The high jumper's vertical velocity at takeoff was 3.14 m/s. The horizontal position of her center of mass at the top of the jump was 5.39 m. The vertical position of her center of mass at the top of the jump was 3.68 m.
To find the vertical velocity at takeoff, we need to determine the vertical component of the resultant velocity. We can use trigonometry to calculate it. Given that the resultant velocity is 4.0 m/s and the angle from the right horizontal is 35°, we can find the vertical velocity using the equation: vertical velocity = resultant velocity * sin(angle). Therefore, the vertical velocity at takeoff is 4.0 m/s * sin(35°) = 3.14 m/s.
To determine the horizontal position of the center of mass at the top of the jump, we need to consider the horizontal displacement. Since there is no horizontal acceleration during the jump, the horizontal displacement can be calculated using the equation: horizontal displacement = horizontal velocity * time. Since the time of flight is the same for the upward and downward motion, we can focus on the upward motion. The initial horizontal velocity is given as 4.0 m/s. Therefore, the horizontal position of the center of mass at the top of the jump is 4.0 m/s * time.
The vertical position of the center of mass at the top of the jump can be found by considering the initial vertical position and the vertical displacement. The vertical displacement is equal to the maximum height reached during the jump. The initial vertical position is given as 1.37 m. Therefore, the vertical position of the center of mass at the top of the jump is 1.37 m + vertical displacement.
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A VW Beetle goes from 0 to 48.0mi/h with an acceleration of +2.35 m/s^2 . (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to 48.0mi/h in 0.600 s. Find the acceleration (in m/s^2) of the dragster. (a) Number Units (b) Number Units
(a) It takes approximately 6.44 seconds for the VW Beetle to reach a speed of 48.0 mi/h.
(b) The acceleration of the top-fuel dragster is approximately 53.33 m/s^2.
(a) To calculate the time it takes for the VW Beetle to reach a speed of 48.0 mi/h, we can use the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given that the initial velocity u is 0 m/s, the final velocity v is 48.0 mi/h, which can be converted to m/s by multiplying it by 0.44704 (since 1 mi/h = 0.44704 m/s), and the acceleration a is +2.35 m/s^2, we can rearrange the equation to solve for time:
v = u + at
48.0 mi/h = 0 + 2.35 m/s^2 * t
Converting 48.0 mi/h to m/s, we get:
48.0 mi/h * 0.44704 m/s = 21.44 m/s
Substituting the values into the equation, we have:
21.44 m/s = 2.35 m/s^2 * t
Solving for t, we find:
t = 21.44 m/s / 2.35 m/s^2 ≈ 6.44 seconds
Therefore, it takes approximately 6.44 seconds for the VW Beetle to reach a speed of 48.0 mi/h.
(b) To find the acceleration of the top-fuel dragster, we can use the same equation of motion v = u + at and rearrange it to solve for acceleration:
a = (v - u) / t
Given that the initial velocity u is 0 m/s, the final velocity v is 48.0 mi/h (which is converted to m/s as explained in part (a)), and the time t is 0.600 s, we can substitute the values into the equation:
a = (48.0 mi/h * 0.44704 m/s - 0 m/s) / 0.600 s
Calculating the expression, we get:
a = 26.92 m/s / 0.600 s ≈ 53.33 m/s^2
Therefore, the acceleration of the top-fuel dragster is approximately 53.33 m/s^2.
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There is a rocket with a mass of 1000 kilograms in space, and 200 kilograms of fuel in the rocket, assuming that there is \( 1.0 \) kilograms of fuel per second The material is transformed into high-s
The momentum principle is the law of physics that applies in the context of rockets, and it can be used to determine how quickly a rocket will accelerate. The momentum principle states that the force exerted on an object is equal to its mass multiplied by its acceleration.
The equation that relates the mass of a rocket, the mass of the fuel, the change in velocity, and the velocity of the exhaust gases is derived from the momentum principle. This equation is known as the Tsiolkovsky rocket equation. It is possible to predict the final velocity of a rocket based on its initial velocity, the mass of the rocket, and the amount of fuel it contains by using this equation.
The final velocity of a rocket is determined by the exhaust velocity of its gases, which is the velocity at which the gases are ejected from the nozzle. The velocity of the exhaust gases can be determined by the temperature and pressure of the gases, as well as the mass of the rocket and the mass of the fuel. As a result, rockets must burn a great deal of fuel to achieve high exhaust velocities.
The rocket equation can be used to determine the amount of fuel required for a rocket to achieve a particular final velocity or reach a particular point in space. By combining the rocket equation with other principles of physics, such as the principles of orbital mechanics, it is possible to design rockets and spacecraft that can travel long distances and reach high speeds.
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A ball is kicked with a speed vo =19.8 m/s and at an angle of θ=38.9∘ with the ground, it hits a wall at the point ×=16.4 m and y=7.68 m. See the figure. How long the ball is in the air before hitting the wall? Use g=9.80 m/s2 if needed. A. 0.395 8. 2.545 C. 1.065 D. 1.325 E. 0.835
The ball is in the air for approximately 1.065 seconds before hitting the wall. Thus, the correct answer is Option C.
To determine how long the ball is in the air before hitting the wall, we can use the projectile motion equations. The horizontal and vertical components of motion can be considered separately.
Given:
Initial speed, vo = 19.8 m/s
Launch angle, θ = 38.9°
Horizontal distance to the wall, x = 16.4 m
Vertical height, y = 7.68 m
Acceleration due to gravity, g = 9.80 m/s²
First, let's calculate the time of flight (total time the ball is in the air) using the vertical component of motion.
The equation for the vertical displacement is given by:
y = vo × t + (1/2)gt²
where
y is the vertical displacement,
vo is the initial vertical velocity,
t is the time, and
g is the acceleration due to gravity.
Rearranging the equation, we have:
t² - (2vo/g)t + (2y/g) = 0
Using the quadratic formula, we can find the time of flight:
t = (-b ± √(b² - 4ac)) / (2a)
where
a = 1
b = -(2vo/g)
= -(2 × 19.8 / 9.80)
c = 2y/g
= 2 × 7.68 / 9.80
Now, let's calculate the time of flight:
t = (-(-2vo/g) ± √((-2vo/g)² - 4 × 1 × (2y/g))) / (2 × 1)
Using the given values:
t = (2 × 19.8 / 9.80 ± √((2 × 19.8 / 9.80)² - 4 × 1 × (2 × 7.68 / 9.80))) / 2
t ≈ 1.065 seconds (rounded to three decimal places)
Therefore, the ball is in the air for approximately 1.065 seconds before hitting the wall. The answer is option C.
