Answer:
Weight in planet X = 0.413 N
Explanation:
Weight = Mass x Acceleration due to gravity.
W = mg
Mass, m = 100 g = 0.1 kg
We have equation of motion s = ut + 0.5 at²
Displacement, s = 10 m
Initial velocity, u = 0 m/s
Time, t = 2.2 s
Substituting
s = ut + 0.5 at²
10 = 0 x 2.2 + 0.5 x a x 2.2²
a = 4.13 m/s²
Acceleration due to gravity, a = 4.13 m/s²
W = mg = 0.1 x 4.13 = 0.413 N
Weight in planet X = 0.413 N
In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.
What is the gravitational acceleration (g)?In physics, gravitational acceleration is the acceleration of an object in free fall within a vacuum (and thus without experiencing drag).
Step 1. Calculate the gravitational acceleration of Planet X.A 100-g (m) ball is released from rest from a height of 10.0 m (s) and it takes 2.2 s (t) for it to reach the ground. We can calculate the gravitational acceleration using the following kinematic equation.
s = 1/2 × g × t²
g = 2 s / t² = 2 (10.0 m) / (2.2 s)² = 4.1 m/s²
Step 2. Calculate the weight (w) of the ball on the surface of Planet X.We will use Newton´s second law of motion.
w = m × g = 0.100 kg × 4.1 m/s² = 0.41 N
In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.
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When four people with a combined mass of 315 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) What is the effective force constant of the springs?
Answer:
3.43 x 10⁵ N/m
Explanation:
M = mass of the car alone = 2000 kg
m = mass of the four people combined = 315 kg
x = compression of the spring when there is car alone
x' = compression of the spring when there is car and four people = x + 0.009
k = spring constant
When there is car, the weight of the car is balanced by the spring force and the force equation is given as
k x = Mg
k x = (2000)(9.8)
k x = 19600 eq-1
When there is car and four people, the weight of the car and four people is balanced by the spring force and the force equation is given as
k x' = (M + m) g
k (x + 0.009) = (2000 + 315) (9.8)
k x + 0.009 k = (2000 + 315) (9.8)
using eq-1
19600 + 0.009 k = 22687
k = 3.43 x 10⁵ N/m
Final answer:
The effective force constant of the springs k = 343000 N/m
Explanation:
To solve for the effective force constant of the springs in a car's suspension system, we can apply Hooke's Law, which is stated as F = -kx. Here, F represents the restoring force supplied by the springs, k is the spring constant or force constant, and x is the displacement of the springs from their equilibrium position.
When the four people with a combined mass of 315 kg sit in the car, their weight causes an additional displacement of 0.90 cm, or 0.009 m.
Since weight acts as the restoring force F, we can calculate it as F = mg, where m is the total mass and g is the acceleration due to gravity (9.80 m/s²). So, F = (315 kg)(9.80 m/s²) = 3087 N.
Using Hooke's Law, we can rearrange the formula to solve for k: k = F/x. Plugging in the values gives us k = 3087 N / 0.009 m, which provides the spring constant for the suspension system.
The effective force constant of the springs k = 343000 N/m
A fire hose ejects a stream of water at an angle of 35.0° above the horizontal. The water leaves the nozzle with a speed of 25.0 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?
To find the distance from the building where the water will hit the highest possible fire, we analyze the horizontal and vertical components of the water's motion. Using the given information, we calculate the horizontal distance using the horizontal velocity and time, and the vertical distance using the vertical velocity and time. By finding the time at which the water reaches its maximum height, we can determine the horizontal distance.
Explanation:A fire hose ejects a stream of water at an angle of 35.0° above the horizontal with a speed of 25.0 m/s. To find the distance from the building where the water will hit the highest possible fire, we can analyze the horizontal and vertical components of the water's motion. The horizontal distance can be calculated using the horizontal velocity and time, while the vertical distance can be determined using the vertical velocity and time. By finding the time at which the water reaches its maximum height, we can calculate the horizontal distance.
Using the given information, we can determine that the initial horizontal velocity is 25.0 m/s * cos(35.0°) = 20.4 m/s, and the vertical velocity is 25.0 m/s * sin(35.0°) = 14.3 m/s. The time it takes for the water to reach its maximum height (where the vertical velocity becomes zero) can be found using the equation vy = v0y - gt, where vy is the vertical velocity, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time. Solving for t, we get t = v0y / g. Plugging in the values, we have t = 14.3 m/s / 9.8 m/s^2 = 1.46 seconds. Finally, we can find the horizontal distance by multiplying the initial horizontal velocity by the time: x = v0x * t = 20.4 m/s * 1.46 s = 29.8 m. Therefore, the fire hose should be located approximately 29.8 meters away from the building to hit the highest possible fire.
