Answer:
From lowest to highest acceleration:
3rd train
2nd train
1st train
Explanation:
The acceleration of an object can be found by using Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
a is the acceleration
F is the net force on the object
m is the mass of the object
We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.
So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.
The acceleration of the three trains can be ranked as follows: the single empty freight car train, the 50 empty freight cars train, and the 50 equally loaded freight cars train.
Explanation:The accelerations of the three trains can be ranked as follows:
The single empty freight car trainThe 50 empty freight cars trainThe 50 equally loaded freight cars trainWhen an equal force is applied to each train, the train with fewer cars will experience a greater acceleration. This is because the force is distributed over fewer cars, resulting in a larger net force acting on each car.
For example, consider two cars:
Single empty freight car train: The force is applied to only one car, resulting in a larger net force per car50 empty freight cars train: The force is distributed over 50 cars, resulting in a smaller net force per carTherefore, the single empty freight car train will have the highest acceleration, followed by the 50 empty freight cars train, and finally the 50 equally loaded freight cars train.
A 1400 kg wrecking ball hangs from a 20-m-long cable. The ball is pulled back until the cable makes an angle of 23.0 ∘ with the vertical. Part A By how much has the gravitational potential energy of the ball changed?
Answer:
21813.46 J
Explanation:
The Potential energy stored at the time it is hanging freely is mgl.
the potential energy stored at the time when it makes a angle 23 degree from the vertical is mglCos 23
Change in gravitational potential energy
U = mgl = mgl cos 23
U = mgl (1 - cos23)
U = 1400 x 9.8 x 20 ( 1 - Cos 23)
U = 21813.46 J
What's the average required power to raise a 150-kg drum to a height of 20 m in a time of 1.0 min? (Answer in kW)
Answer:
Power required, P = 0.49 kW
Explanation:
It is given that,
Mass of drum, m = 150 kg
Height, h = 20 m
Time, t = 1 min = 60 sec
Average power required is given by work done divided by total time taken. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
W = F.d
Here, W = mgh
So, [tex]P=\dfrac{mgh}{t}[/tex]
[tex]P=\dfrac{150\ kg\times 9.8\ m/s^2\times 20\ m}{60\ s}[/tex]
P = 490 watts
or P = 0.49 kW
So, the average power required to raise the drum is 0.49 kilo-watts. Hence, this is the required solution.
Final answer:
The average power required to lift a 150-kg drum to a height of 20 meters in 1 minute is calculated by first finding the work done and then dividing by time. The result is approximately 0.4905 kW.
Explanation:
Calculating Required Power to Lift a Drum
The problem involves finding the average power required to raise a certain weight to a specified height over a defined period of time. The weight in question is a 150-kg drum, the height is 20 meters, and the time is 1 minute (60 seconds).
To calculate power, you need to first calculate the work done, which is the product of force and displacement. Since we are lifting the object vertically, the force is the weight of the object, which is mass (m) times the acceleration due to gravity (g), and the displacement is the height (h). The formula for work (W) is:
W = m × g × h
The average power (P), then, is the work done over the time (t) it takes to do it:
P = W / t
Using the values given:
mass (m) = 150 kg
acceleration due to gravity (g) = 9.81 m/s²
height (h) = 20 m
time (t) = 60 s
First, calculate the work done:
W = 150 kg × 9.81 m/s² × 20 m = 29430 J
Now calculate the power required:
P = 29430 J / 60 s = 490.5 W
To convert this into kilowatts (kW), divide by 1000:
P = 0.4905 kW
Therefore, the average power required to lift the drum is approximately 0.4905 kW.
A 24-g rifle bullet traveling 280 m/s buries itself in a 3.7-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Part A Determine the vertical and horizontal components of the pendulum's maximum displacement.
