A 20.0 kg cannonball is fired from a cannon with a muzzle speed of 100 m/s at an angle of 20.0°. Using conservation of energy, the maximum height reached by the cannonball is approximately 510.2 meters.
A cannon ball weighing 20.0 kg is launched from a cannon with an initial velocity of 100 m/s at an angle of 20.0° above the horizontal.
To determine the maximum height reached by the cannonball using the conservation of energy principle, we consider the conversion of kinetic energy into gravitational potential energy.
Initially, the cannonball has only kinetic energy, given by the equation KE = (1/2)mv², where m is the mass and v is the velocity.
At the highest point of its trajectory, the cannonball has no vertical velocity, meaning it has no kinetic energy but possesses gravitational potential energy, given by the equation PE = mgh, where h is the height and g is the acceleration due to gravity (approximately 9.8 m/s²).
Using the conservation of energy, we equate the initial kinetic energy to the maximum potential energy:
(1/2)mv² = mgh
Canceling the mass and rearranging the equation, we find:
v²/2g = h
Plugging in the given values, we have:
(100²)/(2*9.8) = h
Simplifying the equation, we find:
h ≈ 510.2 m
Therefore, the maximum height reached by the cannonball is approximately 510.2 meters.
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a child on a merry-go-round takes 4.4 s to go around once. what is his angular displacement during a 1.0 s time interval?
The child's angular displacement during a 1.0 s time interval is approximately 1.432 radians.
To determine the angular displacement of the child on the merry-go-round during a 1.0 s time interval, we can use the formula:
Angular Displacement (θ) = Angular Velocity (ω) × Time (t)
The angular velocity (ω) can be calculated by dividing the total angular displacement by the total time taken to complete one revolution.
In this case:
Time taken to go around once (T) = 4.4 s
Angular Velocity (ω) = 2π / T
Angular Velocity (ω) = 2π / 4.4 s ≈ 1.432 radians/s
Now, we can calculate the angular displacement during a 1.0 s time interval:
Angular Displacement (θ) = Angular Velocity (ω) × Time (t)
Angular Displacement (θ) = 1.432 radians/s × 1.0 s
Angular Displacement (θ) ≈ 1.432 radians
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The angular displacement of the child during a 1.0 s time interval is 1.44 radian. The given values are, Time taken by the child to go around once, t = 4.4 s Time interval, t₁ = 1 s
Formula used: Angular displacement (θ) = (2π/t) × t₁. Substitute the given values in the formula, Angular displacement (θ) = (2π/t) × t₁= (2π/4.4) × 1= 1.44 radian. Thus, the angular displacement of the child during a 1.0 s time interval is 1.44 radian.
The change in the angular position of an object or a point in a rotational system is known as angular displacement and it measures the amount and direction of rotation from an initial position to a final position. Angular displacement is an important concept in physics and engineering, as it helps to describe a rotational motion.
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what hall voltage (in mv) is produced by a 0.160 t field applied across a 2.60 cm diameter aorta when blood velocity is 59.0 cm/s?
A 0.160 t field applied across a 2.60 cm diameter aorta when blood velocity is 59.0 cm/s will give Hall voltage of 2.3712 mV.
For calculating this, we know that:
VH = B * d * v * RH
In this instance, the blood flow rate is given as 59.0 cm/s, the magnetic field strength is given as 0.160 T, the aorta diameter is given as 2.60 cm (which we will convert to metres, thus d = 0.026 m), and the magnetic field strength is given as 0.160 T.
Let's assume a value of RH = [tex]3.0 * 10^{-10} m^3/C.[/tex]
VH = (0.160 T) * (0.026 m) * (0.59 m/s) * [tex]3.0 * 10^{-10} m^3/C.[/tex]
VH = 0.0023712 V
Or,
VH = 2.3712 mV
Thus, the Hall voltage produced in the aorta is approximately 2.3712 mV.
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A capacitor is discharged through a 20.0 Ω resistor. The discharge current decreases to 22.0% of its initial value in 1.50 ms.
What is the time constant (in ms) of the RC circuit?
a) 0.33 ms
b) 0.67 ms
c) 1.50 ms
d) 3.75 ms
The time constant (in ms) of the RC circuit is 3.75 ms. Hence, the correct option is (d) 3.75 ms.
The rate of decay of the current in a charging capacitor is proportional to the current in the circuit at that time. Therefore, it takes longer for a larger current to decay than for a smaller current to decay in a charging capacitor.A capacitor is discharged through a 20.0 Ω resistor.
The discharge current decreases to 22.0% of its initial value in 1.50 ms. We can obtain the time constant of the RC circuit using the following formula:$$I=I_{o} e^{-t / \tau}$$Where, I = instantaneous current Io = initial current t = time constant R = resistance of the circuit C = capacitance of the circuit
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The time constant of the RC circuit is approximately 0.674 m s.
