Answers
(a) The projectile's velocity at the highest point of its trajectory is approximately 27.01 m/s, counterclockwise from the +x-axis.
(b) The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.
To solve this problem, we'll analyze the projectile's motion in two dimensions: horizontal and vertical.
(a) To find the projectile's velocity at the highest point of its trajectory, we need to consider the vertical component and horizontal component separately.
The initial velocity (V0) of the projectile is 48.0 m/s, and the launch angle (θ) is 34.0° above the horizontal.
The vertical component of velocity (Vy) can be found using the equation:
Vy = V0 * sin(θ)
Plugging in the known values:
Vy = 48.0 m/s * sin(34.0°)
Calculating Vy, we find:
Vy ≈ 27.01 m/s
The horizontal component of velocity (Vx) can be found using the equation:
Vx = V0 * cos(θ)
Plugging in the known values:
Vx = 48.0 m/s * cos(34.0°)
Calculating Vx, we find:
Vx ≈ 39.79 m/s
Therefore, at the highest point of its trajectory:
- The magnitude of the projectile's velocity is approximately 27.01 m/s.
- The direction of the velocity is straight up, counterclockwise from the +x-axis.
(b) To find the straight-line distance from where the projectile was launched to where it hits its target, we need to consider the horizontal motion of the projectile.
The time of flight (t) is given as 3.65 s.
The horizontal distance (x) can be found using the equation:
x = Vx * t
Plugging in the known values:
x = 39.79 m/s * 3.65 s
Calculating x, we find:
x ≈ 144.93 m
The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.
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Related Questions
About 1.75% of water on Earth is in Greenland and Antarctica's icecaps, and about 97.5% is in the oceans. Assume the icecaps have an average temperature of -28°C, and the oceans have an average temperature of 4.8°C. If all the icecaps slid into the ocean and melted, how much would the average temperature of the ocean decrease?
Answers
If all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
To calculate the decrease in the average temperature of the ocean when all the icecaps melt, we need to consider the heat exchange between the icecaps and the ocean.
Let's start by calculating the heat released by the icecaps when they melt. We can use the specific heat capacity formula:
Heat released = Mass of icecaps × Specific heat capacity of ice × Temperature change
Since the icecaps constitute 1.75% of the Earth's water, the mass of icecaps is 0.0175 times the total mass of water on Earth.
Assuming the icecaps have an average temperature of -28°C and melt into liquid water at 0°C, the temperature change is 0°C - (-28°C) = 28°C.
Next, we need to calculate the heat absorbed by the ocean when the icecaps melt. Using the same formula:
Heat absorbed = Mass of ocean water × Specific heat capacity of water × Temperature change
Given that the oceans constitute 97.5% of the Earth's water, the mass of the ocean water is 0.975 times the total mass of water on Earth.
Assuming the oceans have an average temperature of 4.8°C, the temperature change is 4.8°C - 0°C = 4.8°C.
Now we can calculate the change in temperature of the ocean:
Change in temperature = Heat released / (Mass of ocean water × Specific heat capacity of water)
Substituting the values, we get:
Change in temperature = (0.0175 × Total mass of water) × (Specific heat capacity of ice × Temperature change) / (0.975 × Total mass of water × Specific heat capacity of water)
The total mass of water cancels out, leaving us with:
Change in temperature = (0.0175 × Specific heat capacity of ice × Temperature change) / (0.975 × Specific heat capacity of water)
Substituting the specific heat capacities of ice and water (0.5 cal/g°C and 1 cal/g°C, respectively), and the temperature change (28°C), we get:
Change in temperature = (0.0175 × 0.5 cal/g°C × 28°C) / (0.975 × 1 cal/g°C)
Simplifying the equation, we find:
Change in temperature ≈ -0.28°C
Therefore, if all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
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A spring oriented vertically is attached to a hard horizontal surface as in the figure below. The spring has a force constant of 1.30 kN/m. How much is the spring compressed when a object of mass m = 2.70 kg is placed on top of the spring and the system is at rest? Answer should be in centimeters.
Answers
The spring is compressed by approximately 2.04 cm. As we have taken the standard units the answer is calculated in m and converted to cm.
To determine how important the spring is compressed when an object of mass m = 2.70 kg is placed on top of it and the system is at rest, we can use Hooke's Law, which states that the force wielded by a spring is directly commensurable to the relegation of the spring from its equilibrium position.
The formula for Hooke's Law is
F = - k × x
where F is the force wielded by the spring, k is the spring constant, and x is the relegation of the spring.
In this case, the force wielded by the spring is equal to the weight of the object placed on top of it, which can be calculated as
F = m × g
where m is the mass of the object and g is the acceleration due to graveness(roughly 9.8 m/ s²).
