in the length of the spring is given by :
[tex]x=x'-X[/tex]
[tex]x=3.57\ m-2.45\ m[/tex]
x = 1.12 m
The elastic potential energy of the spring is given by :
[tex]E=\dfrac{1}{2}kx^2[/tex]
[tex]E=\dfrac{1}{2}\times 5.2\times (1.12)^2[/tex]
E = 3.26 J
So, the elastic potential energy stored in the spring is 3.26 joules.
The Elastic Potential Energy stored in the spring mass system
= 3.26 Joule
The Elastic Potential Energy Stored (E) in the spring mass system is given by equation (1)
E = (1/2) [tex]\times K \times x^2[/tex]........(1)
Where K is the Force Constant = 5.2 N/m
and [tex]x[/tex] is the extension or compression of the spring.
Here as the spring length increases so [tex]x[/tex] = Final Length - Initial Length
= (3.57- 2.45) = 1.12 m
E = (1/2) [tex]\times[/tex]5.2 [tex]\times[/tex][tex]1.12^2[/tex] = 3.26144 [tex]\approx[/tex] 3.26 Joule
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a 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what is his recoil velocity? PLEASE HELP ME
Answer:
His recoil velocity is 0.269 m/s to the East
Explanation:
This question is easily solved by using the law of conservation of linear momentum.
The formula for the momentum is
[tex]Mo = mv[/tex]
, where m is the mass and v its speed.
The person + the book are at rest which means their momentum is
[tex]Mo_0=0[/tex]
After the book is released, they both start to move and their combined momentum is
[tex]Mo_f=m_pv_p+m_bv_b[/tex]
Where [tex]m_p, v_p[/tex] are the mass and speed of the person respectively and [tex]m_b, v_b[/tex] are the mass and speed of the book
Knowing that
[tex]m_p=74.9 Kg, m_b=2.44 Kg, v_b=-8.25m/s[/tex] (positive speed is assumed to the right or East), and the total momentum is zero:
[tex]m_pv_p+m_bv_b=0[/tex] =>
[tex]v_p=-\frac{m_bv_b}{m_p} =-\frac{2.44 (-8.25)}{74.9}[/tex]
[tex]v_p=0.269 m/s[/tex]
Since the sign of [tex]v_p[/tex] is positive, it's directed to the East
The recoil velocity of the person after throwing the book is approximately 0.27 m/s to the east, which is the opposite direction of the thrown book.
The recoil velocity of a person after throwing an object can be found by using the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum afterwards, assuming no external forces act on the system. In this case, the person and the book together form a closed system with no external forces since they are on an icy pond, which we can consider frictionless.
The initial momentum of the system is zero because the person is at rest. After the person throws the book, the momentum of the book is mass of the book multiplied by velocity of the book. To find the recoil velocity of the person, we use the formula:
m1×u1 + m2×u2 = m1×v1 + m2×v2,
where m1 and m2 are the masses of the person and book, u1 and u2 are their initial velocities (which are zero), and v1 and v2 are their final velocities. The final velocity of the person (v1) is what we are looking for:
(74.9 kg)(0 m/s) + (2.44 kg)(0 m/s) = (74.9 kg)v1 + (2.44 kg)(-8.25 m/s),
v1 = -(2.44 kg)(-8.25 m/s) / 74.9 kg,
v1 = 0.27 m/s to the east (opposite direction to the book).