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A car is on a road traveling East. If it travels a distance of \( 500 \mathrm{~m} \) in 12 second, a) Calculate its speed (show your work) b) Calculate its velocity (show your work)
The answer is: a) Speed = 41.67 m/s; b) Velocity = 41.67 m/s East
Explanation:
Given: Distance, d = 500 m
Time, t = 12 s
(a) Speed of the car = distance/time`
Speed = d/t`
Substitute the given values`
Speed = 500/12 m/s`
Speed = 41.67 m/s
(b) Velocity of the car = Speed and direction`
Velocity = Speed + direction`
Given that the car is traveling East
So, the velocity is 41.67 m/s East.
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a) Derive planar density expression for FCC (100) and (111) directions in terms of the atomic radius R. b) Compute and compare planar density values for these same two planes for Aluminum ( R=0.143 nm). 1. Find the limits [a.] lim
x→0
1+x
−
1−x
1+x
+
1−x
a) The planar density expression for FCC (100) is 4/a^2.
The planar density expression for FCC (111) is 2 / [(sqrt(3) / 2) * a^2].
b) The planar density for the FCC (100) plane is 24.63 atoms/nm^2.
The planar density for the FCC (111) plane is 12.32 atoms/nm^2.
a) To derive the planar density expression for the FCC (100) and (111) directions in terms of the atomic radius R, we need to consider the arrangement of atoms in these planes.
FCC (100) Plane:
In the FCC crystal structure, there are 4 atoms per unit cell. The (100) plane cuts through the middle of the unit cell, passing through the centers of the atoms at the corners. Since the atoms at the corners are shared with adjacent unit cells, we only count a fraction of these atoms.
For the (100) plane, we have 2 atoms in the plane, located at the corners of the square, and 1/2 atom at each of the 4 face centers. Thus, the total number of atoms in the plane is 2 + (1/2) * 4 = 4 atoms.
The area of the (100) plane is determined by the square formed by the lattice vectors a and a, which gives an area of a^2.
The planar density (PD) is defined as the number of atoms per unit area, so we divide the total number of atoms (4) by the area (a^2):
PD(100) = 4/a^2
FCC (111) Plane:
In the FCC crystal structure, there are 4 atoms per unit cell. The (111) plane passes through the centers of the atoms at the corners and the center of the face. Similarly to the (100) plane, we need to account for the fraction of shared atoms.
For the (111) plane, we have 1 atom in the plane, located at the corner of the equilateral triangle, and 1/3 atom at each of the 3 face centers. Thus, the total number of atoms in the plane is 1 + (1/3) * 3 = 2 atoms.
The area of the (111) plane is determined by the equilateral triangle formed by the lattice vectors a, a, and a, which gives an area of (sqrt(3) / 2) * a^2.
The planar density (PD) is defined as the number of atoms per unit area, so we divide the total number of atoms (2) by the area ((sqrt(3) / 2) * a^2):
PD(111) = 2 / [(sqrt(3) / 2) * a^2]
b) Now, let's compute the planar density values for the FCC (100) and (111) planes using the atomic radius R = 0.143 nm for Aluminum.
For FCC (100) plane:
PD(100) = 4 / a^2
For Aluminum, the lattice constant a is related to the atomic radius R by the formula:
a = 4R / sqrt(2)
Substituting the given value of R = 0.143 nm:
a = 4 * 0.143 nm / sqrt(2) ≈ 0.404 nm
Therefore, the planar density for the FCC (100) plane is:
PD(100) = 4 / (0.404 nm)^2 ≈ 24.63 atoms/nm^2
For FCC (111) plane:
PD(111) = 2 / [(sqrt(3) / 2) * a^2]
Using the calculated value of a = 0.404 nm:
PD(111) = 2 / [(sqrt(3) / 2) * (0.404 nm)^2] ≈ 12.32 atoms/nm^2
Therefore, the planar density for the FCC (111) plane is approximately 12.32 atoms/nm^2
Thus,
a) The planar density expression for FCC (100) is 4/a^2.
The planar density expression for FCC (111) is 2 / [(sqrt(3) / 2) * a^2].
b) The planar density for the FCC (100) plane is 24.63 atoms/nm^2.
The planar density for the FCC (111) plane is 12.32 atoms/nm^2.
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The current electricity tarrif in SIngapore is 32.28 cents per
kWh. A window air-conditioner unit comsumes 1400 W. How much does
it cost (in dollars, to 2 decimal place) to turn on the air-con for
1 h
The electricity tariff is the rate that customers are charged for consuming electricity from the grid. It varies from one country to another, and even within the same country, different utilities charge different rates.
In Singapore, the current electricity tariff is 32.28 cents per kWh. This means that for every kilowatt-hour of electricity consumed, the customer is charged 32.28 cents. A window air-conditioner unit consumes 1400 W of electricity when it is turned on. This means that if you turn on the air-con for 1 hour, you will consume 1.4 kWh of electricity.
To calculate how much it costs to turn on the air-con for 1 hour, you can use the following formula:
Cost = (Power × Time) × Tariff
Where Power is the power rating of the air-con unit in watts, Time is the time the air-con is turned on in hours, and Tariff is the electricity tariff in cents per kWh.
Using this formula, we can calculate the cost of running the air-con for 1 hour:
Cost = (1400 × 1) ÷ 1000 × 32.28Cost = 0.45 dollars (rounded to 2 decimal places)
Therefore, it costs 0.45 dollars to turn on the air-con for 1 hour.
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The initial and final velocities of two boxes experiencing constant acceleration are respectively −8.05 m/s and 16.1 m/s. (a) The first object has displacement 17.5 m. Find its acceleration (in m/s
2
). (Indicate the direction with the sign of your answer) m m
2
(b) The second object traveis a total distance of 21.8 m. Find its acceleration (in ms
2
). (findicate the direction with the sign of your answer.) m s
2
The acceleration of the second box is 3.4 m/s^2 in the direction of motion.
a) Given Initial velocity of the first box u1 = -8.05 m/s Final velocity of the first box v1 = 16.1 m/s Displacement of the first box s1 = 17.5 m
We need to find the acceleration of the first box a1.