To determine the time constant (τ) of an RC circuit, we can use the formula:
τ = RC
Given that the discharge current decreases to 22.0% of its initial value in 1.50 m s, we can calculate the time constant as follows:
The percentage of the initial current remaining after time t is given by the equation:
I(t) =[tex]I_oe^{(-t/\tau)[/tex]
Where:
I(t) = current at time t
I₀ = initial current
e = Euler's number (approximately 2.71828)
t = time
τ = time constant
We are given that the discharge current decreases to 22.0% of its initial value. Therefore, we can set up the following equation:
0.22 =[tex]e^{(-1.50/\tau)[/tex]
To solve for τ, we can take the natural logarithm (ln) of both sides:
ln(0.22) = [tex]\frac{-1.50}{\tau}[/tex]
Rearranging the equation to solve for τ:
τ = [tex]\frac{-1.50 }{ ln(0.22)}[/tex]
Calculating this expression:
τ ≈ 0.674 m s
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What value below has 3 significant digits? a) 4.524(5) kev b) 1.48(4) Mev c) 58 counts d) 69.420 lols Q13: What is the correct count-rate limit of precision for an exactly 24 hour live time count with 4.00% dead time, a count rate of 40.89700 counts/second, and a Fano Factor of 0.1390000? a) 40.897(8) counts/sec b) 40.90(12) counts/sec c) 41.0(5) counts/sec d) 41(5) counts/sec e) Infinite Q14: What kind of detectors have the risk of a wall effect? a) Neutron gas detectors b) All gas detectors c) Neutron semiconductor detectors d) Gamma semiconductor detectors e) Geiger-Müller counters
The value below that has 3 significant digits is: c) 58 counts
In this value, the digits "5" and "8" are considered significant, and the trailing zero does not contribute to the significant figures. The value "58" has two significant digits.
Q13: The correct count-rate limit of precision for an exactly 24 hour live time count with 4.00% dead time, a count rate of 40.89700 counts/second, and a Fano Factor of 0.1390000 is:
b) 40.90(12) counts/sec
The value has 4 significant digits, and the uncertainty is indicated by the value in parentheses. The uncertainty is determined by the count rate's precision and the dead time effect.
Q14: The detectors that have the risk of a wall effect are:
c) Neutron semiconductor detectors
d) Gamma semiconductor detectors
The wall effect refers to the phenomenon where radiation interactions occur near the surface of a detector, leading to reduced sensitivity and accuracy. In the case of neutron and gamma semiconductor detectors, their thin semiconductor material can cause a significant portion of radiation interactions to occur close to the detector surface, resulting in the wall effect.
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please fast.
- 14. A 0.400 kg physics cart is moving with a velocity of 0.22 m/s. This cart collides inelastically with a second stationary cart and the two move off together with a velocity of 0.16 m/s. What was
In an inelastic collision, two or more objects stick together and travel as one unit after the collision. The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on the system, which is also true for an inelastic collision.
As a result, the momentum of the first cart is equal to the combined momentum of the two carts after the collision, since the collision is inelastic. The velocity of the two carts after the collision can be calculated using the conservation of momentum, as follows:0.400 kg x 0.22 m/s + 0 kg x 0 m/s = (0.400 kg + 0 kg) x 0.16 m/s0.088 Ns = 0.064 NsThe total momentum of the system is 0.064 Ns.
The two carts move together after the collision with a velocity of 0.16 m/s. The mass of the second cart is 0 kg, therefore, its initial momentum is 0 Ns. The momentum of the first cart is therefore equal to the total momentum of the system.
The initial momentum of the first cart can be calculated using the following formula:p = mv0.088 Ns = 0.400 kg x v Therefore, the initial velocity of the first cart is:v = p/mv = 0.088 Ns / 0.400 kgv = 0.22 m/s Hence, the initial velocity of the first cart is 0.22 m/s.
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Relative to the ground, a car has a velocity of 17.3 m/s, directed due north. Relative to this car, a truck has a velocity of 23.0 m/s, directed 52.0° north of east. What is the magnitude of the truc
The
magnitude
of the truck's velocity
is approximately 22.783 m/s.
To solve this problem, we can break down the velocities into their x and y components.
The
car's velocity
is directed due north, so its
x-component is 0 m/s and its y-component is 17.3 m/s.
The truck's velocity is directed 52.0° north of east. To find its x and y components, we can use trigonometry. Let's define the
angle
measured counterclockwise from the positive x-axis.
The x-component of the truck's velocity can be found using the cosine function:
cos(52.0°) = adjacent / hypotenuse
cos(52.0°) = x-component / 23.0 m/s
Solving for the x-component:
x-component = 23.0 m/s * cos(52.0°)
x-component ≈ 14.832 m/s
The y-component of the truck's velocity can be found using the sine function:
sin(52.0°) = opposite / hypotenuse
sin(52.0°) = y-component / 23.0 m/s
Solving for the y-component:
y-component = 23.0 m/s * sin(52.0°)
y-component ≈ 17.284 m/s
Now, we can find the magnitude of the truck's velocity by using the
Pythagorean theorem
:
magnitude = √(x-component² + y-component²)
magnitude = √((14.832 m/s)² + (17.284 m/s)²)
magnitude ≈ √(220.01 + 298.436)
magnitude ≈ √518.446
magnitude ≈ 22.783 m/s
Therefore, the magnitude of the truck's
velocity
is approximately 22.783 m/s.