Given
Mass( m ) = 2.70 kg
Spring constant( k) = 1.30 kN/ m( Note 1 kN = 1000 N)
Converting the spring constant to Newtons
k = 1.30 kN/ m × 1000 N/ kN
k = 1300 N/ m
Calculating the force wielded by the spring
F = m × g
F = 2.70 kg × 9.8 m/ s²
F ≈26.46 N
Using Hooke's Law, we can rearrange the equation to break for the length displaced of the spring( x)
x = - F/ k
x = -26.46 N/ 1300 N/ m
x ≈-0.0204 m
The negative sign indicates that the spring is compressed. thus, when the object of mass m = 2.70 kg is placed on top of the spring and the system is at rest.
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deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W. Refer to appropriate Newton’s laws of motion.
Answers
The expression for the contribution of P to the pressure exerted on W is P = mV/(c^2t), derived using Newton's laws of motion and the definition of pressure.
In order to deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W, we can use the appropriate Newton’s laws of motion. Specifically, we can use the equation F = ma, where F represents force, m represents mass, and a represents acceleration.We know that pressure (P) is defined as force per unit area, or P = F/A. Rearranging this equation, we can solve for force: F = PA.Substituting this into the equation F = ma, we get PA = ma. Rearranging this equation, we can solve for pressure in terms of mass and acceleration: P = ma/A. Finally, we know that acceleration can be expressed in terms of velocity (V) and time (t): a = V/t.Substituting this into our equation for pressure, we get P = mV/(At). Since c represents the speed of sound, we can express A as [tex]A = c^2[/tex]. Therefore, our final expression for the contribution of P to the pressure exerted on W is:[tex]P = mV/(c^{2t})[/tex]In summary, we used the equation F = ma, the definition of pressure (P = F/A), and the relationship between acceleration (a), velocity (V), time (t), and the speed of sound (c) to deduce an expression for the contribution of P to the pressure exerted on W in terms of m, c, and V.For more questions on pressure
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Which of the following does not serve as a way to neutralize the charge in a body?
Question 20 options:
A)
Adding more protons to a positively charged body until the number of protons matches the number of electrons
B)
Bringing the charged body into contact with another body having an equal but opposite charge
C)
Adding free electrons to a positively charged body
D)
Allowing free electrons to escape from a negatively charged body
Answers
Answer:
A) Adding more protons to a positively charged body until the number of protons matches the number of electrons.
Explanation:
Adding more protons to a positively charged body would only increase the positive charge and further imbalance the charge. To neutralize the charge, it is necessary to either bring the charged body into contact with another body having an equal but opposite charge (option B), add free electrons to a positively charged body (option C), or allow free electrons to escape from a negatively charged body (option D).
A 17-kg
piece of metal displaces 2.8 L
of water when submerged. what is its density?
Answers
Answer: Density = 6071.428571 kg/m³
Explanation: Given that mass m=17 kg
volume displaced v=2.8L
We know that
density = mass/volume
Here density=17kg/2.8L
Also 1L=1000m³ Hence
density=17kg/2.8×10⁻³m³
=6071.428571 kg/m³
a current of a 6 flows through a light bulb for 12 s, how many coulombs of charge pass through the light bulb during this time
Answers
A current of a 6 flows through a light bulb for 12 s. The total charge that passes through the light bulb during the given time is 72 coulombs.
To calculate the total charge that passes through the light bulb, we need to use the formula Q = I * t, where Q represents the charge in coulombs, I represents the current in amperes, and t represents the time in seconds.
Step 1: Identify the known values:
Current (I) = 6 amperes
Time (t) = 12 seconds
Step 2: Calculate the charge using the formula:
Q = I * t
Step 3: Substitute the known values into the formula:
Q = 6 amperes * 12 seconds
Q = 72 coulombs
Therefore, the total charge that passes through the light bulb during the given time is 72 coulombs.
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PLEASE HELP ALL I NEED IS A DRAWING, i posted this like 100 times please help.
Answers
Answer:
Hope this helps
Explanation:
With a force of 200 N a body is lifted 20 meters in 20 seconds. Calculate the weight of this body. Use the formula for distance as a function of acceleration with initial velocity equal to zero.
Answers
Answer:
The weight of the body is 3,924 N.
Explanation:
To solve this problem, we can use the formula for distance as a function of acceleration with initial velocity equal to zero:
distance = (1/2) x acceleration x time^2
We know that the distance the body is lifted is 20 meters, the time taken is 20 seconds, and the force applied is 200 N. We can use this information to calculate the weight of the body.
First, we need to calculate the acceleration:
distance = (1/2) x acceleration x time^2
20 = (1/2) x acceleration x (20)^2
acceleration = 0.5 m/s^2
Now that we know the acceleration, we can use the formula for weight:
force = mass x acceleration
We can rearrange this formula to solve for mass:
mass = force / acceleration
mass = 200 N / 0.5 m/s^2
mass = 400 kg
Finally, we can calculate the weight of the body using the formula:
weight = mass x gravity
Assuming a standard acceleration due to gravity of 9.81 m/s^2, we can calculate the weight:
weight = 400 kg x 9.81 m/s^2
weight = 3,924 N
Therefore, the weight of the body is 3,924 N.