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The fire hose should be located approximately 59.8 meters from the building from the building to hit a fire at the highest point the water stream can reach.
Projectile Motion in Fire Hose Stream
To determine how far from a building a fire hose should be located to hit the highest possible fire, we need to analyze the projectile motion of the water stream ejected.
The water stream is ejected at an angle of 35.0° above the horizontal with an initial speed of 25.0 m/s. We will assume the water reaches its highest point where its vertical component of velocity becomes zero.
Step-by-Step Calculation
Horizontal and Vertical Components of Velocity:A copper wire has a circular cross section with a radius of 1.25 mm. If the wire carries a current of 3.70 A, find the drift speed in mm/s of the electrons in this wire. (Note: the density of charge carriers (electrons) in a copper wire is n = 8.46 × 1028 electrons/m3)
Answer:
0.0557 mm / s
Explanation:
r = 1.25 mm = 1.25 x 10^-3 m, i = 3.7 A, n = 8.46 x 10^28 per cubic metre,
e = 1.6 x 10^-19 C
Let vd be the drift velocity
use the formula for the current in terms of drift velocity
i = n e A vd
3.7 = 8.46 x 10^28 x 1.6 x 10^-19 x 3.14 x 1.25 x 1.25 x 10^-6 x vd
vd = 5.57 x 10^-5 m/s
vd = 0.0557 mm / s
While doing an experiment you measure the length of an object to be 32.5cm. The measuring device that you are using measures to a millimeter. How can you decrease the error associated with your measurements
Answer and Explanation:
By measuring in millimeter we can decrease the associated error with the measurement because when we measure in smaller unit the measurement is more precise and accurate rather than when we measure in larger unit so when we measure 325 millimeter instead of 32.5 cm then there is chance of less error produced.
What is the magnitude (in Newtons) of the force a 796-uC charge exerts on a 481-nC charge 23.1-cm away? 2 pts Question 5 A proton is released in a uniform electric field, and it experiences an electric force of 6.9x10-15-N toward the south. What is the magnitude (in N/C) of the electric field?
Answer:
(a) 64.58 N
(b) 43125 N/C
Explanation:
(a) q1 = 796 uC = 796 x 10^-6 C, q2 = 481 nC = 481 x 10^-9 C,
r = 23.1 cm = 0.231 m
Force, F = k q1 q2 / r^2
F = (9 x 10^9 x 796 x 10^-6 x 481 x 10^-9) / (0.231)^2
F = 64.58 N
(b) F = 6.9 x 10^-15 N
E = F / q
E = (6.9 x 10^-15) / (1.6 x 10^-19) = 43125 N/C
Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.
Answer:
The correct answer is d. tension pneumothorax.
Explanation:
The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.
Tension pneumothorax is the condition where air that enters the pleural space during inspiration but is unable to exit during expiration, causing a collapse of the lung and compression of surrounding structures.
Explanation:Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called tension pneumothorax. In this condition, there is a buildup of air in the pleural space, leading to a collapse of the lung and compression of surrounding structures. It is a medical emergency that requires immediate treatment.
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Which of these situations would NOT qualify as a perfectly inelastic collision? A. A frog that jumps onto and sits on a lily pad.
B. A bowling ball that knocks pins out of its way as it rolls down a lane
C. A fish, swimming in one direction, that comes upon and swallows another fish swimming towards it
D. A piece of bubble gum that flies through the air and sticks to a golf ball at rest
Answer:
Option (B)
Explanation:
In a perfectly inelastic collision, the two bodies collide each other after the collision.
A. As the frog and the lily pad stick together so it is a perfectly inelastic collision.
B. The bowling ball and the pin do not stick together so it is not an example of perfectly inelastic collision.
C. A fish swallows the another, so it is a perfectly inelastic collision.
D. Bubble gum sticks to the golf ball so it is a perfectly inelastic collision.
An electric heater is rated at 1430 W, a toaster at 890 W, and an electric grill at 1760 W. The three appliances are connected in parallel to a common 120 V circuit. How much total current does this circuit draw?
