Answer:
1.1 sec
Explanation:
m = mass of the box = 8 kg
k = spring constant of the spring = 69 N/m
v = initial speed of the box = 1.5 m/s
t = time period of oscillation of box in contact with the spring
Time period is given as
[tex]t = \pi \sqrt{\frac{m}{k}}[/tex]
Inserting the values
[tex]t = (3.14) \sqrt{\frac{8}{69}}[/tex]
t = 1.1 sec
Starting from rest, a 1.8 x 10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of 0.42 N on it. This force does +2.7 x 10^-4 J of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.
Answer:
a)
1.73 m/s
b)
6.43 x 10⁻⁴ m
Explanation:
m = mass of the flea = 1.8 x 10⁻⁴ kg
v₀ = initial speed of the flea = 0 m/s
v = final speed of the flea
W = work done by the force on the flea = 2.7 x 10⁻⁴ J
Using work-change in kinetic energy, Work done is given as
W = (0.5) m (v² - v₀²)
Inserting the values
2.7 x 10⁻⁴ = (0.5) (1.8 x 10⁻⁴) (v² - 0²)
v = 1.73 m/s
b)
d = distance moved by the flea while pushing off
F = Upward force applied on the flea by ground = 0.42 N
Work done is also given as
W = F d
2.7 x 10⁻⁴ = (0.42) d
d = 6.43 x 10⁻⁴ m
The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many football fields (length=91.4 m) does the shuttle cover in the blink of an eye? The three-toed sloth is the slowest-moving land mammal. On the ground, the sloth moves at an average speed of 0.033 m/s, considerably slower than the giant tortoise have gone relative to the sloth? Do not put units in your answer.
Answer:
It covers distance of 9.15 football fields in the said time.
Explanation:
We know that
[tex]Distance=Speed\times Time[/tex]
Thus distance covered in blinking of eye =
[tex]Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters[/tex]
Thus no of football fields=[tex]\frac{936}{91.4}=9.15Fields[/tex]
A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle of 60° with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, what is the magnitude of the emf induced in the coil?
Explanation:
It is given that,
Length of side of a square, l = 24 cm = 0.24 m
The uniform magnetic field makes an angle of 60° with the plane of the coil.
The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :
[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]
[tex]\dfrac{d\phi}{dt}[/tex] is the rate of change if magnetic flux.
[tex]\phi=BA\ cos\theta[/tex]
[tex]\theta[/tex] is the angle between the magnetic field and the normal to area vector.
[tex]\theta=90-60=30[/tex]
[tex]\epsilon=NA\dfrac{dB}{dt}\times cos30[/tex]
[tex]\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)[/tex]
[tex]\epsilon=0.0598\ T[/tex]
[tex]\epsilon=59.8\ mT[/tex]
or
EMF = 60 mT
So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.
Relative to the ground, a car has a velocity of 15.3 m/s, directed due north. Relative to this car, a truck has a velocity of 22.5 m/s, directed 52.0° north of east. What is the magnitude of the truck's velocity relative to the ground
Answer:
The magnitude of the truck's velocity relative to the ground is 35.82 m/s.
Explanation:
Given that,
Velocity of car relative to ground = 15.3 m/s
Velocity of truck relative to car = 22.5 m/s
We need to calculate the magnitude of the truck's velocity relative to the ground
We need to calculate the x component of the velocity
[tex]v_{x}=22.5\cos\theta[/tex]
[tex]v_{x}=22.5\cos52^{\circ}[/tex]
[tex]v_{x}=13.852\ m/s[/tex]
We need to calculate the y component of the velocity
[tex]v_{y}=15.3+22.5\sin\theta[/tex]
[tex]v_{y}=15.3+22.5\sin52^{\circ}[/tex]
[tex]v_{y}=33.030\ m/s[/tex]
Using Pythagorean theorem
[tex]|v|=\sqrt{v_{x}^2+v_{y}^2}[/tex]
[tex]|v|=\sqrt{(13.852)^2+(33.030)^2}[/tex]
[tex]|v|=35.82\ m/s[/tex]
Hence, The magnitude of the truck's velocity relative to the ground is 35.82 m/s.