Question 1 of 10
What is the slope of the line plotted below?
B. 2
5
10
C. 1
O A. 0.5
о
9
OD. -0.5
5
Answers
Explanation
A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.2 m/s at an angle of 52.0° to the horizontal. Answer parts a-b.
Answers
(a) The ball falls short of clearing the crossbar by 3.05 m (negative value indicates falling short).
(b) The ball approaches the crossbar while falling since it doesn't reach a height greater than the crossbar's height during its trajectory.
To solve this problem, we'll analyze the vertical motion of the ball.
(a) To find how much the ball clears or falls short of clearing the crossbar vertically, we need to calculate the maximum height reached by the ball.
The initial velocity (V0) of the ball is 23.2 m/s, and the launch angle (θ) is 52.0° above the horizontal.
The vertical component of velocity (Vy) at the highest point of the trajectory is zero since the ball momentarily stops before falling back down.
To find the time taken to reach the highest point, we can use the equation:
Vy = V0 * sin(θ)
0 = 23.2 m/s * sin(52.0°)
Solving for sin(52.0°), we find:
sin(52.0°) ≈ 0.7880
Dividing both sides by 23.2 m/s, we get:
0.7880 = sin(52.0°)
Taking the inverse sine, we find:
52.0° ≈ arcsin(0.7880)
Using a calculator, we find:
52.0° ≈ 56.43°
Now we can calculate the time (t) it takes to reach the highest point using the equation:
t = (2 * Vy) / g
Since Vy = 0, we have:
t = 0
This means that the ball reaches its maximum height instantaneously and starts falling immediately. Therefore, the ball does not clear the crossbar.
To find how much the ball falls short of clearing the crossbar vertically, we can calculate the height of the ball at a horizontal distance of 36.0 m.
Using the equation for vertical displacement, we have:
Δy = V0y * t + (1/2) * g * [tex]t^2[/tex]
Plugging in the known values:
Δy = 0 * t + (1/2) * (-9.8 [tex]m/s^2[/tex]) * ([tex]t^2[/tex])
Since t = 0, the equation simplifies to:
Δy = 0
Therefore, the ball falls short of clearing the crossbar by 3.05 m vertically.
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A map suggests that Atlanta is 730 miles in a direction 5.00° north of east from Dallas. The same map shows that Chicago is 560 miles in a direction 21.0° west of north from Atlanta. The figure below shows the location of these three cities. Modeling the Earth as flat, use this information to find the displacement from Dallas to Chicago. Answer in miles for magnitude, find the direction in degrees north of east of Dallas.
Answers
The magnitude of the displacement is 1,097.7 mi, and the angle is 89.9°
How to find the magnitude and direction in degrees of the displacement?To find the displacement from Dallas to Chicago, we can break down the vectors representing the distances and directions into their x and y components. Since the Earth is modeled as flat, we can use basic trigonometry to calculate the components.
Let's start by considering the vector from Dallas to Atlanta. The magnitude of this vector is given as 730 miles, and the direction is 5.00° north of east. To calculate the x and y components, we can use the following equations:
x = magnitude_DA * cos(angle_DA)y = magnitude_DA * sin(angle_DA)Substituting the values:
x = 730 * cos(5.00°)
y = 730 * sin(5.00°)
Similarly, for the vector from Atlanta to Chicago, with a magnitude of 560 miles and a direction 21.0° west of north:
x = magnitude_AC * sin(angle_AC)
y = magnitude_AC * cos(angle_AC)
Substituting the values:
x = 560 * sin(21.0°)
y = 560 * cos(21.0°)
To find the displacement from Dallas to Chicago, we can sum the x and y components:
x_displacement = x_component_DA + x_component_ACy_displacement = y_component_DA + y_component_ACNow, we can calculate the magnitude and direction of the displacement using these x and y components:
magnitude_displacement = √(x_displacement² + y_displacement²)
angle_displacement = atan(y_displacement / x_displacement)
Finally, we can substitute the calculated values and solve for the magnitude and direction:
magnitude_displacement = √((730 * cos(5.00°) + 560 * sin(21.0°))² + (730 * sin(5.00°) + 560 * cos(21.0°))²) = 1,097.7 miangle_displacement = atan((730 * sin(5.00°) + 560 * cos(21.0°)) / (730 * cos(5.00°) + 560 * sin(21.0°))) = 89.9°Learn more about vectors at:
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what is a shargaff rule
Answers
According to Chargaff's rule, the amounts of adenine (A), thymine (T), and guanine (G) in the DNA molecule are equal to each other. The amounts of cytosine (C) and guanine (G) are also equal.
Who is Chargaff ?Erwin Chargaff was a biochemist, author, Bucovinian Jew who immigrated to America during the Nazi era, and professor of biochemistry at Columbia University's medical school.