Answer:
34.01 A
Explanation:
[tex]P_{heater}[/tex] = Power of the electric heater = 1430 W
[tex]P_{toaster}[/tex] = Power of the toaster = 890 W
[tex]P_{grill}[/tex] = Power of the electric grill = 1760 W
V = Voltage of the battery connected in parallel to appliances = 120 volts
Using ohm's law, current drawn by heater is given as
[tex]i_{heater} = \frac{P_{heater}}{V}[/tex]
[tex]i_{heater} = \frac{1430}{120}[/tex]
[tex]i_{heater} [/tex] = 11.92 A
Using ohm's law, current drawn by toaster is given as
[tex]i_{toaster} = \frac{P_{toaster}}{V}[/tex]
[tex]i_{toaster} = \frac{890}{120}[/tex]
[tex]i_{toaster} [/tex] = 7.42 A
Using ohm's law, current drawn by grill is given as
[tex]i_{grill} = \frac{P_{grill}}{V}[/tex]
[tex]i_{grill} = \frac{1760}{120}[/tex]
[tex]i_{grill} [/tex] = 14.67 A
Total circuit drawn is given as
[tex]i_{total} [/tex] = [tex]i_{heater} [/tex] + [tex]i_{toaster} [/tex] + [tex]i_{grill} [/tex]
[tex]i_{total} [/tex] = 11.92 + 7.42 + 14.67
[tex]i_{total} [/tex] = 34.01 A
The cheetah is the fastest land animal, reaching speeds as high as 33 m/s, or roughly 75 mph. What is the kinetic energy of a 65 kg cheetah running at top speed? Express your answer in SI units.
Answer:
Ek= 35.392 kJ or 35.392 10³ J
Explanation:
Formula for Kinetic energy:
Ek= [tex]\frac{1}{2}[/tex] × m × [tex]v^{2}[/tex]
since m= 65kg and v= 33m/s
Ek= [tex]\frac{1}{2}[/tex]× 65×[tex]33^{2}[/tex]
Ek= 35.392 kJ or 35.392 10³ J
Final answer:
The kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is calculated using the formula KE = 1/2 mv², resulting in a kinetic energy of 35442.5 Joules.
Explanation:
The question asks for the kinetic energy of a 65 kg cheetah running at its top speed, which is given to be 33 m/s. To calculate kinetic energy, we use the formula KE = 1/2 mv², where m is the mass of the object (in this case, the cheetah) and v is the velocity of the object. Substituting the given values, we get:
KE = 1/2 × 65 kg × (33 m/s)²
KE = 1/2 × 65 × 1089
KE = 1/2 × 70885
KE = 35442.5 Joules
Therefore, the kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is 35442.5 Joules.
Is the following statement true or false? The time constant of an RC circuit is the time it takes to completely charge or discharge the capacitor. O True False
Answer:
FALSE
Explanation:
As we know that equation for charging of a capacitor is given by
[tex]q = CV(1 - e^{-\frac{t}{\tau}})[/tex]
now at t = 0 we can say there is no charge on the capacitor
now as the time will increase the value of charge will increase on the plates of capacitor
now after t = 1 time constant
we will have
[tex]q = CV(1 - e^{-1})[/tex]
[tex]q = 0.63 CV[/tex]
so after one time constant the capacitor will charged upto 63% of its maximum value.
A gas is equilibrium at T kelvin. If mass of one molecule is m and its component of velocity in x direction is v. Then mean of its v2 is 3kT 2kT 2 kT 3 (4) zero
Answer:
The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
Explanation:
A gas is equilibrium at T kelvin.
Mass = M
We know that,
The average square of the velocity in the x,y and z direction are equal.
[tex]\bar{v}_{x}^2=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}[/tex]
[tex]v_{rms}^2=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}[/tex]
[tex]v_{rms}^2=3v_{x}^2[/tex]
[tex]v_{x}^{2}=\dfrac{v_{rms}^2}{3}[/tex]
Equation of ideal gas
[tex]PV=RT[/tex]
[tex]P=\dfrac{nKT}{V}[/tex]
Here, R = nK
We know that,
[tex]\dfrac{nKT}{V}=\dfrac{nM}{3V}v_{rms}^2[/tex]
[tex]v_{rms}^2=\dfrac{3KT}{M}[/tex]....(I)
Put the value of [tex]v_{rms}^2[/tex] in the equation (I)
[tex]v_{x}^2=\dfrac{1}{3}\times\dfrac{3KT}{M}[/tex]
[tex]v_{x}^2=\dfrac{KT}{M}[/tex]
Hence, The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the station so she accelerates at 1.50 m/s2. What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall? Give your answer relative to an observer who is space walking and therefore does not accelerate with the space station due to the push.
Answer:
[tex]a = 5.83 \times 10^{-4} m/s^2[/tex]
Explanation:
Since the system is in international space station
so here we can say that net force on the system is zero here
so Force by the astronaut on the space station = Force due to space station on boy
so here we know that
mass of boy = 70 kg
acceleration of boy = [tex]1.50 m/s^2[/tex]
now we know that
[tex]F = ma[/tex]
[tex]F = 70(1.50) = 105 N[/tex]
now for the space station will be same as above force
[tex]F = ma[/tex]
[tex]105 = 1.8 \times 10^5 (a)[/tex]
[tex]a = \frac{105}{1.8 \times 10^5}[/tex]
[tex]a = 5.83 \times 10^{-4} m/s^2[/tex]
The magnitude of the acceleration of the space station is mathematically given as
[tex]a = 5.83 * 10^{-4} m/s^2[/tex]
What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall?Question Parameter(s):
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg.