An electron is released from rest in a uniform electric field. The electron accelerates, travelling 6.20 m in 4.50 µs after it is released. What is the magnitude of the electric field in N/C?
Answer:
E= 3.4893 N/C
Explanation:
Given s=6.20 m , t=2.50μs, m=9.11*10^-31 Kg , q= 1.6*10^-19 C
the distance traveled by the electron in time t is
s=ut+0.5at^2
here, u is the initial velocity of the electron, t is time taken and
a is acceleration.
Since the electron is initially at rest u=0
now s=0.5at^2
Therefore a=2s/t^2
also. we know that strength of electric field is
E=ma/q
[tex]E= \frac{2ma}{qt^2}[/tex]
now puting the values we get
[tex]E=\frac{9.11\times 10^-31\times 2\times 6.20}{1.6\times 10^-19\times (4.5\times 10^-6)^2}[/tex]
therefore, E= 3.4865 N/C
The magnitude of the electric field is calculated by first determining the acceleration of the electron and then using the electric force equation to find the electric field. The resulting electric field is 34.8 N/C.
To find the magnitude of the electric field, we first need to calculate the acceleration of the electron. Given that the electron travels a distance of 6.20 m in a time of 4.50 µs (4.50 × 10-6 s), we can use the equations of motion.
Initial velocity, u = 0 (since the electron is released from rest)
Time, t = 4.50 × 10-6 s
Distance, s = 6.20 m
Using the equation of motion: s = ut + 0.5at2
Substitute the values: 6.20 = 0 + 0.5a(4.50 × 10-6)2
6.20 = 0.5a(20.25 × 10-12)
a = 6.20 / (0.5 × 20.25 × 10-12)
a = 6.20 / (10.125 × 10-12)
a = 6.12 × 1011 m/s2
Now, we calculate the electric field using Newton's Second Law, F = ma, and the electric force equation, F = eE, where e is the charge of the electron (1.60 × 10-19 C) and E is the electric field.
ma = eE
(9.11 × 10-31 kg)(6.12 × 1011 m/s2) = (1.60 × 10-19 C)E
(5.57 × 10-19 N) = (1.60 × 10-19 C)E
E = 5.57 × 10-19 N / 1.60 × 10-19 C
E = 3.48 × 101 N/C
Thus, the magnitude of the electric field is 34.8 N/C.
An eagle is flying horizontally at a speed of 3m/s when the fish in her talons wiggles loose and falls into the lake 10m below. Calculate the velocity of the fish relative to the water when it hits the water.
Answer:
Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.
Explanation:
Vertical motion of fish:
Initial speed, u = 0
Acceleration, a = 9.81 m/s²
Displacement, s = 10 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 10 = 196.2
v = 14 m/s
Final vertical speed = 14 m/s
Final horizontal speed = initial horizontal speed = 3 m/s
Final velocity = 3 i - 14 j m/s
Magnitude
[tex]v=\sqrt{3^2+(-14)^2}=14.32m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{-14}{3}\right )=-77.91^0[/tex]
Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.
Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Answer:
[tex]strain = 1.4 \times 10^{-3} [/tex]
Explanation:
As we know by the formula of elasticity that
[tex]E = \frac{stress}{strain}[/tex]
now we have
[tex]E = 110 GPA[/tex]
[tex]E = 110 \times 10^9 Pa[/tex]
Area = 15.2 mm x 19.1 mm
[tex]A = 290.3 \times 10^{-6}[/tex]
now we also know that force is given as
[tex]F = 44500 N[/tex]
here we have
stress = Force / Area
[tex]stress = \frac{44500}{290.3 \times 10^{-6}}[/tex]
[tex]stress = 1.53 \times 10^8 N/m^2[/tex]
now from above formula we have
[tex]strain = \frac{stress}{E}[/tex]
[tex]strain = \frac{1.53 \times 10^8}{110 \times 10^9}[/tex]
[tex]strain = 1.4 \times 10^{-3} [/tex]
The maximum wavelength for photoelectric emission in tungsten is 230 nm. What wavelength of light must be used inorder for electrons with a maximum energy of 1.5 eV to be ejected ?
