You are pushing on a heavy door, trying to slide it open. If your friend stands behind you and helps you push, how have the forces changed?
Question options:
There will be less friction.
The applied force does not change.
The net force will decrease.
The net force applied will increase.
Answer: Option (d) is the correct answer.
Explanation:
Net force is defined as the sum of total number of forces acting on a substance.
For example, when you and your friend are pushing a door in order to slide it open then it means that the net force is sum of force applied by you and force applied by your friend.
Therefore, the force applied earlier was less but it increase when your friend also started applying force.
Thus, we can conclude that if your friend stands behind you and helps you push, then the net force applied will increase.
Calculate the efficiency of an engine in a power plant operating between 40 ° c and 320 °
c. remember that the efficiency is a decimal number less than 1.
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is 150 ft/s. determine the pressure at a stagnation
The pressure at the stagnation point is 149.854 lb/ft².
Explanation:According to Bernoulli's equation, the total pressure at a given point is the sum of the static pressure and the dynamic pressure. The dynamic pressure is given by 0.5 * ρ * v^2, where ρ is the density of the fluid and v is the velocity of the fluid. In this case, the static pressure is -2.0 psi (a vacuum) and the velocity is 150 ft/s. The density of air at sea level is approximately 1.14 kg/m³, which is equivalent to 0.075 lb/ft³. Converting the units, we have -2.0 psi = -18.896 lb/ft². Plugging in the values into Bernoulli's equation, we can calculate the dynamic pressure:
Dynamic pressure = 0.5 * (0.075 lb/ft³) * (150 ft/s)^2 = 168.75 lb/ft².
The total pressure at the stagnation point is equal to the sum of the static pressure and the dynamic pressure. Therefore, the pressure at the stagnation point is:
Total pressure = -18.896 lb/ft² + 168.75 lb/ft² = 149.854 lb/ft².
During an episode of turbulence in an airplane you feel 170 n heavier than usual. if your mass is 73 kg, what are the magnitude and direction of the airplane's acceleration?
Final answer:
The airplane's acceleration that caused the passenger to feel 170 N heavier is 2.33 m/s² upwards.
Explanation:
When a person experiences an increase in weight due to airplane turbulence, it indicates that there's an additional force acting on them because of the airplane's acceleration. To find the magnitude and direction of the airplane's acceleration, we can use Newton's second law of motion (F=ma), where F is the force, m is the mass, and a is the acceleration.
The student feels 170 N heavier, implying that the upward acceleration of the airplane is causing this additional force. Since the student's mass is 73 kg, we can calculate the acceleration using the formula a = F/m, where F is the additional force (170 N) and m is the mass of the student (73 kg). Hence, a = 170 N / 73 kg = 2.33 m/s². This is the magnitude of the acceleration.
The direction of the acceleration is upwards, as it increases the normal force which is felt as an increase in apparent weight.
PLZ HURRY
A tuba makes a low pitched sound and a flute makes a high pitched sound. Which statement accurately describes the wavelengths of the two sounds?
A) The wavelengths of the two sounds are the same.
B) The wavelength of the high pitched sound is not measureable.
C) The wavelength of the low pitched sound is longer than the wavelength of the high pitched sound.
D) The wavelength of the low pitched sound is shorter than the wavelength of the high pitched sound.
A car slams on its brakes, coming to a complete stop in 4.0 s. The car was traveling south at 60.0 mph. Calculate the acceleration.
Answer:
Acceleration, [tex]a=-6.705\ m/s^2[/tex]
Explanation:
It is given that,
Velocity of the car, u = 60 mph = 26.82 m/s
Finally it comes to stop, v = 0
Time taken, t = 4 s
We need to find the acceleration of the car. The change in velocity divided by time is called the acceleration of the object. Mathematically, it is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-26.82}{4}[/tex]
[tex]a=-6.705\ m/s^2[/tex]
Negative sign shows the car is decelerating. So, the acceleration of the car is [tex]6.705\ m/s^2[/tex]. Hence, this is the required solution.
A 2kg particle moving along the x-axis experiences the net force shown at right. The particle’s velocity is 3.0m/s at x = 0m.
A) At what position is the particle moving the fastest?
B) What is the particle’s velocity at x=8m? Show your work
The solid aluminum shaft has a diameter of 50mm and an allowable shear stress of 60 mpa. determine the largest torque t1 that can be applied to the shaft if it is also subjected to the other torsional loadings
Final answer:
The largest torque that can be applied to the solid aluminum shaft is 1.5 MPa.
Explanation:
To determine the largest torque that can be applied to the solid aluminum shaft, we need to consider the allowable shear stress and the diameter of the shaft. The formula to calculate torque is t = rF sin 0, where t is the torque, r is the radius, F is the applied force, and 0 is the angle between the force and the radius. In this case, since the force is perpendicular to the radius, the angle is 90 degrees, so sin 0 = 1.
The diameter of the shaft is given as 50mm, which means the radius is 25mm. We need to convert the radius to meters, so the radius is 25/1000 = 0.025m. The allowable shear stress is given as 60MPa.
Using the formula for torque, t = rF = (0.025m)(60MPa) = 1.5MPa.
Therefore, the largest torque that can be applied to the shaft is 1.5MPa.
Johannes Kepler used math to show that the planets move in perfect circles around the sun.
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