The station pushes off one wall of the station so she accelerates at 1.50 m/s2.
Generally, the equation for the Force is mathematically given as
F = ma
Therefore
F = 70(1.50)
F= 105 N
In conclusion, the space station force
F = ma
105 = 1.8 * 10^5 a
a = 5.83 * 10^{-4} m/s^2
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A force of 36.0 N is required to start a 3.0-kg box moving across a horizontal concrete floor. Part A) What is the coefficient of static friction between the box and the floor? Express your answer using two significant figures. Part B) If the 36.0-N force continues, the box accelerates at 0.54 m/s^2. What is the coefficient of kinetic friction? Express your answer using two significant figures.
Answer:
A) 1.2
B) 1.1
Explanation:
A)
F = force required to start the box moving = 36.0 N
m = mass of the box = 3 kg
[tex]F_{g}[/tex] = weight of the box = mg = 3 x 9.8 = 29.4 N
[tex]F_{n}[/tex] = normal force acting on the box by the floor
normal force acting on the box by the floor is given as
[tex]F_{n}[/tex] = [tex]F_{g}[/tex] = 29.4
[tex]F_{s}[/tex] = Static frictional force = F = 36.0 N
[tex]\mu _{s}[/tex] = Coefficient of static friction
Static frictional force is given as
[tex]F_{s}[/tex] = [tex]\mu _{s}[/tex] [tex]F_{n}[/tex]
36.0 = [tex]\mu _{s}[/tex] (29.4)
[tex]\mu _{s}[/tex] = 1.2
B)
a = acceleration of the box = 0.54 m/s²
F = force applied = 36.0 N
[tex]f_{k}[/tex] = kinetic frictional force
[tex]\mu _{k}[/tex] = Coefficient of kinetic friction
force equation for the motion of the box is given as
F - [tex]f_{k}[/tex] = ma
36.0 - [tex]f_{k}[/tex] = (3) (0.54)
[tex]f_{k}[/tex] = 34.38 N
Coefficient of kinetic friction is given as
[tex]\mu _{k}=\frac{f_{k}}{F_{n}}[/tex]
[tex]\mu _{k}=\frac{34.38}{29.4}[/tex]
[tex]\mu _{k}[/tex] = 1.1
A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 93.1 kg. The mass of the rock is 0.292 kg. Initially the wagon is rolling forward at a speed of 0.456 m/s. Then the person throws the rock with a speed of 15.4 m/s. Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown (a) directly forward in one case and (b) directly backward in another.
Answer:
a) 0.41 m/s
b) 0.51 m/s
Explanation:
(a)
M = total mass of wagon, rider and the rock = 93.1 kg
V = initial velocity of wagon = 0.456 m/s
m = mass of the rock = 0.292 kg
v = velocity of rock after throw = 15.4 m/s
V' = velocity of wagon after rock is thrown
Using conservation of momentum
M V = m v + (M - m) V'
(93.1) (0.456) = (0.292) (15.4) + (93.1 - 0.292) V'
V' = 0.41 m/s
b)
M = total mass of wagon, rider and the rock = 93.1 kg
V = initial velocity of wagon = 0.456 m/s
m = mass of the rock = 0.292 kg
v = velocity of rock after throw = - 15.4 m/s
V' = velocity of wagon after rock is thrown
Using conservation of momentum
M V = m v + (M - m) V'
(93.1) (0.456) = (0.292) (- 15.4) + (93.1 - 0.292) V'
V' = 0.51 m/s
J-s. If your 1400-kg car is parked in an 8.54-m-long garage, what is the uncertainty in its velocity? cm/s the tolerance is +/-2%
Answer:
[tex]\Delta v = 4.41 \times 10^{-37} cm/s[/tex]
Explanation:
As per Heisenberg's uncertainty principle we know that
[tex]\Delta P \times \Delta x = \frac{h}{4\pi}[/tex]
so here we have
[tex]\Delta P = m\Delta v[/tex]
[tex]\Delta x = 8.54 m[/tex]
now from above equation we have
[tex]m\Delta v \times (8.54) = \frac{h}{4\pi}[/tex]
[tex]1400(\Delta v) \times (8.54) = \frac{6.626 \times 10^{-34}}{4\pi}[/tex]
[tex]\Delta v = 4.41 \times 10^{-39} m/s[/tex]
[tex]\Delta v = 4.41 \times 10^{-37} cm/s[/tex]
A proton, traveling with a velocity of 4.5 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.0 × 10−14 N and a direction of due south. What are the magnitude and direction of the magnetic field causing the force?