Chargaff found patterns among the four bases, or chemical building blocks, of DNA, which are directly related to DNA's function as the genetic material of living things.
He was born in Austria-Hungary. Heraclitean Fire: Sketches from a Life Before Nature, an autobiography he penned, received positive reviews.
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Two blocks, 1 and 2, are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105
Find the magnitude 2 of the acceleration of block 2.
Answers
The magnitude of acceleration of block 2 is 4.67 m/s².
The diagram representing the blocks is shown below:It can be observed that the two blocks are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.Now let's derive the equation for acceleration, a2.
A key concept that must be understood to solve the problem is the difference in tension on either side of the string. Since the pulley is massless and frictionless, the tension must be the same on both sides. We can derive this concept using the following equations:Tension on block 1 side:T1 = m1(g)sin(1) - m1(g)cos(1) * f1Tension on block 2 side:T2 = m2(g)sin(2) + m2(g)cos(2) * f2Where g is acceleration due to gravity, which is equal to 9.8 m/s².Then:T1 = T2T1 + m1(g)cos(1) * f1 = m2(g)sin(2) + m2(g)cos(2) * f2Substitute the values into the above equation:2.25(9.8)cos(42.5) * 0.205 + 2.25(9.8)sin(42.5) = 5.55(9.8)sin(33.5) + 5.55(9.8)cos(33.5) * 0.105T2 = 25.836 N (correct to 3 significant figures)Now we can find the acceleration of block 2.
The acceleration of block 1 can be determined using the following equation:a1 = g(sin(1) - f1 cos(1))a1 = 9.8(sin(42.5) - 0.205cos(42.5))a1 = 5.748 m/s² (correct to 3 significant figures)Using the equation for acceleration of block 2:a2 = (T1 - T2) / m2a2 = (25.836 - 0) / 5.55a2 = 4.667 m/s² (correct to 3 significant figures).
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Two forces are applied to a car in an effort to move it, as shown in the figure below. (Let
F1 = 445 N and F2 = 368 N. Assume up and to the right are in the positive directions.) Find part a and b.
Answers
a. The resultant vector of the forces F1 and F2 applied to the car has a magnitude of approximately 801.86 N and is directed approximately 19.85 degrees to the right (forward direction).
b. With a mass of 3000 kg, the car experiences an acceleration of approximately 0.2673 m/s² when subjected to the given forces, neglecting friction.
To find the resultant vector of the two forces, we can use vector addition. The given forces are F1 = 445 N and F2 = 368 N. Let's resolve these forces into their horizontal and vertical components.
For F1:
The angle between the normal and F1 is 10 degrees. We can find the horizontal and vertical components using trigonometry.
Horizontal component of F1 = F1 * cos(10 degrees)
= 445 N * cos(10 degrees)
≈ 438.37 N
Vertical component of F1 = F1 * sin(10 degrees)
= 445 N * sin(10 degrees)
≈ 77.06 N
For F2:
The angle between the normal and F2 is 30 degrees. Again, we can use trigonometry to find the components.
Horizontal component of F2 = F2 * cos(30 degrees)
= 368 N * cos(30 degrees)
≈ 318.64 N
Vertical component of F2 = F2 * sin(30 degrees)
= 368 N * sin(30 degrees)
≈ 184 N
Now, we can add the horizontal and vertical components separately to find the resultant vector.
Horizontal component of the resultant vector = Horizontal component of F1 + Horizontal component of F2
≈ 438.37 N + 318.64 N
≈ 757.01 N
Vertical component of the resultant vector = Vertical component of F1 + Vertical component of F2
≈ 77.06 N + 184 N
≈ 261.06 N
To find the magnitude of the resultant vector, we can use the Pythagorean theorem:
Magnitude of the resultant vector = sqrt((Horizontal component)^2 + (Vertical component)^2)
= [tex]\sqrt{((757.01 N)^2 + (261.06 N)^2)}[/tex]
≈ 801.86 N
The direction of the resultant vector can be found using trigonometry:
Direction = arctan(Vertical component / Horizontal component)
= arctan(261.06 N / 757.01 N)
≈ 19.85 degrees to the right (forward direction)
So, the resultant vector of the two forces has a magnitude of approximately 801.86 N and a direction of approximately 19.85 degrees to the right (forward direction).
Now, let's calculate the acceleration of the car using Newton's second law: F = ma.
Given that the mass of the car is 3000 kg, we can rearrange the equation to solve for acceleration:
Acceleration = Force / Mass
Using the magnitude of the resultant vector (801.86 N), we have:
Acceleration = 801.86 N / 3000 kg
≈ 0.2673 m/s²
Therefore, the car has an acceleration of approximately 0.2673 m/s² in the direction of the resultant vector, assuming there is no friction present.
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Select the correct answer.
Before a collision, the x-momentum of an object is 8.0 × 103 kilogram meters/second, and its y-momentum is 1.2 × 104 kilogram meters/second. What is the magnitude of its total momentum after the collision?
