Answer:
1 km
Explanation:
displacement =velocity ×time
displacement =40m/s ×25s
displacement =1000m equivalent to 1km
Question 1
At one section of a long pipe the velocity of the fluid is 1.6 m/s. At another section of the pipe the diameter is three times greater.
What is the velocity of the fluid at this section?
O 0.533 m/s
○ 4.80 m/s
O Not enough information to tell
O 0.178 m/s
Question 2
Three thermometers are placed in a closed, insulated box and are allowed to reach thermal equilibrium. One is calibrated in
Fahrenheit degrees, one in Celsius degrees, and one in Kelvins. If the Celsius thermometer reads -40 °C the Fahrenheit
thermometer would read -40°F.
True
False
Answer:
Answer 1: The answer is O 0.178 m/s.
Answer 2: True: But in this specific case where the Celsius temperature is -40, the Fahrenheit temperature will also be -40.
So, in short, the answer is:
-40 Celsius is equal to -40 Fahrenheit
1.5 A man with a mass of 90,6 kg walks to the back of a train at a velocity of 1m/s while the train moves at a constant velocity of 36 km/h in a easterly direction 1.5.1 The weight of the man 1.5.2 The velocity of the train in m/s 1.5.3 The resultant velocity of the man 1.5.4 The distance the train has travelled in ten (10) minutes (1) (1) (2) (2)
The weight of the man is approximately 888.6 Newtons.
The velocity of the train is approximately 10 m/s.
The resultant velocity of the man is 11 m/s.
The train has traveled a distance of 6000 meters in ten (10) minutes.
To solve this problem1.5.1 The weight of the man can be calculated using the formula:
Weight = mass * acceleration due to gravity
Given:
Mass of the man (m) = 90.6 kg
Acceleration due to gravity (g) ≈ 9.8 m/s²
Weight of the man = 90.6 kg * 9.8 m/s² ≈ 888.6 N
Therefore, the weight of the man is approximately 888.6 Newtons.
1.5.2 The velocity of the train in m/s can be converted from its given velocity in km/h. Since 1 km/h is equal to 1000/3600 m/s, we can calculate:
Velocity of the train (v) = 36 km/h * (1000/3600) m/s ≈ 10 m/s
Therefore, the velocity of the train is approximately 10 m/s.
1.5.3 By taking into account the relative motion of the man and the train, it is possible to determine the man's final velocity. The resultant velocity will equal the vector sum of the man's walking velocity and the train's velocity because the man is traveling toward the rear of the train, which is moving eastward.
Given:
Velocity of the man relative to the train (v_man) = 1 m/s (backwards)
Resultant velocity of the man (v_resultant) = v_man + v_train
v_resultant = 1 m/s + 10 m/s = 11 m/s
Therefore, the resultant velocity of the man is 11 m/s.
1.5.4 The distance the train has traveled in ten (10) minutes can be calculated using the formula:
Distance = velocity * time
Given:
Time (t) = 10 minutes = 10 * 60 seconds = 600 seconds
Velocity of the train (v) = 10 m/s
Distance = 10 m/s * 600 seconds = 6000 meters
Therefore, the train has traveled a distance of 6000 meters in ten (10) minutes.
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A satellite weighing 5,400 kg is launched into orbit 3.6400 x 107 m above the center of the earth.
The mass of Earth is 6.0 × 1024 kg. The gravitational constant is 6.673 × 10–11 N•m2/kg2.