Answer:
x = 0.95 m
y = 0.166 m
Explanation:
m = mass of the bullet = 24 g = 0.024 kg
v = speed of bullet before collision = 280 m/s
M = mass of the pendulum = 3.7 kg
L = length of the string = 2.8 m
h = height gained by the pendulum after collision
V = speed of the bullet and pendulum combination
Using conservation of momentum
m v = (m + M) V
(0.024) (280) = (0.024 + 3.7) V
V = 1.805 m/s
using conservation of energy
Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination
(m + M) g h = (0.5) (m + M) V²
(9.8) h = (0.5) (1.805)²
h = 0.166 m
y = vertical displacement = h = 0.166 m
x = horizontal displacement
horizontal displacement is given as
x = sqrt(L² - (L - h)²)
x = sqrt(2.8² - (2.8 - 0.166)²)
x = 0.95 m
An object is attached to a spring, the spring is stretched by 23.7 cm in the negative direction, and the object is released, oscillating in simple harmonic motion. After 0.317 s, it is again 23.7 cm from the equilibrium position, having passed through the equilibrium position once in those 0.317 s. Determine the speed of the object after 1.39 seconds. Number cm/s
Answer:83.17 cm/s
Explanation:
Let positive x be negative direction and negative x be positive direction in this question
General equation of motion of SHM is
x=[tex]Asin\left ( \omega_{n}t\right )[/tex] --------1
where [tex]\omega [/tex]is natural frequency of motion given by
[tex]\omega_n[/tex]=[tex]\sqrt{\frac{k}{m}}[/tex]
Where K is spring constant
here A=23.7cm
And it is given it is again at 23.7 from equilibrium position having passed through the equilibrium once.
i.e. it covers this distance in [tex]\frac{T}{2}[/tex] sec
where T is the time period of oscillation i.e. returning to same place after T sec
therefore T=0.634 sec
differentiating equation 1 we get
v=[tex]A\omega_n[/tex]cos[tex]\left ( \omega_{n}t\right )[/tex]
and [tex]T\times \omega_n[/tex]=[tex]2\pi[/tex]
[tex]\omega_n[/tex]=9.911rad/s
[tex]v[/tex]=[tex]23.7\times 9.911cos\left (789.261\degree\right)[/tex]
v=83.17 cm/s
Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.55 μC charge and flies due west at a speed of 685 m/s over the Earth's south magnetic pole, where the 8.00 x 10^-5 T magnetic field points straight up. What is the magnitude of the magnetic force on the plane, in newtons?
Answer:
3.014 x 10⁻⁸ N
Explanation:
q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C
v = speed of the jet = 685 m/s
B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T
θ = angle between the magnetic field and direction of motion = 90
magnitude of the magnetic force is given as
F = q v B Sinθ
F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90
F = 3.014 x 10⁻⁸ N
Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M = 1kg and radius R = 1m about an axis perpendicular to the hoop’s plane at an edge. (Express your answer in units of kg*m^2).
Answer:
2 kg m^2
Explanation:
M = 1 kg, R = 1 m
The moment of inertia of the hoop about its axis perpendicular to its plane is
I = M R^2
The moment of inertia of the hoop about its edge perpendicular to it splane is given by the use of parallel axis theorem
I' = I + M x (distance between two axes)^2
I' = I + M R^2
I' = M R^2 + M R^2
I' = 2 M R^2
I' = 2 x 1 x 1 x 1 = 2 kg m^2
The moment of inertia of a hoop about an axis perpendicular to its plane at an edge is calculated using the Parallel-Axis Theorem and for a hoop with mass M = 1kg and radius R = 1m, it is 2 kg*m².
Calculation of the Moment of Inertia for a Hoop
To find the moment of inertia of a hoop with mass M = 1kg and radius R = 1m about an axis perpendicular to the hoop's plane at an edge, we use the Parallel-Axis Theorem. The moment of inertia of a hoop about its central axis (through its center, perpendicular to the plane) is MR². According to the Parallel-Axis Theorem, the moment of inertia about an axis parallel to this but passing through the edge of the hoop is given by I = MR₂ + MR₂ (because the distance from the central axis to the outer edge of the hoop is R). Thus, the moment of inertia for the hoop about the edge is 2MR₂ which simplifies to 2 * 1kg * (1m)² = 2 kg*m².
When an electron enters a magnetic field, it will accelerate up the field. True OR False
Answer:
True
Explanation:
The force on the electron when it enters in a magnetic field is given by
F = q ( v x B)
F = -e x V x B x Sin∅
here, F is the force vector, B be the magnetic field vector and v be the velocity vector.
If the angle between the velocity vector and the magnetic field vector is 0 degree, then force is zero.
When the electrons enters in the magnetic field at any arbitrary angle, it experiences a force and hence it accelerate up.
A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 ? What is the magnitude of the induced current?
Answer:
0.5 A
Explanation:
N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm
The induced emf is given by
e = - N dФ/dt
Where, dФ/dt is the rate of change of magnetic flux.
