Answer:
1. downward
2. to the left
3. downward and to the left
Explanation:
This is gotten by using vector law of triangle Addition which states that If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3rd side of that triangle taken in opposite order.
If a current flowing through a lightbulb is 0.75 ampere and the voltage difference across the lightbulb is 120 volts, how much resistance does the light bulb have
Answer: [tex]160 \Omega[/tex]
Explanation:
According to Ohm's law:
[tex]V=R.I[/tex]
Where:
[tex]V=120 V[/tex] is the voltage difference across the light bulb
[tex]R[/tex] is the resistance of the light bulb (the value we want to find)
[tex]I=0.75 A[/tex] is the electric current
Isolating [tex]R[/tex]:
[tex]R=\frac{V}{I}[/tex]
[tex]R=\frac{120 V}{0.75 A}[/tex]
Finally:
[tex]R=160 \Omega[/tex] This is the resistance of the light bulb
Final answer:
To calculate the resistance of a lightbulb with a current of 0.75 amperes and a voltage difference of 120 volts, use Ohm's law. The formula is R = V / I, which yields a resistance of 160 ohms for the lightbulb.
Explanation:
Calculating the Resistance of a Lightbulb
The student has asked how to calculate the resistance of a lightbulb when a current of 0.75 amperes flows through it and the voltage difference across it is 120 volts. To find the resistance, we can use Ohm's law, which states that the resistance (R) of a circuit is equal to the voltage (V) across it divided by the current (I) flowing through it, which can be written as:
R = V / I
Plugging in the given values, we get:
R = 120 V / 0.75 A
R = 160 ohms
Therefore, the resistance of the lightbulb is 160 ohms.
In a clear, coherent, paragraph-length answer, explain why the speed of the apple half way between the release and the maximum height is not half the initial speed. Include a qualitative justification for the speed being greater or less than half the initial speed. You may use no equations to justify your conclusion
Answer:
the speed of the apple half way between the release and the maximum height is not half the initial speed due to the acceleration due to gravity numerically equal to 9.81 m/s2. An acceleration in the direction of motion of a body will cause its velocity to increase, thus, this acceleration due to gravity increases the velocity of the apple so that it will be at its maximum speed before hitting the ground.
The value of this acceleration implies that there is an increase of 9.8 m/s in its speed for every one seconds of fall. This means that at 1 sec when it starts to fall, the speed will be 9.8 m/s. By two seconds it becomes 19.62 m/s. It is clear to see that the velocity halfway will be more than its initial velocity.
A circular swimming pool has a diameter of 18 m. The circular side of the pool is 3 m high, and the depth of the water is 1.5 m. (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m 3 kg/m^3.)How much work (in Joules) is required to: i. pump all of the water over the side? ii. pump all of the water out of an outlet 2m over the side?
To calculate the work required to pump water out of the swimming pool, physics principles such as the formula for work, volume of a cylinder, and weight of water are used to derive the necessary values for the complete calculation.
Explanation:The question requires the use of physics concepts to calculate work done in pumping water out of a swimming pool. Since we are dealing with the force of gravity, density of water, height to which water is raised, and volume of water, we must use the formula for work done (work = force x distance) alongside formulas for the volume of a cylinder (Volume = πr^2h) and the weight of the water (weight = mass x gravity).
To solve the problem, first calculate the volume of water in the pool using the volume formula for a cylinder, taking into account the water's depth. Then, calculate the mass by multiplying the volume by the density of water. The weight of the water is found by multiplying the mass by the acceleration due to gravity. Finally, calculate the work done by multiplying the weight of the water by the height to which it needs to be lifted, considering both scenarios: pumping over the side and pumping through an outlet 2 meters over the side.
Final answer:
Calculation of work required to pump water out of a circular swimming pool at different heights.
Explanation:
The work required to pump all the water over the side of the pool is:
Work = mgh = density x volume x gravity x height = 1000 kg/m³ x (pi x (9m)² x 1.5m) x 9.8 m/s² x 3m = 1222540 J
For pumping all water out of the outlet 2m over the side:
Work = mgh = density x volume x gravity x height = 1000 kg/m³ x (pi x (9m)²x 1.5m) x 9.8 m/s² x 5m = 2037567 J
What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffraction grating has 700 lines per mm?
Answer:
The wavelength is 3500 nm.
Explanation:
d= [tex]\frac{1}{700 lines per mm} = 0.007mm = 7000 nm[/tex]
n= 1
θ= 30°
λ= unknown
Solution:
d sinθ = nλ
λ = [tex]\frac{7000 nm sin 30}{1}[/tex]
λ = 3500 nm
If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself. ะเ50% Part (a) If you had a nic that you would have to use?
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.
a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?
b) What type of electromagnetic radiation would this be?
Given Information:
Wavelength = λ = 1.5×10⁻¹⁰ m
Required Information:
a) Frequency = f = ?
b) Type of electromagnetic radiation = ?