The gravitational force of Earth on the satellite is ___
Group of answer choices
9.1 x 10^4
1.6 x 10^3
2.1 x 10^6
Answer:
[tex]\tt F=1.63*10^3 N[/tex]
Explanation:
Gravitational force is defined as the force of attraction between two objects with mass. It is a fundamental force of nature, and it is what keeps us on the ground and what keeps the planets in orbit around the Sun.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
For the Question:
We can use the following formula to calculate the gravitational force between the Earth and the satellite:
[tex]\boxed{\tt F =\frac{ G * M * m }{ r^2}}[/tex]
Where:
F is the gravitational force
G is the gravitational constant[tex]\tt (6.673 * 10^{-11} Nm^2/kg^2)[/tex]
M is the mass of the Earth [tex]\tt (6.0 * 10^24 kg)[/tex]
m is the mass of the satellite[tex]\tt (5,400 kg)[/tex]
r is the distance between the satellite and the center of the Earth [tex]\tt (3.6400 * 10^7 m)[/tex]
Plugging in these values, we get the following:
[tex]\tt F = \frac{6.673 * 10^{-11} * 6.0 * 10^{24}* 5,400 }{ (3.6400 * 10^7 )^2}[/tex]
[tex]\tt F=1.63*10^3 N[/tex]
Therefore, answer is [tex]\tt F=1.63*10^3 N[/tex]
The temperature of a aluminum bar rises by 10.0°C when it absorbs 4.73 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of aluminum from these data. Answer is in kJ/kg · °C.
Answer:
Certainly! We can use the formula:
q = mcΔT
where q is the amount of heat absorbed, m is the mass of the aluminum bar, c is the specific heat capacity of aluminum, and ΔT is the change in temperature.
Substituting the given values, we get:
4.73 kJ = (0.525 kg) x c x (10.0°C)
Solving for c, we get:
c = 0.901 kJ/kg · °C
Therefore, the specific heat of aluminum is 0.901 kJ/kg · °C.
Explanation:
What is Moral subjectivism?
Answer:
What Is Moral Subjectivism? Moral subjectivism is based on an individual person's perspective of what is right or wrong. An individual can decide for themselves that they approve or disapprove of a certain behavior, and that is what determines if the behavior is right or wrong.
A 66 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m
. What is the effective spring constant of the spring system in the taptap?
Enter the spring constant numerically in newtons per meter using two significant figures
Explanation:
You want N/m
N = 66 * 9.81
m = 2.3 x 10^-2 m
66* 9.81 / 2.3 x 10^-2 = 28150 = 28 000 N/m to two S D
¿Cuál es el trabajo neto en J que se necesita para acelerar un auto de 1500 kg de 55 m/s a 65 m/s?
What is the net work in J required to accelerate a 1500 kg car from 55 m/s to 65 m/s?
The net work done (in J) required to accelerate a 1500 kg car from 55 m/s to 65 m/s is 3127500 J
How do i determine the net work done?First, we shall obtain the initial kinetic energy. Details below:
Mass (m) = 1500 Kginitial velocity (u) = 55 m/sInitial kinetic energy (KE₁) =?KE₁ = ½mu²
= ½ × 1500 × 55²
= 41250 J
Next, we shall final kinetic energy. Details below:
Mass (m) = 1500 KgFinal velocity (v) = 65 m/sFinal kinetic energy (KE₂) =?KE₂ = ½mv²
= ½ × 1500 × 65²
= 3168750 J
Finally, we shall determine the net work done. Details below:
Initial kinetic energy (KE₁) = 41250 JFinal kinetic energy (KE₂) = 3168750 JNet work done (W) =?W = KE₂ - KE₁
= 3168750 - 41250
= 3127500 J
Thus, the net work done is 3127500 J
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Acceleration
Quiz Active
1 2 3 4 5 6 7 8 9 10
Study the motion map shown. Some of the vectors have been circled.
X
What do the circled vectors represent?
distance
speed
velocity
acceleration
Mark this and return
Save and Exit
Next
TIME REMAINING
29:41
e:
Submit
The circled vectors represents acceleration.
The last option is correct.
How do we explain?
We see in the first motion diagram the length of velocity vector is increasing this shows that the velocity is increasing in the magnitude with time so this is an accelerated motion in which a uniform acceleration must be in the same direction of velocity must be there.
We also notice in the second motion diagram the length of velocity vector is decreasing with time which shows the velocity is decreasing me magnitude which shows a constant deceleration and the direction of acceleration must be opposite to that of velocity.