Ф = B A
dФ/dt = A dB/dt
so,
e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V
negative sign shows the direction of magnetic field.
induced current, i = induced emf / resistance = 0.2 / 0.4 = 0.5 A
In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. Find the rotational kinetic energy of the pitcher’s arm given its moment of inertia is 0.720 kg m2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder.
The rotational kinetic energy of the pitcher's arm when throwing a softball at a speed of 139 km/h is calculated using the equation for rotational kinetic energy, and taking moment of inertia and angular velocity into account. The angular velocity is inferred from the linear speed of the ball and the distance it leaves the pitcher's hand from the pivot at the shoulder. The rotational kinetic energy is found to be approximately 1491 Joules.
Explanation:This question pertains to the rotational kinetic energy of the pitcher's arm during the act of throwing a softball. By definition, the kinetic energy associated with rotational motion (rotational kinetic energy) can be given by the equation: K_rot = 0.5 * I * ω² where 'I' is the moment of inertia and 'ω' is the angular velocity.
To calculate the angular velocity 'ω', we can infer it from the linear speed of the ball when it leaves the pitcher's hand, as 'ω = v/r', v = 139 km/h = 38.6 m/s, and r = 0.600 m (distance ball leaves hand from pivot at shoulder). Therefore, 'ω' is approximately 64.4 rad/s.
Substituting 'I' and 'ω' into the equation above, we get K_rot = 0.5 * 0.720 kg*m² * (64.4 rad/s)² = 1491 Joules, which is the rotational kinetic energy of the pitcher's arm.
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A 3.0 kg ball tied to the end of a 50 cm long string being swung in a circle in a horizontal plane at constant speed. If the tension of the string is 3.8 N, what is the speed of the ball?
Question 31 options:
A)
1.2 m/s
B)
1.0 m/s
C)
1.4 m/s
D)
0.8 m/s
Answer:
The speed of the ball is 0.8 m/s.
(D) is correct option.
Explanation:
Given that,
mass of ball = 3.0 kg
length = 50 cm
We need to calculate the speed of the ball
Using relation between the centripetal force and tension
[tex]\dfrac{mv^2}{r}=T[/tex]
[tex]v^2=\dfrac{T\times r}{m}[/tex]
Where, m = mass
T = tension
r = radius
Put the value into the formula
[tex]v=\sqrt{\dfrac{3.8\times50\times10^{-2}}{3.0}}[/tex]
[tex]v=0.8\ m/s[/tex]
Hence, The speed of the ball is 0.8 m/s.
The resistivity of gold is 2.44×10-8 Ω•m at room temperature. A gold wire that is 1.8 mm in diameter and 40 cm long carries a current of 860 mA. What is the electric field in the wire?
To find the electric field in the wire, we first find the cross-sectional area of the wire and then substitute the given values and the calculated area into the formula for electric field. The electric field in the gold wire is approximately 8.24 kN/C.
Explanation:The question asks for the electric field in a gold wire of given dimensions and current. We'll solve this using the formula for electric field (E) in terms of resistivity (ρ),current (I) and area (A), which is given by E = ρI/A.
First, we need to find the cross-sectional area (A) of the wire using the formula for the area of a circle, since the wire is cylindrical. The area of a circle is given by A = π*(d/2)^2, where d is the diameter. Substituting d = 1.8 mm or 1.8 * 10^-3 m, we find A ≈ 2.54 * 10^-6 m^2.
Next, we need to substitute the known values into the formula E = ρI/A. Using the given values ρ = 2.44 *10^-8 Ω•m, I = 860 mA or 860 * 10^-3 A, and the calculated A ≈ 2.54 * 10^-6 m^2, we find E ≈ 8.24 * 10^3 N/C or 8.24 kN/C.
So, the electric field in the wire is approximately 8.24 kN/C.
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The electric field within the gold wire can be calculated utilizing resistivity formula, determining the cross-sectional area of the wire, and then computing the current density. After these intermediate calculations, the ultimate electric field in the wire is found to be approximately 72 mV/m.
Explanation:
The electric field in the gold wire can be calculated by using the formula for
resistivity: ρ=E/J where
ρ (rho) is the resistivity of the material,
E is the electric field, and
J is the current density.
Before we apply this formula, let's calculate the cross-sectional area of the wire (A) using the formula A=πr², where r is the radius of the wire. The diameter is given as 1.8mm, therefore
r=0.9mm = 0.9x10^-3 m.
