An airplane on a runway accelerates at 4.0 meters/second2 for 28.0 seconds before takeoff. How far does the plane travel on the runway before takeoff?
Answers
the plane travels a distance of 1568 meters on the runway before takeoff. To find the distance traveled by the airplane before takeoff, we can use the following kinematic equation:
Where:
- d is the distance traveled.
- is the initial velocity, which is 0 m/s since the airplane starts from rest.
- a is the acceleration, which is 4.0 m/s².
- t is the time, which is 28.0 seconds.
Plugging these values into the equation:
Calculating this:
So, the plane travels a distance of 1568 meters on the runway before takeoff.
To know more about distance
#SPJ12
x28
==
112
I believe that is the answer.
Related Questions
An observer on Earth measures the length of a spacecraft travelling at a speed of 0.700c to be 78.0 m long. Determine the proper length of the spacecraft.
A fan uses 7A when connected to a 110V line. If electric power costs 10 cents per kilowatt-hour in this location, what is the cost of running the fan for 13 hours?
What makes us human?
Select all of the following that describes momentum, p [mark all correct answers] a. momentum is the product of mass and acceleration b. momentum is the product of mass and speed c. momentum is the product of mass and velocity d. momentum is a scalar quantity e. momentum is a vector quantity f. momentum has unit of newton (N) g. momentum has unit of kgm/s h. momentum has unit of kgm2/s2 i. momentum has unit of joule (J) j. momentum is one-half the product of mass and the square of the object's speed
b.T1 is ..... M1g.
c. T3 is ..... m1g + M2g
d.T1 is ..... T2
e.The magnitude of the acceleration of M2 is ..... the magnitude of the acceleration on m1.
f. T1 + T2 is ..... T3
Answers
Answer:
a. center of mass acceleration supposed to be acceleration due to gravity, 9.81 m/s^2,
b. T1 = 9.81m1 N; c. T3 =9.81(M1+M2) N; d. T3-T1, e. (T3-T1)/M2; f. (M1+M2)T3/M3
Explanation:
Final answer:
In this frictionless, massless pulley system, the center of mass accelerates downward with an acceleration equal to the acceleration due to gravity. The tension in the string connected to mass M1 is equal to M1g, and the tension in the string connected to mass M2 is equal to m1g + M2g. The magnitudes of the accelerations of M1 and M2 are equal, and the sum of the tensions T1 and T2 is equal to the tension T3.
Explanation:
a. The center of mass accelerates: When considering the system as a whole, the acceleration of the center of mass is determined by the net external force acting on the system. In this case, the only external force is the force due to gravity. Therefore, the center of mass accelerates downward with an acceleration equal to g, the acceleration due to gravity.
b. T1 is equal to M1g: The tension in the string connected to mass M1 is equal to the weight of M1, which is given by the formula T1 = M1g.
c. T3 is equal to m1g + M2g: The tension in the string connected to mass M2 is equal to the sum of the weights of M1 and M2, which is given by the formula T3 = m1g + M2g.
d. T1 is equal to T2: Since the pulley is assumed to be frictionless and massless, the tension in the string connected to mass M1 is the same as the tension in the string connected to mass M2.
e. The magnitude of the acceleration of M2 is equal to the magnitude of the acceleration on M1: This is due to the constraint imposed by the tension in the string. Since the tension in the string connecting M1 and M2 is the same, their accelerations must also be the same.
f. T1 + T2 is equal to T3: The sum of the tensions T1 and T2 is equal to the tension T3, as the total force acting on mass M2 is equal to the sum of the individual tensions.
Learn more about Dynamics here:
#SPJ12
Answers
Inertia is determined by the mass of the object. greater the mass an object has, greater will be its inertia. smaller the mass an object has, smaller will be its inertia. Greater mass enables an object to resist change in its state of motion or rest while on the other hand smaller mass object can be easily made to change its state of motion or rest.
Answers
d = -2 m
a = - 9.81 m/s^2
t = ?
Formula
d = vi*t + 1/2 a t^2
- 2 = 6m/s *t - .5 * 9.81 * t^2
Put the - 2 on the right.
4.905 t^2 + 6 m/s t +2 = 0
a = -4.905 m/s^2
b = 6 m/s
c = 2 m
t = [- b +/- square root(b^2 - 4ac)]/(2a)
t = [ -6 +/- square root (6^2 - 4*(-4.905)*2)] / (- 4.905*2)
t = [-6 +/- 8.674) / - 9.81
t = -14.674/-9.81
t = 1.69 seconds.
cooled air
Answers
I also think it's hot air, but I'm pretty sure it's because hot air rises. I'm not 100% sure so I'm sorry if it wrong :)
b. a loop
c. weather
d. climate zones
Answers
Answer:
The answer is B. a loop hopefully this helps!
a. 10N
b. 100N
c. 200N
d. 98N
Answers
The smallest value of the force that will make the block not to slide down is 10 N.
We'll begin by calculating the normal reaction. This can be obtained as follow:
- Mass (m) = 8 Kg
- Acceleration due to gravity (g) = 10 m/s²
- Normal reaction (N) =?
N = mg
N = 8 × 10
N = 80 N
Finally, we shall determine the frictional force.
- Coefficient of friction (μ) = 0.4
- Normal reaction (N) = 80 N
- Frictional Force (F) =?
F = μN
F = 0.4 × 80
F = 32 N
Since the frictional force is 32 N, therefore, a force lesser than the frictional force will make the blocknot to slide down.
From the options given above, only option A has a force that is lesserthan the frictional force.
Therefore, the correct answer to the question is Option A. 10 N
Learn more about frictional force:
Final answer:
The smallest value of the force that will not slide the 8.0 kg block down the wall is 31.36 N.
Explanation:
To determine the smallest value of the force such that the 8.0 kg block will not slide down the wall, we need to consider the static friction between the block and the wall. The formula for static friction is fs = μs * N, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which is mg = 8.0 kg * 9.8 m/s^2 = 78.4 N. Therefore, the smallest value of the force is equal to the maximum static friction force, which can be calculated as fs = 0.4 * 78.4 N = 31.36 N. So the correct answer is 31.36 N.
Learn more about Static Friction here:
#SPJ3