Answer:
180.04 nm
Explanation:
λ₀ = maximum wavelength for photoelectric emission in tungsten = 230 x 10⁻⁹ m
E₀ = maximum energy of ejected electron = 1.5 eV = 1.5 x 1.6 x 10⁻¹⁹ J
λ = wavelength of light used = ?
Using conservation of energy
Energy of the light used = Maximum energy required for photoelectric emission + Energy of ejected electron
[tex]\frac{hc}{\lambda }=\frac{hc}{\lambda_{o} } + E_{o}[/tex]
[tex]\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{\lambda }=\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{230 \times 10^{-9} } + 1.5 \times 1.6 \times 10^{-19}[/tex]
λ = 180.04 x 10⁻⁹ m
λ = 180.04 nm
The world energy consumption was about 6*10^22 J. How much area must a parallel plate capacitor need to store this energy Assume we maintain the capacitor at delta V= 5 volts for safety reasons, and have a plate separation distance of 1 meter. 5 *10^32 m^2 5 *10^61 m^2 9.10^8 m^2 4*10^14 m^2 2 *10^23 m^2
Answer:
[tex]A = 5 \times 10^{32} m^2[/tex]
Explanation:
As we know that the energy stored in the capacitor is given as
[tex]Q = \frac{1}{2}CV^2[/tex]
here we know that
[tex]Q = 6 \times 10^{22} J[/tex]
also we know that
[tex]V = 5 Volts[/tex]
now we have
[tex]6 \times 10^{22} = \frac{1}{2}C(5^2)[/tex]
[tex]C = 4.8 \times 10^{21} F[/tex]
now we know the formula of capacitance
[tex]C = \frac{\epsilon_0 A}{d}[/tex]
[tex]4.8 \times 10^{21} = \frac{(8.85 \times 10^{-12})(A)}{1}[/tex]
[tex]A = 5 \times 10^{32} m^2[/tex]
Can the resistors in an "unbalanced" Wheatstone bridge circuit be treated as a combination of series and/or parallel resistors? What about a "balanced bridge?
Answer:
Explanation:
The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.
In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.
Here, In the balanced wheat stone bridge
R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.
Resistors in a Wheatstone bridge can be treated as combinations of series and/or parallel resistors for simplification in both balanced and unbalanced bridges. A balanced bridge allows separate treatment of two parallel branches, while unbalanced requires careful analysis. Not all resistor networks can be simplified into series or parallel models.
Explanation:In a Wheatstone bridge, the resistors can indeed be treated as combinations of series and/or parallel resistors when aiming to simplify calculations or understand the behavior of the circuit. For an unbalanced bridge, resistors are not in simple series or parallel arrangements with respect to the entire circuit due to the bridge not being in equilibrium. However, within certain parts of the bridge, resistors may appear to be in series or parallel with each other. In a balanced bridge, where the bridge is in a state of equilibrium and the central voltmeter reads zero, the two arms of the bridge can be treated separately as two parallel voltage dividers, because no current flows through the meter, effectively decoupling the two parallel branches.
Attempting to simplify a complex resistor network encountered in bridges can indeed be done by identifying and replacing series and parallel resistor combinations step by step until a single resistance value is found. However, this approach cannot always be applied to any arbitrary combination of resistors. Some configurations might contain elements that cannot be reduced to mere parallel or series connections, usually because they form more intricate networks, such as bridges or loops not separable into simpler series or parallel sections.