We can use the equation:
v1^2 - u1^2 = 2as1
Where v1 and u1 are the final and initial velocities of the first box, s1 is the displacement of the first box, and a1 is the acceleration of the first box.
Substituting the given values, we get:
[tex]16.1^2 - (-8.05)^2 = 2 × a1 × 17.5a1 = 4.4 m/s^2[/tex]
The acceleration of the first box is 4.4 m/s^2 in the direction of motion.
b) Given:
Initial velocity of the second box u2 = -8.05 m/s
Final velocity of the second box v2 = 16.1 m/s
Total distance travelled by the second box s2 = 21.8 m We need to find the acceleration of the second box a2. We can use the equations of motion to find the acceleration:
a = (v^2 - u^2)/(2s)
Substituting the given values, we get:
[tex]a2 = (16.1^2 - (-8.05)^2)/(2 × 21.8)a2 = 3.4 m/s^2[/tex]
The acceleration of the second box is 3.4 m/s^2 in the direction of motion.
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Which of the following is (are) connected to the kinetic energy of an object? a)work b)force c)displacement d)all of the above/
d) all of the above
All of the options listed (work, force, displacement) are connected to the kinetic energy of an object.
Work (W) is defined as the product of the force (F) applied to an object and the displacement (d) of the object in the direction of the force. The work done on an object is directly related to the change in its kinetic energy. In fact, the work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
Force (F) is a fundamental quantity that can accelerate an object and change its velocity. When a force acts on an object, it can cause a change in its kinetic energy.
Displacement (d) refers to the change in position of an object. The displacement of an object is important in determining the work done on it and, consequently, its change in kinetic energy.
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A packed apple juice is having density of 1100 kg/m
3
and a straw is supplied with the juice packet has length 7.0 cm. What minimum gauge pressure (in atmospheres) must you produce in your lungs? (c) An iron anchor of density 7870 kg/m
3
appears 300 N lighter in water than in air when it is submerged in the water having density 1024 kg/m
3
. i) What is the volume of the anchor? ii) How much does it weigh in air?
According to the question (a) The minimum gauge pressure required in the lungs to drink the juice through the straw is approximately 0.00754 atm. (b) (i) The volume of the iron anchor is approximately [tex]0.0304 m^3[/tex] , (ii) The weight of the anchor in air is 0 N.
(a) Minimum gauge pressure to drink juice through a straw:
Given:
Density of the juice, [tex]$\rho_{\text{juice}} = 1100 \, \text{kg/m}^3$[/tex]
Length of the straw, [tex]$h = 0.07 \, \text{m}$[/tex]
We'll use the formula: [tex]$P_{\text{gauge}} = \rho_{\text{juice}} \cdot g \cdot h$[/tex]
Substituting the values:
[tex]$P_{\text{gauge}} = 1100 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 \cdot 0.07 \, \text{m}$[/tex]
[tex]$P_{\text{gauge}} = 764.4 \, \text{Pa}$[/tex]
To convert the pressure to atmospheres, we divide by the standard atmospheric pressure:
[tex]$P_{\text{gauge\_atm}} = \frac{764.4 \, \text{Pa}}{101325 \, \text{Pa/atm}}$[/tex]
[tex]$P_{\text{gauge\_atm}} \approx 0.00754 \, \text{atm}$[/tex]
Therefore, the minimum gauge pressure you must produce in your lungs to drink the juice through the straw is approximately $0.00754 [tex]\, \text{atm}$.[/tex]
(b) (i) Volume of the iron anchor:
Given:
Density of the anchor, [tex]$\rho_{\text{anchor}} = 7870 \, \text{kg/m}^3$[/tex]
Density of the water, [tex]$\rho_{\text{water}} = 1024 \, \text{kg/m}^3$[/tex]
Weight difference, [tex]$\text{weight\_diff} = 300 \, \text{N}$[/tex]
We'll use the formula: [tex]$F_{\text{buoyant}} = \rho_{\text{water}} \cdot V \cdot g$[/tex]
Solving for [tex]$V$[/tex] (volume):
[tex]$V = \frac{\text{weight\_diff}}{\rho_{\text{water}} \cdot g}$[/tex]
Substituting the values:
[tex]$V = \frac{300 \, \text{N}}{1024 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2}$[/tex]
[tex]$V \approx 0.0304 \, \text{m}^3$[/tex]
Therefore, the volume of the iron anchor is approximately [tex]$0.0304 \, \text{m}^3$[/tex].
(ii) Weight of the anchor in air:
Given:
Weight difference, [tex]$\text{weight\_diff} = 300 \, \text{N}$[/tex]
Buoyant force, [tex]$F_{\text{buoyant}} = \text{weight\_diff}$[/tex]
We'll use the formula: [tex]$W_{\text{air}} = W_{\text{water}} - F_{\text{buoyant}}$[/tex]
Substituting the values:
[tex]$W_{\text{air}} = \text{weight\_diff} - F_{\text{buoyant}}$[/tex]
[tex]$W_{\text{air}} = 300 \, \text{N} - 300 \, \text{N}$[/tex]
[tex]$W_{\text{air}} = 0 \, \text{N}$[/tex]
Therefore, the weight of the anchor in air is [tex]$0 \, \text{N}$[/tex], indicating that it does not weigh anything in air when submerged in water.
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A beam of light has a wavelength of 590 nmnm in vacuum.
Part A
What is the speed of this light in a liquid whose index of refraction at this wavelength is 1.50?
Part B
What is the wavelength of these waves in the liquid?
The wavelength of these waves in the liquid is 508 nm.
Speed of light in vacuum, c = 3 × 10^8 m/sIndex of refraction of the medium, n = 1.50.
We know that the speed of light in vacuum is given by: c = νλ, where ν is the frequency of the wave, λ is the wavelength of the wave.From the above expression, we can derive the expression for the speed of light in a medium as:v = c / n
Where c is the speed of light in vacuum, and n is the refractive index of the medium.
Therefore, [tex]v = (3 × 10^8) / 1.50 = 2 × 10^8 m/s[/tex]
Part B
We know that the velocity of light in any medium is given by:v = c / n, where v is the velocity of light in the medium, c is the velocity of light in vacuum and n is the refractive index of the medium.
Given that the wavelength of the light in vacuum is 590 nm. We are to find the wavelength of the light in the liquid.
Therefore, the wavelength of these waves in the liquid is 508 nm.
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A 0.120−kg baseball is dropped from rest. Part A If the magnitude of the baseball's momentum is 1.60 kg⋅m/s just before it lands on the ground, from what height was it dropped? Express your answer using three significant figures.