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. A 0.140 kg baseball is pitched toward home plate at 30.0 m/s.
The batter hits the ball back (opposite direction) to the pitcher at
44.0 m/s. Assume that towards home plate is positive. What is
the change in momentum for the ball?
The change in momentum for the baseball, which is hit back in the opposite direction by the batter, is -10.36 kg·m/s. This change in momentum is obtained by subtracting the initial momentum of 4.2 kg·m/s from the final momentum of -6.16 kg·m/s. The negative sign indicates the opposite direction of the momentum.
To find the change in momentum for the baseball, we can use the formula:
Change in momentum = Final momentum - Initial momentum
Momentum is defined as the product of mass and velocity.
Given data:
Mass of the baseball (m) = 0.140 kg
Initial velocity of the baseball ([tex]v_i_n_i_t_i_a_l)[/tex] = 30.0 m/s
Final velocity of the baseball ([tex]v_f_i_n_a_l_[/tex]) = -44.0 m/s (negative sign indicates opposite direction)
To calculate the initial momentum, we multiply the mass by the initial velocity:
Initial momentum = m * [tex]v_i_n_i_t_i_a_l[/tex] = 0.140 kg * 30.0 m/s = 4.2 kg·m/s
To calculate the final momentum, we multiply the mass by the final velocity:
Final momentum = m * [tex]v_f_i_n_a_l_[/tex] = 0.140 kg * (-44.0 m/s) = -6.16 kg·m/s
Now we can find the change in momentum:
Change in momentum = Final momentum - Initial momentum
Change in momentum = (-6.16 kg·m/s) - (4.2 kg·m/s)
Change in momentum = -10.36 kg·m/s
Therefore, the change in momentum for the baseball is -10.36 kg·m/s. The negative sign indicates a change in direction, as the ball is hit back in the opposite direction.
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A simple harmonic oscillator (SHO) has a spring constant of 280 N/m, Total energy of 150 J, and a mass of 4.00 kg. What is its
maximum velocity?
Numerical answer is in units of m/s
The maximum velocity is 8.66 m/s²
As we know, simple harmonic motion refers to a to-and-fro motion in a periodic manner and spring constant refers to the force required to stretch or compress a spring.
The spring constant for a simple harmonic oscillator is given as 280 N/m, the total energy is 150 J and the mass is 4 kgs. We have to find the maximum velocity of the given simple harmonic motion.
We know that Energy = force x perpendicular distance
In an SHM, energy is in the form of Kinetic Energy. Hence, we use the formula for kinetic energy.
To find the maximum velocity, we will apply the formula for kinetic energy.
Since Kinetic Energy = 1/2 mass x velocity²
Therefore, 150 = 1/2 mass x velocity² ; velocity = 8.66 m/s²
Hence, the maximum velocity for the given system of SHM is 8.66 m/s²
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Suppose a distant world with surface gravity of 6.32 m/s^2 has an atmospheric pressure of 9.00 ✕ 10^4 Pa at the surface. Answer parts a-c.
The Force ≈ [tex]1.13 * 10^6 N[/tex]
The Weight ≈ [tex]1.66 * 10^5 N[/tex]
Pressure ≈ 6.32 × 10⁴Pa
How to solve for the force(a)
Force = Pressure × Area
Force = (9.00 × 10⁴ Pa) × (π × (2.00 m)²)
Force ≈ [tex]1.13 * 10^6 N[/tex]
(b)
Weight = Density × Volume × g
Weight = (415 kg/m³) × (π × (2.00 m)² × 10.0 m) × (6.32 m/s²)
Weight ≈ 1.66 × 10⁵ N
(c)
Pressure = Pressure at the surface + Density × g × depth
Pressure = [tex](9.00 * 10^4 Pa) + (415 kg/m^3)* (6.32 m/^2)* (10.0 m)[/tex]
Pressure ≈ 6.32 × 10⁴Pa
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On March 21_occurs where in the length of the day and night are equal
On March 21, an equinox occurs where the length of the day and night are equal. An equinox happens when the Earth's axis is not tilted towards or away from the sun. During an equinox, the sun's rays are equally distributed across the Earth's surface.
