Answer:
a) 1.34 Volts
b) 3.08 W
c) 2.68 W
Explanation:
Given:
Emf of the cell, E = 1.54 V
current, i = 2.0 A
internal resistance, r = 0.100Ω
(a) Terminal voltage (V) = E - v
where,
v is the potential difference across the resistance 'r'
now,
according to the Ohm's Law, we have
v = i × r
substituting the values in the above equation we get
v = 2.0 × 0.100 = 0.2 Volts
thus,
Terminal voltage (V) = (1.54 - 0.2) = 1.34 V
(b) Now, the Total power (P) is given as
P = E × i = (1.54 × 2.0) = 3.08 W
(c) Power into its load = [terminal voltage, v] * i
= (1.34 × 2.0) = 2.68 W
Final answer:
Terminal voltage of a carbon-zinc cell is 1.34 V, producing 2.68 W of electrical power, with 2.68 W going to the load.
Explanation:
Terminal voltage: The formula to calculate terminal voltage is V = EMF - I * r, where V is the terminal voltage, EMF is the electromotive force, I is the current, and r is the internal resistance of the cell. So, V = 1.54 V - 2.00 A * 0.100 Ω = 1.34 V.
Electrical power: The electrical power produced by the cell is given by P = V * I, where P is power, V is voltage, and I is current. Substituting the values, P = 1.34 V * 2.00 A = 2.68 W.
Power to load: The power delivered to the load is equal to the voltage supplied to the load times the current flowing through it. Hence, the power to the load is P = V_load * I = 1.34 V * 2.00 A = 2.68 W.
. The inner and outer surfaces of a 4-m × 7-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m-K are maintained at temperatures of 20 °C and 5 °C, respectively. Determine the rate of heat transfer through the wall, in W.
Answer:
[tex]\frac{dQ}{dt} = 966 W[/tex]
Explanation:
As we know that the rate of heat transfer due to temperature difference is given by the formula
[tex]\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}[/tex]
here we know that
[tex]K = 0.69 W/m-K[/tex]
A = 4 m x 7 m
thickness = 30 cm
temperature difference is given as
[tex]\Delta T = 20 - 5 = 15 ^oC[/tex]
now we have
[tex]\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}[/tex]
[tex]\frac{dQ}{dt} = 966 W[/tex]
A 8.0 N force acts on a 0.30-kg object for 0.20 seconds. By how much does the object's momentum change (in kg-m/s)? (NEVER include units in the answer to a numerical question.)
Answer:
The change in momentum of an object is 1.6 kg-m/s.
Explanation:
It is given that,
Mass of the object, m = 0.3 kg
Force acting on it, F = 8 N
Time taken, t = 0.2 s
We need to find the change in momentum of an object. The change in momentum is equal to the impulse imparted on an object i.e.
[tex]\Delta p=J=F.\Delta t[/tex]
[tex]\Delta p=8\ N\times 0.2\ s[/tex]
[tex]\Delta p=1.6\ kg-m/s[/tex]
So, the change in momentum of an object is 1.6 kg-m/s. Hence, this is the required solution.
The change in the momentum of the object is 1.6 kg-m/s.
What is change in momentum?Change in momentum is the product of force acting on a body and time
To calculate the change in momentum of the object, we use the formula below
Formula:
ΔM = Ft...............Equation 1Where:
ΔM = change in momentum of the objectm = Force acting on the objectt = time.From the question,
Given:
F = 8 Nt = 0.20 secondsSubstitute these values into equation 1
ΔM = (8×0.2)ΔM = 1.6 kg-m/sHence, The change in the momentum of the object is 1.6 kg-m/s.
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Give an example of an elastic and an inelastic collision:; State whether or not momentum and kinetic energy are conserved in each case.
Answer:
Elastic collision is that collision in which two objects are collides and separate out without and deformation in shape and size
example: if a block connected with ideal spring will collide with another block in which spring will compress during the collision and then again come to its natural state.
Inelastic collision in that in which two objects will collide and then stick with each other after collision and there is permanent deformation in the objects.
example : if two cars will collide and stuck together after collision and move together with same speed
So here in both examples and both type of collision there is no external force during collision of two objects so here total momentum of the system will remains conserved.