Substituting this into the formula gives us
A=π(0.9x10^-3)²≈2.54x10^-6 m².
Now we calculate current density: J=I/A. where I is current, given as
860mA=860x10^-3 A.
Substituting values of I and A into the formula gives
J=860x10^-3/2.54x10^-6≈3.39x10^8 A/m².
Finally, the electric field E=ρ/J=2.44x10^-8/3.39x10^8 ≈ 7.2x10^-2 V/m or 72mV/m.
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Two circular rods, one steel and the other copper, are both 0.780 m long and 1.50 cm in diameter. Each is subjected to a force with magnitude 4350 N that compresses the rod. What is the difference in the length of the two rods when compressed?
Answer:
The difference in the length of the two rods when compressed is [tex]5.4\times10^{-5}\ m[/tex].
Explanation:
Given that,
Length = 0.780 m
Diameter = 1.50 cm
Force = 4350 N
(a). For steel rod
We know ,
The young modulus for steel rod
[tex]Y=2\times10^{11}[/tex]
Using formula of young modulus
[tex]e_{s}=\dfrac{Fl}{AY}[/tex]
[tex]e_{s}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times2\times10^{11}}[/tex]
[tex]e_{s}=9.6\times10^{-5}\ m[/tex]
(b). For copper rod
We know ,
The young modulus for steel rod
[tex]Y=1.1\times10^{11}[/tex]
Using formula of young modulus
[tex]e_{c}=\dfrac{Fl}{AY}[/tex]
[tex]e_{c}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times1.1\times10^{11}}[/tex]
[tex]e_{c}=1.5\times10^{-4}\ m[/tex]
The difference in the length of the two rods when compressed is
[tex]difference\ in\ length=e_{c}-e_{s}[/tex]
[tex]difference\ in\ length=1.5\times10^{-4}-9.6\times10^{-5}[/tex]
[tex]difference\ in\ length =5.4\times10^{-5}\ m[/tex]
Hence, The difference in the length of the two rods when compressed is [tex]5.4\times10^{-5}\ m[/tex].
Given that a fluid at 260°F has a kinematic viscosity of 145 mm^2/s, determine its kinematic viscosity in SUS at 260°F.
Answer:
kinematic viscosity in SUS is = 671.64 SUS
Explanation:
given data
kinetic viscosity = 145 mm^2/s
we know
1 mm = 0.1 cm
so kinetic viscosity in cm is [tex]\nu =145 (0.1)^{2} =1.45 cm^{2}/s[/tex]
other unit of kinetic viscosity is centistokes
[tex]1 cm^{2}/s = 100 cst[/tex]
so 1.45 cm^2/s will be 145 cst
if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS
Two ocean liners, each with a mass of 42,000 metric tons, are moving on parallel courses, 93 m apart. What is the magnitude of the acceleration of one of the liners toward the other due to their mutual gravitational attraction? Treat the ships as particles.
Answer:
0.324×10⁻³ m/s²
Explanation:
G=Gravitational constant=6.67408×10⁻⁸ m³/kg s
m₁=m₂=Mass of the ships=42000×10³ kg
r=Distance between the ships=93 m
The ships are considered particles
From Newtons Law of Universal Gravitation
[tex]F=G\frac {m_1m_2}{r^2}\\Here\ m_1=m_2\ hence\ m_1\times m_2=m_1^2\\\Rightarrow F=G\frac {m_1^2}{r^2}\\\Rightarrow F=\frac{6.67408\times 10^{-8}\times (42000\times 10^3)^2}{93^2}\\\Rightarrow F=13612.0674\ N\\[/tex]
[tex]F=ma\\\Rightarrow 13612.0674=42000\times 10^3a\\\Rightarrow a=0.324\times 10^{-3}\ m/s^2[/tex]
∴Acceleration of one of the liners toward the other due to their mutual gravitational attraction is 0.324×10⁻³
A gas expands from an initial volume of 0.040 m^3 and an initial pressure of 210 kPa to a final volume of 0.065 m^3 while its temperature is kept constant. How much work is done by the system?
Answer:
286
Explanation:
p1v1/T1=p2v2/T2
then subtitude your values
T2=760*0.65*273/210*0.040
T2=135/8.4=16+273=289
The pressure-volume work done by the system during the expansion is 5.25 x 10³J.
