Answer:
The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Explanation:
Given that,
Weight Fg = mg
Acceleration = a
Tension = T
Drag force = Fa
Vertical force = L
We need to find the correct relationships
Using balance equation
In horizontally,
The acceleration is a
...(I)
In vertically,
No acceleration
Put the value of mg
....(II)
Hence, The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Answer:
Earth would continue moving by uniform motion, with constant velocity, in a straight line
Explanation:
The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:
"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"
This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.
In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.
Answer:
Explanation:
Given that
re= 46 cm
Vp= 180 m/s
We know that
So
Now by putting the all given values in the questions
So the average electric field is .
Explanation:it is beause they are sharper and also have less surface area and therefore more pressure
Answer:
Explanation:
During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.
Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,
The work done is given by the friction force and the distance traveled,
Where ![\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)](https://tex.z-dn.net/?f=%20%5Cmu_k%20%5B%2F%20tex%5D%20is%20the%20coefficient%20of%20kinetic%20friction%3C%2Fp%3E%3Cp%3EN%20is%20the%20normal%20force%20previously%20found%20d%20is%20the%20distance%20traveled%2C%3C%2Fp%3E%3Cp%3EReplacing%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW_f%20%3D%20%280.80%29%28441%29%280.42%29)
The thermal energy released through the work done is,
