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pentagon
2 days ago
14

A weather balloon travels upward for 6 km whi le the wind blows it 10 km north and 8 km east. Approximately what is its final di

splacement from its initial position?
Physics
1 answer:
Maru [1K]2 days ago
6 0

The final displacement is calculated as √(6² + 10² + 8²).

This results in √(36 + 100 + 64).

Ultimately, the final displacement is √200.

In conclusion, the final displacement is 14.14 km (up and to the northeast of the launch)

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a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,
inna [974]
Definamos h como la distancia que hay desde el borde del pozo hasta la superficie del agua (en metros).

Consideremos la gravedad g como 9.8 m/s² y despreciemos la resistencia del aire.

La velocidad inicial vertical del guijarro es nula.
Ya que el guijarro impacta el agua tras 1.5 segundos, entonces:
h = 0.5 * (9.8 m/s²) * (1.5 s)² = 11.025 m

Resultado: 11.025 m
7 0
16 days ago
Read 2 more answers
For a projectile, which of the following quantities are constant during the flight: x, y, vx, vy, v, ax, ay? Check all that appl
Yuliya22 [1153]

Response:

C. vx

F. ax

G. ay

Clarification:

The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.

Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.

The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.

5 0
7 days ago
A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a
Softa [904]

The overall force acting on the vehicle is zero

Explanation:

Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

  • The weight component acting perpendicular to the slope, mgcos \theta, directed into the slope
  • The normal force N, directed outward from the slope

Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

  • The weight component aligned with the slope, mgsin \theta, directed down the slope
  • The frictional force F_f, acting up the slope

The car moves at a constant speed in this direction, indicating that its acceleration is zero.

a=0

Thus, according to Newton's second law,

F=ma

implying the net force is zero:

F=0

Learn more about slopes and friction:

5 0
11 hours ago
A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What
kicyunya [1011]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = 60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s

s = Distance traveled = 123\ ft=\dfrac{123}{3.281}=37.4885705578\ m

\theta = Incline = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

Friction coefficient

\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561

mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

The calculated stopping distance is 83.1946504051 m

6 0
15 days ago
A string is stretched by two equal but opposite forces f newton each what is tension in string
Maru [1038]

The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.

Explanation:

If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.

When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.

Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.

8 0
4 days ago
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