a. To determine the rock's velocity after 5 seconds on each planet, we can use the formula:
velocity = initial velocity + acceleration × time
On Mars:
Acceleration on Mars = 3.71 m/s² (downward)
Initial velocity on Mars (when dropped) = 0 m/s
Time = 5 s
velocity on Mars = 0 m/s + 3.71 m/s² × 5 s
velocity on Mars = 18.55 m/s (downward)
On Earth:
Acceleration on Earth = 9.8 m/s² (downward)
Initial velocity on Earth (when dropped) = 0 m/s
Time = 5 s
velocity on Earth = 0 m/s + 9.8 m/s² × 5 s
velocity on Earth = 49 m/s (downward)
Therefore, the rock's velocity 5 seconds after being dropped on Mars is -18.55 m/s (downward), and on Earth, it is -49 m/s (downward).
b. To determine how the position changes after 5 seconds on Mars, we can use the formula:
position = initial position + initial velocity × time + 0.5 × acceleration × time²
On Mars:
Acceleration on Mars = 3.71 m/s² (downward)
Initial velocity on Mars (when dropped) = 0 m/s
Initial position on Mars = 0 m (assuming we dropped the rock from the surface)
Time = 5 s
position change on Mars = 0 m + 0 m/s × 5 s + 0.5 × 3.71 m/s² × (5 s)²
position change on Mars = 45.875 m (downward)
Therefore, if you drop a rock on Mars and measure the position change after 5 seconds, it would be 45.875 meters downward.
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A- The instantaneous position "y" of an object that is moving at a constant (or uniform) acceleration can be expressed in function of the time "t", its initial velocity vo, its initial position yo, and its acceleration "a". Give that expression with the units of all symbols
The instantaneous position of an object that is moving at a constant acceleration can be expressed as y = yo + vot + 1/2at², where y, yo, vo, a, and t have units of meters, meters per second, meters per second squared, and seconds, respectively.
The expression for the instantaneous position "y" of an object that is moving at a constant (or uniform) acceleration can be expressed in function of the time "t", its initial velocity vo, its initial position yo, and its acceleration "a".
The expression is:y = yo + vot + 1/2at² Where y = instantaneous position of the object yo = initial position of the object vo = initial velocity of the object a = acceleration of the object t = time taken by the object to reach the instantaneous position.
The units of all the symbols are:y: meters (m)yo: meters (m)vo: meters per second (m/s)a: meters per second squared (m/s²)t: seconds (s)
Therefore, the instantaneous position of an object that is moving at a constant acceleration can be expressed as y = yo + vot + 1/2at², where y, yo, vo, a, and t have units of meters, meters per second, meters per second squared, and seconds.
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Two cars collide head-on and stick together.
Car A, with a mass of 2000 kg, was initially
moving at a velocity of 10 m/s to the east. Car
B, with an unknown mass, was initially at rest.
After the collision, both cars move together at
a velocity of 5 m/s to the west. What is the
mass of Car B?
The mass of Car B is 2000 kg.
Given details:
Car A, mass = 2000 kg Initial velocity of Car A = 10 m/s (East)
Car B, mass = unknown Initial velocity of Car B = 0 m/s
After the collision, both cars move together
Velocity of both cars = 5 m/s (West)
Let the mass of Car B be m kg.
Using the law of conservation of momentum, the total momentum before the collision = total momentum after the collision.
So,2000 kg × 10 m/s + 0 kg × 0 m/s = (2000 + m) kg × 5 m/s (to the West)
20000 kg·m/s = (2000 + m) kg × 5 m/s (to the West)
20000 kg·m/s = 10000 kg·m/s + 5m kg·m/s
20000 kg·m/s - 10000 kg·m/s = 5m kg·m/s
10000 kg·m/s = 5m kg·m/sm
= 10000/5 kgm
= 2000 kg
Therefore, the mass of Car B is 2000 kg.
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When a potential difference of $12 \mathrm{~V}$ is applied to a wire $7.2 \mathrm{~mm}$ long and $0.34 \mathrm{~mm}$ in diameter the result is an electric current of $2.1 \mathrm{~A}$. What is the resistivity of the wire? Express your answer using two significant figures. In the units omega* $\mathrm{m}$.
The resistivity of the wire is 7.2 × 10^-7 Ωm (two significant figures)
The given parameters are:
Potential difference $V = 12$V,
length $l = 7.2$mm,
diameter $d = 0.34$mm and
electric current $I = 2.1$A.
Resistance of the wire is given by the formula as,
R = (ρl)/A
where R is resistance, ρ is resistivity of wire, l is length of wire, and A is the cross-sectional area of the wire.
Area of the wire,
A = (πd²)/4
Resistance can also be given as,
R = V/I
Combining both the equations,
R = V/I = (ρl)/A
We can rearrange this equation to find ρ as:ρ = (RA)/l
Given values: V = 12V, l = 7.2 mm, d = 0.34 mm and I = 2.1 A
The diameter of the wire is 0.34 mm so the radius is 0.17 mm or 0.00017 m.
Diameter of wire = 0.34 mm => radius of wire = 0.17 mm = 0.17 × 10^-3 m
Cross-sectional area of wire:
A = (πd²)/4 = (π(0.34 × 10^-3 m)²)/4 = 9.04 × 10^-8 m²
Resistance of wire:
R = V/I = 12/2.1 = 5.71 Ω
Resistivity of wire:
ρ = (RA)/l = (5.71 Ω × 9.04 × 10^-8 m²)/7.2 × 10^-3 m = 7.15 × 10^-7 Ωm≈ 7.2 × 10^-7 Ωm
Therefore, the resistivity of the wire is 7.2 × 10^-7 Ωm (two significant figures).
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The momentum of a small object moving in a straight line is 24 kg.m/s and its kinetic energy is 98 J. What is the mass of the object? 3 kg 4 kg b kig 8 kg
According to the question, the mass of the object is 24 kg.
To find the mass of the object, we can use the formula for kinetic energy and momentum.
The formula for kinetic energy is given by:
Kinetic Energy = (1/2) * mass * velocity^2
The formula for momentum is given by:
Momentum = mass * velocity
Given that the momentum is 24 kg.m/s and the kinetic energy is 98 J, we can set up the following equations:
24 kg.m/s = mass * velocity
98 J = (1/2) * mass * velocity^2
Since we are given the momentum and kinetic energy, we can solve for the velocity using the equation for momentum.
velocity = momentum / mass
Substituting the value of momentum and mass, we have:
24 kg.m/s = mass * (momentum / mass)
Simplifying, we get:
24 kg.m/s = momentum
Therefore, the mass of the object is 24 kg.