Answer:
a) Frequency = f = 2×10¹⁸ Hz
b) Type of electromagnetic radiation = X-rays
Explanation:
a) The frequency of the electromagnetic radiation is given by
f = c/ λ
Where λ is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s
f = 3×10⁸/1.5×10⁻¹⁰
f = 2×10¹⁸ Hz
Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.
b)
The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz
The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays
To observe atomic-sized features, we would need to use forms of radiation with similar wavelengths to the atom's size, such as X-rays. This is because observable detail with electromagnetic radiation is limited by wavelength; therefore, visible light cannot detect atoms due to their much smaller size.
To observe features that are around the size of atoms with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself. In the area of physics, this concept is a key part of the study of wave optics.
If we imagine atoms to be about 1.5 x 10^-10 m (or 0.1 nm) in size, because wavelength and size must be on similar scales, we'd need to use forms of radiation with similar wavelengths.
This falls within the range of X-rays on the electromagnetic spectrum. Therefore, to observe atomic-sized features, an appropriate method would be X-ray microscopy or other forms of X-ray imaging.
It's important to note that visible light can never detect individual atoms because atoms are much smaller than visible light's wavelength, reinforcing the concept that observable detail is limited by wavelength.
For more such questions on electromagnetic, click on:
https://brainly.com/question/32167648
#SPJ6
A circular loop of wire with area A Bz of B By of B Determine the component Bz B . Determine the component By Bx of B? k . Determine the component Bx j , and k^ i , j^ A , i^ I , A U = - Mu - B is negative. The magnitude of the magnetic field is B0=15D/IA Determine the vector magnetic moment of the current loop. Express your answer in terms of the variables I D is a positive constant, and for this orientation of the loop the magnetic potential energy, where B vector is given by -z -direction toward the origin, a current I is circulating clockwise around the loop. The torque produced by an external magnetic field B? z -axis looking in the xy -plane. As viewed along the z A lies in the xy
Find the given attachment
A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center. The turntable is turned on and 0.8 s later it has an angular speed of 45 rpm. Assume the angular acceleration is constant and determine the following quantities for the ant 0.4 s after the turntable has been turned on. Express all quantities using appropriate mks units.
Complete Question
The complete is shown on the first uploaded image
Answer:
[tex]\alpha = 5.89 rad/s^2[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
[tex]v= 0.212m/s[/tex]
[tex]a_t= 0.5301 m/s[/tex]
[tex]a_r = 0.499 m/s[/tex]
[tex]a = 0.7279 m/s[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
Explanation:
From the question we are told that
mass of the ant is [tex]m_a = 0.8g = \frac{0.8}{1000} = 0.00018kg[/tex]
The distance from the center is [tex]d = 9cm = \frac{9}{100} = 0.09m[/tex]
The angular speed is [tex]w = 45rpm = 45 * \frac{2 \pi }{60} = 1.5 \pi[/tex]
The time taken to attain angular acceleration of 45rpm [tex]t_1 = 0.8s[/tex]
The time taken is [tex]t_2 = 0.4 s[/tex]
The angular acceleration is mathematically represented as
[tex]\alpha = \frac{w}{t}[/tex]
[tex]= \frac{1.5}{0.8}[/tex]
[tex]\alpha = 5.89 rad/s^2[/tex]
The angular velocity at time t= 0.4s is mathematically represented as
[tex]w__{0.4s}} = \alpha * t_2[/tex] Recall angular acceleration is constant
[tex]= 5.89 * 0.4[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
The linear velocity is mathematically represented as
[tex]v = w__{t_2}} * r[/tex]
[tex]= 2.36 * 0.09[/tex]
[tex]v= 0.212m/s[/tex]
The tangential acceleration is mathematically represented as
[tex]a_{t} = \alpha * r[/tex]
[tex]= 5.89 * 0.09[/tex]
[tex]a_t= 0.5301 m/s[/tex]
The radial acceleration is mathematically represented as
[tex]a_r = \frac{v^2}{r}[/tex]
[tex]= \frac{0.212^2}{0.09}[/tex]
[tex]a_r = 0.499 m/s[/tex]
The resultant velocity is mathematically represented as
[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]
[tex]= \sqrt{0.53^2 + 0.499^2}[/tex]
[tex]a = 0.7279 m/s[/tex]
The net force is mathematically represented as
[tex]F_{net} = 0.0008 * 0.7279[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
The left end of a long glass rod 7.00 cmcm in diameter has a convex hemispherical surface 3.50 cmcm in radius. The refractive index of the glass is 1.60. The glass rod is immersed in oil (nn = 1.45). An object placed to the left of the rod on the rod's axis is to be imaged 1.15 mm inside the rod.
How far from the left end of the rod must the object be located to form the image?
Answer:
n1/p + n2/q = (n2-n1)/R
1.45/p + 1.6/1.15 = (1.6 - 1.45)/0.035 (m)
1.45/p + 1.39 = 4.28
p = 0.50 m (50 cm)
A 325-turn circular-loop coil 9.4 cm in diameter is initially aligned so that its axis is parallel to Earth’s magnetic field. In 2.48 ms the coil is flipped so that its axis is perpendicular to Earth’s magnetic field. If an average voltage of 0.17 V is thereby induced in the coil, what is the value of Earth’s magnetic field at that location? Answer in units of µT.