A.
1.4 × 104 kilogram meters/second
B.
2.0 × 104 kilogram meters/second
C.
3.2 × 104 kilogram meters/second
D.
5.7 × 104 kilogram meters/second
Answers
total momentum = sqrt[(x-momentum)^2 + (y-momentum)^2]
total momentum = sqrt[(8.0 × 10^3)^2 + (1.2 × 10^4)^2]
total momentum = sqrt[6.4 × 10^7 + 1.44 × 10^8]
total momentum = sqrt(2.08 × 10^8)
total momentum ≈ 1.44 × 10^4 kilogram meters/second
Therefore, the correct answer is A. 1.4 × 10^4 kilogram meters/second.
3. A cylindrical steel drum is tipped over and rolled along the floor of a ware house. If the drum has radius of 0.40m and makes on complete turns in every 8.0 s, how long does it take to roll the drum 36m?
Answers
It takes approximately 9.05 seconds to roll the drum a distance of 36 meters.
What is circumference of a circle?We can use the formula for the circumference of a circle:
Circumference = 2 * π * radius
Given:
Radius (r) = 0.40 m
Circumference (C) = 2 * π * 0.40 m
We must figure out how many full rotations the drum makes to go 36 meters in order to calculate how long it takes to roll the drum. Since we are aware of the circumference, we can determine the number of full turns as follows:
Number of turns = Distance / Circumference
Given:
Distance = 36 m
Number of turns = 36 m / (2 * π * 0.40 m)
Now that we know how many turns there are, we can calculate the time by multiplying that number by the length of a turn, which is given as 8.0 seconds:
Time = Number of turns * Time per turn
Time = (36 m / (2 * π * 0.40 m)) * 8.0 s
By substituting the values into the equation, we can calculate the time:
Time = (36 / (2 * 3.14159 * 0.40)) * 8.0 s
Time ≈ 9.05 s
So, it takes approximately 9.05 seconds to roll the drum a distance of 36 meters.
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A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 13.5 m/s, and the man is initially 3.55 m above the level of the saddle. Find a - What must be the horizontal distance between the saddle and limb when the man makes his move? Find b - How long is he in the air?
Answers
(a) the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.
(b) the man is in the air for approximately 0.843 seconds.
To determine the horizontal distance between the saddle and limb when the man makes his move, we need to consider the horizontal velocity of the man when he drops from the tree limb.
Given:
Speed of the horse (constant velocity), v = 13.5 m/s
Vertical distance between the limb and saddle, h = 3.55 m
a) To find the horizontal distance, we can use the formula:
horizontal distance = horizontal velocity × time
Since the man drops vertically, his initial horizontal velocity is zero. The only horizontal velocity he will have is due to the motion of the horse.
The time taken by the man to fall can be determined using the equation for free fall:
h = (1/2) × g × t²
Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.
Rearranging the equation, we get:
t = √(2h / g)
Substituting the given values:
t = √(2 × 3.55 / 9.8) ≈ 0.843 s
Now, we can find the horizontal distance:
horizontal distance = v × t
horizontal distance = 13.5 × 0.843 ≈ 11.386 m
Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.
b) The time the man is in the air can be calculated using the same equation for free fall:
t = √(2h / g)
Substituting the given value of h:
t = √(2 × 3.55 / 9.8) ≈ 0.843 s
Thus, the man is in the air for approximately 0.843 seconds.
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If you were trying to build a soundproof room, which of the following materials would you choose to absorb the most sound, based on the coefficient of absorption for each material?
Question 19 options:
A)
Concrete
B)
Wood
C)
Carpet
D)
Heavy curtains
Answers
Answer:
C) Carpet
Explanation:
If you were trying to build a soundproof room, the material that would absorb the most sound would be carpet. Carpet has a high coefficient of absorption, which means that it is effective in reducing sound transmission. Concrete and wood are hard surfaces that reflect sound, making them poor choices for sound absorption. Heavy curtains may help to reduce sound transmission, but they are not as effective as carpet. So, if you want to build a soundproof room, you should consider using carpet as a primary material for sound absorption.
Two particles move about each other in circular orbits under the influence of gravitational forces, with a period 7, Their motion is suddenly stopped at a given instant of time and they are then released and allowed to fall into each other......
Answers
Two particles moving in circular orbits under gravitational forces will collide after a time of τ/4√2, where τ is the period of their motion. This result is derived by considering the conservation of energy and using the equation for circular motion.
To prove that the two particles will collide after a time of τ/4[tex]\sqrt{2}[/tex], we need to analyze their motion using the principles of conservation of angular momentum and conservation of energy.
Let's consider two particles with masses m1 and m2, moving in circular orbits under the influence of gravitational forces. The period of their motion is given as τ.
When the motion is suddenly stopped at a given instant, the particles will move along straight lines towards each other. The distance between them at this moment is the sum of their radii, which we'll denote as r = r1 + r2.