Answer:
Magnetic field, B = 0.11 i (in k direction)
Explanation:
It is given that,
Velocity of the proton, [tex]v=4.5\times 10^6\ m/s[/tex]
Magnetic force, [tex]F=8\times 10^{-14}\ N[/tex] (due south)
The magnetic force acting on the electron is given by :
[tex]F=qvB[/tex]
(-j) = (i) × (B)
[tex]B=\dfrac{F}{qv}[/tex]
q is the charge on proton
[tex]B=\dfrac{8\times 10^{-14}\ N}{1.6\times 10^{-19}\ C\times 4.5\times 10^6\ m/s}[/tex]
B = 0.11
So, the magnitude of 0.11 T is acting on the proton in and is acting directed upward above the plane. Hence, this is the required solution.
Final answer:
To calculate the magnitude of the magnetic field that causes a force on a moving proton, we use the rearranged Lorentz force law, with inputs of proton charge, velocity, and the force experienced. The direction of the magnetic field is determined by the right-hand rule, given the directions of the force and proton's velocity.
Explanation:
The student's question involves the effect of a magnetic field on a moving charged particle, specifically a proton, which falls under the subject of Physics. The Lorentz force law states that the force (F) on a charged particle is directly proportional to the charge (q), the velocity (v), and the magnetic field (B), and is given by the equation F = q(v x B), where v x B denotes the cross product of the velocity vector and the magnetic field vector. The force is perpendicular to both the velocity and the magnetic field.
To find the magnitude of the magnetic field, we can rearrange this equation to B = F / (q * v * sin(θ)), where θ is the angle between the velocity of the proton and the direction of the magnetic field. In the case where force is maximum, the angle is 90 degrees, and the sine of 90 degrees is 1. Given the maximum magnetic force (8.0 × 10⁻¹⁴ N), the charge of a proton (approximately 1.6 × 10⁻¹⁹ C), and the velocity of the proton (4.5 × 10⁶ m/s), we can calculate the magnitude of the magnetic field. As the force experienced by the proton is directed due south and the velocity is due east, the magnetic field must be pointing downwards to produce a force in that direction according to the right-hand rule.
a football of mass 0.38 kg is traveling towards a boy at 2.8ms-1. the boy kicks the ball with a force of 21N, reversing its direction. if the time of imapct between the football and boy’s foot is 250ms. what is the new speed of trhe ball?
Answer:
Final speed in reverse direction will be 11 m/s
Explanation:
As per impulse momentum theorem we know that
Change in momentum of the object is product of force and time
so we have
[tex]P_f - P_i = F\Delta t[/tex]
now we have
[tex]P_f = mv_f[/tex]
[tex]P_i = mv_i[/tex]
[tex]\Delta t = 250 ms[/tex]
F = 21 N
now we have
[tex]0.38(v_f) - 0.38(-2.8) = 21(250\times 10^{-3})[/tex]
[tex]v_f = 13.82 - 2.8[/tex]
[tex]v_f = 11 m/s[/tex]
Define electric potential and inductance using sentences an elementary student could understand. Do not use math
Answer and Explanation:
1). Electric Potential:
Electric Potential can be defined as the amount of work required to move a unit charge from infinity or reference point to a particular or specific point inside the electric field (without any accelerated motion of charge particle). The other terms for electric potential are 'electrostatic potential, potential difference(if a charge is moved from one point to another inside the field) or electric field potential.
2). Inductance:
It can be define as that property of electric circuits due to which it opposes any change in the voltage under the influence of changing magnetic field.
We can also say that the property of an electric conductor due to which it opposes any change in electric current passing through it. Any change in electric current results in the production of magnetic field which when varied results in the generation of Electro Motive Force (EMF) which according to Lenz law is produced in a way that it opposes its cause of production.
Answer:
this is my basic understanding o
. Electric Potential:
Electric Potential can be defined as the amount of work required to move a unit charge from infinity or reference point to a particular or specific point inside the electric field (without any accelerated motion of charge particle). The other terms for electric potential are 'electrostatic potential, potential difference(if a charge is moved from one point to another inside the field) or electric field potential.
2). Inductance:
It can be define as that property of electric circuits due to which it opposes any change in the voltage under the influence of changing magnetic field.
We can also say that the property of an electric conductor due to which it opposes any change in electric current passing through it. Any change in electric current results in the production of magnetic field which when varied results in the generation of Electro Motive Force (EMF) which according to Lenz law is produced in a way that it opposes its cause of production.