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An electric dipole consists of charges +e and -e separated by 0.56 nm. It is in an electric field of strength 5.50 106 N/C. The (imaginary) line connecting the two charges forms a 31° angle with the electric field.
(a) What is the net force on the electric dipole?
N
(b) What is the net torque on the electric dipole?
N·m
For the given electric dipole with charges +e and -e separated by 0.56 nm, and in an electric field of strength 5.50 x 10^6 N/C, we calculated the net force and net torque on the dipole.
(a) To find the net force on the electric dipole, we need to consider the forces acting on each charge of the dipole due to the electric field. The force on a charge in an electric field is given by the formula F = qE, where F is the force, q is the charge, and E is the electric field strength.
For the positive charge (+e), the force is in the direction of the electric field, so F_+e = +e * E. Since the electric field strength and charge are given, we can calculate F_+e = (+e) * (5.50 * 10^6 N/C).
For the negative charge (-e), the force is in the opposite direction of the electric field, so F_-e = -e * E. Thus, F_-e = (-e) * (5.50 * 10^6 N/C).
The net force on the dipole is the vector sum of these forces:
Net force = F_+e + F_-e.
(b) To find the net torque on the electric dipole, we can use the formula τ = p * E * sin(θ), where τ is the torque, p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field.
The dipole moment (p) of an electric dipole is given by p = q * d, where q is the magnitude of either charge and d is the separation between the charges.
We can calculate the dipole moment: p = (+e) * (0.56 nm).
Finally, the net torque on the dipole is given by:
Net torque = p * E * sin(θ).
Given:
Charge magnitude: e = 1.6 x 10^-19 C
Separation between charges: d = 0.56 nm = 0.56 x 10^-9 m
Electric field strength: E = 5.50 x 10^6 N/C
Angle between dipole moment and electric field: θ = 31°
(a) Calculating the net force:
F_+e = (+e) * E
F_+e = (1.6 x 10^-19 C) * (5.50 x 10^6 N/C)
F_+e = 8.8 x 10^-13 N
F_-e = (-e) * E
F_-e = (-1.6 x 10^-19 C) * (5.50 x 10^6 N/C)
F_-e = -8.8 x 10^-13 N
Net force = F_+e + F_-e
Net force = (8.8 x 10^-13 N) + (-8.8 x 10^-13 N)
Net force = 0 N
Therefore, the net force on the electric dipole is 0 N.
(b) Calculating the net torque:
Dipole moment: p = q * d
p = (e) * (0.56 x 10^-9 m)
p = 0.896 x 10^-19 C·m
Net torque = p * E * sin(θ)
Net torque = (0.896 x 10^-19 C·m) * (5.50 x 10^6 N/C) * sin(31°)
Net torque ≈ 2.56 x 10^-19 N·m
Therefore, the net torque on the electric dipole is approximately 2.56 x 10^-19 N·m.
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Wey Lidento calertme fiefs 5. The rotor of an electric generator is rotating at 200rpm when the motor is turn off. Due to frictional effects. the angular deceleration of the rotor after it is turned off is α=−0.015ωrad/s
2
, where ω is the angular velocity in rad/s. How many revolutions does the rotor turn after the motor turned off? How long does it take the motor moved off?
The number of revolutions the rotor turns after the motor is turned off can be calculated by finding the change in angular velocity and converting it to the number of revolutions.
The initial angular velocity is 200 rpm, which can be converted to rad/s by multiplying by 2π/60 to get approximately 20.94 rad/s. The final angular velocity is 0 rad/s since the rotor stops rotating.
The change in angular velocity is Δω = ωf - ωi = 0 - 20.94 = -20.94 rad/s.
To find the number of revolutions, we divide the change in angular velocity by 2π, since one revolution is equal to 2π radians:
Number of revolutions = Δω / (2π) = -20.94 / (2π) ≈ -3.34 revolutions.
Therefore, the rotor turns approximately -3.34 revolutions after the motor is turned off. The negative sign indicates that the rotor rotates in the opposite direction.
The time it takes for the motor to come to a stop can be determined by using the formula:
Δω = αt
Rearranging the equation, we have:
t = Δω / α
Substituting the values, we get:
t = (-20.94 rad/s) / (-0.015 rad/s²) ≈ 1396 seconds.
Therefore, it takes approximately 1396 seconds for the motor to come to a stop.
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Three point charges are arranged in a triangle as shown in the figure. - Point charge Q
1
has a charge of 5.36nC. - Point charge Q
2
has a charge of 5.72nC. - Point charge Q
3
has a charge of 2.25nC. - Point charges Q
1
and Q
2
are separated by a distance D
12
of 0.146 m. - Point charges Q
2
and Q
3
are separated by a distance D
23
of 0.525 m. - Point charges Q
1
and Q
3
are separated by a distance D
13
of 0.578 m. What is the electrostatic potential energy U
tot
of this configuration of charges? Assume that zero potential energy corresponds to all the charges being infinitely far apart. U
tot
= J
The electrostatic potential energy (Utot) of the configuration of charges is calculated using the formula Utot = k * (Q1 * Q2 / D12 + Q2 * Q3 / D23 + Q1 * Q3 / D13).
To calculate the electrostatic potential energy (Utot) of this configuration of charges, we can use the formula:
Utot = k * (Q1 * Q2 / D12 + Q2 * Q3 / D23 + Q1 * Q3 / D13)
Where:
- Utot is the total electrostatic potential energy
- k is the electrostatic constant, approximately equal to 8.99 x 10^9 N m^2/C^2
- Q1, Q2, and Q3 are the charges of the point charges in coulombs
- D12, D23, and D13 are the distances between the point charges in meters
Substituting the given values:
Q1 = 5.36 nC = 5.36 x 10^-9 C
Q2 = 5.72 nC = 5.72 x 10^-9 C
Q3 = 2.25 nC = 2.25 x 10^-9 C
D12 = 0.146 m
D23 = 0.525 m
D13 = 0.578 m
k = 8.99 x 10^9 N m^2/C^2
Calculating the potential energy:
Utot = (8.99 x 10^9 N m^2/C^2) * [(5.36 x 10^-9 C * 5.72 x 10^-9 C) / 0.146 m + (5.72 x 10^-9 C * 2.25 x 10^-9 C) / 0.525 m + (5.36 x 10^-9 C * 2.25 x 10^-9 C) / 0.578 m]
Evaluating the expression gives the value of Utot in joules.