But since the shape will change in inelastic collision so kinetic energy is not conserved in inelastic collision while in elastic collision kinetic energy is conserved
Final answer:
An example of an elastic collision is two colliding billiard balls, with both momentum and kinetic energy conserved, while an inelastic collision, such as a car crash, conserves momentum but not kinetic energy.
Explanation:
An example of an elastic collision is a collision between two billiard balls where they bounce off of each other without any deformation or generation of heat. In this case, both momentum and kinetic energy are conserved. Momentum conservation means the total momentum before the collision is equal to the total momentum after the collision. Similarly, kinetic energy conservation implies that the total kinetic energy before the collision is the same as the total kinetic energy after the collision.
In contrast, an example of an inelastic collision is a car crash, where the cars collide and deform, often sticking together. In this type of collision, while again momentum is conserved, kinetic energy is not conserved because some of it has been transformed into other forms of energy, such as heat or sound. The reduction in internal kinetic energy is a hallmark of inelastic collisions.
The conservation of momentum and kinetic energy together allow us to calculate the final velocities of objects in one-dimensional, two-body collisions. However, in inelastic collisions, energy is not conserved due to the conversion of kinetic energy to other energy forms.
Find an expression for the current in a circuit where the resistance is 8 Ω, the inductance is 4 H, a battery gives a constant voltage of 32 V, and the switch is turned on when t
Answer:
[tex]I(t)=4[1-e^{-2t}][/tex]
Explanation:
For a LR circuit as shown the current at any time t in the circuit is given by
[tex]I(t)=\frac{V}{R}[1-e^{\frac{-Rt}{L}}][/tex]
where
'V' is the voltage
'R' is resistance in the circuit
'L' is the inductance of the circuit
't' is time after circuit is turned on
Applying the given values we get
[tex]I(t)=\frac{32}{8}[1-e^{\frac{-8t}{4}}]\\\\I(t)=4[1-e^{-2t}][/tex]
Two 30 uC charges lie on the x-axis, one at the origin and the other at 9 m. A third point is located at 27 m. What is the potential at this third point relative to infinity? (The value of k is 9.0 x10^9 N-m^2/C^2.) A) 750 V B) 2500 V C) 2000 V D) 3000 V
Answer:
25000 V
Explanation:
The formula for potential is
V = Kq/r
Potential at B due to the charge placed at origin O
V1 = K q / OB
[tex]V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}[/tex]
V1 = 10000 V
Potential at B due to the charge placed at A
V2 = K q / AB
[tex]V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}[/tex]
V2 = 15000 V
Total potential at B
V = V1 + V2 = 10000 + 15000 = 25000 V
A car of mass 1689-kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is essentially elastic. If the velocity of the truck is 17 km/h (in the same direction as the car's initial velocity) after the collision, what was the initial speed of the car?
The question is about determining the initial speed of a car involved in an elastic collision with a parked truck, using the laws of conservation of momentum.
The subject of this question is Physics, and it involves the principles of conservation of momentum and elastic collisions. We are given masses for two vehicles (a car and a truck) and the post-collision speed of one of the vehicles (the truck). To find the car's initial speed, we use the equations for an elastic collision, which ensures that both momentum and kinetic energy are conserved.
For an elastic collision, the conservation of momentum is given by:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
where m1 and m2 are the masses of the car and truck respectively; v1 and v2 are the initial velocities of the car and truck respectively (since the truck is initially parked, v2 = 0); v1' and v2' are the final velocities of the car and truck respectively.
To find the final velocity v1' of the car, we must use the conservation of kinetic energy along with the above equation. However, we only need the conservation of momentum equation solved for v1 (initial speed of the car) since v2' (final velocity of the truck) is given.
After plugging in the given values (car's mass 1689 kg, truck's mass 2000 kg, truck's final velocity 17 km/h), the equation is solved to find the car's initial velocity.
Using a simple pendulum, we can measure. (a) Acceleration along a circle (b) Acceleration due to gravity (c) Frequency (d) None
Answer:
Option B is the correct answer.
Explanation:
Period of simple pendulum is given by the expression [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex], where l is the length of pendulum and g is the acceleration due to gravity value.
So if we have a simple pendulum with length l we can find its period. Using the above formula we can calculate acceleration due to gravity value of that place.
So using simple pendulum we can measure acceleration due to gravity value.