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a lens has a refractive power of -1.50. what is its focal length?
It has been determined that the focal length of the lens is -0.6667 m.
Given: The refractive power of a lens is -1.50We are supposed to find the focal length of the given lens
Solution:The formula to find the focal length of a lens is given by:1/f = (n-1) (1/R1 - 1/R2)
Given: Refractive power (P) = -1.50
As we know that, P = 1/f (Where f is the focal length)
Hence, -1.50 = 1/fOr, f = -1/1.5= -0.6667 m
Therefore, the focal length of the given lens is -0.6667 m.
From the above calculations, it has been determined that the focal length of the lens is -0.6667 m.
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what is the best definition of relativistic thought according to perry
Relativistic thought refers to the recognition that our perceptions and beliefs are influenced by our experiences, upbringing, and cultural and social environments, according to Perry.
It suggests that reality is subjectively constructed rather than objectively discovered, and that what is "true" or "right" for one person or group may not be for another. Relativistic thinking entails a degree of tolerance for opposing viewpoints and a willingness to engage in dialogue rather than debate or dismiss opposing perspectives. Instead of seeing things in black and white, relativistic thought acknowledges the nuances and complexity of human experience and acknowledges that there may be multiple valid perspectives on any given issue.
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A 0.200-kg object is attached to a spring that has a force constant of 95.0 N/m. The object is pulled 7.00 cm to the right of equilibrium and released from rest to slide on a horizontal, frictionless table. Calculate the maximum speed Umas of the object. Upis m/y Find the location x of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up. m
The maximum speed of the object is Umas = 1.516 m/s. The location of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up is x = 6.97 cm..
To find the maximum speed of the object, we can use the concept of mechanical energy conservation. At the maximum speed, all the potential energy stored in the spring is converted into kinetic energy.
The potential energy stored in the spring is given by:
Potential energy (PE) = (1/2)kx²
Where:
k = force constant of the spring = 95.0 N/m
x = displacement of the object from equilibrium = 7.00 cm = 0.0700 m (converted to meters)
Substituting the values into the equation:
PE = (1/2)(95.0 N/m)(0.0700 m)²
PE ≈ 0.230 Joules
At the maximum speed, all the potential energy is converted into kinetic energy:
Kinetic energy (KE) = 0.230 Joules
The kinetic energy is given by:
KE = (1/2)mv²
Where:
m = mass of the object = 0.200 kg
v = maximum speed of the object (Umas)
Substituting the values into the equation:
0.230 Joules = (1/2)(0.200 kg)v²
v² = (0.230 Joules) * (2/0.200 kg)
v² = 2.30 Joules/kg
v ≈ 1.516 m/s
Therefore, the maximum speed of the object is Umas ≈ 1.516 m/s.
To find the location of the object relative to equilibrium when it has one-third of the maximum speed, we can use the concept of energy conservation again. At this point, the kinetic energy is one-third of the maximum kinetic energy.
KE = (1/2)mv²
(1/3)KE = (1/6)mv²
Substituting the values into the equation:
(1/3)(0.230 Joules) = (1/6)(0.200 kg)v²
0.077 Joules = (0.0333 kg)v²
v² = 2.311 Joules/kg
v ≈ 1.519 m/s
Now, we need to find the displacement x of the object from equilibrium at this velocity. We can use the formula for the potential energy stored in the spring:
PE = (1/2)kx²
Rearranging the equation:
x² = (2PE) / k
x² = (2 * 0.230 Joules) / 95.0 N/m
x² ≈ 0.004842 m²
x ≈ ±0.0697 m
Since the object is moving to the right, the displacement x will be positive:
x ≈ 0.0697 m
Converting this to centimeters:
x ≈ 6.97 cm
Therefore, the location of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up is x ≈ 6.97 cm.
The maximum speed of the object is Umas ≈ 1.516 m/s. The location of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up is x ≈ 6.97 cm.
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The displacement of a wave traveling in the negative y-direction
is D(y,t)=(9.0cm)sin(45y+70t+π)D(y,t)=(9.0cm)sin(45y+70t+π), where
y is in m and t is in s.
What is the frequency of this wave?
Wh
The displacement of a wave traveling in the negative y-direction depends on the amplitude and frequency of the wave.
The displacement of a wave traveling in the negative y-direction is a combination of factors. The first factor is the amplitude, which is the maximum distance that a particle moves from its rest position as a wave passes through it. The second factor is the frequency, which is the number of waves that pass a fixed point in a given amount of time. The displacement of a wave is given by the formula y = A sin(kx - ωt + ϕ), where A is the amplitude, k is the wave number, x is the position, ω is the angular frequency, t is the time, and ϕ is the phase constant. This formula shows that the displacement depends on the amplitude and frequency of the wave.
These variables have the same fundamental meaning for waves. In any case, it is useful to word the definitions in a more unambiguous manner that applies straightforwardly to waves: Amplitude is the distance between the wave's maximum displacement and its resting position. Frequency is the number of waves that pass by a particular point every second.
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