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The electrons that produce the picture in a TV set are accelerated by a very large electric force as they pass through a small region in the neck of the picture tube. This region is 1.9 cm in length, and the electrons enter with a speed of 1×10
5
m/s and leave with a speed of 2.5×10
6
m/s. What is their acceleration over this 1.9 cm length? Answer in units of m/s
2
. 005 (part 2 of 2) 10.0 points How long is the electron in the accelerating region? Answer in units of s. 006 (part 1 of 2) 10.0 points A car traveling 29mi/h accelerates uniformly for 7.5 s, covering 501ft in this time. What was its acceleration? Round your answer to the nearest 100th place. Answer in units of ft/s
2
. 007 (part 2 of 2 ) 10.0 points What is the final velocity at this time? Answer in units of ft/s. 00810.0 points A 1000 kg automobile enters a freeway on-ramp at 20 m/s and accelerates uniformly up to 40 m/s in 10 seconds. How far does the automobile travel during that time? 1. 500 m 2. 600 m 3. 200 m 4. 400 m 5. None of these 6. 100 m 7. 300 m 009 (part 1 of 4) 10.0 points A young woman named Kathy Kool buys a sports car that can accelerate at the rate of 5.95 m/s
2
. She decides to test the car by drag racing with another speedster, Stan Speedy.
The acceleration of electrons is [tex]1.23 * 10^{15} m/s^2[/tex].
The time of electrons in the accelerating region is [tex]2.02 * 10^{-11} s.[/tex]
Part 1: Acceleration of Electrons To calculate the acceleration of electrons, we will use the kinematic equation:
v² = u² + 2as
where:
v = final velocity
= [tex]2.5 * 10^6 m/s[/tex]
u = initial velocity
= [tex]1 * 10^5 m/s[/tex]
a, acceleration = ?
s = displacement
= 1.9 cms
= 0.019 m
Substituting the values in the above equation,
v² - u² = [tex]2as(2.5 * 10^6)^2 - (1 * 10^5)^2[/tex]
[tex]= 2 \times a \times 0.019a[/tex]
[tex]=1.23 * 10^{15} m/s^2[/tex]
Therefore, the acceleration of electrons is [tex]1.23 * 10^{15} m/s^2[/tex].
Part 2: Time of Electrons in Accelerating Region
To calculate the time of electrons in the accelerating region, we will use the kinematic equation:
v = u + at
where:
v = final velocity
= [tex]2.5 * 10^6 m/s[/tex]
u = initial velocity
= [tex]1 * 10^5 m/s[/tex]
a = acceleration
= [tex]1.23 * 10^{15} m/s^2[/tex]
t, time = ?s
Substituting the values in the above equation,
[tex]2.5 \times 10^6 m/s =1 * 10^5 m/s + 1.23 * 10^{15} m/s^2 * t[/tex]
= 2.02 × 10^-11 s
Therefore, the time of electrons in the accelerating region is [tex]2.02 * 10^{-11} s.[/tex]
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The magnitude of vector
A
is 7.0 m. It points in a direction which makes an angle of 145
∘
measured counterdockwise from the positive x-axis. (a) What is the x component of the vector −4.5
A
? in (b) What is the y component of the vector −4.5
A
? (c) What is the magnitude of the vector −4.5
A
^
?
Answer: (a) -18.05 m, (b) -25.81 m, (c) 31.5 m.
Given data:
The magnitude of vector A is 7.0 m and the angle of the vector with positive x-axis is 145°.
Parts (a) and (b) require us to determine the horizontal and vertical components of the vector -4.5 A.
The minus sign indicates that this vector is in the opposite direction of the vector A.
Therefore, the magnitude of the vector -4.5 A is-4.5A
= -4.5 x 7.0
= -31.5 m
(a) x component of the vector −4.5A = -4.5A cos θ
where cos θ = cos(145°)
= -0.5736 (from the cosine table)
-4.5A cos θ = -31.5 m x (-0.5736)
= 18.05 m (toward left side is considered negative)
(b) y component of the vector −4.5A = -4.5A sin θ
where sin θ = sin(145°)
= 0.8192 (from the sine table)
-4.5A sin θ = -31.5 m x 0.8192
= -25.81 m (upward direction is considered positive)
(c) The magnitude of the vector −4.5 A^ is the same as the magnitude of 4.5 A, which is 31.5 m.
Answer: (a) -18.05 m,
(b) -25.81 m,
(c) 31.5 m.
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what is the gravitational force on a 20 kg satellite
The gravitational force on a 20 kg satellite 150 km above the Earth's surface is about 3000 N.
Here's how to solve for it:
According to Newton's law of gravitation,
the force of attraction between two objects with masses m1 and m2 separated by a distance r is:
F = G(m1m2)/r^2
where G is the gravitational constant (6.674 x 10^-11 Nm^2/kg^2).
In this case, one of the objects is the Earth, which has a mass of about 5.97 x 10^24 kg. The other object is the satellite with a mass of 20 kg. The distance between them is the radius of the Earth plus the altitude of the satellite, which we'll assume is 150 km above the Earth's surface.
So the distance between the centers of the Earth and the satellite is:
r = 6,371 km + 150 km= 6,521 km= 6,521,000 m
Substituting these values into the formula:
F = G(m1m2)/r^2F = (6.674 x 10^-11 Nm^2/kg^2)(20 kg)(5.97 x 10^24 kg)/(6,521,000 m)^2F = 2.99 x 10^3 N ≈ 3000 N
the gravitational force on a 20 kg satellite 150 km above the Earth's surface is about 3000 N.
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What is true about the speed of sound in air?A. It increases as the temperature risesB. It does not depend on the temperatureC. It decreases as the temperature increasesD.The speed of sound is always constant
The correct option is A. It increases as the temperature rises. The speed of sound in air is directly proportional to the temperature of the air.
The correct option is A. The speed of sound in air is directly proportional to the temperature of the air. As the temperature rises, the speed of sound in air also increases. This relationship is described by the formula:
v = √(γ * R * T)
where:
v is the speed of sound,
γ is the adiabatic index (a constant),
R is the gas constant for air,
T is the temperature of the air.
Since the temperature is in the square root term, any increase in temperature will result in an increase in the speed of sound. Conversely, a decrease in temperature would result in a decrease in the speed of sound.
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3. (6 pts total) A crate rests on a rough inclined plane, angled at 38.0° above the horizontal. The crate has a mass of 11.5 kg, and the coefficient of static friction between the crate and the inclined plane is 0.292 A force F is exerted in the direction up the plane and parallel to the slope in order to keep the cate in plane.
a) (6 pts) Draw a free -body diagram of the situation,assuming the crate is not moving . Be sure to label all forces (Hint: be carefful with the diection of fiction force) b)(3 pts) What minimum force,directed parallel to the slope up the plane,is required to keep the crate from the siding the plane.