Explanation:
Two workers are sliding 390 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 200 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Answer:
The coefficient of kinetic friction [tex]\mu= 0.16989[/tex]
Explanation:
From Newton's second law
[tex]\sum\overset{\rightarrow}{F}=m\cdot\overset{\rightarrow}{a}[/tex]
If the velocity is constant, that means the summation of all forces must be equal to zero. Draw the free-body diagram to obtain the sums of forces in x and y. It must include the Friction Force, in the opposite direction of the displacement, the weight ([tex]W=mg=390*9.81=3825.9N[/tex]), the Normal Force, which is the is the consequence of Newton's third law and the forces from the two workers.
The sum in y is:
[tex]\sum F_{y}=F_{N}-3825.9=0[/tex]
Solving for the [tex]F_{N}[/tex]:
[tex]F_{N}=$ $3825.\,\allowbreak9N[/tex]
The sum in x is:
[tex]\sum F_{x}=450+200-F_{f}=0[/tex]
Solving for the [tex]F_{f}[/tex]:
[tex]$F_{f}=650.0N[/tex]
The formula of the magnitude of the Friction force is
[tex]F_{f}=\mu F_{N}[/tex]
That means the coefficient of friction is:
[tex]\mu=\frac{F_{f}}{F_{N}}=\frac{650.0}{3825.\,\allowbreak9}=\allowbreak0.16989[/tex]
An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV = ΔVmax sin(2πft) and the source frequency is 16.9 Hz, at what time t will the current flowing in this circuit be 55.0% of the peak current?
0.005366 seconds is the time will the current flowing in this circuit be 55.0% of the peak current.
In a simple AC circuit with a resistor connected to an AC voltage source, the current flowing through the circuit can be calculated using Ohm's law:
I(t)=ΔV(t)/R
Where: I(t) is the current at time
ΔV(t) is the instantaneous voltage at time t
R is the resistance of the resistor.
Given that the source voltage is [tex]\bigtriangleup V=\bigtriangleup V_{max}sin(2 \pi ft)[/tex], and you want to find the time t at which the current is 55% of the peak current, you need to find the time when [tex]I(t)=0.55\times I_{peak}.[/tex]
Find the peak current [tex]I_{peak}[/tex].
The peak current corresponds to the current when the source voltage is at its peak value [tex]\bigtriangleup V_{max}[/tex] .
Using Ohm's law:
[tex]I_{peak}=\frac{\bigtriangleup V_{max}}{R}[/tex]
The time t when the current I(t) is 55% of the peak current:
[tex]I(t)=0.55 \times I_{peak}[/tex]
Substitute the expressions for I(t) and [tex]I_{peak}[/tex].
[tex]\frac{\bigtriangleup V(t)}{R}=0.55 \times \frac{\bigtriangleup V_{max}}{R}[/tex]
2πft=arcsin(0.55)
t=arcsin(0.55)/2πf
t= 0.5716/2×3.14×16.9
t=0.5716/106.132
t=0.0053s
Hence, at approximately 0.005366 seconds after the start of the AC cycle (or 5.366 ms 5.366ms), the current flowing in the circuit will be 55.0% of the peak current.
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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?
Answer:
The distance is 11 m.
Explanation:
Given that,
Friction coefficient = 0.24
Time = 3.0 s
Initial velocity = 0
We need to calculate the acceleration
Using newton's second law
[tex]F = ma[/tex]...(I)
Using formula of friction force
[tex]F= \mu m g[/tex]....(II)
Put the value of F in the equation (II) from equation (I)
[tex]ma=\mu mg[/tex]....(III)
[tex]a = \mu g[/tex]
Put the value in the equation (III)
[tex]a=0.24\times9.8[/tex]
[tex]a=2.352\ m/s^2[/tex]
We need to calculate the distance,
Using equation of motion
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
[tex]s=0+\dfrac{1}{2}2.352\times(3.0)^2[/tex]
[tex]s=10.584\ m\ approx\ 11\ m[/tex]
Hence, The distance is 11 m.
The truck can travel a maximum distance of 10.584 meters in 3.0 seconds without the box sliding.
To determine the maximum distance the truck can travel in 3.0 s without the box sliding, we start by calculating the maximum acceleration the box can experience before it starts to slide.
This is given by the static friction force, which is the product of the coefficient of static friction and the normal force (N). Assuming the box’s weight is W = mg, the normal force is N = mg.
The maximum static friction force (f) is:
f = μ N = μ mg
Given μs = 0.24, we need to ensure:
f > ma
Therefore, 0.24mg = ma, solving for a gives us:
a = 0.24g = 0.24 × 9.8 m/s² ≈ 2.352 m/s²
Next, we need to calculate the maximum distance using the kinematic equation for constant acceleration:
d = ut + (1/2)at²
Where:
u = initial velocity = 0a = 2.352 m/s²t = 3.0 sSubstituting these values, we get:
d = 0 + (1/2) × 2.352 m/s² × (3.0 s)² = 0.5 × 2.352 × 9 = 10.584 m
Thus, the maximum distance the truck can travel without the box sliding is approximately 10.584 meters.