Option B is the correct answer.
f a vector has magnitude V_1 = 46 N, and direction theta_1 = 30 degree Another vector V_2 = 63 N, and direction theta_2 = 151 degree. Calculate the reference angle of the vector V_1 + V_2. Round your answer such that it has no decimal.
Answer:
105.8 degree from + X axis
Explanation:
V1 = 46 N at 30 degree
V2 = 63 N at 151 degree
Write the vector form
V1 = 46 (Cos 30 i + Sin 30 j) = 39.84 i + 23 j
V2 = 63 (Cos 151 i + Sin 151 j) = - 55 i + 30.54 j
The resultant of V1 and V2 is given by
V = V1 + V2 = 39.84 i + 23 j - 55 i + 30.54 j = - 15.16 i + 53.54 j
The angle made by resultant of V1 and V2 with X axis is
tan∅ = 53.54 / (- 15.16) = - 3.53
∅ = 105.8 degree from + X axis.
A particle that has a mass of 5.0 kg is acted on by a force of 4.5 N. The acceleration of the particle is: a 0.9 m/s^2 b 1.1 m/s^2 c. 2.9 m/s^2 d. 22.5 m/s^2
Answer:
a) [tex]a = 0.9 m/s^2[/tex]
Explanation:
As per Newton's 2nd law we know that net force on an object of mass "m" is given by the formula
[tex]F = ma[/tex]
so net force is the product of mass and acceleration
here we know that
mass = 5 kg
net force on the particle is 4.5 N
so from above equation we have
[tex]4.5 = 5 \times a[/tex]
so the acceleration is given as
[tex]a = \frac{4.5}{5} = 0.9 m/s^2[/tex]
A 30 resistor is connected in parallel with a 6.0 Q resistor. This combination is connected in series with a 4.00 resistor. The resistons are connected to a 12. volt battery. How much power is dissipated in the 3.0 resistor? A) 7.7 W E) 5.3 W DO 12 W B) 2.7 W C) 6 W
Answer:
Power dissipated in 3 ohms resistor is 5.32 watts
Explanation:
Resistor 1, R₁ = 3 ohms
Resistor 2, R₂ = 6 ohms
Resistor 3, R₃ = 4 ohms
Voltage source, V = 12 V
We need to find the power dissipated in the 3 ohms resistor. Firstly, we will find the equivalent resistance of R₁ and R₂.
[tex]\dfrac{1}{R'}=\dfrac{1}{R_1}+\dfrac{1}{R_2}[/tex]
[tex]\dfrac{1}{R'}=\dfrac{1}{3}+\dfrac{1}{6}[/tex]
R' = 2 ohms
Now R' is connected in series with R₃. Their equivalent is given by :
[tex]R_{eq}=R'+R_3[/tex]
[tex]R_{eq}=2+4[/tex]
[tex]R_{eq}=6\ ohms[/tex]
Total current flowing through the circuit, [tex]I=\dfrac{12}{6}=2\ A[/tex]
Voltage across R', [tex]V'=IR'=2\times 2=4\ V[/tex]
The voltage across R₁ and R₂ is 4 volts as they are connected in parallel. So, current across 3 ohm resistor is,
[tex]I=\dfrac{4}{3}=1.33\ A[/tex]
Power dissipated is given by, P = I × V
[tex]P=1.33\ A\times 4\ \Omega[/tex]
P = 5.32 watts
So, 5.32 watt of power is dissipated in 3 ohms resistor. Hence, this is the required solution.
A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find the coordinates of the point where the net electric field strength due to these charges is zero.
Answer:
The coordinates of the point is (0,0.55).
Explanation:
Given that,
First charge [tex]q_{1}=9\times10^{-6}\ C[/tex] at origin
Second charge [tex]q_{2}=6\times10^{-6}\ C[/tex]
Second charge at point P = (0,1)
We assume that,
The net electric field between the charges is zero at mid point.
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}[/tex]
[tex]\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}[/tex]
[tex]\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1[/tex]
[tex]\dfrac{1}{d}=1.82[/tex]
[tex]d=\dfrac{1}{1.82}[/tex]
[tex]d=0.55\ m[/tex]
Hence, The coordinates of the point is (0,0.55).
It is a physical field occupied by a charged particle on another particle in its surrounding. The coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).
What is an electric field?It is a physical field occupied by a charged particle on another particle in its surrounding.