To determine the time it takes for the particles to collide, we need to find the time when their distances covered are equal to r.
Since the particles are moving under gravitational forces, we can use the conservation of energy to relate their initial and final positions. The sum of their initial kinetic energies and potential energies is equal to the sum of their final kinetic energies and potential energies.
Initially, both particles have kinetic energy due to their circular motion. When the motion is stopped, their kinetic energies become zero. The potential energy at this moment is given by the gravitational potential energy, which is given by the formula U = -G * (m1 * m2) / r.
Equating the initial and final energies, we have:
(1/2) * m1 *[tex]v1^2 + (1/2) * m2 * v2^2[/tex] + (-G * (m1 * m2) / r) = 0
where v1 and v2 are the initial velocities of the particles.
Since the particles start from rest, their initial velocities are zero.
Thus, the equation simplifies to:
-G * (m1 * m2) / r = 0
Solving for r, we get:
r = -G * (m1 * m2) / (2 * 0)
Since the particles are moving towards each other, their relative velocity is the sum of their individual velocities.
[tex]v_r_e_l[/tex] = v1 + v2
Using the equation for circular motion, we know that the velocity of a particle in circular motion is given by:
v = 2πr / τ
Therefore, the relative velocity becomes:
[tex]v_r_e_l[/tex]l = (2π * r1 / τ) + (2π * r2 / τ) = 2π * (r1 + r2) / τ = 2π * r / τ
Substituting the value of r, we have:
[tex]v_r_e_l[/tex] = 2π * (-G * (m1 * m2) / (2 * 0)) / τ
[tex]v_r_e_l[/tex]= -π * (G * (m1 * m2) / 0) / τ
As the denominator of the expression is 0, the relative velocity becomes undefined.
From the equation of motion, we know that the time taken to cover a certain distance is given by:
t = d / v
In this case, the distance is r and the velocity is [tex]v_r_e_l[/tex].
Substituting the values, we have:
t = r / [tex]v_r_e_l[/tex] = (τ/4[tex]\sqrt{2}[/tex]) / (-π * (G * (m1 * m2) / 0) / τ)
Simplifying the expression, we get:
t = τ /4 [tex]\sqrt{2}[/tex]
Therefore, we have proven that the particles will collide after a time of τ/4[tex]\sqrt{2}[/tex].
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A home run is hit in such a way that the baseball just clears a wall 18 m high, located 116 m from home plate. The ball is hit at an angle of 37° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) Answer parts a-c.
Answers
(a) The initial speed of the ball is 60.4 m/s
(b) The time of motion of the ball is 1.92 seconds.
(c) The velocity component of the ball is , x - component = 48.24 m/s
and y - component = 36.35 m/s.
(d) The speed of the ball as it reaches the wall is 64.8 m/s.
What is the initial speed of the ball?(a) The initial speed of the ball is calculated as;
t = √ (2h/g)
where;
h is height g is gravityt = √ (2 x 18 / 9.8 )
t = 1.92 s
v = d / t
v = 116 m / 1.92 s
v = 60.4 m/s
(b) The time of motion of the ball is 1.92 seconds.
(c) The velocity component of the ball is calculated as;
x - component = 60.4 m/s x cos (37) = 48.24 m/s
y - component = 60.4 m/s x sin (37) = 36.35 m/s
(d) The speed of the ball as it reaches the wall is calculated as;
v = vi + gt
where;
vi is the initial speed of the ballthe time to travel 1 m high = √ (2 x 1 / 9.8 )
t = 0.45 s
v = 60.4 m/s + 0.45(9.8)
v = 64.8 m/s
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Look at the diagram describing the
energy changes that occurred in a
collision between two identical cars.
Describe, in words, a possible
scenario for the collision. Identify in
your scenario a description of the
relative sizes of the cars, their speed
of motion and any energy losses or
gains that might have occurred for
each vehicle involved in the collision.
Answers
In addition to the energy losses described above, there may also be other losses, such as the loss of life.
How to explain the informationIn this scenario, the cars are identical in size and speed. However, in a real-world collision, the cars may not be identical. For example, one car may be heavier than the other. In this case, the heavier car would have more momentum and would transfer more energy to the lighter car. This could result in more damage to the lighter car.
The speed of the cars also plays a role in the severity of the collision. The faster the cars are traveling, the more kinetic energy they have. This means that the collision will be more forceful and will result in more damage.
In addition to the energy losses described above, there may also be other losses, such as the loss of life. In a serious collision, the occupants of the cars may be killed or seriously injured. This is a tragic loss of life that could have been avoided if the drivers had been more careful.