In conclusion, while more complex connections of resistors in circuits like the Wheatstone bridge can often be broken down into combinations of series and parallel, this is not universally the case for all resistor networks. In certain scenarios, specific techniques or theorems such as Kirchhoff's laws might be required to analyze the circuit effectively.
The strength of the Earth’s magnetic field B at the equator is approximately equal to 5 × 10−5 T. The force on a charge q moving in a direction perpendicular to a magnetic field is given by F = q v B, where v is the speed of the particle. The direction of the force is given by the right-hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 4 × 10−9 C. If you are at the equator and driving west at a speed of 60 m/s, what is the strength of the magnetic force on your head due to the Earth’s magnetic field? Answer in units of N
Answer:
1.2 x 10⁻¹¹ N
Explanation:
B = magnitude of magnetic field at the equator = 5 x 10⁻⁵ T
q = magnitude of charge acquired by the head = 4 x 10⁻⁹ C
v = speed of driving at the equator = 60 m/s
Magnitude of magnetic force on the head at the equator is given as
F = q v B
Inserting the above values in the equation
F = (4 x 10⁻⁹) (60) (5 x 10⁻⁵)
F = 1.2 x 10⁻¹¹ N
The activation energy for the uncatalyzed decomposition of hydrogen peroxide at 20°C is 75.3 kJ/mol. In the presence of the enzyme catalase, the activation energy is reduced to 29.3 kJ/mol. Use the following form of the Arrhenius equation, RT ln1k1/k22 5 Ea2 2 Ea1 , to calculate how much larger the rate constant of the catalyzed reaction is.
Answer:
The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex] times.
Explanation:
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
[tex]\log K=\log A-\frac{Ea}{2.303\times RT}[/tex]
The expression used with catalyst and without catalyst is,
[tex]\log K_1=\log A-\frac{Ea_1}{2.303\times RT}[/tex]...(1)
[tex]\log K_2=\log A-\frac{Ea_2}{2.303\times RT}[/tex]...(2)
On subtracting (2) from (1)
[tex]\log \frac{K_2}{K_1}=\frac{Ea_1-Ea_2}{2.303RT}[/tex]
where,
[tex]K_2[/tex] = rate of reaction with catalyst
[tex]K_1[/tex] = rate of reaction without catalyst
[tex]Ea_2[/tex] = activation energy with catalyst = 29.3 kJ/mol = 29300 J/mol
[tex]Ea_1[/tex] = activation energy without catalyst = 75.3 kJ/mol=75300 J/mol
R = gas constant =8.314 J /mol K
T = temperature = [tex]20^oC=273+20=293K[/tex]
Now on substituting all the values in the above formula, we get
[tex]\log \frac{K_2}{K_1}=\frac{75300 kJ/mol-29300 kJ/mol}{2.303\times 8.314 J/mol K\times 293}=1.58\times 10^{8}[/tex]
The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex] times.
4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with 5m/s in the same direction. They suffer an elastic collision and block B is now moving with 8m/s. What is the speed of block A after collision? 10m/s
8m/s
5m/s
13m/s
18m/s
Answer:
Option A is the correct answer.
Explanation:
Here momentum is conserved.
That is [tex]\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}[/tex]
Substituting values
[tex]3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s[/tex]
Speed of block A after collision = 10 m/s
Option A is the correct answer.
Answer:
8m/s
Explanation:
Suppose you apply a force of 40 N to a 0.25-meter-long wrench attached to a bolt in a direction perpendicular to the bolt. Determine the magnitude of the torque when the force is applied at an angle of 135degrees to the wrench.