The following data are given as
q₁ is the first charge at the origin
q₂ is the Second charge
point of the Second charge is P = (0,1)
As we know the net electric field between the charges is zero at the midpoint.
The relation of the electric field with the distance between the charged particle is given by the formula
[tex]\rm E=\frac{Kq}{d^2} \\\\0=\frac{K\times9\times10^{-6}}{d^2} +\frac{K\times6\times10^{-6}}{(1-d)^2} \\\\\frac{1-d}{d} =\sqrt{\frac{6}{9} } \\\\\frac{1}{d} =\sqrt{\frac{{6} }{9} } +1\\\\\rm\frac{1}{d} =1.82\\\\\rm d=0.55m[/tex]
Hence the coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).
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A power plant burns coal at 830 K, and it exhaust to air at 288 K. If it runs at the Carnot efficiency, how much heat must it exhaust into the air to produce 230,000 J of work?
Answer:
122241.02 J
Explanation:
Work Done = 230000 J
[tex]T_H=\text {High Temperature Reservoir}=830\ K\\T_L=\text {Low Temperature Reservoir}=288\ K\\\text{In case of Carnot Cycle}\\\text{Efficiency}=\eta\\\eta=1-\frac{T_L}{T_H}\\\Rightarrow \eta =1-\frac{288}{830}\\\Rightarrow \eta=0.65\\[/tex]
[tex]\eta=\frac{\text{Work}}{\text{Heat}}\\\Rightarrow \eta=\frac{W}{Q_H}\\\\\Rightarrow 0.65=\frac{230000}{Q_H}\\\Rightarrow Q_H=\frac{230000}{0.65}\\\Rightarrow Q_H=352214.02[/tex]
[tex]Q_L=Q_H-W\\\Rightarrow Q_L=352214.02-230000\\\Rightarrow Q_L=122241.02\ J[/tex]
∴Heat exhausted into the air to produce 230,000 J of work is 122241.02 Joule
Correct Answer:
122000 J
A toy car has a kinetic energy of 8 J. What is its kinetic energy after a frictional force of 0.5 N has acted on it for 6 m?
Answer:
Final kinetic energy, [tex]KE_f=5\ J[/tex]
Explanation:
It is given that,
Kinetic energy of toy car, [tex]KE_i = 8\ J[/tex]
Frictional force, F = 0.5 N
Distance, d = 6 m
We need to find the kinetic energy after this frictional force has acted on it. We know that frictional force is an opposing force. The work done by the toy car is given by :
[tex]W=F\times d[/tex]
[tex]W=-0.5\ N\times 6\ m[/tex]
W = -3 J
We know that the work done is equal to the change in kinetic energy. Let [tex]KE_f[/tex] is the final kinetic energy.
[tex]W=KE_f-KE_i[/tex]
[tex]-3=KE_f-8[/tex]
[tex]KE_f=5\ J[/tex]
So, the final kinetic energy is the toy car is 5 J. Hence, this is the required solution.
A 0.380 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 28.0 N/m force constant that is at its equilibrium length. A 0.0600 kg wad of Play-Doh is thrown horizontally at the block with a speed of 2.60 m/s and sticks to it. Determine the amount in centimeters by which the Play-Doh-block system compresses the spring.
Initial momentum of the Play-doh: 0.0600 x 2.60 = 0.156 kg/m/s
Total mass of the block and play-doh: 0.38 + 0.0600 = 0.44 kg.
Final momentum is mass x velocity = 0.44v
V = Initial momentum / mass
V = 0.156 / 0.44 = 0.3545 m/s
Work done by spring is equal to the Kinetic enrgy.
Work Done by spring = 1/2 *28.0 * distance^2 = 14 * d^2
KE = 1/2 * 0.44* 0.3545^2
set to equal each other:
14 * d^2 = 0.22 *0.12567
Solve for d:
d = √(0.22*0.12567)/14
d = 0.44 meters = 4.4cm
Answer:
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The block rests at a distance of 2 m from the center of the platform. If the coefficient of static friction between the block and the platform is ms = 0.3, determine the maximum speed which the block can attain before it begins to slip. Assume the angular motion of the disk is slowly increasing.