The word equinox is derived from the Latin words "aequus" and "nox," which means "equal night."
The equinox occurs twice a year, around March 21 and September 21. During an equinox, the duration of day and night is equal across the entire world.
It means that every place on the planet experiences almost the same amount of daylight and darkness.
On March 21, the vernal equinox, marks the beginning of spring in the Northern Hemisphere, while in the Southern Hemisphere, it marks the beginning of autumn.
The equinox has been considered a significant day in many cultures throughout history. In many cultures, the equinox is celebrated as a time of renewal, rebirth, and fertility. It's also been considered a day of balance, where darkness and light are equal, and the world is in harmony.
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In a water pistol, a piston drives water through a larger tube of radius 1.30 cm into a smaller tube of radius 1.10 mm as in the figure below. Answer parts a-f.
It takes 0.47 seconds for water to travel from the nozzle to the ground when the water pistol is fired horizontally.
What is the time it takes for water to travel from the nozzle to the ground?We will denote the height of the water pistol above the ground as h and the initial velocity of water exiting the nozzle as v2. Assuming negligible air resistance, we will analyze the vertical motion of the water droplets.
The vertical displacement of the water droplets is calculated using equation: h = (1/2) * g * t^2.
Rearranging equation, we solve for time:
t = sqrt(2h / g).
Given data:
Height h = 1.10 m and the acceleration due to gravity g = 9.8 m/s^2, we get:t = sqrt(2 * 1.10 / 9.8)
t = 0.47380354147
t = 0.47 seconds.
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A steam turbine receives steam with a velocity of 28 m/s, specific enthalpy 3000 kJ/kg at a rate of 3500 kg per hour. The steam leaves the turbine with a specific enthalpy of 2200 kJ/kg at 180 m/s. Calculate the turbine output, neglecting losses.
A steam turbine receives steam with a velocity of 28 m/s, specific enthalpy 3000kJ/kg at a rate of 3500 kg per hour. The steam leaves the turbine with a specific enthalpy of 2200kJ/kg at 180 m/s then turbine output is 777.76kW.
To get the turbine output, we must first compute the change in specific enthalpy (h) and mass flow rate ().
Assume that the inlet steam velocity (v1) is 28 m/s.
Specific enthalpy at the inlet (h1) = 3000 kJ/kg
()=3500kg/h mass flow rate
2200 kJ/kg outlet specific enthalpy (h2)
v2 (outlet steam velocity) = 180 m/s
To begin, convert the mass flow rate from kg/h to kg/s as follows: =
[tex]3500 kg/h (1 h/3600 s) = 0.9722 kg/s[/tex]
The change in specific enthalpy (h) can then be calculated:
3000kJ/kg-2200kJ/kg=800kJ/kgh=h1-h2
The following formula can be used to compute the turbine output (P):
[tex]P = ṁ * Δh[/tex]
Substituting P=0.9722kg/s*800kJ/kg=777.76kJ/sork W
As a result, ignoring losses, the turbine output is roughly 777.76kW
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Joe is painting the floor of his basement using a paint roller. The roller has a mass of 2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor, Joe applies a force F = 16 N directed at an angle of 35° as shown. Ignoring the mass of the roller handle, what is the magnitude of the angular acceleration of the roller?
The magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².
To find the magnitude of the angular acceleration of the roller, we can use the rotational analog of Newton's second law: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
First, let's calculate the moment of inertia of the roller. The moment of inertia of a solid cylinder rotating about its central axis is given by the formula: I = (1/2)mr², where m is the mass and r is the radius.
