Since the car is at rest, the force experienced by car will be normal face(exerted by surface to which car is in contact) and weight of the car.
As the car is at rest, net force on the car should be zero.
Answer:
Hope it helps..
Explanation:
Let n be the number of the vernier scale division which coincides with the main scale division. Rotate the vernier caliper 90° and repeat the steps 4 and 5 for measuring the internal diameter in perpendicular direction. To measure the depth, find the total reading and zero correction.
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I think that by "Classical physics" is meant low speed things. By low speed, I think is meant speed far below very roughly half the speed of light, so that Relativistic, special or general, effects can be ignored. Or at least it is hoped that they can be ignored.
Fire extinguishers and rockets get propelled by forcing out large amounts of material (gases under very high pressure) through a nozzle, and the RECOIL from that propels something forward. So, if the action is the ejection of material, the reaction (recoil) is the ejector moving along the same line in the other direction. And that's an example of Newton's third law.
Given a propulsion system, the magnitude of the force recoiling on the ejector will change the momentum of the ejector, often written as the equation F=ma where F is the force, m is the mass being accelerated, and a being the acceleration.
Just as something will stay still until it is moved - inertia - so once set in uniform motion in a straight line, the thing will continue in that motion, theoretically for ever or until something alters its momentum. Newton's first law is to the effect of "every body continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force". Which, I think, is where the concept of inertia stems from.
I think that the above mostly tcuches on the 3 laws.Any more help needed, please ask.
Answer:
Maybe A is the correct answer
Answer:
R = 103.7 N, 31.6° above x-axis
Explanation:
First we find the x components of all the forces:
F1x = F1 Cos 60°
F1x = (100 N)(Cos 60°)
F1x = 50 N
F2x = F2 Cos 140°
F2x = (200 N)(Cos 140°)
F2x = -153.2 N
F3x = F3 Cos 320°
F3x = (250 N)(Cos 320°)
F3x = 191.5 N
So, the x component of resultant will be the sum of the x component of each force:
Rx = F1x + F2x + F3x
Rx = 50 N - 153.2 N + 191.5 N
Rx = 88.3 N
Now we find the y components of all the forces:
F1y = F1 Sin 60°
F1y = (100 N)(Sin 60°)
F1y = 86.6 N
F2y = F2 Sin 140°
F2y = (200 N)(Sin 140°)
F2y = 128.5 N
F3y = F3 Sin 320°
F3y = (250 N)(Sin 320°)
F3y = -160.7 N
So, the y component of resultant will be the sum of the y component of each force:
Ry = F1y + F2y + F3y
Ry = 86.6 N + 128.5 N - 160.7 N
Ry = 54.4 N
Hence, the magnitude of resultant force will be:
|R| = √(Rx² + Ry²)
|R| = √[(88.3 N)² + (54.4 N)²]
|R| = √10756.25 N²
|R| = 103.7 N
And the direction θ will be:
θ = tan⁻¹(Ry/Rx)
θ = tan⁻¹(54.4/88.3)
θ = 31.6° above x-axis
Hence, the resultant vector will be:
<u>R = 103.7 N, 31.6° above x-axis</u>