Answer:
[tex]B = 187\ \mu T[/tex]
Explanation:
Given,
Number of turns, N = 325
Diameter of coil, d = 9.4 cm
time,t = 2.48 m s
average voltage,ε = 0.17 V
Earth magnetic field, B = ?
We know that
[tex]\epsilon = \dfrac{NBA}{t}[/tex]
[tex]0.17= \dfrac{325\times B \times \pi (0.047)^2}{2.48\times 10^{-3}}[/tex]
[tex]B = 187 \times 10^{-6}\ T[/tex]
[tex]B = 187\ \mu T[/tex]
Magnetic field of the earth is equal to [tex]B = 187\ \mu T[/tex]
A cylinder with a piston contains 0.300 molmol of oxygen at 2.50×105 PaPa and 350 KK . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.
Part A
Find the work done by the gas during the initial expansion.
--
Part B
Find the heat added to the gas during the initial expansion.
--
Part C
Find internal-energy change of the gas during the initial expansion.
--
Part D
Find the work done during the final cooling;
--
Part E
Find the heat added during the final cooling;
--
Part F
Find the internal-energy change during the final cooling;
--
Part G
Find the internal-energy change during the isothermal compression.
Answer:
Explanation:
find the solution below
The work, heat, and internal energy changes are calculated for different stages of the gas expansion and cooling processes in a cylinder with a piston containing oxygen as an ideal gas.
Explanation:Part A: The work done by the gas during the initial expansion can be calculated using the formula:
Work = Pressure * Change in Volume
Since the expansion is isobaric (constant pressure) and the volume doubles, the change in volume is 2 times the initial volume. Therefore, the work done is 2 times the initial pressure times the initial volume.
Part B: The heat added to the gas during the initial expansion can be calculated using the formula:
Heat = Pressure * Change in Volume, since the expansion is isobaric.
Part C: The internal energy change of the gas during the initial expansion is equal to the heat added minus the work done.
Part D: The work done during the final cooling is zero because the process is isochoric (constant volume) and no work is done.
Part E: The heat added during the final cooling can be calculated using the equation:
Heat = Change in Internal Energy, since no work is done.
Part F: The internal energy change during the final cooling is equal to the negative of the heat added.
Part G: The internal energy change during the isothermal compression is also equal to the negative of the heat added, as no work is done during an isothermal process.
Learn more about work, heat, internal energy here:https://brainly.com/question/34205246
#SPJ3
A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and land at ground level on the other side. He has done this stunt many times and approaches it with confidence. Unfortunately, the motor-cycle engine dies just as he starts up the ramp. He is going 11 m/s at that instant, and the rolling friction of his rubber tires (coefficient 0.02) is not negligible. Does he survive, or does he become croco-dile food? Justify your answer by calculating the distance he travels through the air after leaving the end of the ramp
Answer:
He becomes a croco-dile food
Explanation:
From the question we are told that
The height is h = 2.0 m
The angle is [tex]\theta = 20^o[/tex]
The distance is [tex]w = 10m[/tex]
The speed is [tex]u = 11 m/s[/tex]
The coefficient of static friction is [tex]\mu = 0.02[/tex]
At equilibrium the forces acting on the motorcycle are mathematically represented as
[tex]ma = mgsin \theta + F_f[/tex]
where [tex]F_f[/tex] is the frictional force mathematically represented as
[tex]F_f =\mu F_x =\mu mgcos \theta[/tex]
where [tex]F_x[/tex] is the horizontal component of the force
substituting into the equation
[tex]ma = mgsin \theta + \mu mg cos \theta[/tex]
[tex]ma =mg (sin \theta + \mu cos \theta )[/tex]
making a the subject of the formula
[tex]a = g(sin \theta = \mu cos \theta )[/tex]
substituting values
[tex]a = 9.8 (sin(20) + (0.02 ) cos (20 ))[/tex]
[tex]= 3.54 m/s^2[/tex]
Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as
[tex]sin \theta = \frac{h}{l}[/tex]
making [tex]l[/tex] the subject
[tex]l = \frac{h}{sin \theta }[/tex]
substituting values
[tex]l = \frac{2}{sin (20)}[/tex]
[tex]l = 5.85m[/tex]
Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as
[tex]v^2 =u^2 + 2 (-a)l[/tex]
The negative a means it is moving against gravity
substituting values
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
[tex]v= \sqrt{79.582}[/tex]
[tex]= 8.92m/s[/tex]
The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is
Initial velocity along the x-axis which is mathematically evaluated as
[tex]v_x = vcos 20^o[/tex]
substituting values
[tex]v_x = 8.92 * cos (20)[/tex]
[tex]= 8.38 m/s[/tex]
Initial velocity along the y-axis which is mathematically evaluated as
[tex]v_y = vsin\theta[/tex]
substituting values
[tex]v_y = 8.90 sin (20)[/tex]
[tex]= 3.05 m/s[/tex]
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2} a_y t^2[/tex]
where [tex]y_o[/tex] is the initial height,
[tex]u_y[/tex] is the initial velocity in the y-axis
[tex]a_y[/tex] is the initial acceleration in the y axis with a constant value of ([tex]g = 9.8 m/s^2[/tex])
at the y= 0 which is when the height above ground is zero
Substituting values
[tex]0 = 2 + (3.05)t - 0.5 (9.8)t^2[/tex]
The negative sign is because the acceleration is moving against the motion
[tex]-(4.9)t^2 + (2.79)t + 2m = 0[/tex]
Solving using quadratic formula
[tex]\frac{-b \pm \sqrt{b^2 -4ac} }{2a}[/tex]
substituting values
[tex]\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}[/tex]
[tex]t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}[/tex]
[tex]t = -0.39s \ or \ t = 1.02s[/tex]
since in this case time cannot be negative
[tex]t = 1.02s[/tex]
At this time the position the motorcycle along the x-axis is mathematically evaluated as
[tex]x = u_x t[/tex]
x [tex]=8.38 *1.02[/tex]
[tex]x =8.54m[/tex]
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool
If the horizontal distance x covered by the motorcyclist is greater than or equal to 10 meters, he survives; otherwise, he does not, From the given calculations, x = 8.21 m, so he becomes crocodile food.