Answer:
v = 2.43 m/s
Explanation:
As we know that block is placed at rest at distance
d = 2 m
so here the centripetal force on the block to move in the circle is due to static friction force
now when block is just going to slide over the disc
then the friction force is maximum static friction which is given as
[tex]\mu_s mg = \frac{mv^2}{R}[/tex]
now we have
[tex]v = \sqrt{\mu_s Rg}[/tex]
now plug in the values in the above equation
[tex]v = \sqrt{0.3(2)(9.81)}[/tex]
[tex]v = 2.43 m/s[/tex]
To determine the maximum speed at which the block can attain before it begins to slip, calculate the force of static friction using the given coefficient and weight of the block. Then, use the equation fs = m(a+g) to determine the maximum speed.
Explanation:To determine the maximum speed at which the block can attain before it begins to slip, we need to calculate the force of static friction. The formula for static friction is fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is the weight of the block, which is mg. The maximum force of static friction is therefore μs(mg). Since the block is on a horizontal surface, the normal force is equal to the weight of the block, and the maximum force of static friction is μs(mg). We can calculate μs using the given coefficient of static friction and determine the maximum speed using the equation fs = m(a+g), where fs is the force of static friction, m is the mass of the block, a is the acceleration, and g is the acceleration due to gravity.
Plug in the given values: ms = 0.3, m = (mass of the block), a = 2 m/s², and g = 9.8 m/s².
Then, solve for the maximum speed by rearranging the equation to isolate v: v = sqrt((fs - m*g)/m).
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An object moves with simple harmonic motion. If the amplitude and the period are both doubled, the object's maximum speed is 1. quartered 2. halved 3. quadrupled. 4. doubled 5. unchanged
Answer:
The object's maximum speed remains unchanged.
Explanation:
The speed of a particle in SHM is given by :
[tex]v(t)=A\omega\ sin(\omega t)[/tex]
Maximum speed is, [tex]v_{max}=A\omega[/tex]
If A' = 2A and T' = 2T
[tex]v'_{max}=(2A)\dfrac{2\pi}{2T}[/tex]
[tex]v'_{max}=(A)\dfrac{2\pi}{T}[/tex]
[tex]v'_{max}=v_{max}[/tex]
So, the maximum speed of the object remains the same i.e. it remains unchanged. Hence, this is the required solution.
If the amplitude and period of an object's simple harmonic motion are both doubled, the object's maximum speed will be halved.
Explanation:When an object moves with simple harmonic motion, doubling the amplitude and period will affect various properties of the motion. In this case, if the amplitude and the period are both doubled, the object's maximum speed will be halved. This means that the object will reach its maximum velocity at a slower rate compared to its original motion.
To understand why this happens, it's important to know that in simple harmonic motion, the maximum speed occurs when the object passes through equilibrium. Doubling the period means that the object will take twice as long to complete one full cycle, which results in a decrease in its maximum speed by a factor of 2.
A 9800-N automobile is pushed along a level road by four students who apply a total forward force of 600 N. Neglecting friction, the acceleration of the automobile is: a) 0.0 m /s^2 b) 0.54 m/s^2 c) 0.6 m/s^2 d) 9.8 m/s^2
Answer:
c). [tex]a = 0.60 m/s^2[/tex]
Explanation:
As we know that weight of the automobile is given here
so weight = mass times gravity
[tex]W = mg[/tex]
[tex]9800 = m(9.8)[/tex]
[tex]m = 1000 kg[/tex]
now from Newton's law
[tex]F = ma[/tex]
[tex]600 = 1000 a[/tex]
[tex]a = \frac{600}{1000}[/tex]
[tex]a = 0.60 m/s^2[/tex]
The question is about finding the acceleration of a car with a given weight and applied force. To do this, the weight of the car is first converted to mass, then using Newton's Second Law (F = ma), we can determine the acceleration. The calculated acceleration of the car is 0.6 m/s^2.
Explanation:The subject matter of this question is related to physics, specifically Newton's Second Law of Motion. Noting that the force, mass, and acceleration of the automobile are connected by the formula F = ma (force equals mass times acceleration), the given problem can be solved. First, we need to convert the weight of the automobile (9800 N) into mass, as weight is the product of mass and gravity (9.8 m /s^2). Therefore, the mass of the car is 1000 kg.