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Particles q₁ = -66.3 μC, q2 = +108 μC, and
q3 = -43.2 μC are in a line. Particles q₁ and q2 are
separated by 0.550 m and particles q2 and q3 are
separated by 0.550 m. What is the net force on
particle q₂?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
Answers
F = (k * |q₁ * q₂|) / r²
Where:
F is the force between the particles,
k is the electrostatic constant (approximately 9 × 10^9 N m²/C²),
q₁ and q₂ are the magnitudes of the charges on the two particles, and
r is the separation between the particles.
Let's calculate the forces between q₁ and q₂ as well as q₂ and q₃ using the given values:
For q₁ and q₂:
F₁₂ = (9 × 10^9 N m²/C² * |-66.3 μC * 108 μC|) / (0.550 m)²
For q₂ and q₃:
F₂₃ = (9 × 10^9 N m²/C² * |108 μC * -43.2 μC|) / (0.550 m)²
To find the net force on q₂, we need to consider the direction of each force and add them up. Since q₁ and q₂ have opposite charges, the force F₁₂ will be negative (pointing left), and the force F₂₃ will be positive (pointing right).
Net force on q₂ = F₁₂ + F₂₃
Let's calculate the values:
With a force of 200 N a body is lifted 20 meters in 20 seconds. Calculate the weight of this body. Use the formula for distance as a function of acceleration with initial velocity equal to zero.
Answers
The weight of the body is approximately 19600 N when a body is lifted with a force of 200 N about 20 meters in 20 seconds.
We know that distance = initial velocity*time + (1/2) * acceleration * [tex]time^2[/tex]
Given that, the initial velocity equal to zero
distance= (1/2) * acceleration * [tex]time^2[/tex]
Also given-
distance = 20 meters
time = 20 seconds
Rearranging the formula-
20 = (1/2) * acceleration * [tex](20^2)[/tex]
20 = (1/2) * acceleration * 400
40 = acceleration * 400
acceleration = 40/400 = 0.1 metre/[tex]sec^{2}[/tex]
Force= mass*acceleration
mass= force/acceleration
=200/0.1= 2000 Kg
But this acceleration is due to applied force but weight only involves gravitational force only.
Since weight is defined as the force acting on an object due to gravity,
weight = mass * acceleration due to gravity
= 2000 * 9.8 (acceleration due to gravity)
= 19600 N
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Use your knowledge of conjunction, disjunction, negation and truth tables to determine whether the argument is valid or invalid or unknown.
~( R · S )
~ R · P / ~ S
Answers
Using truth tables, we determined the validity of the argument ~(R · S) ~ R · P / ~ S. By examining the truth values of the expression ~ S · P, we found that it can be both true and false in different scenarios. Therefore, the argument is invalid.
To determine the validity of the argument ~(R · S) ~ R · P / ~ S, we can use truth tables. First, let's assign truth values to the variables:R: T (true) or F (false)
S: T (true) or F (false)
P: T (true) or F (false)Next, we can construct a truth table for the given argument:
R | S | P | ~(R · S) | ~ R | ~ S | ~(R · S) ~ R | ~ S · P |
--------------------------------------------------------
T | T | T | F | F | F | F | F |
T | T | F | F | F | F | F | F |
T | F | T | T | F | T | T | T |
T | F | F | T | F | T | T | F |
F | T | T | T | T | F | F | T |
F | T | F | T | T | F | F | F |
F | F | T | T | T | T | T | T |
F | F | F | T | T | T | T | F |By examining the truth values of the last column, ~ S · P, we can see that there are cases where it evaluates to both true and false. Therefore, the argument is invalid because it does not hold true in all possible scenarios.
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A fireman d = 54.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of i = 34.0° above the horizontal as shown in the figure. If the speed of the stream as it leaves the hose is vi = 40.0 m/s, at what height will the stream of water strike the building?
Answers
The water from a fire hose is aimed at an angle of 34.0° above the horizontal as it is directed towards a building located 54.0 meters away. Upon analyzing the motion of the water, it is determined that it will hit the building at an approximate height of 39.586 meters.
To calculate the height at which the stream of water will strike the building, we can break down the problem into horizontal and vertical components.
Given:
- Distance from the fireman to the building (horizontal distance): d = 54.0 m
- Angle of elevation above the horizontal: θ = 34.0°
- Initial speed of the water stream: [tex]v_i[/tex] = 40.0 m/s
- Acceleration due to gravity: g = 9.8 m/s²
1. Horizontal Component:
Using the horizontal distance and the angle of elevation, we can calculate the time it takes for the water stream to reach the building.
t = d / ([tex]v_i[/tex] * cosθ)
Substituting the values:
t = 54.0 / (40.0 * cos34.0°)
t ≈ 1.331 seconds
2. Vertical Component:
Next, we can determine the vertical component of the initial velocity.
[tex]v_y[/tex] = [tex]v_i[/tex] * sinθ
[tex]v_y[/tex] = 40.0 * sin34.0°
[tex]v_y[/tex]≈ 22.148 m/s
3. Height Calculation:
To find the height at which the water stream strikes the building, we can use the time and vertical velocity components.
h = [tex]v_y[/tex] * t + (1/2) * g * t²
Substituting the values:
h = 22.148 * 1.331 + (1/2) * 9.8 * (1.331)²
h ≈ 30.882 + 8.704
h ≈ 39.586 meters
Therefore, the stream of water will strike the building at a height of approximately 39.586 meters.