An electric field of 4.0 μV/m is induced at a point 2.0 cm from the axis of a long solenoid (radius = 3.0 cm, 800 turns/m). At what rate is the current in the solenoid changing at this instant?
Answer:
The rate of current in the solenoid is 0.398 A/s
Explanation:
Given that,
Electric field [tex]E = 4.0\ \mu V/m[/tex]
Distance = 2.0 cm
Radius = 3.0 cm
Number of turns per unit length = 800
We need to calculate the rate of current
Using formula of electric field for solenoid
[tex]E = \dfrac{x}{2}\mu_{0}n\dfrac{dI}{dt}[/tex]
Where, x = distance
n = number of turns per unit length
E = electric field
r = radius
Put the value into the formula
[tex]4.0\times10^{-6}=\dfrac{2.0\times10^{-2}}{2}\times4\pi\times10^{-7}\times800\times\dfrac{dI}{dt}[/tex]
[tex]\dfrac{dI}{dt}=\dfrac{4.0\times10^{-6}\times2}{2.0\times10^{-2}\times4\pi\times10^{-7}\times800}[/tex]
[tex]\dfrac{dI}{dt}=0.397\ A/s[/tex]
Hence, The rate of current in the solenoid is 0.398 A/s.
During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket has an angular speed of 12.0 rad/s. If the distance between the top of the racket and the shoulder is 1.30 m, find the magnitude of the total acceleration of the top of the racket.
Final answer:
The magnitude of the total acceleration of the top of the racket during the serve, considering both tangential and radial components, is approximately 270.5 m/s^2.
Explanation:
To find the magnitude of the total acceleration of the top of the racket, we need to consider both tangential and radial (centripetal) acceleration components. The tangential acceleration (at) is directly provided by the angular acceleration (α), and is calculated by multiplying the angular acceleration by the radius (r), so at = α × r. Here, α = 150 rad/s2 and r = 1.30 m, giving at = 195 m/s2. The radial acceleration (ar), also known as centripetal acceleration, depends on the angular speed (ω) and the radius (r), and is calculated with the formula ar = ω2 × r. Given ω = 12.0 rad/s and r = 1.30 m, we find ar = 187.2 m/s2. Finally, the total acceleration (a) is the square root of the sum of the squares of at and ar, resulting in a = √(at2 + ar2). This gives us a = √(1952 + 187.22) m/s2, and by calculating, we find the magnitude of the total acceleration to be approximately 270.5 m/s2.
A man is holding a 6.0-kg (weight = 59 N) dumbbell at arm's length, a distance of 0.56 mfrom his shoulder. What is the torque on the shoulder joint from the weight of the dumbbell if thearm is held at 15° above the horizontal? On the picture, draw the lever arm for this force
Answer:
[tex]\vec \tau = 31.9 N[/tex]
Explanation:
As we know that torque due to a force is given by the formula
[tex]\vec \tau = \vec r \times \vec F[/tex]
here we know that force is exerted due to weight of the mass hold in his hand
so we have
F = mg = 59 N
Now Lever arm is the perpendicular distance on the line of action of force from the axis about which the system is rotated
so here we can say
[tex]r = Lcos\theta [/tex]
[tex]r = 0.56 cos15 = 0.54 m[/tex]
now we have
[tex]\vec \tau = (0.54)(59)[/tex]
[tex]\vec \tau = 31.9 Nm[/tex]
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subjected to an axial load of 200 N. Determine the deflection of the spring. C = 80 kN/mm2.
Answer:
The deflection of the spring is 34.56 mm.
Explanation:
Given that,
Diameter = 10 mm
Number of turns = 10
[tex]Radius_{mean} = 60\ mm[/tex]
[tex]Diameter_{mean} = 120\ mm[/tex]
Load = 200 N
We need to calculate the deflection
Using formula of deflection
[tex]\delta=\dfrac{8pD^3n}{Cd^4}[/tex]
Put the value into the formula
[tex]\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}[/tex]
[tex]\delta =34.56\ mm[/tex]
Hence, The deflection of the spring is 34.56 mm.
You heat a circular steel sheet of diameter 50.0 cm in an oven. What will be the radius for 50 K increase in temperature?