Using this converted mass and the applied force (600 N) in the formula, we can calculate the acceleration. The calculation goes as follows: 600N = 1000 kg * a, where a is the unknown acceleration we're trying to find. When you solve for 'a', you get a = 0.6 m/s^2. Hence, the acceleration of the 9800-N automobile when a forward force of 600N is applied, neglecting friction, is option c) 0.6 m/s^2.
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In getting ready to slum-dunk the ball, a basketball player starts from rest and sprints to a speed of 6.0 m/s in 1.5s. Assuming that the player accelerates uniformly, determine the distance he runs.
Answer:
Distance ran by basketball player = 4.5 m
Explanation:
We have equation of motion, v = u+at
Substituting
6 = 0 + a x 1.5
a = 4m/s²
Now we need to find distance traveled, we have the equation of motion
v² = u² + 2as
Substituting
6² = 0² + 2 x 4 x s
8 s = 36
s = 4.5 m
Distance ran by basketball player = 4.5 m
Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.
Answer:
Inductive reactance is 125.7 Ω
Explanation:
It is given that,
Inductance, [tex]L=50\ mH=0.05\ H[/tex]
Voltage source, V = 15 volt
Frequency, f = 400 Hz
The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :
[tex]X_L=2\pi fL[/tex]
[tex]X_L=2\pi \times 400\ Hz\times 0.05\ H[/tex]
[tex]X_L=125.66\ \Omega[/tex]
[tex]X_L=125.7\ \Omega[/tex]
So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.
To calculate the inductive reactance for a 50 mH inductor with a 400 Hz source, apply the formula XL = 2πfL, resulting in an inductive reactance of 125.6 ohms.
To determine the inductive reactance of a 50 mH inductor across a 15 volt, 400 Hz source, we use the formula for inductive reactance, which is:
L = 2πfL
where:
XL is the inductive reactance
f is the frequency of the AC source (400 Hz in this case)
L is the inductance of the coil (50 mH or 0.050 H)
Substituting the given values:
XL = 2π x 400 Hz x 0.050 H
XL = 125.6 Ω
Therefore, the inductive reactance is 125.6 ohms (Ω).
Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium
Answer:
A
Explanation:
Iron and gadlinium are both very easily made into magnetic substances. Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but
Aluminum by itself is not able to be magnetized.
Trey races his bicycle for 192m. A wheel of his bicycle turns 48 times as the bicycle travels this distance. What is the diameter of the wheel? Use the value 3.14 for π. Round your answer to the nearest tenth. Do not round any intermediate steps.
Answer:
1.27 m
Explanation:
Distance = 192 m
number of rotations = 48
Distance traveled in one rotation = 2 x π x r
Where, r be the radius of wheel.
so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r
It is equal to the distance traveled.
192 = 48 x 2 x 3.14 x r
r = 0.637 m
diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m
Answer:
1.29 meters
Explanation:
In order to calculate this you first have to calculate the circumference of the wheel, as you know to travel 192 meters the wheel did 48 revolutions, that means that it took 48 circumferences of the wheel to cover 192 meters, we first have to divide the 192 meters by the circumference:
Circumference:192 m/48=4m
So to know the diameter we just need to divide the circumference by pi:
diameter=circumference/pi
Diameter=4 meters/3.14
Diameter= 1.2861 meters.
How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is connected to a 2.00-V battery, what charge is stored?
Answer:
(a) 45 micro coulomb
(b) 6 micro Coulomb
Explanation:
C = 3 micro Farad = 3 x 10^-6 Farad
V = 15 V
(a) q = C x V
where, q be the charge.
q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb
(b)
V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad
q = C x V
where, q be the charge.
q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb
A ball thrown straight upwards with an initial velocity of 15 m/s. Calculate how long the ball will remain in the air and the maximum height the ball reaches?
Answer:
The ball remain in air for 3.06 seconds.
Maximum height reached = 22.94 m
Explanation:
We have equation of motion v = u + at
At maximum height, final velocity, v =0 m/s
Initial velocity = 15 m/s
acceleration = -9.81 m/s²
Substituting
0 = 15 -9.81 t
t = 1.53 s
Time of flight = 2 x 1.53 = 3.06 s
The ball remain in air for 3.06 seconds.
We also have equation of motion v² = u² + 2as
At maximum height, final velocity, v =0 m/s
Initial velocity = 15 m/s
acceleration = -9.81 m/s²
Substituting
0² = 15² - 2 x 9.81 x s
s = 22.94 m
Maximum height reached = 22.94 m
a wave travels in a string at 58 m/s. a second string of 10% greater linear density has the same tension applied as in the first string. what will be the resulting wave speed in the second string
Answer:
The speed of wave in the second string is 55.3 m/s.