The minimum force, directed parallel to the slope up the plane, required to keep the crate from sliding down the plane is 40.449 N.
When a crate rests on a rough inclined plane at an angle of 38.0° above the horizontal, a force is needed to keep the crate from sliding down the slope. The force F required in the direction up the plane and parallel to the slope to keep the crate from sliding is calculated as follows:
Calculate the frictional force acting against the crate. The frictional force (Ff) acting against the crate is calculated using the following formula:
Ff = (coefficient of static friction) x (normal force)
N = (mass of the crate) x (acceleration due to gravity)
N = 11.5 kg × 9.81 m/s²
N = 112.715 N
The normal force acting against the crate is:
cos θ = (adjacent/hypotenuse)
cos 38.0° = (N/112.715)
N = 86.042 N
The frictional force is:
Ff = 0.292 x 86.042
Ff = 25.102 N
The force F required to keep the crate from sliding is:
F = Ff/sin θ
F = 25.102/sin 38.0°
F = 40.449 N
Therefore, the minimum force, directed parallel to the slope up the plane, required to keep the crate from sliding down the plane is 40.449 N.
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A proton and an electron are moving du yeast in a constant electric field that also points due east. The electric field has a magnitude of 2.0×10
4
N/C Determine the magnitude of the acceleration of the proton and the electron.
Therefore, the magnitude of acceleration of the proton is 1.9 × 10¹³ m/s² and the magnitude of acceleration of the electron is 3.5 × 10¹⁴ m/s²
Given
Magnitude of electric field = 2.0 × 10⁴ N/C.
The acceleration of a proton in an electric field is given by
F = ma
Where F = qE,
where E is the electric field strength and q is the charge of the proton.
Substituting, F = ma ⟹ qE = ma,
Thus, a = qE/m
As the mass of the proton is given as 1.67 × 10⁻²⁷ kg and the charge as 1.6 × 10⁻¹⁹ C,
we have a = (1.6 × 10⁻¹⁹ C)(2.0 × 10⁴ N/C) / (1.67 × 10⁻²⁷ kg) = 1.9 × 10¹³ m/s²
Similarly, the acceleration of an electron in an electric field is given by F = ma
Where F = qE,
where E is the electric field strength and q is the charge of the electron.
Substituting, F = ma ⟹ qE = ma,
Thus, a = qE/m
As the mass of the electron is given as 9.1 × 10⁻³¹ kg and the charge as -1.6 × 10⁻¹⁹ C (negative because it is an electron),
we have
a = (-1.6 × 10⁻¹⁹ C)(2.0 × 10⁴ N/C) / (9.1 × 10⁻³¹ kg)
= -3.5 × 10¹⁴ m/s² (because it is opposite to the direction of the electric field).
Therefore, the magnitude of acceleration of the proton is 1.9 × 10¹³ m/s² and
the magnitude of acceleration of the electron is 3.5 × 10¹⁴ m/s².
Hence, a = (1.9 × 10¹³ m/s²) (for proton)
a = (3.5 × 10¹⁴ m/s²) (for electron)
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A spherical shell of radius 6.5 m is placed in a uniform electric field with magnitude 6450 N/C. Find the total electric flux through the shell. Answer in units of N⋅m
2
/C 00210.0 points A (3.2 m by 3.2 m) square base pyramid with height of 5.91 m is placed in a uniform vertical electric field of 74.3 N/C. The pyramid encloses no charge. Calculate the electric flux which goes out through one of the pyramid's four slanted surfaces. Answer in units of Nm
2
/C.
The flux through one of the pyramid's four slanted surfaces is given by:Φ = 794.98 N⋅m².
Radius of the shell, r = 6.5 mm
Magnitude of the uniform electric field, E = 6450 N/C
We know that the electric flux (Φ) through a closed surface is given by the formula:
Φ = ∫ E · dA
The electric flux Φ is proportional to the charge enclosed in the surface. Since there is no charge enclosed in the spherical shell, the electric flux through the spherical shell is zero.
Now, let's calculate the electric flux through one of the pyramid's four slanted surfaces.
Given:
Electric field, E = 74.3 N/C
The area of one of the slanted surfaces of the square pyramid is given by the formula:
A = (1/2) × base × height
The slant height of the pyramid can be calculated using the Pythagorean theorem.
Let the length of the base be x. Then the slant height is given by:
l² = (x/2)² + h²
Where h is the height of the pyramid.
Let the slant height be l. Then the area of one of the slanted surfaces is given by:
A = (1/2) × x × l
The total flux through the slanted surface is given by:
Φ = EA
The electric field E and the area A are perpendicular to each other. Therefore, we don't have to worry about the dot product. Thus,
Φ = EA = (74.3 N/C) × (1/2) × 3.2 m × l = (118.88 Nm²/C) × l
We need to find the value of l. Using the Pythagorean theorem,
l² = (3.2/2)² + 5.91²
l = 6.68 m
Thus, the flux through one of the pyramid's four slanted surfaces is given by:
Φ = (118.88 Nm²/C) × (6.68 m)
Φ = 794.98 N⋅m²/C
Answer: 794.98 N⋅m²/C.
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A crate of mass m is going down a frictionless inclined plane. The plane is at angle θ with respect to horizontal. Given m=1.3( kg),θ=56 (deg), g=9.81 m/s
∧
2. (Give your answers as numbers with 3 significant figures) Find the magnitude of the normal force that the plane exerts on the crate (in N) Find the magnitude of the crate's acceleration (in m/s
∧
2 ) A mechanical system is given as in figure. Two block of masses m1 and m2 are connected by a massless string. The pully's mass is also negligible. Assume the system can move with no friction. Given: m1=6.2 (kg),m2=1.6( kg),g=9.81 m/s
∧
2. (Give your answers as numbers with 3 significant figures) Find the magnitude of the acceleration of m1 (in m/s
∧
2 ) Find the tension in the string (in N)
1. The magnitude of the normal force that the plane exerts on the crate is 6.02 N.
2. The tension in the string is 19.7 N.
1. To find the magnitude of the normal force that the plane exerts on the crate (in N), we have to use the formula;
Fnet = ma
Where
Fnet is the net force,
m is the mass
a is the acceleration.