Answer:
R = 50.016 cm
Explanation:
Coefficient of thermal expansion for steel is given as
[tex]\alpha = 13 \times 10^{-6} per ^0C[/tex]
now as we know that the change in the length due to thermal expansion depends of change in temperature and initial length of the object
so here we will have
[tex]\Delta R = R_o \alpha \Delta T[/tex]
here we know that
[tex]R_o = \frac{50}{2} cm = 25 cm[/tex]
[tex]\Delta T = 50 K[/tex]
now we will have
[tex]\Delta R = (25)(13 \times 10^{-6})(50)[/tex]
[tex]\Delta R = 0.016 cm[/tex]
so final radius of the disc will be R = 50.016 cm
An electron and a proton are each placed at rest in an electric field of 500 N/C. Calculate the speed (and indicate the direction) of each particle 54.0 ns after being released. electron m/s O in the same direction as the field O in a direction opposite to the field proton m/s O in a direction opposite to the field O in the same direction as the field
Answer:
For proton: 2592 m/s In the same direction of electric field.
For electron: 4752000 m/s In the opposite direction of electric field.
Explanation:
E = 500 N/C, t = 54 ns = 54 x 10^-9 s,
Acceleration = Force /mass
Acceleration of proton, ap = q E / mp
ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2
Acceleration of electron, ae = q E / me
ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2
For proton:
u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s
use first equation of motion
v = u + at
vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.
For electron:
u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s
use first equation of motion
v = u + at
vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.
What will be the new volume when 128 mL of gas at 20.0°C is heated to 40.0°C while pressure remains unchanged? What will be the new volume when 128 mL of gas at 20.0°C is heated to 40.0°C while pressure remains unchanged? 64.0 mL 256 mL 120. mL 137 mL 128 mL
The answer will be 137 mL
Optics (Lens) Problem A lens forms an image of an object. The object is 16.0cm from the lens. The image is 12cm from the lens on the same side as the object. a. What is the focal length of the lens? Is the lens converging or diverging? b. If the object is 8.5mm tall, how tall is the image? Is it erect or inverted? c. Draw a principal ray diagram.
Answer:
Part a)
[tex]f = -48 cm[/tex]
Since focal length is negative so its a diverging lens
Part b)
[tex]h_i = 6.375 mm[/tex]
Since the magnification is position for diverging lens so it is ERECT
Explanation:
Part a)
As we know by lens formula
[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]
[tex]-\frac{1}{12cm} + \frac{1}{16} = \frac{1}{f}[/tex]
[tex]f = -48 cm[/tex]
Part b)
Since focal length is negative so its a diverging lens
Part c)
As we know that
[tex]\frac{h_i}{h_o} =\frac{d_i}{d_o}[/tex]
[tex]\frac{h_i}{8.5 mm} = \frac{12}{16}[/tex]
[tex]h_i = 6.375 mm[/tex]
Part d)
Since the magnification is position for diverging lens so it is ERECT
Part e)
A thin conducing plate 2.3 m on a side is given a total charge of −20.0 µC. (Assume the upward direction is positive.) (a) What is the electric field (in N/C) 1.0 cm above the plate? (Indicate the direction with the sign of your answer.)
Answer:
The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Distance = 2.3 m
Charge [tex]q= -20.0\ \muC[/tex]
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{\sigma}{2\epsilon_{0}}[/tex]
Where, [tex]\sigma=\dfrac{Q}{A}[/tex]
Q = charge
A = area
Put the value into the formula
[tex]E=\dfrac{Q}{2A\epsilon_{0}}[/tex]
[tex]E=\dfrac{-20.0\times10^{-6}}{2\times2.3\times2.3\times8.85\times10^{-12}}[/tex]
[tex]E=-213599.91\ N/C[/tex]
[tex]E=-2.14\times10^{5}\ N/C[/tex]
Negative sign shows the direction of the electric field.
The direction of electric field is toward the plates.
Hence, The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Final answer:
The electric field 1.0 cm above a thin conducting plate with a side length of 2.3 m and a total charge of -20.0 µC is -213,559.32 N/C, indicating a downward direction.
Explanation:
A thin conducting plate 2.3 m on a side is given a total charge of -20.0 µC. To find the electric field 1.0 cm above the plate, we use the concept of surface charge density (σ) and the formula for the electric field near an infinite plane sheet of charge, E = σ / (2ε0), where ε0 is the vacuum permittivity (ε0 = 8.85 x [tex]10^-^1^0[/tex] C2/N·m²). The surface charge density is σ = Q / A, with Q being the charge and A the area of the plate.
The area of the plate, A, is 2.3 m × 2.3 m = 5.29 m2. Therefore, σ = (-20.0 x 10-6 C) / (5.29 m2) = -3.78 x [tex]10^-^6[/tex] C/m². Substituting σ into the electric field equation yields E = -3.78 x [tex]10^-^6[/tex] C/m² / (2 × 8.85 x [tex]10^-^1^2[/tex] C2/N·m2) = -213,559.32 N/C, indicating the direction is downward due to the negative sign.