Explanation:
Given that,
Speed of wave in first string= 58 m/s
We need to calculate the wave speed
Using formula of speed for first string
[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]...(I)
For second string
[tex]v_{2}=\sqrt{\dfrac{T}{\mu_{2}}}[/tex]...(II)
Divided equation (II) by equation (I)
[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\dfrac{T}{\mu_{2}}}{\dfrac{T}{\mu_{1}}}}[/tex]
Here, Tension is same in both string
So,
[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{\mu_{2}}}[/tex]
The linear density of the second string
[tex]\mu_{2}=\mu_{1}+\dfrac{10}{100}\mu_{1}[/tex]
[tex]\mu_{2}=\dfrac{110}{100}\mu_{1}[/tex]
[tex]\mu_{2}=1.1\mu_{1}[/tex]
Now, Put the value of linear density of second string
[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{1.1\mu_{1}}}[/tex]
[tex]v_{2}=v_{1}\times\sqrt{\dfrac{1}{1.1}}[/tex]
[tex]v_{2}=58\times\sqrt{\dfrac{1}{1.1}}[/tex]
[tex]v_{2}=55.3\ m/s[/tex]
Hence, The speed of wave in the second string is 55.3 m/s.
Final answer:
The resulting wave speed in the second string with a 10% greater linear density and the same tension will be the same as the wave speed in the first string.
Explanation:
The wave speed in a string can be determined by the tension and linear mass density of the string. In this case, the wave speed in the first string is 58 m/s. To find the resulting wave speed in the second string with a 10% greater linear density and the same tension, we can use the formula for wave speed:
v = √(T/μ)
Let's assume the linear mass density of the first string is μ and the linear mass density of the second string is 1.1μ. Since the tension is the same in both strings, we have:
v1 = v2
√(T/μ) = √(T/(1.1μ))
Cross multiplying and simplifying, we get:
Tμ = 1.1T
μ = 1.1
So, the resulting wave speed in the second string will be 58 m/s. Therefore, the wave speed remains the same.
A boy drags a 100 N sled up a 20 slope at constant velocity. If the coefficient of friction between sled and hill is 0.20, what force must he exert at an angle of 35 with respect to the hill?
Answer:
[tex]F = 56.75 N[/tex]
Explanation:
As per the free body diagram of the box we can say that the applied force must be greater than or equal to the sum of friction force and component of the weight of the box along the hill
So we can say that
[tex]F_{net} = F cos35 - (mgsin20 + F_f)[/tex]
on the other side the force perpendicular to the plane must be balanced so that it remains in equilibrium in that direction
so we can say that
[tex]F_n + Fsin35 = mgcos20[/tex]
now we will have
[tex]F_f = \mu F_n[/tex]
[tex]F_f = (0.20)(mg cos20 - Fsin35)[/tex]
now we have
[tex]0 = Fcos35 - mg sin20 - (0.20)(mg cos20 - Fsin35)[/tex]
[tex]F(cos35 + 0.20 sin35) = mg sin20 + 0.20 mgcos20[/tex]
[tex]F = \frac{mg sin20 + 0.20 mgcos20}{(cos35 + 0.20 sin35)}[/tex]
[tex]F = \frac{100(sin20 + 0.20 cos20)}{(cos35 + 0.20 sin35)}[/tex]
[tex]F = 56.75 N[/tex]
A proton moves through a region containing a uniform electric field given by E with arrow = 54.0 ĵ V/m and a uniform magnetic field B with arrow = (0.200 î + 0.300 ĵ + 0.400 k) T. Determine the acceleration of the proton when it has a velocity v with arrow = 170 î m/s.