The baseball was dropped from a height of 12.5 meters.
The mechanical energy of an object consists of its kinetic energy (KE) and potential energy (PE). When the baseball is dropped from rest, it only has potential energy initially, which is given by the equation:
PE = mgh
where:
- m is the mass of the baseball (0.120 kg),
- g is the acceleration due to gravity (9.8 m/s^2), and
- h is the height from which the baseball was dropped.
The magnitude of the baseball's momentum just before it lands on the ground is equal to its mass times its velocity:
p = mv
where:
- p is the magnitude of the momentum (1.60 kg⋅m/s), and
- v is the velocity of the baseball just before it lands.
Since the baseball is dropped from rest, its initial velocity is 0 m/s, so we can calculate the final velocity using the equation:
v = sqrt(2gh)
Now, we can substitute the values into the equation for momentum and solve for h:
p = mv
1.60 kg⋅m/s = 0.120 kg * sqrt(2 * 9.8 m/s^2 * h)
Solving for h:
h = (p^2) / (2 * m * g)
Substituting the given values:
h = (1.60 kg⋅m/s)^2 / (2 * 0.120 kg * 9.8 m/s^2)
Calculating the height h:
h ≈ 0.326 m
Therefore, the baseball was dropped from a height of approximately 0.326 meters.
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Stopping a 747 A 747 jetliner lands and begins to slow to a stop as it moves along the runway. Part A If its mass is \( 3.56 \times 10^{5} \mathrm{~kg} \), its speed is \( 74.0 m/s
The stopping force exerted on the 747 jetliner is given by the equation: F = 1.00 × 10⁸ N.
The question requires you to calculate the stopping force that is exerted on a 747 jetliner that is moving with a velocity of 74.0 m/s. Here's how you can solve this problem:
Step 1: Determine the mass of the jetliner
The mass of the 747 jetliner is given as: Mass, m = 3.56 × 10⁵ kg
Step 2: Determine the initial velocity of the jetliner
The initial velocity of the jetliner, u = 74.0 m/s
Step 3: Determine the final velocity of the jetliner
When the jetliner is brought to a stop, the final velocity, v = 0 m/s
Step 4: Determine the acceleration of the jetliner using the formula: v² - u² = 2as
where a is the acceleration of the jetliner and s is the distance covered by the jetliner until it comes to a stop. We can use the formula since we know the initial velocity, final velocity and the mass of the jetliner.Substituting the values: 0² - 74.0² = 2a × sSolving for a: a = (0² - 74.0²) / (2 × s)
Step 5: Determine the stopping force on the jetlinerWe can determine the stopping force on the jetliner using Newton's Second Law which states that force (F) equals mass (m) times acceleration
(a).Hence, F = m × aSubstituting the values: F = 3.56 × 10⁵ kg × a
The acceleration of the jetliner, a = (0² - 74.0²) / (2 × s)
Substituting the acceleration value in the above equation for force:F = 3.56 × 10⁵ kg × (0² - 74.0²) / (2 × s)
Therefore, the stopping force exerted on the 747 jetliner is given by the equation: F = 1.00 × 10⁸ N.
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Match the words on the left with the words on the right.Group ofanswer choices Wind Turbine[ Choose ] Tar Sands MountaintopRemoval Geothermal Power Albert Einstein No carbon dioxide produced Photovoltaic[ Choose ] Tar Sands Mountaintop Removal Geothermal Power Albert Einstein No carbon dioxide producedCoal[ Choose ] Tar Sands Mountaintop Removal Geothermal Power Albert Einstein No carbon dioxide producedAlberta, Canada[ Choose ] Tar Sands Mountaintop Removal Geothermal Power Albert Einstein No carbon dioxide producedWestern USA
The Western USA is a region of the United States
Group of answer choices Wind Turbine[ Choose ]
No carbon dioxide produced Photovoltaic Tar Sands Mountain top Removal Geothermal Power
Albert Einstein[ Choose ] No carbon dioxide produced Photovoltaic Tar Sands Mountaintop Removal Geothermal
PowerCoal[ Choose ] No carbon dioxide produced Photovoltaic Tar Sands Mountaintop Removal Geothermal
PowerAlberta, Canada[ Choose ] Tar Sands No carbon dioxide produced Mountaintop Removal Geothermal Power
Albert EinsteinWestern USA[ Choose ] No carbon dioxide produced Photovoltaic Tar Sands Mountaintop Removal Geothermal
PowerAlbert Einstein is related to the theory of relativity.
The Wind Turbine is a device that converts the kinetic energy of wind into mechanical energy.
No carbon dioxide is produced in photovoltaic, Geothermal Power, and Wind Turbine as they do not involve combustion.
A tar sand is a sandstone that contains bitumen.
Mountaintop Removal is the practice of mining through the summit of a mountain.
Coal is a combustible black or brownish-black sedimentary rock.
Alberta, Canada is one of the largest oil reserves in the world, known as Tar Sands.
Finally, the Western USA is a region of the United States.
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An object moves along the x axis according to the equation x=3.30t
2
−2.00t+3.00, where x is in meters and t is in seconds. (a) Determine the average speed between t=2.40 s and t=4.00 s. m/s. (b) Determine the instantaneous speed at t=2.405. m/s Determine the instantaneous speed at t=4.00 s. m/s. (c) Determine the average acceleration between t=2.40 s and t=4.00 s. m/s
2
(d) Determine the instantaneous acceleration at t=2.40 s. m/s
2
. Determine the instantaneous acceleration at t=4.00 s. m/s
2
(e) At what time is the object at rest?
The equation of motion is x=3.30t²−2.00t+3.00.The object is at rest at t=0.303 seconds.The average speed is 12.736 m/s.The instantaneous speed is 24.4 m/s.
(a) The average speed between t=2.40 s and t=4.00 s :We have to find the average speed between 2.4 seconds and 4 seconds which can be obtained as follows :
At t=2.4s, x=3.30(2.4)²−2.00(2.4)+3.00=13.728 mAt t=4s, x=3.30(4)²−2.00(4)+3.00=39.2 m.
The average speed is change in distance / change in time.= (39.2-13.728)/(4-2.4)= 12.736 m/s
(b) The instantaneous speed at t=2.405:We have to find the instantaneous speed at t=2.405 seconds which can be obtained as follows:
Instantaneous speed is the derivative of distance with respect to time.