According to question, the height of motorcycle daredevil is h = 2.0 m, the angle of the ramp is θ = 20°, the distance of the wide pool is w = 10 m, the speed of the motorcycle daredevil is u = 11 m/s, and the coefficient of static friction between the ramp and the tyres of the motorcycle daredevil is μ = 0.02.
At equilibrium, the forces acting on the motorcycle are equated as:
ma = mgsinθ + f
here, m = mass of the motorcycle, a is the acceleration of the motorcycle, g is the acceelration due to gravity (g = 9.8 m/s²), and f is the frictional force between the ramp and the tyres of the motorcycle daredevil mathematically represented using normal reaction N as
f =μ N = μ mgcosθ
substituting into the equation
ma = mg sinθ + μ mg cosθ
ma =mg (sinθ + μ cosθ)
⇒ a = g(sinθ + μ cosθ)
⇒ a = 9.8 (sin(20°) + (0.02) cos (20°))
⇒ a = 3.54 m/s²
Now, for the ramp, we can say that:
[tex]sin \theta = \frac{h}{l}[/tex]
[tex]\therefore l = \frac{h}{sin \theta}[/tex]
substituting the above values, we get:
[tex]l = \frac{2}{sin20^{\circ}}[/tex]
or, l = 5.85m
Using Newton equation of motion, the final velocity of the motorcycle at the end of the ramp, can be given as:
[tex]v^2 =u^2 + 2 (-a)l[/tex]
As the motorcycle is moving against gravity, the acceleration (a) is taken negative. From the above values, we get:
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
⇒ [tex]v= \sqrt{79.582} \hspace{0.8 mm} m/s[/tex]
or, v = 8.92m/s
The motion of the motorcycle is a projectile motion. The initial velocity at the beginning of the pool (end of ramp) is composed of two component which are the x, and the y components respectively.
Initial velocity along the x-axis which is [tex]v_x[/tex] = v cos 20°
substituting values, we get:
[tex]v_x[/tex] = 8.92 cos (20°)
or, v = 8.38 m/s
Initial velocity along the y-axis which [tex]v_y[/tex] = v sin 20°
substituting values, we get:
[tex]v_y[/tex] = 8.90 sin (20)
[tex]v_y[/tex] = 3.05 m/s
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2}a_y t^2[/tex]
where y₀ is the initial height, [tex]u_y[/tex] is the initial velocity in the y-axis, and [tex]a_y[/tex] is the initial acceleration in the y-axis with a constant value of (g = 9.8 m/s^2)
At y = 0
0 = 2 + (3.05) t - 0.5 (9.8) t²
The negative sign is because the acceleration is moving against the motion:
- (4.9) t² + (2.79) t + 2m = 0
To solve for t, we can use the quadratic formula:
[tex]t = \frac{- b \hspace{0.5} \pm \sqrt{(b^2 - 4ac)}}{2a}[/tex]
In this case, a = 4.9, b = -2.79, and c = -2.
Plugging in these values, we get:
[tex]t = \frac{(2.79 \pm \sqrt{(-2.79)^2 - 4(4.9)(-2)}}{2(4.9)}[/tex]
Solving for both possible values of t, we get:
t ≈ [tex]\frac{(2.79 + 6.858)}{9.8}[/tex] ≈ 0.98 seconds
t ≈ [tex]\frac{(2.79 - 6.858)}{9.8}[/tex] ≈ -0.41 seconds
Since time cannot be negative,
∴ t ≈ 0.98 seconds
Hence displacement along the x-coordinate will be:
x = [tex]u_x[/tex] t
⇒ x =8.38 × 0.98 m
x = 8.21 m
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool.