What is done by pressure-volume work?The work that is done when a fluid is compressed or expanded is known as pressure-volume work. Pressure-volume work occurs whenever there is a change in volume but the outside pressure stays the same.
Here,
The pressure of the system of gas, P = 210 kPa
Initial volume of the system of gas, V₁ = 0.04 m³
Final volume of the system of gas, V₂ = 0.065 m³
The expression for the pressure-volume work done by a system of gas is given by,
Work done,
W = PΔV
W = P(V₂ - V₁)
Applying the values of P, V₁ and V₂,
W = 210 x 10³(0.065 - 0.04)
W = 210 x 10³x 0.025
W = 5.25 x 10³J
Hence,
The pressure-volume work done by the system during the expansion is 5.25 x 10³J.
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Mario, a hockey player, is skating due south at a speed of 4.79 m/s relative to the ice. A teammate passes the puck to him. The puck has a speed of 9.13 m/s and is moving in a direction of 17.8 ° west of south, relative to the ice. What are (a) the magnitude and (b) direction (relative to due south) of the puck's velocity, as observed by Mario
Mario's velocity vector relative to the ice is
[tex]\vec v_{M/I}=-4.79\,\vec\jmath[/tex]
(note that all velocities mentioned here are given in m/s)
The puck is moving in a direction of 17.8 degrees west of south, or 252.2 degrees counterclockwise relative to east. Its velocity vector relative to the ice is then
[tex]\vec v_{P/I}=9.13(\cos252.2^\circ\,\vec\imath+\sin252.2^\circ\,\vec\jmath)=-2.79\,\vec\imath-8.69\,\vec\jmath[/tex]
The velocity of the puch relative to Mario is
[tex]\vec v_{P/M}=\vec v_{P/I}+\vec v_{I/M}=\vec v_{P/I}-\vec v_{M/I}[/tex]
[tex]\vec v_{P/M}=-2.79\,\vec\imath-3.90\,\vec\jmath[/tex]
Then, relative to Mario,
a. the puck is traveling at a speed of [tex]\boxed{\|\vec v_{P/M}\|=4.80}[/tex], and
b. is moving in a direction [tex]\theta[/tex] such that
[tex]\tan\theta=\dfrac{-3.90}{4.80}\implies\theta=-126^\circ[/tex]
which is about [tex]\boxed{35.6^\circ}[/tex] west of south.
The magnitude of the puck's velocity as observed by Mario is 4.9 m/s and the direction is 35.6 degrees west of south.
Explanation:When addressing this problem, we're dealing with relative velocity in two dimensions. Since we need to find the puck's velocity as seen by Mario, we should use the principle that the relative velocity of the puck with respect to Mario is the vector difference between the velocity of the puck and Mario.
Given that Mario is travelling south at 4.79 m/s and the puck is travelling at an angle of 17.8 degrees west of south at a speed of 9.13 m/s, we can break the puck's velocity into southward and westward components. The southward component is 9.13 m/s × cos(17.8°) = 8.74 m/s and the westward component is 9.13 m/s × sin(17.8°) = 2.84 m/s.
The relative southward velocity of the puck with respect to Mario is 8.74 m/s - 4.79 m/s = 3.95 m/s. The westward component is unchanged since Mario has no westward velocity. So, the puck appears to Mario to be moving southwest at an angle of arctan(2.84 m/s / 3.95 m/s) = 35.6 degrees west of south.
The magnitude is given by the Pythagorean theorem, sqrt((2.84 m/s)² + (3.95 m/s)²) = 4.9 m/s.
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Water vapor can react reversibly with solid carbon to yield a mixture of hydrogen gas and carbon monoxide. Suppose you continually add more water vapor to the reaction. In what direction does this shift the equilibrium?
Answer:
Product side
Explanation:
When water vapor reacts reversibly with solid carbon to yield a mixture of hydrogen gas and carbon monoxide and we continually add more water vapor to the reaction the equilibrium of the reaction shifts to the product side.
Because gaseous water is reactant that appears in the reaction quotient expression.
[tex]H_{2}O+ C_{s}\leftrightharpoons H_{2}_{g}+ CO[/tex]
When we add more water vapor to the reaction the product formation is increased. The reaction goes in forward direction affecting the equilibrium.