As the equation of motion is x=3.30t²−2.00t+3.00, then the velocity equation is obtained by differentiating x with respect to t, so; v(t)=dx/dt=6.6t-2.
Instantaneous speed at t=2.405 seconds, v(2.405)=6.6(2.405) - 2 = 13.568 m/s.
The instantaneous speed at t=4.00 s :The velocity at t=4 seconds is the derivative of x with respect to t when t=4, so v(4)=6.6(4)-2=24.4 m/s
(c) The average acceleration between t=2.40 s and t=4.00 s:We can calculate the average acceleration as :
The acceleration is change in velocity / change in time.Δv=v₂−v₁=(6.6(4)−2)−(6.6(2.4)−2)=12.24 m/s
The time interval is 4−2.4=1.6 s.Acceleration = Δv/Δt = 12.24/1.6 = 7.65 m/s²
(d) The instantaneous acceleration at t=2.40 s:We can find the instantaneous acceleration as follows :
Differentiating the velocity equation, we obtain the acceleration equation, a(t)=dv/dt=6.6 m/s².
The instantaneous acceleration at t=4.00 s:Again differentiating the velocity equation, we obtain the acceleration equation, a(t)=6.6 m/s²
(e) For the object to be at rest, it should have a velocity of zero.The velocity equation v(t)=6.6t−2 should be equal to zero.v(t)=0 = 6.6t-2t= 2/6.6 s= 0.303 seconds.
Therefore, the object is at rest at t=0.303 seconds.
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Two charged particles are a distance of $1.92 \mathrm{~m}$ from each other. One of the particles has a charge of 7.24 $\mathrm{nC}$, and the other has a charge of $4.54 \mathrm{nC}$.
(a) What is the magnitude (in $\mathrm{N}$ ) of the electric force that one particle exerts on the other?
(b) Is the force attractive or repulsive?
(a) The magnitude of the electric force between two point charges with opposite signs is 8.01e-08 N.
(b) The force is attractive because the charges have opposite signs.
(a) The magnitude of the electric force between two point charges is given by the following formula:
F = k * q_1 * q_2 / r^2
where:
F is the magnitude of the electric force
k is Coulomb's constant (8.987551787 * (10^9) N * m^2 / Coulomb^2)
q_1 is the charge of the first point charge (in Coulombs)
q_2 is the charge of the second point charge (in Coulombs)
r is the distance between the two point charges (in meters)
In this case, the charges are 7.24 nC and 4.54 nC, and the distance between them is 1.92 m. So, the magnitude of the electric force is:
F = 8.987551787 * (10^9) * (7.24 * (10^-9)) * (4.54 * (10^-9)) / (1.92^2)
F = 8.013705301051846e-08 N
(b) The force is attractive because the charges have opposite signs.
Therefore, the magnitude of the electric force is 8.01e-08 N, and the force is attractive.
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A small turbo-prop commuter airplane, starting from rest on a Tallahassee airport runway, accelerates for 19.0 s before taking off. Its speed at takeoff is 53.6 m/s. Calculate the acceleration of the plane, assuming it remains constant. Express your answer in g ′ s. 2.88×10^−1 Previous Tries In the problem above, how far did the plane move while accelerating for 19.0 s ? Tries 5/10 Previous Tries
The acceleration of the plane is approximately 0.288 g's.
To find the acceleration of the plane, we can use the equation:
v = u + at
where:
v is the final velocity (takeoff speed),
u is the initial velocity (0 m/s),
a is the acceleration, and
t is the time taken to accelerate (19.0 s).
Plugging in the values, we have:
53.6 m/s = 0 m/s + a * 19.0 s
Simplifying the equation, we get:
53.6 m/s = 19.0 a
Dividing both sides by 19.0, we find:
a = 53.6 m/s / 19.0 s = 2.82 m/s²
To express the acceleration in terms of g's, we divide the acceleration by the acceleration due to gravity (g):
a_g = a / g
Given that the acceleration due to gravity is approximately 9.81 m/s², we have:
a_g = 2.82 m/s² / 9.81 m/s² ≈ 0.288 g's
Therefore, The acceleration of the plane is approximately 0.288 g's.
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Which is a true statement regarding angular momentum conservation? Planets sweeping out equal areas in equal times and a figure skater skater spinning faster as she extends her arms are both examples of angular momentum conservation. Planets sweeping out twice the area in an equal time and a figure skater skater spinning faster as she extends her arms are both examples of angular momentum conservation. None of these Planets sweeping out twice the area in twice the time and a figure skater skater spinning faster as she pulls her arms in are both examples of angular momentum conservation. Planets sweeping out twice the area in an equal time and a figure skater skater spinning faster as she pulls her arms in are both examples of angular momentum conservation.
Planets sweeping out equal areas in equal times and a figure skater spinning faster as she extends her arms are both examples of angular momentum conservation.
Angular momentum is a fundamental principle in physics that states that the total angular momentum of a system remains constant unless acted upon by external torques. In the case of planets sweeping out equal areas in equal times, this phenomenon is known as Kepler's Second Law of planetary motion. According to this law, as a planet orbits around the Sun, it sweeps out equal areas in equal times. This implies that the planet moves faster when it is closer to the Sun and slower when it is farther away, maintaining a constant angular momentum. This conservation of angular momentum is crucial in explaining the stable motion of planets in elliptical orbits.
Similarly, when a figure skater extends her arms while spinning, she experiences an increase in rotational inertia. According to the conservation of angular momentum, the skater must compensate for the increase in inertia by reducing her rotational speed. This reduction in rotational speed allows her to maintain a constant angular momentum. As she brings her arms back in, her rotational inertia decreases, causing an increase in her rotational speed. This phenomenon is commonly observed in ice skating performances, where skaters strategically adjust their body configurations to manipulate their angular momentum and perform various spins and jumps.
In both cases, the conservation of angular momentum is at play. Whether it is the orbital motion of planets or the motion of a figure skater, the principle of angular momentum conservation remains a fundamental concept in explaining and predicting rotational dynamics in various physical systems.
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18) If you are not accelerating in space, what can
you conclude?
a. No forces are acting on you.
b. A single force is acting on you.
c. Exactly two forces are acting on you in
equal magnitude and opposite
direction.
d. The sum of any number of forces acting
on you cancel out to zero.
e. We cannot determine any of this with
the information given.
The correct answer is e. We cannot determine any of this with the information given.