Answer:
acceleration is 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k ) m/s
Explanation:
given data
uniform electric field E = 54.0 ĵ V/m
uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T
velocity v = 170 î m/s.
to find out
acceleration
solution
we know magnetic force for proton is
i.e = e (velocity × uniform magnetic field)
magnetic force = e (170 î × (0.200 î + 0.300 ĵ + 0.400 k) )
magnetic force = e (- 68 ĵ + 51 k ) ..................1
and now for electric force for proton i.e
= uniform electric field × e ĵ
electric force = e (54 ĵ ) ............2
so net force will be add magnetic force + electric force
from equation 1 and 2
e (- 68 ĵ + 51 k ) + e (54 ĵ )
e (- 14 ĵ + 51 k )
so the acceleration (a) for proton will be
net force = mass × acceleration
a = e (- 14 ĵ + 51 k ) / 1.6 [tex]10^{-19}[/tex] / 1.67 × [tex]10^{-27}[/tex]
acceleration = 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k ) m/s
The acceleration of the proton can be determined using the equation F = qvBsinθ. Plug in the given values to calculate the acceleration.
Explanation:The acceleration of the proton can be determined using the equation F = qvBsinθ, where F is the force, q is the charge of the proton, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since the question does not provide the angle, we assume that θ = 0. Therefore, the acceleration of the proton is given by:
a = (q * v * B) / m
where m is the mass of the proton. Plugging in the given values, we can calculate the acceleration.
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A machine gun fires 33 g bullets at a speed of 342 m/s. The gun fires 115 bullets/min.What is the average force the shooter must exert to keep the gun from moving?
Final answer:
The required force is approximately 21.63 newtons.
Explanation:
The question involves determining the average force that a shooter must exert to keep a gun from moving while it fires bullets. To find this force, we can use the concept of impulse, which relates force, time, and change in momentum.
The gun fires 115 bullets per minute, which is 115/60 bullets per second. Multiplying the number of bullets per second by the mass and speed of each bullet gives the total momentum per second, which is the same as the force when mass is in kg and speed is in m/s.
First, we find the momentum of each bullet:
Bullet speed (v) = 342 m/s
F ≈ 21.63 N
The shooter must exert an average force of approximately 21.63 newtons to keep the gun from moving.
Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?
Answer:
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
At the time of collision velocity of ball one is descending.
Explanation:
Velocity of ball 1 = 146 ft/sec = 44.50m/s
The balls are to collide at an altitude of 234 ft
H = 234 ft = 71.32 m
We have equation of motion
v² = u² + 2as
v² = 44.50² + 2 x (-9.81) x 71.32
v = ±24.10 m/s.
Time for each velocity can be calculated using equation of motion
v = u + at
24.10 = 44.50 - 9.81 t , t = 2.07 s
-24.10 = 44.50 - 9.81 t , t = 6.99 s
Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.
So at the time of collision velocity of ball one is descending.
The collision occurs at t = 6.99 s.
Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.
Height traveled by ball 2 = 71.32 m
We need to find velocity
We have
s = ut + 0.5 at²
71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²
u = 38.21 m/s = 125.36 ft/s
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
Answer:
v2=139 ft
Explanation:
First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.
h_1=v_1*t-(a*t/2)
We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.
t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s
Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.
Solving a similar equation, but this time for v2 to obtain the result.
h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft
A puck sliding on ice approaches a 20 degrees ramp at 10 m/s with coefficient of friction 0.2. How high will it go up the ramp along the ramp’s surface?
Answer:
it will go up along the inclined plane by d = 9.62 m
Explanation:
As we know that puck is moving upward along the slide
then the net force opposite to its speed is given as
[tex]F_{net} = - mgsin\theta - \mu mgcos\theta[/tex]
so here deceleration is given as
[tex]a = -g(sin\theta + \mu cos\theta)[/tex]
now plug in all values in it
[tex]a = -9.81(sin20 + 0.2cos20)[/tex]
[tex]a = -5.2 m/s^2[/tex]
now the distance covered by the puck along the plane is given as
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 10^2 = 2(-5.2)d[/tex]
[tex]d = 9.62 m[/tex]
A vertical block-spring system on earth has a period of 6.0 s. What is the period of this same system on the moon where the acceleration due to gravity is roughly 1/6 that of earth? A)1.0s B)2.4s C)6.0s D)15s E)36s
The period of the vertical block-spring system on the moon is approximately 2.45 seconds.
Explanation:The period of a vertical block-spring system is determined by the square root of the ratio of the mass of the block to the spring constant. Since the mass and spring constant remain the same, the period will only be affected by the acceleration due to gravity.
On the moon, where the acceleration due to gravity is about 1/6 that of Earth, the period of the system will be sqrt(6), or approximately 2.45 seconds.
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