A solid iron cylinder weighs 600 N. It has a density of 7860 kg/m^3. (a) Calculate the cylinder's volume. (b) How much would the cylinder weigh if it is completely submerged in water?
Answer:
7.66 (litres); 523.66 (N).
Explanation:
all the details are provided in the attached picture. Note, P_i means 'Weight of iron', ρ_i means "Density of iron', ρ_w means 'Density of water', F_f means 'Arhimede force'.
The answers are marked with red colour.
What is log2 8 ? [watch out for the different base]
Answer:
[tex]log_28=3[/tex]
Explanation:
To calculate the value of [tex]log_28[/tex]
Also,Some relation for log values:
[tex]loga^m= mloga[/tex] -----------------------------------------------------1
[tex]log_aa= 1[/tex]---------------------------------------------------------------2
Also, [tex]2^3= 8[/tex]
Applying in the question as:
[tex]log_28=log_2(2^3)[/tex]
Using Relation-1 mentioned above:
[tex]log_28=3\times log_2(2)[/tex]
Using Relation-2 mentioned above:
[tex]log_28=3\times 1[/tex]
Thus,
[tex]log_28=3[/tex]
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 × 108 m. Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is 5.60 × 1010 m from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.
Answer:
(a):The time what it takes his voice reach the earth via radio waves is t1= 1.28 seconds.
(b): The time what it takes his voice reach from Mars to the earth via radio waves is t2= 186.66 seconds.
Explanation:
d1= 3.85 *10⁸ m
d2= 5.6 *10¹⁰ m
C= 300,000,000 m/s
t1= d1/C
t1= 1.28 s
t2= d2/C
t2= 186.66 s
An object with a mass of 7.8 kg is pulled on a horizontal surface by a horizontal pull of 50 N to the right. The friction force on this object is 30 N to the left. What is the acceleration of the object? 0.16 m/s^2
10 m/s^2
2.6 m/s^2
6.4 m/s^2
Answer:
2.6 m/s²
Explanation:
m = mass of the object = 7.8 kg
Consider the right direction as positive and left direction as negative
F = horizontal applied force towards right = 50 N
f = frictional force acting towards left = - 30 N
a = acceleration
Force equation for the motion of the object is given as
F + f = ma
50 - 30 = (7.8) a
a = 2.6 m/s²
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.00 s.
Answer:
The net force acting on this object is 180.89 N.
Explanation:
Given that,
Mass = 3.00 kg
Coordinate of position of [tex]x= 5t^3+1[/tex]
Coordinate of position of [tex]y=3t^2+2[/tex]
Time = 2.00 s
We need to calculate the acceleration
[tex]a = \dfrac{d^2x}{dt^2}[/tex]
For x coordinates
[tex]x=5t^3+1[/tex]
On differentiate w.r.to t
[tex]\dfrac{dx}{dt}=15t^2+0[/tex]
On differentiate again w.r.to t
[tex]\dfrac{d^2x}{dt^2}=30t[/tex]
The acceleration in x axis at 2 sec
[tex]a = 60i[/tex]
For y coordinates
[tex]y=3t^2+2[/tex]
On differentiate w.r.to t
[tex]\dfrac{dy}{dt}=6t+0[/tex]
On differentiate again w.r.to t
[tex]\dfrac{d^2y}{dt^2}=6[/tex]
The acceleration in y axis at 2 sec
[tex]a = 6j[/tex]
The acceleration is
[tex]a=60i+6j[/tex]
We need to calculate the net force
[tex]F = ma[/tex]
[tex]F = 3.00\times(60i+6j)[/tex]
[tex]F=180i+18j[/tex]
The magnitude of the force
[tex]|F|=\sqrt{(180)^2+(18)^2}[/tex]
[tex]|F|=180.89\ N[/tex]
Hence, The net force acting on this object is 180.89 N.
To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.
Explanation:To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. The x-component of the force can be found by taking the derivative of the x-coordinate equation with respect to time and multiplying it by the mass of the object. The same process can be applied to find the y-component of the force. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.
Starting with the x-coordinate equation, x = 5t³ - 1, taking its derivative gives dx/dt = 15t². Multiplying the derivative by the mass of the object (3.00 kg) gives the x-component of the force: Fx = 3.00 kg * 15t².
Similarly, for the y-coordinate equation y = 3t² + 2, taking its derivative gives dy/dt = 6t. Multiplying the derivative by the mass of the object (3.00 kg) gives the y-component of the force: Fy = 3.00 kg * 6t.