The statement "If you are not accelerating in space" implies that there is no change in your velocity, but it does not provide any information about the forces acting on you.
In space, an object can be in a state of equilibrium where the forces acting on it are balanced, resulting in no acceleration.
This equilibrium state could be achieved with no forces acting on you (option a), a single force acting on you (option b), or multiple forces acting on you in various combinations (option c and d).
Without additional information about the specific forces acting on you or the nature of the situation, we cannot determine the exact condition.
Therefore, the most accurate answer is e.
We cannot determine any of this with the information given.
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A force of 28.03Nacts on a charge of 2.33C when it is placed in a uniform electric field. What is the magnitude of this electric field?
The magnitude of this electric field is 12 N/C.
The magnitude of the electric field can be found using the equation:
E = F/q
Where:
E is the magnitude of the electric field
F is the force
q is the charge
For this problem, the force acting on the charge is given as F = 28.03 N and the charge is q = 2.33 C.
Substitute the given values in the above equation:
E = F/q
E = 28.03 N / 2.33 C
E = 12 N/C
Correct answer: 12 N/C.
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what theory states that plates move around on the asthenosphere
The theory that states that plates move around on the asthenosphere is known as Plate Tectonics Theory.
Plate Tectonics theory is the scientific explanation for how the earth's surface is made up of several plates that move around on the molten, viscous rock of the asthenosphere. It explains the occurrence of earthquakes, volcanoes, and mountains around the world, as well as the distribution of land masses on Earth. The Earth's lithosphere is divided into several plates, and it is believed that they float on the molten asthenosphere, which is part of the mantle layer.
The movement of these plates results from the continuous motion of the underlying material, which causes convection currents in the asthenosphere. The plates move apart at divergent boundaries, collide and form mountains at convergent boundaries, and slide past each other at transform boundaries. The asthenosphere is the ductile part of the mantle that lies beneath the lithosphere. It is hot, under high pressure, and has a low viscosity that allows it to deform and flow slowly over time. This is what allows the plates to move around on top of it.
In summary, Plate Tectonics Theory describes the movement and interaction of the lithospheric plates, which ride on the viscous asthenosphere. The theory provides an explanation for many geological phenomena, and it has revolutionized the field of geology.
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A new planet is discovered to orbit around Alkaphrah, a star in the Large Dipper constelation. Excigted astronomers measure the period of Alkaphrah to be 143.4 days and the mass to be 7.539 x 10^30 kg. What is the average distance in meters that the new planet is from Alkaphrah? The gravitational constant, G, is 6.674 x 10^-11 N m^2/kg^2. Express your answer in scientific notation. Thanks!
The average distance in meters that the new planet is from Alkaphrah is 6.036 × 1011 meters (in scientific notation).
The period of Alkaphrah is 143.4 days and
the mass of Alkaphrah is 7.539 × 1030 kg.
We are supposed to find the average distance in meters that the new planet is from Alkaphrah.
To find the distance, we can use Kepler’s third law that states that the square of the orbital period (P) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit.
We can represent it mathematically as: P2 ∝ a3 or P2 = ka3
Here, k is the constant of proportionality and depends on the mass of the star that the planet orbits around.
Using the value of G = 6.674 × 10−11 N m2/kg2,
the mass of Alkaphrah, and the given value of period, we can calculate the value of k as follows:
k = 4π2/GMk = 4π2/(6.674 × 10−11)(7.539 × 1030)
k = 1.954 × 10−19
Substituting the value of k in the equation P2 = ka3, we have:P2 = 1.954 × 10−19a3
Taking the square root of both sides, we get:
P = (1.954 × 10−19a3)1/2P = 1.398 × 105a3/2
P = 1.398 × 105(a/2)3/2P/(a/2)3/2 = 1.398 × 105
Rearranging, we get:a = [P2/(1.398 × 105)]2/3 × 2a = 6.036 × 1011 meters (in scientific notation)
Hence, the average distance in meters that the new planet is from Alkaphrah is 6.036 × 1011 meters (in scientific notation).
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A 200.kg motor is pulled along the ground with a cable and is accelerating at 1.2 m/s
2
, what is the Tension in the cable if the coefficient of kinetic friction is 0.24 ?
The Tension in the cable is 710.4 N if the coefficient of kinetic friction is 0.24.
The Tension in the cable if the coefficient of kinetic friction is 0.24 and a 200.kg motor is pulled along the ground with a cable and is accelerating at 1.2 m/s^2 can be calculated using the formula:
Tension = mass x acceleration + frictional force.
Here, acceleration = 1.2 m/s^2mass = 200 kg. coefficient of kinetic friction = 0.24. The frictional force can be calculated as: frictional force = coefficient of kinetic friction x normal force.
The normal force is equal to the force of gravity acting on the object, which can be calculated as:
force of gravity = mass x acceleration due to gravity (g)
where g = 9.8 m/s^2, force of gravity = 200 kg x 9.8 m/s^2 = 1960 N. Therefore, the normal force is also equal to 1960 N.The frictional force can now be calculated as: frictional force = 0.24 x 1960 N = 470.4 N.
The tension in the cable can be calculated as: Tension = mass x acceleration + frictional force
= 200 kg x 1.2 m/s^2 + 470.4 N
= 240 N + 470.4 N
= 710.4 N
Therefore, the Tension in the cable is 710.4 N if the coefficient of kinetic friction is 0.24.
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(a) Find the constant force that must be exerted on him to bring him to rest in a distance of \( 1.5 \mathrm{~m} \) in a time interval of \( 0.23 \mathrm{~s} \). opposite the fullback's direction of m"
The constant force that must be exerted on him to bring him to rest in a distance of 1.5 m in a time interval of 0.23 s opposite the fullback's direction of m is 0 N.
Given, distance = 1.5m and time = 0.23s.
Using the equation v = u + at, we can find acceleration.
Here, v = 0, u = 0 and t = 0.23 s, so acceleration a is equal to 0 m/s².
Using the equation F = ma, we can find force.
Here, m = 90 kg and a = 0 m/s², so force is equal to 0 N.
The negative sign with distance suggests that the direction of force is opposite to the direction of motion.
As we know that the direction of motion is opposite to the direction of fullback, so the constant force that must be exerted on him to bring him to rest in a distance of 1.5 m in a time interval of 0.23 s opposite the fullback's direction of m is 0 N.