The Michelson-Morley experiment a) confirmed that time dilation occurs. b) proved that length contraction occurs. c) verified the conservation of momentum in inertial reference frames. d) supported the relationship between mass and energy. e) indicated that the speed of light is the same in all inertial reference frames.
Answer:
e) indicated that the speed of light is the same in all inertial reference frames.
Explanation:
In 18th century, many scientists believed that the light just like air and water needs a medium to travel. They called this medium aether. They believed that even the space is not empty and filled with aether.
Michelson and Morley tried to prove the presence and speed of this aether through an interference experiment in 1887. They made an interferometer in which light was emitted at various angles with respect to the supposed aether. Both along the flow and against the flow to see the difference in the speed of light. But they did not find no major difference and thus it became the first proof to disprove the theory of aether.
It thus proved that the speed of light remains same in all inertial frames.
Also, it became a base for the special theory of relativity by Einstein.
the triceps muscle in the back of the upper arm extends the forearm. this muscle in a professional boxer exerts a force of 2.00×103 n with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 121 rad/s2. what is the moment of inertia of the boxer's forearm?
Answer: 0.512 kgm²
Explanation:
Given
Force, F = 2*10^3 N
Angular acceleration, α = 121 rad/s²
Lever arm, r(⊥) = 3.1 cm = 3.1*10^-2 m
τ = r(⊥) * F
Also,
τ = Iα
Using the first equation, we have
τ = r(⊥) * F
τ = 0.031 * 2*10^3
τ = 62 Nm
Now we calculate for the inertia using the second equation
τ = Iα, making I subject of formula, we have
I = τ / α, on substituting, we have
I = 62 / 121
I = 0.512 kgm²
Thus, the moment of inertia of the boxers forearm is 0.512 kgm²
Albert is piloting his spaceship, heading east with a speed of 0.92 cc relative to Earth. Albert's ship sends a light beam in the forward (eastward) direction, which travels away from his ship at a speed cc. Meanwhile, Isaac is piloting his ship in the westward direction, also at 0.92 cc, toward Albert's ship
With what speed does Isaac see Albert's light beam pass his ship?
Answer:
He sees the light as 1c
Explanation:
According to relativity, the speed of light is the same in all inertial frame of reference.
If we were to add the velocities as applicable to a normal moving bodies, the relative speed of the light beam will exceed c which will break relativistic law since nothing can go past the speed of light.
You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor.What frequency is required?
Answer:
Frequency required will be 2421.127 kHz
Explanation:
We have given inductance [tex]L=0.450H=0.45\times 10^{-3}H[/tex]
Current in the inductor [tex]i=1.90mA=1.90\times 10^{-3}A[/tex]
Voltage v = 13 volt
Inductive reactance of the circuit [tex]X_l=\frac{v}{i}[/tex]
[tex]X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm[/tex]
We know that
[tex]X_l=\omega L=2\pi fL[/tex]
[tex]2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10[/tex]
f = 2421.127 kHz
A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the body is pulled 1 m to the left, stretching the spring, and set in motion with an initial velocity of 20 m/s to the left. (a) Find x(t) in the form Upper C cosine (omega 0 t minus alpha ). (b) Find the amplitude and the period of motion of the body.
The problem refers to an object undergoing simple harmonic motion, initially displaced and set in motion. The displacement and the period of motion can be calculated using information of spring constant, mass, initial velocity, and principles of oscillations.
Explanation:The question refers to a classic setup for simple harmonic motion (SHM), where a body of known mass is attached to a spring with a known force constant. The force constant is the ratio of the force applied to the spring and the resulting extension; we can calculate it as k = F/x = 9N/0.2m = 45 N/m. This accounts for the spring's stiffness.
As SHM is in effect here, the displacement of the body from the equilibrium position can be represented by the equation x(t) = A cos(ωt + α). In this case, the body is initially displaced 1m to the left (A = 1m) and given an initial velocity to the left (which determines α).
Angular frequency ω can be determined by the formula √(k/m), where k is the spring constant and m is the mass of the body (noting to convert grams to kilograms for SI units). Hence, ω = √(45 N/m / 0.2 kg) = 15 rad/s.
On the other hand, initial velocity is related to α by the equation v = -Aωsinα, thus α can be calculated as such. Here, it is worth mentioning that the amplitude is the maximum displacement (1m) and the period of motion, given by T = 2π/ω, refers to the time taken for one complete oscillation.
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
#SPJ3
3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed and acceleration of her swing. She has a device that measures the linear velocity of the bat during the swing. Her bat is 0.85 m long. On her first swing, the linear velocity of the bat just prior to the swing is 0 m/s, and 0.15 s later, just prior to ball contact is 16 m/s. What is the average angular acceleration of the end of her bat during the swing?