Using the Pythagorean theorem, the magnitude of the net force Fnet can be calculated as: Fnet = sqrt(Fx² + Fy²). Substituting the values into the equation will give you the magnitude of the net force at t = 2.00 s.
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C is fixed to the x axis at x = +2.0 m. Find the location of the place(s) along the x axis where the electric field due to these two charges is zero.
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}[/tex]
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}[/tex]
x = - 1.4 m
Electric field is zero due to positive point charge Q1 and negative Q2 along the x axis is at the location of 3.62 meters.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Given information-
The charge of the point 1 is [tex]2.5\times10^{-5} \rm C[/tex].
The charge of the point 2 is [tex]-5.0\times10^{-6} \rm C[/tex].
The distance between the point 1 and point 2 is 2 meters away from the x axis.
Let the position of the point 1 is at x.
Thus the electric force on point Q1 is,
[tex]F_1=\dfrac{kQ1}{x^2}[/tex]
As the distance between the point Q1 and point Q2 is 2 meters away from the x axis. Thus the position of it should be at (x+2). The electric force on point 2
[tex]F_2=\dfrac{kQ1}{(x+2)^2}[/tex]
As the force of two is equal and opposite thus,
[tex]\dfrac{kQ1}{x^2}=\dfrac{kQ2}{(x+2)^2}\\\dfrac{2.5\times10^{-5}}{x^2}=\dfrac{-5\times10^{-6}}{(x+2)^2}\\x=1.62[/tex]
Thus the position of point 2 is,
[tex]p_2=2+1.62\\p_2=3.62\rm m[/tex]
Thus, the location of the place along the x axis where the electric field due to these two charges is zero is 3.62 meters.
Learn more about electric field here;
https://brainly.com/question/14372859
An electron is released from rest in a uniform electric field. The electron accelerates, travelling 5.50 m in 4.00 µs after it is released. What is the magnitude of the electric field in N/C?
Answer:
3.91 N/C
Explanation:
u = 0, s = 5.50 m, t = 4 us = 4 x 10^-6 s
Let a be the acceleration.
Use second equation of motion
s = u t + 1/2 a t^2
5.5 = 0 + 1/2 a (4 x 10^-6)^2
a = 6.875 x 10^11 m/s^2
F = m a
The electrostatic force, Fe = q E
Where E be the strength of electric field.
So, q E = m a
E = m a / q
E = (9.1 x 10^-31 x 6.875 x 10^11) / ( 1.6 x 10^-19)
E = 3.91 N/C
While on a sailboat at anchor, you notice that 15 waves pass its bow every minute. The waves have a speed of 6.0 m/s . Part A What is the distance between two adjacent wave crests
Answer:
Distance between two adjacent wave crests = 24m
Explanation:
Distance= speed × time
Distance traveled by waves in 60 seconds (15 crests)= 15 × distance
15 × distance = 6,0 (meters/second) × 60 seconds
distance = (360 meters) / 15 = 24 meters (between two adyacent waves)
. The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration?
Answer:
1.59 m/s^2, 65.2°
Explanation:
F1 = 390 N North
F2 = 180 N east
m = 270 kg
Net force is the vector sum of both the forces.
[tex]F = \sqrt{F_{1}^{2}+F_{2}^{2}}[/tex]
[tex]F = \sqrt{390^{2}+180^{2}}[/tex]
F = 429.53 N
Direction of force
tan∅ = F1 / F2 = 390 / 180 = 2.1667
∅ = 65.2°
The direction of acceleration is same as the direction of net force.
The magnitude of acceleration is
a = F / m = 429.53 / 270 = 1.59 m/s^2
Final answer:
The magnitude of the sailboat's acceleration is 1.59 m/s², and the direction is 25 degrees east of north. This is calculated using Newton's second law and vector addition of the orthogonal forces exerted by the wind and water.
Explanation:
To calculate the magnitude and direction of the sailboat's acceleration, we need to use Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Here, we combine the forces exerted by the wind and the water to obtain the net force. The net force can be calculated using vector addition, where the force due to the wind (390 N north) and the force due to the water (180 N east) are treated as orthogonal vectors.