As per the question, the crate is moving down the inclined plane without friction. Hence the force acting on the crate is only the gravitational force. Therefore, the formula can be simplified to;
Fg = ma
Where
Fg is the force due to gravity acting on the crate,
m is the mass of the crate
a is the acceleration of the crate
Using the above formula, we can find the acceleration of the crate.
a = gsinθ
a = (9.81)(sin56°)
a = 7.94 m/s²
To find the normal force acting on the crate, we have to use the formula;
Fn = mgcosθ
Fn = (1.3)(9.81)(cos56°)
Fn = 6.02 N
Thus, the magnitude of the normal force that the plane exerts on the crate is 6.02 N.
2. We know that the two blocks are connected by a massless string. Therefore, the tension in the string is the same for both blocks.
Using the following formula for each block separately;
m1g − T = m1a
m2g + T = m2a
Here,
g is the acceleration due to gravity
T is the tension in the string
Adding the two equations gives us;
m1g − m2g = (m1 + m2)a
Simplifying further;
a = g(m1 − m2) / (m1 + m2)
a = (9.81)(6.2 − 1.6) / (6.2 + 1.6)
a = 6.16 m/s²
Thus, the magnitude of the acceleration of m1 is 6.16 m/s².
To find the tension in the string, we can use any of the two equations we got for each block separately. Using the first equation;
m1g − T = m1a
T = m1(g − a)
T = 6.2(9.81 − 6.16)
T = 19.7 N
Thus, the tension in the string is 19.7 N.
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a basketball olayer jumps straight up at 4 m/s. 0.3 seconds later he shoots the ball with a velocity of 6 m/s up and 7m/s east relative to him. what is the velocity of ball relative to ground as player shoots the ball
The velocity of the ball relative to the ground as player shoots the ball is obtained as ,Initial velocity of the basketball player = 4 m/s
Time taken by the player to shoot the ball after jumping up = 0.3 sVelocity of the ball w.r.t basketball player
= Velocity of the ball – Velocity of the player
= 6 m/s up + 7 m/s east – 4 m/s up
= 2 m/s up + 7 m/s eastNow,Velocity of the basketball player w.r.t the ground
= 0 m/s (since there is no movement in the horizontal direction)
Applying relative velocity formula, Relative velocity of the ball w.r.t the ground = Velocity of the ball w.r.t basketball player + Velocity of the basketball player w.r.t ground= (2 m/s up + 7 m/s east) + 0 m/s
= 2 m/s up + 7 m/s eastHence, the velocity of the ball relative to the ground as player shoots the ball is 2 m/s up and 7 m/s east.
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A cylinder of length L has a circular cross section radius R. The Volume of the cylinder is 15.0±0.5 cm3 and the length is 20 ± 0.1 cm. Calculate the radius of the cylinder with its uncertainty.
The radius of the cylinder with uncertainty is found to be 0.488 ± 0.01 cm. This can be understood by propagation of errors.
Volume of the cylinder, V = 15.0 ± 0.5 cm³
Length of the cylinder, L = 20 ± 0.1 cm.
Circular cross-section radius, r =?
We know, Volume of the cylinder = πr²L
`V=π r²L`
r²= V/(πL)
Taking the square root of both the sides, `r = √(V/(πL))`
Let's calculate the uncertainty of the radius,
`r = √(V/(πL))` = √15/(π×20) = 0.488 cm
`Δ(r)/r = 1/2 × [Δ(V)/V + Δ(L)/L]`
`Δ(r) = r/2 × [Δ(V)/V + Δ(L)/L]`
Substitute the values of V, L, and Δ(V) and Δ(L)
`Δ(r) = r/2 × [0.5/15 + 0.1/20]`=`r/2 × (0.033 + 0.005)`=`r/2 × 0.038`=`0.019r`
Now substitute the value of r,
`Δ(r) = 0.019 × 0.488 = 0.0092 ≈ 0.01` (rounded to 2 significant figures)Therefore, the radius of the cylinder with uncertainty is `0.488 ± 0.01 cm`.
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You are deadlifting 1,130 N. What is the net force needed to accelerate the weights upwards at 1.6 m/s2?
To determine the net force required to accelerate the weights upwards at 1.6 m/s², we can use Newton's second law of motion, which states that the net force (F_net) acting on an object is equal to the product of its mass (m) and its acceleration (a).
Given:
Force due to deadlifting (F_deadlift) = 1,130 N
Acceleration (a) = 1.6 m/s²
To calculate the net force, we need to know the mass (m) of the weights being lifted. Without this information, we cannot provide an exact value for the net force.
However, we can discuss the relationship between force, mass, and acceleration.
According to Newton's second law, the net force is directly proportional to the acceleration and the mass of the object.
This means that as the mass increases, the net force required to achieve the same acceleration also increases. Conversely, if the mass decreases, a smaller net force is needed to achieve the desired acceleration.
In the case of deadlifting, the net force required to accelerate the weights upwards will depend on the mass of the weights. The greater the mass, the larger the net force needed to produce the desired acceleration.
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A puck is moving on an air hockey table. Relative to an x,y coordinate system at time t−0 s, the x components of the puck's initial velocity and acceleration are v
0x
=+2.5 m/s and a
x
=+2.2 m/s
2
. The y components of the puck's initial velocity and acceleration are v
O
=+9.0 m/s and a
y
=−8.7 m/s
2
. Find (a) the magnitude v and (b) the direction θ of the puck's velocity at a time of t=0.50 s. Specify the direction relative to the +x axis. (a) v= (b) θ= degrees the +x axis
(a) The magnitude of the puck's velocity at t = 0.50 s is approximately 9.34 m/s.
(b) The direction of the puck's velocity at t = 0.50 s, relative to the +x axis, is approximately 75.96 degrees.
To find the magnitude and direction of the puck's velocity at t = 0.50 s, we can use the equations of motion.
(a) To find the magnitude of the velocity, we can use the Pythagorean theorem:
v = √(v_x^2 + v_y^2)
Given:
v_0x = +2.5 m/s (initial x-component velocity)
v_0y = +9.0 m/s (initial y-component velocity)
Using the given values, we can calculate the magnitude of the velocity:
v = √(v_0x^2 + v_0y^2)
v = √((+2.5 m/s)^2 + (+9.0 m/s)^2)
v = √(6.25 m^2/s^2 + 81.0 m^2/s^2)
v = √(87.25 m^2/s^2)
v ≈ 9.34 m/s
Therefore, the magnitude of the velocity at t = 0.50 s is approximately 9.34 m/s.
