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Vinil7 [7]
3 years ago
6

To get a feeling for inertial forces discuss the familiar cases of accelerating in a car in a straight line while increasing or

decreasing speed and turning the wheel to change direction. What direction do you feel a force in these scenarios and how does the strength of that force change if you either hit gas/brake harder or turn sharper?
Physics
1 answer:
inn [45]3 years ago
7 0

Answer:

Explanation:

When we accelerate in a car on a straight path we tend to lean backward because our lower body part which is directly in contact with the seat of the car gets accelerated along with it but the upper the upper body experiences this force  later on due to its own inertia. This force is accordance with Newton's second law of motion and is proportional to the rate of change of momentum of the upper body part.

Conversely we lean forward while the speed decreases and the same phenomenon happens in the opposite direction.

While changing direction in car the upper body remains in its position due to inertia but the lower body being firmly in contact with the car gets along in the direction of the car, seems that it makes the upper body lean in the opposite direction of the turn.

On abrupt change in the state of motion the force experienced is also intense in accordance with the Newton's second law of motion.

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Trumpeter A holds a B-flat note on the trumpet for a long time. Person C is running towards the trumpeter at a constant velocity
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You didn't mention it, but the trumpeter herself has to be standing still.

<span>Person C, the one running towards the trumpeter, hears a pitch
that is higher than B-flat.  (A)

Person B, the one running away from the trumpeter, hears a pitch
that is lower than B-flat.

Person D, the one standing still the whole time, hears the B-flat.</span>
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3 years ago
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Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 14.0 m/s toward the east and the
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4 0
3 years ago
A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the
pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

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6 0
2 years ago
A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points
Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

(c) B-A = 15i -( -5j) = 15i + 5j

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

5 0
3 years ago
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6 0
3 years ago
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