Answer:
Torque = 7.07 N.m
Explanation:
It is given that,
Force acting on the wrench, F = 40 N
Length of wrench, l = 0.25 meters
It is attached perpendicular to the bolt such that the force is applied at an angle of 135 degrees to the wrench. The formula for torque is given by :
[tex]\tau=r\times F[/tex]
[tex]\tau=rF\ sin\theta[/tex]
[tex]\tau=0.25\ m\times 40\ N\ sin(135)[/tex]
[tex]\tau=7.07\ N.m[/tex]
So, the magnitude of torque applied to the wrench is 7.07 N-m. Hence, this is the required solution.
To calculate the magnitude of the torque, use the formula τ = r * F * sin(θ), where r is the lever arm length (0.25 m), F is the force applied (40 N), and θ is the angle between force and lever arm (135°). The sine of 135° provides the necessary component of the force that contributes to the torque.
Explanation:The question deals with the concept of torque in physics, particularly how torque is influenced by the angle at which a force is applied. Torque (τ) is the product of the force (F) applied, the distance (r) from the pivot point to the point where the force is applied, and the sine of the angle (θ) between the force vector and the lever arm, which can be represented as τ = r * F * sin(θ). Given that a force of 40 N is applied to the 0.25-meter-long wrench at a 135-degree angle to the wrench, the magnitude of the torque can be calculated using this formula.
Using the provided equation:
Torque = radius * force * sin(angle)Torque= 0.25 m * 40 N * sin(135°) = 0.25 m * 40 N * sin(135°)
Here, sin(135°) is a positive value since 135° is in the second quadrant where sine values are positive. It is important to note that the angle must be converted to radians or the correct sine value must be used if the calculator is set to degrees. The calculated torque will have the unit of Newton-meters (N.m).
The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and the walls thickness is 1.3 cm. Assume steady state conditions exist. Determine the temperature gradient in units of K/m and degree C/m. If the temperature gradient were larger what happens to the heat flux and why? Comment on the direction of heat flux given a negative temperature gradient and again for the case of a positive temperature gradient.
Answer:
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards
Explanation:
As we know that heat flux is given by the formula
[tex]q^n = K\frac{dT}{dx}[/tex]
here we know that
K = thermal conductivity
[tex]\frac{dT}{dx}[/tex] = temperature gradient
now we know that
[tex]q^n = 11 W/m^2[/tex]
also we know that
K = 1.7 W/mK
now we have
[tex]11 = 1.7 \frac{dT}{dx}[/tex]
so temperature gradient is given as
[tex]\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m [/tex]
also in other unit it will be same
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards
Salt water has greater density than fresh water. A boat floats in both fresh water and in salt 12) water. Where is the buoyant force greater on the boat?
Final answer:
The buoyant force on a boat is the same in both salt water and fresh water, as it depends on the weight of the fluid displaced, not on the fluid's density.
Explanation:
According to the principle discovered by Archimedes, a boat experiences buoyant force which is equal to the weight of the water it displaces. This buoyant force is not dependent on the weight of the boat but rather on the density of the fluid and the volume of the fluid displaced.
As the density of salt water is higher than that of fresh water, a boat will displace less volume of salt water to float. Therefore, the weight of the salt water displaced is equal to the weight of fresh water displaced when the boat floats.
This means that the buoyant force on the boat is the same in both salt water and fresh water.
Name a common product produced by blow molding.
Parts made from blow molding are plastic, hollow, and thin-walled, such as bottles and containers that are available in a variety of shapes and sizes. Small products may include bottles for water, liquid soap, shampoo, motor oil, and milk, while larger containers include plastic drums, tubs, and storage tanks.
A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its nose into the snow to catch its meal. If a fox jumps up to a height of 81 cm , calculate the speed at which the fox leaves the snow and the amount of time the fox is in the air. Ignore air resistance.