Answer:
The average angular acceleration is [tex]\alpha =125.487 rad /s^2[/tex]
Explanation:
From the question we are told that
From the question we are told that
The length of the bat is [tex]l = 0.85m[/tex] \
The initial linear velocity is [tex]u = 0 m/s[/tex]
The time is [tex]t = 0.15s[/tex]
The velocity at t is [tex]v = 16 m/s[/tex]
Generally average angular acceleration is mathematically represented as
[tex]\alpha = \frac{w_f - w_o}{t}[/tex]
Where [tex]w_f[/tex] is the finial angular velocity which is mathematically evaluated as
[tex]w_f = \frac{v}{l}[/tex]
[tex]w_f = \frac{16}{0.85}[/tex]
[tex]= 18.823 rad/s[/tex]
and [tex]w_o[/tex] is the initial angular velocity which is zero since initial linear velocity is zero
So
[tex]\alpha = \frac{18.823 - 0}{0.15}[/tex]
[tex]\alpha =125.487 rad /s^2[/tex]
Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their mass ratios were 1:4 and their respective speeds as they approached the intersection were both 13 m/s, find the magnitude and direction of the final velocity of the wreck.
Answer with Explanation:
Let mass of one vehicle =m
Mass of other vehicle=m'
[tex]\frac{m}{m'}=\frac{1}{4}[/tex]
[tex]m'=4m[/tex]
Velocity of one vehicle=[tex]v=13 im/s[/tex]
Velocity of other vehicle=[tex]v'=13jm/s[/tex]
In x- direction
By law of conservation of momentum
[tex]mv+0=(m+m')V_x[/tex]
[tex]13m=(m+4m)V_x[/tex]
[tex]13m=5mV_x[/tex]
[tex]V_x=\frac{13}{5}[/tex]
In y- direction
By law of conservation of momentum
[tex]0+m'v'=(m+m')V_y[/tex]
[tex]4m(13)=5mV_y[/tex]
[tex]V_y=\frac{52m}{5m}=\frac{52}{5}[/tex]
Magnitude of velocity of the wreck,V=[tex]\sqrt{V^2_x+V^2_y}=\sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}=10.72 m/s[/tex]
Direction:[tex]\theta=tan^{-1}(\frac{V_y}{V_x})[/tex]
[tex]\theta=tan^{-1}(\frac{\frac{52}{5}}{\frac{13}{5}})=75.96^{\circ}[/tex]
A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 48 feet across at its opening and 4 feet deep at its center, where should the receiver be placed
The receiver of the satellite dish, which is shaped like a paraboloid, should be positioned 9 feet from the base along the axis of symmetry for optimal signal reception.
Explanation:This question is related to the properties of a parabolic dish (or paraboloid). The satellite dish is a 3D shape formed by rotating a parabola along its axis of symmetry. For a parabola, the focus is located a certain distance from the vertex, and this distance is defined by the depth of the dish.
The formula for the focus (F) in a parabolic mirror/dish is given by: F = D^2/16d where D is the diameter of the dish (48ft in this case) and d is the depth of the dish (4ft in this case). By substitifying these values, we obtain F = 48^2/(16*4) = 144/16 = 9 feet.
Therefore, to ensure the strongest reception of signals, the receiver of the satellite dish should be positioned 9 feet from the base along the axis of symmetry.
Learn more about Parabolic Dish here:https://brainly.com/question/32342363
#SPJ12
Three nested coffee filters fall, quickly reaching a terminal speed of ~ 1.7 m/s. If the air resistance force on a single falling coffee filter was F1, how large is the air resistance force on the three falling filters? 1) F1 2) 3*F1 3) (1/3)*F1 4) not enough information
Final answer:
The air resistance force on three nested coffee filters falling is 3*F1.
Explanation:
The air resistance force on three nested coffee filters falling is 3*F1.
When the coffee filters fall, they quickly reach a terminal speed of approximately 1.7 m/s. Terminal velocity occurs when the air resistance force equals the weight of the object. Since the filters are nested, the air resistance force on a single falling filter is F1. Therefore, assuming the air resistance force is the same for each filter, the air resistance force on the three falling filters is 3*F1.
A physical pendulum consists of a vertical board of of mass 6.12 kg, length 191 cm, and width 12 cm hanging from a horizontal, frictionless axle. A bullet of mass 143 g and a purely horizontal speed v impacts the pendulum at the bottom edge of the board. The board then makes a complete circle.
(a) If the bullet passed through the board, reducing its speed by a factor of 1/3, what is the minimum speed of the bullet?
(b) If the bullet embedded itself in the board, what is the minimum speed of the bullet?
Note (hint from instructor): It appears you may use Conservation of Energy to find angular speed. For potential energy: use center of mass. For kinetic energy, use impulse. Then, to find initial speed, use conservation of angular momentum.
This question involves applying the principles of Conservation of Energy and Angular Momentum to calculate the minimum speed of a bullet impacting a physical pendulum. The solution requires us to consider different scenarios: when the bullet passes through and when it embeds itself into the board.
Explanation:This problem involves the concepts of Conservation of Energy and Conservation of Angular Momentum.