(b) To find the direction θ of the velocity, we can use the inverse tangent function:
θ = tan^(-1)(v_y / v_x)
Using the given values, we can calculate the direction:
θ = tan^(-1)(v_0y / v_0x)
θ = tan^(-1)(+9.0 m/s / +2.5 m/s)
θ ≈ 75.96 degrees
Therefore, the direction of the velocity at t = 0.50 s relative to the +x axis is approximately 75.96 degrees.
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A square parallel plate capacitor having an area of 200 cm2 has a separation distance of 2.4 mm. Determine its capacitance when a dielectric material of constant k = 2 is inserted between its plates.
a.148pF
b.1476pF
c.7.2mF
d.15nF
The capacitance of the capacitor with the dielectric material inserted is approximately 148 pF. The correct answer is option a) 148 pF.
To determine the capacitance of the square parallel plate capacitor with a dielectric material inserted between its plates, we can use the formula:
C = (ε₀ * k * A) / d
where C is the capacitance, ε₀ is the vacuum permittivity (8.85 x 10^-12 F/m), k is the dielectric constant, A is the area of the plates, and d is the separation distance between the plates.
Given:
Area (A) = 200 cm² = 200 x 10^-4 m²
Separation distance (d) = 2.4 mm = 2.4 x 10^-3 m
Dielectric constant (k) = 2
Substituting the values into the formula:
C = (8.85 x 10^-12 F/m * 2 * 200 x 10^-4 m²) / (2.4 x 10^-3 m)
Simplifying the expression:
C = 148 pF
Therefore, the capacitance of the capacitor with the dielectric material inserted is approximately 148 pF. The correct answer is option a) 148 pF.
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A long, nonconducting, solid cylinder of radius 5.4 cm has a nonuniform volume charge density p that is a function of radial distance r from the cylinder axis: p = Ar². For A = 2.1 μC/m5, what is the magnitude of the electric field at (a) r = 1.6 cm and (b) r = 11 cm.
The magnitude of the electric field at a radial distance of 3.0 cm from the cylinder is (dependent on the numerical integration calculation), and at a radial distance of 5.0 cm, the magnitude of the electric field is (dependent on the numerical integration calculation).
the magnitude of the electric field at a radial distance from the cylinder, we can use Gauss's Law and the symmetry of the problem. Since the cylinder has cylindrical symmetry, we can consider a Gaussian surface in the shape of a cylinder centered on the axis of the cylinder.
The electric field outside the cylinder will only depend on the charge enclosed within the Gaussian surface. For a Gaussian surface at a radial distance r, the charge enclosed can be calculated by integrating the volume charge density over the volume of the cylinder.
The magnitude of the electric field (E) at a radial distance r is given by:
E = (1 / (4πε₀)) * (Q_enclosed / r²),
where ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 C²/(N·m²)) and Q_enclosed is the charge enclosed within the Gaussian surface.
a) For d = 3.0 cm:
The radial distance r is 3.0 cm = 0.03 m.
To find the charge enclosed, we integrate the volume charge density over the volume of the cylinder:
Q_enclosed = ∫ρ dV = ∫(Aor²) dV,
where the integral is taken over the volume of the cylinder.
The volume element dV for a cylindrical coordinate system is given by: dV = r dr dθ dz.
Since the cylinder is long and nonconducting, the integration over the θ and z directions gives the length of the cylinder and can be ignored.
Integrating over r from 0 to r, we have:
Q_enclosed = ∫(Aor²) r dr = A ∫(r³) dr.
Evaluating the integral, we get:
Q_enclosed = A * (1/4) * r⁴.
Substituting the given values, A = 2.5 μC/m⁵ and r = 0.03 m, we can calculate Q_enclosed.
Finally, we can calculate the magnitude of the electric field using the formula E = (1 / (4πε₀)) * (Q_enclosed / r²).
b) For d = 5.0 cm:
The radial distance r is 5.0 cm = 0.05 m. Using the same procedure as above, we can calculate the charge enclosed Q_enclosed for this distance and then calculate the magnitude of the electric field E
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child on a sled is sliding some snow-covered hills starting to rest at location A. in picture three different locations are marked with the letter A, B, and C Rank from the highest to lowest kinetic energy of the child
The three locations marked as A, B, and C, have to be ranked from the highest to the lowest kinetic energy of the child. Kinetic energy is defined as the energy that a moving object possesses due to its motion. As the child on a sled is sliding down the snow-covered hills, it's gaining kinetic energy due to its motion.
The higher the velocity of the sled, the higher will be the kinetic energy possessed by the sled. Hence, the ranking of the three locations from highest to lowest kinetic energy will be as follows: B > C > A Here's why: Location B is at the bottom of the hill, where the sled will have the highest velocity and, therefore, the highest kinetic energy. Hence, it ranks first in terms of kinetic energy. Location C is somewhere in between the top and bottom of the hill.
At this location, the sled will have less velocity than at location B but more than at location A. Hence, the kinetic energy of the child on a sled will be lower than at location B but higher than at location A. Therefore, location C ranks second in terms of kinetic energy. Location A is at the top of the hill, where the sled is starting to slide.
At this point, the sled is at rest, which means that the velocity of the sled is zero, and, therefore, the kinetic energy of the sled is also zero. Hence, location A ranks last in terms of kinetic energy.
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A standard 1 kilogram weight is a cylinder 42.0 mm in height and 44.5 mm in diameter. What is the density of the material?
kg/m3
A standard 1-kilogram weight is a cylinder 42.0 mm in height and 44.5 mm in diameter. The density of the material is approx [tex]30,368.75 kg/m^3[/tex]
For calculating the density of the material, need to find the mass and volume of the weight. The mass of the weight is given as 1 kilogram.
For finding the volume, need to use the dimensions of the weight. The weight is in the shape of a cylinder, so use the formula for the volume of a cylinder:
[tex]V = \pi * r^2 * h[/tex]
where r is the radius and h is the height.
The given diameter is 44.5 mm, which means the radius is half of that, so
r = 44.5 mm / 2 = 22.25 mm.
Convert the radius to meters by dividing it by 1000:
r = 22.25 mm / 1000 = 0.02225 m.
The height is given as 42.0 mm, which is equivalent to 0.0420 m.
Now calculate the volume:
[tex]V = \pi * (0.02225 m)^2 * 0.0420 m = 0.0000329 m^3[/tex]
Finally, calculate the density by dividing the mass by the volume:
[tex]density = 1 kg / 0.0000329 m^3 \approx 30,395.1365 kg/m^3[/tex].