Answer:
4 m/s
0.82 s
Explanation:
h = height to which the fox jumps = 81 cm = 0.81 m
v₀ = speed at which the fox leaves the snow
v = speed of the fox at highest point = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
Using the kinematics equation
v² = v₀² + 2 a h
0² = v₀² + 2 (- 9.8) (0.81)
v₀ = 4 m/s
t = amount of time in air while going up
Using the equation
v = v₀ + a t
0 = 4 + (- 9.8) t
t = 0.41 s
T = Total time
Total time is given as
T = 2 t
T = 2 (0.41)
T = 0.82 s
The speed at which the fox leaves the snow is approximately 3.987 m/s. The fox is in the air for approximately 0.407 seconds.
Explanation:To calculate the speed at which the fox leaves the snow, we can use the concept of vertical motion and 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 displacement. Since the fox jumps straight up, the initial velocity is 0 m/s and the displacement is 81 cm (or 0.81 m). Assuming the acceleration due to gravity is 9.8 m/s^2, we can now calculate the final velocity:
v^2 = u^2 + 2as
v^2 = 0^2 + 2(9.8)(0.81)
v^2 = 15.876
v = √15.876
v ≈ 3.987 m/s
The time the fox is in the air can be calculated using the equation v = u + at, where t is the time. Again, the initial velocity is 0 m/s and the acceleration due to gravity is 9.8 m/s^2. Plugging in these values, we have:
v = u + at
3.987 = 0 + (9.8)t
3.987 = 9.8t
t = 3.987/9.8
t ≈ 0.407 s
A solenoid having N turns and carrying a current of 2.000 A has a length of 34 00 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mT what is the value of N? (μo = 4π x10^-7 T. m/A) A) 2318 B) 1218 C) 860.0 D) 3183 E) 1591
Answer:
B) 1218
Explanation:
N = Total number of turns in the solenoid
L = length of the solenoid = 34.00 cm = 0.34 m
B = magnetic field at the center of the solenoid = 9 mT = 9 x 10⁻³ T
i = current carried by the solenoid = 2.000 A
Magnetic field at the center of the solenoid is given as
[tex]B = \frac{\mu _{o}N i}{L}[/tex]
[tex]9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}[/tex]
N = 1218
The value of N is about B) 1218
[tex]\texttt{ }[/tex]
Further explanationLet's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:
[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]
B = magnetic field strength from current carrying wire (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
d = distance (m)
[tex]\texttt{ }[/tex]
[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]
B = magnetic field strength at the center of the solenoid (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
N = number of turns
L = length of solenoid (m)
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
Current = I = 2000 A
Length = L = 34.00 cm = 0.34 m
Magnetic field strength = B = 9000 mT = 9 T
Permeability of free space = μo = 4π × 10⁻⁷ T.m/A
Asked:
Number of turns = N = ?
Solution:
[tex]B = \mu_o \frac{I N}{L}}[/tex]
[tex]\frac{I N}{L} = B \div \mu_o[/tex]
[tex]IN = BL \div \mu_o[/tex]
[tex]N = BL \div (\mu_o I)[/tex]
[tex]N = ( 9 \times 0.34 ) \div ( 4 \pi \times 10^{-7} \times 2000 )[/tex]
[tex]\boxed {N \approx 1218}[/tex]
[tex]\texttt{ }[/tex]
Learn moreTemporary and Permanent Magnet : https://brainly.com/question/9966993The three resistors : https://brainly.com/question/9503202A series circuit : https://brainly.com/question/1518810Compare and contrast a series and parallel circuit : https://brainly.com/question/539204[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Magnetic Field
Two points are on a disk turning at constant angular velocity. One point is on the rim and the other halfway between the rim and the axis. Which point moves the greater distance in a given time?
Answer:
The point on the rim
Explanation:
All the points on the disk travels at the same angular speed [tex]\omega[/tex], since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the distance of the point from the centre of the disk
As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.
An electron is released from rest in a uniform electric field of 418 N/C near a particle detector. The electron arrives at the detector with a speed of 3.70 x 106 m/s (a) What was the uniform acceleration of the electron? (Enter the magnitude.) m/s (b) How long did the electron take to reach the detector? (c) What distance was traveled by the electron? cm (d) What is the kinetic energy of the electron when it reaches the detector?
