A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?
To solve this problem it is necessary to apply to the concepts related to energy conservation. For this purpose we will consider potential energy and kinetic energy as the energies linked to the body. The final kinetic energy is null since everything is converted into potential energy, therefore
Potential Energy can be defined as,
Kinetic Energy can be defined as,
Now for Conservation of Energy,
Therefore the highets position the car reaches above the bottom of the hill is 40.02m
A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes
T1 = 499.9N, T2 = 865.8N, T3 = 1000N
To find the tensions we need to find the vertical and horizontal components of T1 and T2
T1x = T1 cos60⁰, T1y = T1 sin60⁰
Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰
For the forces to be in equilibrium,
the sum of vertical forces must be zero and the sum of horizontal forces must also be zero
Sum of Fx = 0
That is, T1x - T2x=0
NB: T2x is being subtracted because T1x and T2x are in opposite directions
T1 cos60⁰ - T2 cos30⁰ = 0
0.866T1 - 0.5T2 = 0 ............ (1)
Sum of Fy = 0
T1y + T2y - 1000 = 0
T1 sin60⁰ + T2 sin30⁰ - 1000 = 0
NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.
0.5T1 - 0.866T2 - 1000 = 0 ........(2)
make T1 the subject
T1 = 0.5T2/0.866
Substitute T1 into (2)
0.5 (0.5T2/0.866) - 0.866T2 = 1000
(0.25/0.866)T2 - 0.866T2 = 1000
0.289T2 - 0.866T2 = 1000
1.155T2 = 1000
T2 = 865.8N
Then T1 = 0.5 x 865.8 / 0.866
T1 = 499.9N
T3 = 1000N
NB: The weight of the bag is the Tension above the rope, which is T3
A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in- dependently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 28 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular
t = 39.60 s
Let's take a careful look at this interesting exercise.
In the first case the two motors apply the force in the same direction
F = m a₀
a₀ = F / m
with this acceleration it takes t = 28s to travel a distance, starting from rest
x = v₀ t + ½ a t²
x = ½ a₀ t²
t² = 2x / a₀
28² = 2x /a₀ (1)
in a second case the two motors apply perpendicular forces
we can analyze this situation as two independent movements, one in each direction
in the direction of axis a, there is a motor so its force is F/2
the acceleration on this axis is
a = F/2m
a = a₀ / 2
so if we use the distance equation
x = v₀ t + ½ a t²
as part of rest v₀ = 0
x = ½ (a₀ / 2) t²
let's clear the time
t² = (2x / a₀) 2
we substitute the let of equation 1
t² = 28² 2
t = 28 √2
t = 39.60 s
A satellite travels around the Earth in a circular orbit. What is true about the forces acting in this situation? A. The resultant force is the same direction as the satellite’s acceleration. B. The gravitational force acting on the satellite is negligible. C. There is no resultant force on the satellite relative to the Earth. D. The satellite does not exert any force on the Earth.
A. The resultant force in the same direction as the satellite’s acceleration.
Launching a satellite in the space and then placing it in orbit around the Earth is a complicated process but at the very basic level it works on simple principles. Gravitational force pulls the satellite towards Earth whereas it acceleration pushes it in straight line.
The resultant force of gravity and acceleration makes the satellite remain in orbit around the Earth. It is condition of free fall where the gravity is making the satellite fall towards Earth but the acceleration doesn't allow it and keeps it in orbit.
In a circular orbit around the Earth, the resultant force acting on a satellite is in the same direction as its acceleration.
In a satellite orbiting the Earth in a circular orbit, there are several forces at play. The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in its orbit. The centripetal force acts towards the center of the circular orbit, while the satellite's acceleration is directed towards the center as well. Therefore, option A is correct: the resultant force is in the same direction as the satellite's acceleration.
The gravitational force acting on the satellite is not negligible; in fact, it is crucial in providing the necessary centripetal force. Therefore, option B is incorrect.
Option C is incorrect as well. There is a resultant force acting on the satellite relative to the Earth, which is responsible for keeping the satellite in its circular orbit.
Lastly, option D is also incorrect. According to Newton's third law of motion, the satellite exerts an equal and opposite force on the Earth, keeping the Earth and the satellite in orbit around their common center of mass.
Learn more about Satellites in Circular Orbit here:
Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.What is the frequency of the fundamental wave on the guitar string?
Fundamental frequency= 174.5 hz
mass per unit length==0.00427
Now calculating velocity v=
Distance between two nodes is 0.7 m.
Plugging these values into to calculate frequency
f = =174.5 hz
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.