Given:
Mass of the roller (m) = 2.4 kg
Radius of the roller (r) = 3.8 cm = 0.038 m
Moment of inertia (I) = (1/2) * 2.4 kg * (0.038 m)² = 0.0021744 kg·m²
Next, we need to calculate the torque (τ) applied to the roller. Torque is given by the formula: τ = rFsin(θ), where r is the distance from the axis of rotation to the point of application of the force, F is the magnitude of the force, and θ is the angle between the force and the line connecting the axis of rotation and the point of application.
Given:
Force applied (F) = 16 N
Angle (θ) = 35°
Distance from the axis of rotation to the point of application (r) is equal to the radius of the roller, so r = 0.038 m.
Torque (τ) = (0.038 m) * (16 N) * sin(35°) = 0.2366 N·m
Now, we can use the equation τ = Iα and solve for the angular acceleration (α):
0.2366 N·m = (0.0021744 kg·m²) * α
α = 0.2366 N·m / 0.0021744 kg·m² ≈ 108.8 rad/s²
Therefore, the magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².
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Joe is painting the floor of his basement using a paint roller. The roller has a mass of 2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor, Joe applies a force F = 16 N directed at an angle of 35° as shown. Ignoring the mass of the roller handle, what is the magnitude of the angular acceleration of the roller?
The magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
The magnitude of the angular acceleration of the roller can be determined using the torque equation and Newton's second law for rotational motion.
Step 1: Calculate the moment of inertia of the roller.
The moment of inertia (I) of a solid cylinder is given by the formula I = (1/2) * m * r^2, where m is the mass of the object and r is the radius.
In this case, the mass of the roller is 2.4 kg and the radius is 0.038 m.
So, I = (1/2) * 2.4 kg * (0.038 m)^2.
Step 2: Calculate the torque applied to the roller.
Torque (τ) is equal to the force (F) applied multiplied by the perpendicular distance (r) from the axis of rotation.
In this case, the force applied by Joe is 16 N and the distance is equal to the radius of the roller, 0.038 m.
So, τ = F * r.
Step 3: Use the torque equation.
The torque applied to the roller causes an angular acceleration (α) according to the equation τ = I * α.
Rearranging the equation, we get α = τ / I.
Step 4: Substitute the values into the equation.
Using the values we calculated earlier, we can substitute them into the equation α = τ / I.
α = (16 N * 0.038 m) / [(1/2) * 2.4 kg * (0.038 m)^2].
Step 5: Calculate the magnitude of the angular acceleration.
Evaluating the expression, we find that the magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
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Someone goes to lift a crate that is resting on the bottom of the pool filled with water (density of water is 1000 kg/m^3). While
still submerged, only 310 N is required to lift the crate. The crate is shaped like a cube with sides of 0.25 m. What is the density of
the cube? Numerical answer is assumed to be in units of kg/m^3
Answer:
the density of the cube is approximately 2016.07 kg/m^3.
Explanation:
The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Let's first calculate the weight of the crate:
mass of crate = density * volume = density * (side length)^3 = density * 0.25^3 = 0.015625 * density
weight of crate = mass of crate * gravity = 0.015625 * density * 9.81 = 0.1530875 * density
where gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.
Since the crate is submerged in water, the buoyant force acting on it is:
buoyant force = weight of water displaced = density of water * volume of water displaced * gravity
The volume of water displaced is equal to the volume of the cube, which is 0.25^3 = 0.015625 m^3. Therefore, the buoyant force is:
buoyant force = 1000 kg/m^3 * 0.015625 m^3 * 9.81 m/s^2 = 1.534453125 N
According to the problem, it takes 310 N to lift the crate while it is still submerged. This means that the net force acting on the crate is:
net force = lifting force - buoyant force = 310 N - 1.534453125 N = 308.465546875 N
This net force is equal to the weight of the crate:
net force = weight of crate = 0.1530875 * density
Therefore, we can solve for the density of the crate:
density = net force / 0.1530875 = 308.465546875 / 0.1530875 = 2016.06666667 kg/m^3
Rounding to the nearest hundredth, we get:
density ≈ 2016.07 kg/m^3
Therefore, the density of the cube is approximately 2016.07 kg/m^3