Answer:
a) 7.35 x 10¹³ m/s²
b) 5.03 x 10⁻⁸ sec
c) 9.3 cm
d) 6.23 x 10⁻¹⁸ J
Explanation:
E = magnitude of electric field = 418 N/C
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of the electron = 9.1 x 10⁻³¹ kg
a)
acceleration of the electron is given as
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}[/tex]
a = 7.35 x 10¹³ m/s²
b)
v = final velocity of the electron = 3.70 x 10⁶ m/s
v₀ = initial velocity of the electron = 0 m/s
t = time taken
Using the equation
v = v₀ + at
3.70 x 10⁶ = 0 + (7.35 x 10¹³) t
t = 5.03 x 10⁻⁸ sec
c)
d = distance traveled by the electron
using the equation
d = v₀ t + (0.5) at²
d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²
d = 0.093 m
d = 9.3 cm
d)
Kinetic energy of the electron is given as
KE = (0.5) m v²
KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²
KE = 6.23 x 10⁻¹⁸ J
A person with mass of 68kg rides in an elevator that has an upward acceleration of 1.5 m/s2. Draw 1 body diagram of the person and determine the magnitude of the force of the elevator floor ree on the person?
Answer:
N = 768.4 N
Explanation:
As per given FBD we can see that the person inside the elevator have two forces on it
1) Normal force upwards
2) weight downwards
Now from Newton's law of motion we can say
[tex]F_{net} = ma[/tex]
[tex]N - mg = ma[/tex]
[tex]N = mg + ma[/tex]
now plug in all values in it
[tex]N = 68(1.5) + 68(9.8)[/tex]
[tex]N = 768.4 N[/tex]
he "escape velocity from Earth (the speed required to escape Earth's gravity) is 2.5 x 10 miles per hour. What is this speed in m/s? (1 mile 1609 m)
Answer: [tex]11.17\ \text{ m/s}[/tex]
Explanation:
Given : The escape velocity : [tex]v=2.5\times10\text{ miles per hour}[/tex]
We know that 1 mile = 1609 meters (approx)
and 1 hour= 3600 seconds
To convert escape velocity 2.5 x 10 miles per hour into m/s , we need to multiply it by 1609.34 and divide it by 3600.
Thus, the escape velocity in m/s is given by :-
[tex]v=2.5\times10\times\dfrac{1609}{3600}\\\\=11.1736111111\approx11.17\text{ m/s}[/tex]
Hence, the speed in m/s = 11.17
According to the quantum theory of radiation,
A. radiation goes through only one of the two slits in the double-slit experiment with radiation.
B. an electromagnetic field can have only certain values of total energy and no other values.
C. electromagnetic fields are really made of particles.
D. electromagnetic fields must vibrate only at one particular frequency.
E. the total quantity of energy that an electromagnetic field can contain must be less than a certain maximum value
Answer:
Option B
Explanation:
According to quantum theory of radiation electromagnetic field or electromagnetic radiation( like light) produce by accelerated charge object and the quantum of EM radiations is photon which has discrete energy. So, EM field can have only certain values of total energy and no other value due the discrete nature of the energy of photon. Hence option B is correct
(c) If η = 60% and TC = 40°F, what is TH, in °F?
The force of attraction between a star of mass M and a planet of mass m (where т «М) is: ЗЫ? a F where is the angular momentum of the planet and a,b are both positive constants. [Note: black hole, in the non- this does approximate the force of attraction between a planet and a relativistic limit, with a = GMm. (a) 15 points Under what conditions is a stable circular orbit possible? Give the radius of the stable circular orbit in terms of the given parameters (M, m,a, b, l). function of a (b) 15 pointsWhat is the smallest radius possible for any circular orbit as a and b, allowing for arbitrary l? (Hint: this occurs in the limit of very large .) Is this circular orbit stable or unstable? (c) [10 points] If the planet travels in a slightly non-circular orbit about a stable radius, find expression for the angular frequency of small radial oscillations
Answer:
2b2t
Explanation:
2b2t
1/32nd of a radioactive element's atoms remain in a lunar rock sample. How many half lives old is the rock?