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A lambda particle (Λ) decays into a proton and a pion, Λ→p+π, and it is observed that the proton is left at rest. (a) What is the energy of the pion? (b) What was the energy of the original Λ ? (The masses involved are m
A
=1116,m
p
=938, and m
π
=140, all in MeV/c
2
. As is almost always the case, your best procedure is to solve the problem algebraically, in terms of the symbols m
A
,m
p
,m
π
, and, only at the end, to put in numbers.)
Eπ = mπc² (energy of the pion)
EΛ = Ep + (mπc²) + mpc² (energy of the original Λ particle)
(a) To find the energy of the pion (Eπ), we can use conservation of energy. Since the proton is left at rest, the initial energy of the Λ particle (EΛ) is equal to the energy of the pion (Eπ). We can express this as:
EΛ = Eπ
The total energy of a particle can be calculated using the formula:
E = mc²
where m is the mass of the particle and c is the speed of light.
For the pion, its mass is given as mπ.
Therefore, we have:
Eπ = mπc²
(b) To find the energy of the original Λ particle (EΛ), we need to consider the conservation of energy. Initially, the Λ particle has an energy of EΛ, and after the decay, the proton is left at rest, which means its energy is the rest energy:
EΛ = Ep + Eπ + rest energy
The rest energy of a particle can be calculated using the formula:
rest energy = mc²
where m is the mass of the particle and c is the speed of light.
For the proton, its mass is given as mp.
Therefore, we have:
EΛ = Ep + Eπ + mpc²
Substituting the expression for Eπ from part (a), we get:
EΛ = Ep + (mπc²) + mpc²
Now, to find the values of Ep and EΛ, we need to consider the conservation of momentum. Since the proton is left at rest, the total momentum before and after the decay must be zero:
Initial momentum = Final momentum
mΛvΛ = mpvp + mπvπ
Since the proton is left at rest, its velocity vp is 0.
Therefore, we have:
mΛvΛ = mπvπ
Solving for vΛ, we get:
vΛ = (mπ/mΛ) vπ
Now, we can express the energy of the Λ particle in terms of the pion velocity:
EΛ = (mΛc²) / √(1 - (vΛ/c)²)
Substituting the expression for vΛ, we get:
EΛ = (mΛc²) / √(1 - ((mπ/mΛ) vπ / c)²)
Finally, substituting the values for the masses and simplifying the expression gives the energy of the original Λ particle.
Note: The actual numerical value depends on the specific values of the masses provided in MeV/c².
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7. A diver steps off a diving board and hits the surface of the water in the swimming pool 1.5 s later. If the instant they enter the water, the diver starts to decelerate at 16 m/s
2
, how deep do they go into the water before stopping? Ignore air resistance. Hint: Stepping off the diving board means that they will start with zero velocity and go straight down into the water. (All options are given as negative because they are how far below the surface of the water.) x
2
−4.05m x
2
=−6.75m x
2
=−11m x
2
=−14.7m x
2
=−18m
The diver goes 12 meters below the surface of the water before coming to a stop.
To solve this problem, we can use the equations of motion for uniformly decelerated motion. We know that the initial velocity (u) is zero because the diver starts with zero velocity when stepping off the diving board. The acceleration (a) is -16 m/s² because the diver decelerates at a constant rate of 16 m/s². We need to find the displacement (x) of the diver when they come to a stop.
The equation for displacement in uniformly decelerated motion is given by:
x = ut + (1/2)at²
Since the initial velocity (u) is zero, the equation simplifies to:
x = (1/2)at²
Plugging in the values:
x = (1/2)(-16 m/s²)(1.5 s)²
x = -12 m
Therefore, the diver goes 12 meters below the surface of the water before coming to a stop.
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A particle of charge +4.1μC is released from rest at the point x=54 cm on an x axis. The particle begins to move due to the presence of a charge Q that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 11 cm if (a)Q Q=+18μC and (b)Q=−18μC? (a) Number Units (b) Number Units
(a) The kinetic energy of the particle is 6.34 J when Q = +18 μC. (b) The kinetic energy of the particle when Q = +18 μC is 6.34 J, and when Q = -18 μC is 7.75 J.
(a) When Q = +18 μC
A particle of charge +4.1μC is released from rest at the point x = 54 cm on an x-axis.
The particle begins to move due to the presence of a charge Q that remains fixed at the origin.
As we know that, the potential energy and kinetic energy of a particle that is acted upon by an electrostatic force are given as:
PE = qV
KE = (1/2)*mv²
where, q is the charge of the particle,
V is the potential energy,
m is the mass of the particle,
v is the velocity of the particle.
Initially, the particle is at rest. So, the initial kinetic energy is zero.
At x = 11 cm:
Displacement, `d` = x - x₀ = 11 - 54 = -43 cm
Here, the displacement is negative because the particle moves toward the origin.
Distance, `r` = |d| = 43 cm
Distance between the point where the charge is present and where the particle is located, `r₁` = 54 cm
Charge of the particle, `q` = +4.1 μC
Charge of the charge at the origin, `Q₁` = +18 μC
Charge of the charge at the origin, `Q₂` = -18 μC
For electrostatic potential, we can use the following formula:
V = (1/4πε₀)Q/r
where ε₀ is the permittivity of free space.
For calculating the kinetic energy, we have to calculate the potential energy at `x = 54 cm` and `x = 11 cm`.
Then, we have to find the difference between them to get the work done due to electrostatic force, and finally, we can calculate the kinetic energy of the particle.
So, let's calculate the electrostatic potential energy of the particle at `x = 54 cm` and `x = 11 cm`.
The potential at x = 54 cm,
`V₁` = (1/4πε₀)Q/r₁
= (9 × 10⁹ N⁻¹ m² C⁻²) × (18 × 10⁻⁶ C) / (0.54 m)
≈ 5.56 × 10⁵ V
The potential at x = 11 cm,
`V₂` = (1/4πε₀)Q/r = (9 × 10⁹ N⁻¹ m² C⁻²) × (18 × 10⁻⁶ C) / (0.11 m) ≈ 1.64 × 10⁷ V
Now, let's calculate the work done by the electrostatic force,
`W` = q(V₂ - V₁) ≈ (4.1 × 10⁻⁶ C) × (1.64 × 10⁷ V - 5.56 × 10⁵ V) ≈ 6.34 J
Now, the kinetic energy of the particle,`KE` = W= 6.34 J
Therefore, the kinetic energy of the particle is 6.34 J when Q = +18 μC.