(a) If the bullet passes through the board, we consider that the board moves by gaining kinetic energy (KE) from the bullet’s impact. The bullet's final speed is (1/3)*v, so the KE lost by the bullet will be (4/9)*mv². This energy is used for the board to perform a complete circle, which means it acquires a potential energy of (2*M*g*H), where H is the height when the board makes a complete circle (H=L/2). Equating these energies (4/9)*mv²-2*M*g*H = 0, we solve for v.
(b) If the bullet embeds itself, the system's final angular momentum equals its initial, which gives (1/2*M*L)v = (M+m)Lω. Then we equate the initial kinetic energy and potential energy during the complete circle, getting (1/2)(M+m)L²ω² = (M+m)gL. Solving for v will give us the minimum speed.
Learn more about Physics of Angular Momentum here:https://brainly.com/question/29716949
#SPJ12
PLEASE HELP! It’s urgent... and please show your work!!
1. Calculate the following, and express the answer in scientific notation with
the correct number of significant figures:
(0.82 +0.042)(4.4 x 10°)
a) 3.8 x 10
b) 3.78 x 10
c) 3.784 x 103
d) 3784
The correct result of the given operation expressed in scientific notation and with the appropriate number of significant figures should be 3.4 x 10^3. However, there are no matching options provided. A possible mistake might exist in the question.
Explanation:To solve the problem, you should first add the two numbers in the brackets together. This gives us 0.82 + 0.042 = 0.862. Next, we need to multiply this result by the number outside the brackets, which is 4.4 x 103. Thus, our equation becomes 0.862 x 4.4 x 103 = 3.3928 x 103.
The result needs to be expressed in scientific notation and with the correct number of significant figures. The least precise number in our calculation is 4.4 (which has two significant figures), so our final answer should also have two significant figures. Therefore, we round 3.3928 x 103 to 3.4 x 103.
However, none of the options provided matches this result. Please double-check the question as there might be a typo.
Learn more about Scientific Notation here:https://brainly.com/question/2005529
#SPJ2
First, add 0.82 + 0.042 = 0.862. Second, multiply the sum 0.862 x 4.4 = 3.7928. Finally, round the result to two significant figures in scientific notation giving the answer as 3.8 x 10 or just 3.8.
Explanation:To solve this problem, first, we need to add the numbers in the parentheses, then multiply the sum by 4.4 x 10⁰, which is just 4.4 since any number raised to the power of zero equals one.
Step 1: Add 0.82 + 0.042 = 0.862
Step 2: Multiply 0.862 x 4.4 = 3.7928
Lastly, we need to express the answer in scientific notation with the correct number of significant figures. As the question gives the numbers 0.82 and 0.042 with two and three significant figures respectively, we should express our final answer to two significant figures, as you always round to the fewest number of significant figures.
So, round 3.7928 to two significant figures, gives you 3.8 x 10⁰, which can alternatively be written as just 3.8, making the correct answer (a) 3.8 x 10.
Learn more about Scientific Notation here:https://brainly.com/question/2005529
#SPJ2
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration the gravitational pull of the moon?
Answer:
The magnitude of the acceleration of earth due to the gravitational pull of earth is a = Gm/r^2
Where r = the center to center distance between the earth and the moon,
m = mass of the moon, and,
G is the gravity constant.
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
[tex]a_e = \frac{Gm}{r^2}[/tex]
Explanation:
We assume that:
M to represent the mass of the earth
m to equally represent the mass of the moon
r should be the distance between the center of the earth to the center of the moon.
Then;
the expression for the gravitational force can be written as:
[tex]F = \frac{GMm}{r^2}[/tex]
Where [tex]a_e[/tex] is the acceleration produced by the earth; then:
[tex]F =M *a_e[/tex]
Then:
[tex]M*a_e = \frac{GMm}{r^2}[/tex]
[tex]a_e = \frac{GMm}{Mr^2}[/tex]
[tex]a_e = \frac{Gm}{r^2}[/tex]
Therefore, the magnitude of the acceleration of the earth due to the gravitational pull of the moon [tex]a_e = \frac{Gm}{r^2}[/tex]
A ball is thrown upward from a height of 256 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Find the attachments for complete solution
The function giving the height of the ball at time t is s(t) = 96t - 16t^2 + 256. The ball takes 6 seconds to reach the ground and achieves a maximum height of 400 feet.
Explanation:The physics problem you've posed can be approached by first understanding that the height of the ball (s(t)) at time t can be found by integrating the velocity function, due to v(t) being the derivative of s(t). So, integrating v(t)=96-32t with respect to t gives s(t) = 96t - 16t^2 + 256, where 256 is the constant of integration corresponding to the initial height from the ground.
To find out how long will the ball take to reach the ground, we solve the function s(t)=0, yielding t = 6 seconds as the answer, as that's when the ball hits the ground. Now, for the maximum height reached by the ball, it's where the ball has a velocity of zero before it begins its descent, i.e. v(t)=0. Solving this gives t=3 seconds, substituting this back into the s(t) equation gives the maximum reached height as 400 feet.