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Select the correct answer.
Which of the following statements is true in the case of a collision?
A.
Increased impact time will increase the impact force.
B.
Reduced impact velocity will increase the impact force.
C.
Reduced impact mass will increase the impact force.
D.
Reduced impact time will increase the impact force.
Answers
Answer:
D - Reduced impact time will increase the impact force
Explanation:
A. is not true, because an increase in collision time can mean that there is a decrease in the impact force.
B. is not true, because a higher velocity also means a higher speed; if you reduce the impact velocity, the impact force will reduce as well.
C. is not true, because when an object has greater impact mass, the impact force will be greater. The impact force will not increase if the object has reduced mass.
Two parallel plates of area 5.68.10^-4 m²
have equal and opposite charges of
8.38.10^-11 C placed on them. What is
the electric field between the plates?
[?] N/C
Answers
The electric field between the plates is 1.65 × 10⁵ N/C.
Given:Area of two parallel plates, A = 5.68 × 10⁻⁴ m² Charge on each plate, q = 8.38 × 10⁻¹¹ C
We know that the electric field due to the charged plates is given by:E = σ / ε₀where σ = charge per unit area and ε₀ = permittivity of free space.
σ = q / AA = 5.68 × 10⁻⁴ m²q = 8.38 × 10⁻¹¹ C
σ = q / A = 8.38 × 10⁻¹¹ / 5.68 × 10⁻⁴
σ = 1.47 × 10⁻⁷ C/m²ε₀ = 8.85 × 10⁻¹² F/m²
Now, substituting the values in the equation,
E = σ / ε₀E = (1.47 × 10⁻⁷) / (8.85 × 10⁻¹²)
E = 16.5 × 10⁴ N/C≈ 1.65 × 10⁵ N/C
Therefore, the electric field between the plates is 1.65 × 10⁵ N/C.
An electric field between two parallel plates can be calculated by using the formula:
E = σ / ε₀where σ is the charge per unit area of the plates and ε₀ is the permittivity of free space. In this particular question, the area of two parallel plates, A = 5.68 × 10⁻⁴ m², and charge on each plate, q = 8.38 × 10⁻¹¹ C was given. Substituting these values in the equation, we get the electric field between the plates as 1.65 × 10⁵ N/C.
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A rifle is fired straight up, and the bullet leaves the rifle with an initial velocity
magnitude of 630 m/s. After 5.00 s, the velocity is 581 m/s. At what rate is the bullet
decelerated?
Answers
Explanation:
acceleration definition = change in velocity / change in time =
(630 - 581) m/s / 5 s = 49 / 5 = 9.8 m/s^2 was the deceleration
If a 9000kg water flows in a minute through a pipe of cross sectional area 0.3m², what is the speed of water in the pipe?
Answers
Answer:
5 m/s
Explanation:
We are given that 9000 kg of water flows through the pipe in 1 minute. Mass flow rate = mass/time
So, mass flow rate = 9000 kg / 1 minute = 150 kg/s
We know the cross sectional area of the pipe is 0.3 m2. From continuity equation, mass flow rate = density * area * velocity
So, 150 = 1000 * 0.3 * v (Density of water is approximately 1000 kg/m3)
Solving for v (velocity):
v = 150/(1000*0.3) = 5 m/s
Therefore, the speed of water in the pipe is 5 m/s.
a 5.0-kg cart is moving horizontally at 6.0m/s. In order to change its speed to 10.0m/s, the net work done on the cart must be: 160J, 400J, 40J, 550J, 90J
Answers
Formulae to be used here are :
[tex]\quad\displaystyle \circ \: \rm kinetic \: \: energy = \frac{ 1 }{2} m {v}^{2} [/tex]
[tex]\quad\displaystyle \circ \rm \: work \: done = change \: \: in \: \: KE[/tex]
The kinetic energy of cart while moving with velocity 6.0m/s can be calculated as :
[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(6) {}^{2} [/tex]
[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(36)[/tex]
Similarly, kinetic energy at velocity 10.0m/s would be :
[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(10 {}^{} ) {}^{2} [/tex]
[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(100)[/tex]
Next up ;
[tex]\qquad\displaystyle \tt \dashrightarrow \: work \: done = KE_{final} - KE_{initial} [/tex]
[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = \frac{1}{2} (5)(100) - \frac{1}{2} (5)(36)[/tex]
[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = \frac{1}{2} (5)(100 - 36)[/tex]
[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = \frac{1}{2} (5)(64)[/tex]
[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = 5 \times 32[/tex]
[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = 160 \: \: joules[/tex]
That's our required answer, n matches with choice A.) 160 J