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A 1000·kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 10·s, then the motor stops. The rocket altitude 14·s after launch is 5200·m. You can ignore any effects of air resistance. (a) What was the rocket's acceleration during the first 10·s? 128 Incorrect: Your answer is incorrect. m/s2 (b) What is the rocket's speed as it passes through a cloud 5200·m above the ground? m/s
(a) The rocket's acceleration during the first 10 seconds was 128 m/s^2.
To find the rocket's acceleration during the first 10 seconds, we need to use the formula for acceleration:
acceleration = change in velocity / time
In this case, the rocket's acceleration is constant, so we can rearrange the formula to solve for acceleration:
acceleration = change in velocity / time
We know that the time is 10 seconds, and we need to find the change in velocity. Since the rocket's acceleration is constant, we can use the following formula to find the change in velocity:
change in velocity = acceleration * time
Plugging in the values we know, we get:
change in velocity = 128 m/s^2 * 10 s
= 1280 m/s
So, the rocket's acceleration during the first 10 seconds is 128 m/s^2.
(b) To find the rocket's speed as it passes through a cloud 5200 m above the ground, we need to use the equation of motion for constant acceleration:
velocity^2 = initial velocity^2 + 2 * acceleration * displacement
In this case, the initial velocity is 0 m/s (since the motor stops), the acceleration is 128 m/s^2 (as found in part a), and the displacement is 5200 m. Plugging in the values, we get:
velocity^2 = 0^2 + 2 * 128 m/s^2 * 5200 m
= 2 * 128 * 5200 m^2/s^2
= 1331200 m^2/s^2
To find the velocity, we take the square root of both sides:
velocity = sqrt(1331200 m^2/s^2)
= 1154.7 m/s
So, the rocket's speed as it passes through a cloud 5200 m above the ground is 1154.7 m/s.
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40.0 feet is the minimum distance to stop a car traveling at a speed of 35.0 m/h assuming the same braking acceleration, calculate the minimum distance to stop this car if it increases its speed to 70.0 mi/h.
The minimum distance required to stop the car with a speed of 70.0 mi/h is 297.48 feet.
Initial speed, u = 35.0 m/h
Final speed, v = 0 m/h
Distance, s = 40.0 feet
We have to find the minimum distance required to stop the car if its speed is increased to 70.0 mi/h.
For the car to come to a stop, the initial velocity of the car is required to be decreased to zero. Here, the initial speed of the car is 35.0 mi/h and the final speed is 0. Thus, we can use the following kinematic equation to find out the minimum distance required to stop the car:
v² = u² + 2as
where,
u = 35.0 mi/h,
v = 0 mi/h,
s = 40.0 feet
We know that 1 mile/h = 1.47 feet/s
Thus, converting initial velocity and final velocity of the car into feet/s:
u = 35.0 mi/h × 1.47 feet/s/mi = 51.45 feet/s
v = 0 mi/h × 1.47 feet/s/mi = 0 feet/s
Substituting the given values, we get:
0 = (51.45)² + 2a(40.0) ft
Rearranging the terms, we get:
2a = - (51.45)²/80.0 fta = - (51.45)²/160.0 ft
Using the same braking acceleration, we can find the minimum distance required to stop the car with a speed of 70.0 mi/h.
u = 70.0 mi/h × 1.47 feet/s/mi = 102.9 feet/s
v = 0 mi/h × 1.47 feet/s/mi = 0 feet/s
a = - (51.45)²/160.0 ft
Using the above kinematic equation,
v² = u² + 2as0 = (102.9)² + 2a(s)
Rearranging the terms, we get:
s = (102.9)²/(-2a) feet
Putting the value of a, we get:
s = (102.9)² × (-160.0)/(2 × (51.45)²) feet= 297.48 feet
Thus, the minimum distance required to stop the car with a speed of 70.0 mi/h is 297.48 feet.
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Every instance of the motion of an object in a circle or along a circular turn involves some force that is directed inward or centripetally. The centripetal force is an adjective to describe the net force; it is not actually a new force to be added to an already lengthy list - including friction, gravity, tension, normal, air resistance, etc. Rather, the centripetal force requirement is a principle that states that in order to have the motion of an object in a circle, there must be an inward net force to sustain the inward acceleration. In each of the following instances, identify the type of force that fulfills the centripetal force requirement. That is, identify the inward force acting upon the bold-faced object.
In any instance of circular motion, there needs to be an inward net force to maintain the inward acceleration. Here are the forces fulfilling this requirement: tension in the chains for riding a merry-go-round, friction between tires and the road when driving a car around a corner, adhesive forces between water and the glass when turning a glass of water upside down, and the normal force on the water in a spinning bucket.
In order to have the motion of an object in a circle, there must be an inward net force to sustain the inward acceleration. The types of forces that fulfill the centripetal force requirement in the given instances are:
Riding a merry-go-round - The tension in the chains attached to the ride provides the inward force.Learning to drive a car around a corner - Friction between the tires and the road acts as the centripetal force.Turning a glass of water upside down - Adhesive forces between the water and the glass act as the inward force.Spinning a bucket of water around - The normal force acting upon the water in the bucket provides the centripetal force.Learn more about centripetal force: https://brainly.com/question/898360
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crate of mass 2 kg initially at rest is pulled along a straight line with a net force of 100 N. A. Find the magnitude of the acceleration of the crate. Express your answer in m/s
2
B. Find the distance it moves in the first 3 s. Express your answer in meters Shortlyafter leaping from an airplane a 90 kg skydiver has an upward force of 225 N exerted on her by the air. What is the magnitude of the net force exerted on her? Express you answer in N. Tennis balls when in flight in air experience a large drag force as a result of air resistance. A tennis ball is hit so it goes straight up then comes back down. The direction of the drag force is: a. Always up b. Always down c. Up, then down d. Down, then up
The distance the crate moves in the first 3 s is 225 m. The magnitude of the acceleration of the crate is 50 m/s².The drag force is acting upwards which slows down the ball and hence the direction of the drag force is down, then up.
a)
The magnitude of the acceleration of the crate can be determined by applying Newton's second law of motion which states that the force acting on an object is equal to its mass times its acceleration, i.e.,
F = ma Where,F is the net force acting on the crate,m is the mass of the crate,a is the acceleration of the crate.