Learn more about Projectile Motion here:https://brainly.com/question/29545516
#SPJ11
g If the primary coil of wire on a transformer is kept the same and the number of turns of wire on the secondary is increased, how will this affect the voltage observed at the secondary? There is no voltage measured on the secondary coil since it is only connected to a resistor and not connected to a battery. The voltage will stay the same. The voltage will increase. The voltage will decrease.
Answer:
The voltage will increase.
Explanation:
Increason the number of coil in the secondary will increase the voltage in a transformer.
A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is connected to a battery producing a current of 4.0 A in the wire. (a) What is the magnitude of the magnetic dipole mo- ment of this device? (b) At what axial distance d will the mag- netic field have the magnitude 5.0 mT (approximately one-tenth that of Earth’s magnetic field)?
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
When 450-nm light is incident normally on a certain double-slit system the number of interference maxima within the central diffraction maximum is 5. When 900-nm light is incident on the same slit system the number is: a. 9b. 2c. 3d. 10e. 5
Answer:
Explanation:
width of central diffraction maxima = 2 λD / d₁
λ is wave length of light , D is screen distance and d₁ is slit width
width of each interference fringe = λD / d₂ , d₂ is slit separation.
No of interference fringe in central diffraction fringe
= width of central diffraction maxima / width of each interference fringe
= 2 λD / d₁ x λDd₂ / λD
No = 2 d₂ / d₁
No = 5
5 = 2 d₂ / d₁
Since this number does not depend upon wavelength so it will remain the same
No of required fringe will be 5 .
right option
e ) 5.
When 900-nm light is incident on the same slit system the number is 5.
The given parameters:
First wavelength of the incident light, λ₁ = 450 nmThe number of interference maxima, n = 5Second wavelength, λ₂ = 900 nmThe number of interference fringe in central diffraction fringe is calculated as follows;
[tex]n = \frac{width \ of \ central \ diffraction\ maxima}{ width \ of \ each \ interference\ fringe}\\\\n = \frac{2\pi \lambda D/d_1}{2\pi \lambda D/d_2} \\\\n = \frac{2d_2}{d_1}[/tex]
The number of number of interference fringe is independent of the wavelength.
Thus, when 900-nm light is incident on the same slit system the number is 5.
Learn more about number of interference fringe here: https://brainly.com/question/4449144
A strip of copper 130 µm thick and 4.40 mm wide is placed in a uniform magnetic field of magnitude B = 0.79 T, that is perpendicular to the strip. A current i = 26 A is then sent through the strip such that a Hall potential difference V appears across the width. Calculate V. (The number of charge carriers per unit volume for copper is 8.47 × 1028 electrons/m3.)
Answer:
V = 1.1658 × [tex]10^{-5}[/tex] V
Explanation:
given data
strip of copper thick = 130 µm
strip of copper wide = 4.40 mm
uniform magnetic field of magnitude B = 0.79 T
current i = 26 A
number of charge carriers per unit volume = 8.47 × [tex]10^{28}[/tex] electrons/m³
solution
we know that number density is express as
n = \frac{Bi}{Vle} ...............1
B is uniform magnetic field and i is current and V is hall potential difference and l is thickness and e is electron charge 1.6 × [tex]10^{-19}[/tex] C
so V will be as
V = \frac{iB}{nle} .....................2
so put here value and we get V
V = [tex]\frac{26 \times 0.79}{8.47\times 10^{28}\times 130\times10^{-6}\times1.6 \times10^{-19}}[/tex]
V = 1.1658 × [tex]10^{-5}[/tex] V
Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
Answer:
check the diagram in the attachment below.
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
Explanation:
question
Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
ANSWER;
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
Answer:
the potential difference across the membrane of this single segment is zero.
ΔVmem=0
Explanation:
When the switch is closed, the capacitor ([tex]C_{mem}\\[/tex]) will be short. It means the current will not flow through the resistor([tex]R_{mem[/tex]). Rather current will only flow through the capacitor as there is no resistance. Due to zero resistance, the voltage across the capacitor will also be zero. So, the potential difference across the membrane of this single segment is zero.
ΔVmem=0
An electron, moving south, enters a uniform magnetic field. Because of this uniform magnetic field, the electron curves upward. We can conclude that the magnetic field must have a component A. upward; B. downward; C. toward the north; D. toward the east; E. toward the south; F. toward the west.
Answer:
Towards the west
Explanation:
Magnetic force is the interaction between a moving charged particle and a magnetic field.
Magnetic force is given as
F = q (V × B)
Where F is the magnetic force
q is the charge
V is the velocity
B is the magnetic field
V×B means the cross product of the velocity and the magnetic field
NOTE:
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
So, if the electron is moving southward, then, it implies that the velocity of it motion is southward, so the electron is in the positive z-direction
Also, the electron is curved upward due to the magnetic field, this implies that the force field is directed up in the positive y direction.
Then,
V = V•k
F = F•j
Then, apply the theorem
F •j = q ( V•k × B•x)
Let x be the unknown
From vector k×i =j.
This shows that x = i
Then, the magnetic field point in the direction of positive x axis, which is towards the west
You can as well use the Fleming right hand rule
The thumb represent force
The index finger represent velocity
The middle finger represent field