The net force (Fnet) is the vector sum of the two individual forces. We calculate the net force using the Pythagorean theorem:
Fnet = √Fwind² + Fwater² = √390² + 180² = √152100 + 32400 = √184500 N
So, the magnitude of the net force is approximately 429.53 N. To find the acceleration (a), we use the formula:
a = Fnet / m
Substituting the mass of the sailboat (270 kg) and the net force, we get:
a = 429.53 N / 270 kg = 1.59 m/s²
To determine the direction of the acceleration, we take the arctangent of the ratio of the forces. Since arctan(180/390) equals approximately 25 degrees, this is the angle east of north.
The sailboat's acceleration has a magnitude of 1.59 m/s² and a direction of 25 degrees east of north.
beam of light, traveling in air(the index of refraction for air is 1), strikes the surface of mineral oil at an ngle of 30° with the normal to the surface.( the index of refraction for this mineral oil is 1.38) What is the ngle of refraction? What is the speed of the light traveling in mineral oil?
Answer:
2.17 x 10^8 m/s
Explanation:
Angle of incidence, i = 30 degree, refractive index of mineral oil, n = 1.38
Let r be the angle of refraction.
By use of Snell's law
n = Sin i / Sin r
Sin r = Sin i / n
Sin r = Sin 30 / 1.38
Sin r = 0.3623
r = 21.25 degree
Let the speed of light in oil be v.
By the definition of refractive index
n = c / v
Where c be the speed of light
v = c / n
v = ( 3 x 10^8) / 1.38
v = 2.17 x 10^8 m/s
A 1200 W microwave oven transforms 1.8 x10^5 J of energy while reheating some food. Calculate how long the food was in the microwave. Answer in minutes.
Answer:
2.5 min
Explanation:
Hello
by definition the power is the amount of work done per unit of time. in this case, we know the total power and the work developed
[tex]1 Watt= \frac{Joule }{sec}[/tex]
Let
[tex]P=1200 W\\E=1.8 *10^{5}\\X=\frac{1.8 *10^{5} j }{1200 W}\\X= unknown\ time\ (sec) \\\\we\ need\ the\ answer\ in\ minutes,\\ 1\ min=60\ sec\\\\x=150 sec \\150\ sec*(\frac{1 min}{ 60 sec})=2.5 min\\[/tex]
hence , the data is not altered, we did it like this to eliminate the sec units.
Answer 2.5 min
I hope it helps
The food was in the 1200 W microwave for 150 seconds, which is equivalent to 2.5 minutes.
To calculate how long the food was in the microwave, we use the formula for power, which is the rate at which work is done or energy is transferred:
Power (P) = Energy (E) / Time (t)
First, rearrange this formula to solve for time (t):
t = E / P
Plugging in the given values:
So:
t = (1.8 times 10⁵J) / (1200 W)
Calculate this to find the time in seconds, and then convert it to minutes as follows:
t = 150 seconds
Divide by 60 to get minutes:
t = 2.5 minutes
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If the ramp is inclined at 29.9 o with respect to the ground, what is the component of her acceleration that is parallel to the ground?
Answer:
1.7 m/s²
Explanation:
d = length of the ramp = 13.5 m
v₀ = initial speed of the skateboarder = 0 m/s
v = final speed of the skateboarder = 7.37 m/s
a = acceleration
Using the equation
v² = v₀² + 2 a d
7.37² = 0² + 2 a (13.5)
a = 2.01 m/s²
θ = angle of the incline relative to ground = 29.9
a' = Component of acceleration parallel to the ground
Component of acceleration parallel to the ground is given as
a' = a Cosθ
a' = 2.01 Cos29.9
a' = 1.7 m/s²
What is the minimum index of refraction of a clear material if a minimum thickness of 121 nm , when laid on glass, is needed to reduce reflection to nearly zero when light of 675 nm is incident normally upon it? Assume that the film has an index less than that of the glass.
Answer:
[tex]\mu = 1.39[/tex]
Explanation:
Since the reflection is nearly zero intensity
so here we will say that the reflected light must show destructive interference
so here we have
[tex]path \: difference = \frac{\lambda}{2}[/tex]
[tex]2 \mu t = \frac{\lambda}{2}[/tex]
[tex]t = \frac{\lambda}{4\mu}[/tex]
here we have
[tex]\lambda = 675 nm[/tex]
t = 121 nm
now from above equation we have
[tex]\mu = \frac{675 nm}{4(121 nm)}[/tex]
[tex]\mu = 1.39[/tex]