Case (b):
When Q = -18 μC:
The potential at x = 54 cm,
`V₁` = (1/4πε₀)Q/r₁ = (9 × 10⁹ N⁻¹ m² C⁻²) × (-18 × 10⁻⁶ C) / (0.54 m) ≈ -5.56 × 10⁵ V
The potential at x = 11 cm,
`V₂` = (1/4πε₀)Q/r = (9 × 10⁹ N⁻¹ m² C⁻²) × (-18 × 10⁻⁶ C) / (0.11 m) ≈ -1.64 × 10⁷ V
Now, let's calculate the work done by the electrostatic force,
`W` = q(V₂ - V₁) ≈ (4.1 × 10⁻⁶ C) × (-1.64 × 10⁷ V - (-5.56 × 10⁵ V)) ≈ -7.75 J
Now, the kinetic energy of the particle,
`KE` = |W|= 7.75 J
Therefore, the kinetic energy of the particle is 7.75 J when Q = -18 μC.
So, the kinetic energy of the particle when Q = +18 μC is 6.34 J, and when Q = -18 μC is 7.75 J.
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1. Suppose that your friend goes out for a run and comes back 2 hours later. Their fitness watch says that they traveled a distance of 5 miles in that time. Determine their speed, in units of mph. Were they really running that whole time? How do you know? 2. Use a navigation app to determine how many miles you traveled to get to school today. Then estimate how long it took you to get to school today. Use these numbers to determine estimates for your travel speed, in both mph and m/s. 3. Driving down the road, you notice that you go from being about 300 feet in front of a traffic light to about 300 feet past it in 1 second of time. About how fast are you traveling, in both m/s and mph ? 4. Further down the road, you notice that you go from being about 600 feet in front of a traffic light to about 200 feet in front of it in 1.5 seconds of time. About how fast are you traveling now, in both m/s and mph ?
1. Speed = distance/time = 5 miles / 2 hours = 2.5 miles/hour. We do not know if they were running the whole time because speed is a scalar quantity that only tells us how fast they were moving, not how they were moving.
2. The distance from home to school is needed to find speed. If distance traveled to school is x miles and time taken to travel to school is t, then average speed = distance/time = x/t mph and x/t × 1609.34/t m/s.
3.The average speed of the car is the total distance divided by the total time. The distance that the car covers in 1 second is 600 + 300 = 900 ft. So, the speed of the car is (900 feet / 1 sec) / 5280 feet/mile = 0.17 miles/sec or 60 × 0.17 = 10.2 miles/hour or 900/3600 = 0.25 meters/second.
4. For the second part, the initial position of the car is 600 feet and the final position is 200 feet, which means the distance traveled is 400 feet. The time taken to travel that distance is 1.5 seconds. So, the speed of the car is 400 feet / (1.5 seconds × 5280 feet/mile) = 400/7920 = 0.05 miles/second or 18 miles/hour. In meters per second, the speed of the car is 0.05 miles/sec × 1609.34 meters/mile = 80.47 m/s.
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Lift is a pressure caused by the fluid in a direction horizontal to the direction of travel of the body. True False
Lift is not a pressure caused by the fluid in a horizontal direction to the body's direction of travel is False.
Lift is a force perpendicular to the direction of fluid flow.
Lift is the force generated on an object, such as an airplane wing, when it is in motion through a fluid (such as air).
The generation of lift is primarily due to the difference in pressure on the upper and lower surfaces of the wing.
As a wing moves through the fluid, the shape and angle of the wing cause the air to move faster over the top surface, resulting in lower pressure, and slower under the bottom surface, resulting in higher pressure.
The pressure difference between the upper and lower surfaces of the wing creates an upward force perpendicular to the direction of the fluid flow.
This upward force is known as lift and is responsible for enabling flight and supporting the weight of the aircraft.
Therefore, the statement that lift is a pressure caused by the fluid in a horizontal direction to the body's direction of travel is false.
Lift is a force perpendicular to the direction of fluid flow.
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A derrick boat approaches a two-mile marker 100 m ahead at a velocity of 32.5 m/s. The pilot reduces the throttle, slowing the boat with a constant acceleration of −2.90 m/s2.(a) How long (in s) does it take the boat to reach the marker? ____ s.(b) What is the velocity (in m/s) of the boat when it reaches the marker? (Indicate the direction with the sign of your answer.) ____ m/s.
To solve this problem, we can use the equations of motion. It takes approximately 11.21 seconds for the boat to reach the marker. The velocity of the boat when it reaches the marker is approximately -21.82 m/s.
To solve this problem, we can use the equations of motion. Let's denote the initial velocity of the boat as v0, the final velocity as vf, the acceleration as a, and the displacement as d.
(a) We need to find the time it takes for the boat to reach the marker. We can use the equation:
vf = v0 + at
Given:
v0 = 32.5 m/s
a = -2.90 m/s^2
d = 100 m
To find the time, we rearrange the equation:
t = (vf - v0) / a
Substituting the given values, we have:
t = (0 - 32.5) / (-2.90)
Simplifying the expression, we find:
t ≈ 11.21 s
Therefore, it takes approximately 11.21 seconds for the boat to reach the marker.
(b) To find the velocity of the boat when it reaches the marker, we can use the equation:
vf^2 = v0^2 + 2ad
Substituting the given values, we have:
vf^2 = (32.5)^2 + 2(-2.90)(100)
Simplifying the expression, we find:
vf^2 = 1056.25 - 580
vf^2 ≈ 476.25
Taking the square root of both sides, we find:
vf ≈ 21.82 m/s
Since the boat is slowing down, the velocity is negative. Therefore, the velocity of the boat when it reaches the marker is approximately -21.82 m/s.
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A capacitor with plates of area A that are separated by a distance D is connected to a battery, and charged to a charge Q. The battery is left in contact with the plates, while the plates are pressed closer together. What happens to the electric field E inside the plates of the capacitor?
A capacitor with plates of area A that are separated by a distance D is connected to a battery, and charged to a charge Q. The battery is left in contact with the plates, while the plates are pressed closer together.
In a capacitor, electric field E inside the plates is directly proportional to the voltage V across the plates and inversely proportional to the separation distance d between them. That is, E = V / d. Capacitance C of a capacitor with plates of area.
A separated by distance D is given by C = εA / D where ε is the permittivity of the medium between the plates.Now, if the plates of the capacitor are pressed closer together to distance d' (d' < d), the capacitance of the capacitor will increase to C' = εA / d'.
Hence, for a constant charge Q, the voltage V across the capacitor plates will increase to V' = Q / C' = Qd' / εA. As the voltage across the capacitor plates increases, the electric field E inside the plates will also increase, the electric field E inside the plates of the capacitor will increase when the plates are pressed closer together.
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