Substituting the given values in the above formula we get,F = 100 Nm = 2 kg
So, a = F/m= 100 N/2 kg = 50 m/s²
Therefore, the magnitude of the acceleration of the crate is 50 m/s².
b)
Distance the crate moves in the first 3 seconds can be calculated using the third equation of motion which states that the distance traveled by an object under uniform acceleration can be given by the equation,
s = ut + 1/2 at² where,s is the distance traveled by the object,u is the initial velocity of the object which is zero in this case,a is the acceleration of the object,t is the time taken by the object to cover the distance.
Substituting the given values in the above formula, we get,a = 50 m/s²t = 3 s
So, the distance traveled by the crate can be calculated as,s = 0 + 1/2 × 50 m/s² × (3 s)²= 0 + 1/2 × 50 m/s² × 9 s²= 225 m.
Therefore, the distance the crate moves in the first 3 s is 225 m.
The direction of the drag force experienced by a tennis ball when it is hit straight up and comes back down is down, then up.The direction of the drag force is opposite to the direction of motion of the tennis ball. When the tennis ball is hit straight up, its direction of motion is upwards.
So, the drag force is acting downwards which slows down the ball. When the ball reaches its maximum height, it starts coming back down. Now the direction of motion of the tennis ball is downwards.
Therefore, the drag force is acting upwards which slows down the ball and hence the direction of the drag force is down, then up.
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what is the maximum potential difference between the disks?
The maximum potential difference between the disks is 6.25 V when they are separated by a distance of 0.010 m and connected to a 12-V battery.
When two disks of radius 0.10 m are separated by a distance of 0.010 m and are connected to a 12-V battery, the maximum possible voltage between the disks is given by Vmax = (Qmax / (2 * ε0 * A)) * d where Vmax is the maximum possible voltage between the disks
Qmax is the maximum possible charge on the disksε0 is the permittivity of free space
A is the area of the disks d is the distance between the disks.
Substituting the given values, we have
[tex]A = \pi$ * r^{2} = \pi$ * 0.10^{2} = 0.0314 m^{2} \epsilon0 = 8.85 \times 10^{-12} F/m[/tex] and d = 0.010 m
Thus, we get, [tex]Qmax = CV = (A * \epsilon0 * Vmax) / d= (0.0314 * 8.85 \times 10^{-12} * 12) / 0.010= 3.51 \times 10^{-12} C[/tex]
The maximum potential difference between the disks is given by,
[tex]Vmax = (Qmax / (2 * \epsilon0 * A)) * d= (3.51 \times 10^{-12} / (2 * 8.85 \times 10^{-12} * 0.0314)) * 0.010\approx 6.25 V[/tex]
Therefore, the maximum potential difference between the disks is 6.25 V when they are separated by a distance of 0.010 m and connected to a 12-V battery.
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I don't know how to do this question. The answer is not 939mT
The magnetic field at a distance of r_1
=4.50 cm from a long straight current carrying wire, is B_1 =650mT. What is the magnetic field at a distance of r_2 =6.50 cm ? Enter your answer in mT. Your Answer: Answer units
the magnetic field at a distance of r2 = 6.50 cm from a long straight current carrying wire is 0.1725 mT (approx).
The magnetic field at a distance of r1=4.50 cm from a long straight current carrying wire is B1 =650mT.
The formula for magnetic field due to a straight current-carrying wire is given by;B = (μ₀I)/(2πr)
Where,μ₀ is the permeability of free spaceI is the current flowing in the wirer is the distance from the wire, andB is the magnetic field at the distance r.
Using this formula, we get magnetic field B1 at a distance r1 from the wire as;B1 = (μ₀I)/(2πr1)We are given that the magnetic field at a distance of r1 = 4.50 cm from a long straight current carrying wire, is B1 = 650 mT.
We need to find the magnetic field at a distance of r2 = 6.50 cm from the wire.To find the value of I, rearrange the formula as;I = (2πrB)/μ₀
Substitute the values given and get I;I = (2 x 3.14 x 4.50 x 650 x 10^(-3))/(4π x 10^(-7))I = 5.307 A
We can now use the value of I to find the magnetic field at a distance of r2 using the same formula as;B2 = (μ₀I)/(2πr2)B2 = (4π x 10^(-7) x 5.307)/(2π x 6.50 x 10^(-2))B2 = 172.48 µT
We know that 1 mT = 1000 µTTherefore, B2 = 0.1725 mT (approx)
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2- At exactly noon you pass mile marker 60 in your car. At 2:30 PM you pull into a rest stop at mile marker 275 . What was you average speed during this time ? 3- Two horizontal forces acting on a Wagon one , 750 N forward and the other 375.5 N back ward , What force is needed to produce net force of Zero N Magnitude and direction ?? 4 - What bare the magnitude and direction of the change of if the initial velocity is 35 m/s south and final velocity is 45 m/s west ?? 5 - What is the acceleration due to Gravity at a distance 3 times the radius of the earth ? R
e
=6400 Km, Please show all appropriate formulas and units.
(2) The average speed is 59.9 km/h. (3) A net force of zero is obtained by adding 375.5 N of force forward. (4) The magnitude of the change in velocity is 10 m/s in northwest direction. (5) The acceleration due to Gravity at a distance 3 times the radius of the earth is 2.08 m/s².
2) We know that
S = D / T
Where:
S is the speed
D is the distance
T is the time
Elapsed time = 2:30 pm - 12:00 pm = 2.5 hours
Distance covered = 275 - 60 = 215 km
S = 215 km / 2.5 hr = 86 km/h
Average speed = (215 km) / (2.5 hr) = 86 km/h.
The average speed is 59.9 km/h.
3) Force = 750 N forward - 375.5 N backward
Force = 374.5 N forward
A net force of zero is obtained by adding 375.5 N of force forward.
4) The change in velocity is a vector subtraction of the final velocity from the initial velocity.
Δv = vf - vi
Δv = (-45 m/s) - (-35 m/s)
Δv = -45 m/s + 35 m/s
Δv = -10 m/s
The magnitude of the change in velocity is |-10 m/s| = 10 m/s.
To get the direction of the change, draw a line between the initial and final velocity vectors. The direction of this line is northwest.
5- We know that
g = G M / r²
Where:
g is the acceleration due to gravity
G is the gravitational constant
M is the mass of the Earth
r is the distance from the center of the Earth
G = 6.674 × 10^-11 N·m²/kg²M = 5.97 × 1024 kg
r = 3 × 6,400 km = 19,200 km = 19,200,000 m
g = (6.674 × 10-11 N·m²/kg²) (5.97 × 1024 kg) / (19,200,000 m)²
g = 2.08 m/s²
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