A. 1
B. 2
C. 3
D. 4
E. 5
Answer:
E. 5
Explanation:
N₀ = initial total number of radioactive elements number
N = Number of atoms of radioactive element after "n" half lives = N₀ /32
n = number of half lives
Number of atoms of radioactive element after "n" half lives is given as
[tex]N = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]
inserting the values
[tex]\frac{N_{0}}{32} = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]
[tex]\frac{1}{32} = \left ( \frac{1}{2} \right )^{n}[/tex]
n = 5
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall. A) How far from the wall is she?B) What is the period of her up and down motion?
If ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.0 m/s .
A. The kayaker is 11 meters away from the wall.
B. The period of the kayaker's up and down motion is 5.5 seconds.
What is the distance?
A) The distance from the kayaker to the wall is:
Distance = 1/2 × Wavelength
Distance = 1/2 × 22 m
Distance = 11 m
Therefore the kayaker is 11 meters away from the wall.
B) The period can be calculated using the formula:
Period = 1 / Frequency
Use this formula:
Frequency = Velocity / Wavelength
Frequency = 4.0 m/s / 22 m
Frequency ≈ 0.182 Hz
Period:
Period = 1 / 0.182 Hz
Period ≈ 5.49 seconds
Period ≈ 5.5 seconds
Therefore, the period of the kayaker's up and down motion is approximately 5.5 seconds.
The elementary particle called a muon is unstable and decays in about 2.20μs2.20μs , as observed in its rest frame, into an electron, a neutrino, and an antineutrino. What lifetime do you observe for muons approaching you at 0.9270.927 the speed of light?
Answer:
5.865 μs
Explanation:
t₀ = Time taken to decay a muon = 2.20 μs
c = Speed of Light in vacuum = 3×10⁸ m/s
v = Velocity of muon = 0.927 c
t = Lifetime observed
Time dilation
[tex]t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-\frac{(0.927c)^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-0.927^2}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{0.140671}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{0.3750}\\\Rightarrow t=5.865\times 10^{-6}\ seconds[/tex]
∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs
The acceleration of a particle is given by a = 6t - 28, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at t = 0 is s0= -8 m, and the initial velocity is v0 = 6 m/s. After you have the general expressions, evaluate these expressions at the indicated times.
Answers:
At t = 5.8 s, s = m, v = m/s
At t = 12.7 s, s = m, v = m/s
Answer:
Velocity, V = 3t²- 28t+6
Displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = -249.05 m
v = -55.48 m/s
At t = 12.7 s
s = -141.48 m
v = 134.27 m/s
Explanation:
We have acceleration of a particle is given by a = 6t - 28
Velocity
[tex]v=\int adt=\int (6t - 28)dt=3t^2-28t+C[/tex]
At t = 0 we have v₀ = 6 m/s
v₀ = 6 = 3 x 0 ²-28 x 0+C
C = 6
So velocity, V = 3t² - 28t+6
Displacement
[tex]s=\int vdt=\int (3t^2-28t+6)dt=t^3-14t^2+6t+C[/tex]
At t = 0 we have s₀ = -8 m
s₀ = -8 = 0³ + 14 x 0² + 6 x 0 + C
C = -8
So displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = 5.8³ - 14 x 5.8²+6 x 5.8 - 8 = -249.05 m
v = 3 x 5.8² - 28 x 5.8 + 6 = -55.48 m/s
At t = 12.7 s
s = 12.7³ - 14 x 12.7²+6 x 12.7 - 8 = -141.48 m
v = 3 x 12.7² - 28 x 12.7 + 6 = 134.27 m/s