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3 years ago

For a snowboard jumper in the air, what force or forces will be most important for modeling the motion?

Physics

1 answer:

Reptile [31]3 years ago

4 0

Answer: Gravitational force and drag force

Explanation:

For a snowboard jumper in the air, two forces would be acting. One in the downward direction- the gravitational pull and second in the opposite direction to the motion, the drag force due to air. If the snowboard jumper jumps in the air at a certain angle with the horizontal. The forces are written as the sum of vertical and horizontal components. Hence, for the modeling the motion, gravitational force and drag force are important,

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A weather station records the wind as blowing from the northeast at 12 km/h.

weqwewe [10]

Answer:

Explanation:

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3 years ago

The rate at which an object changes position is that object's ______.

Alexxandr [17]

The correct answer is: Velocity

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4 years ago

1. Define the following in the case of a concave mirror.

insens350 [35]

Answer:

Principal focus of a concave mirror. The principal focus of a concave mirror is a point on its principal axis to which all the light rays are parallel and close to the axis converge after reflection from the concave mirror.

Focal length of a concave mirror. The focal length of a concave mirror is the distance between its pole and the principal focus

The reflecting surface of a spherical mirror forms a part of a sphere. The centre of this sphere. This point is called the centre of curvature of the spherical mirror. Center of curvature can also be defined as the point in the centre of the sphere from which the mirror was sliced. It is represented by the letter C. Please note that the centre of curvature lies outside the mirror's reflecting surface. The centre of curvature of a concave mirror lies in front of it. However, it lies behind the mirror in case of a convex mirror.

If a concave mirror were thought of as being a slice of a sphere, then there would be a line passing through the center of the sphere and attaching to the mirror in the exact center of the mirror. This line is known as the principal axis.

5 0

4 years ago

. Egbert is in a rowboat four miles from the nearest point on Egbert a straight shoreline. He wishes to reach a house 12 miles f

swat32

Answer:

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

Explanation:

Given:

Distance from the shore, $s=4\ mi$

distance of house from the shore, $h=12\ mi$

speed of rowing, $v_r=3\ mi.hr^{-1}$

speed of walking, $v_w=4\ mi.hr^{-1}$

For the least amount of time to reach the house one must row at the nearest point on the shore and then walk from there.

Now the time taken to reach the shore:

$t_s=\frac{s}{v_s}$

$t_s=\frac{4}{3}\ mi.hr^{-1}$

Time taken in walking to house from the shore:

$t_h=\frac{h}{v_w}$

$t_h=\frac{12}{4}$

$t_h=3\ mi.hr^{-1}$

Therefore total time taken:

$t=t_h+t_s$

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

6 0

3 years ago

A pair of fuzzy dice is hanging by a string from your rearview mirror. while you are accelerating from a stoplight to 28 m/s in

Lerok [7]

We can use the formula of motion in physics (2nd law od newton) in this problem:
x direction: Fsin ∅ = ma
y direction: Fcos ∅ -mg = 0
∅ is equal to sin ∅ / cos ∅ or x/y
tan ∅ = ma / mg = a /g

Applying acceleration formula:
v = vo + at ; 28 = 0 + 6a ; a = 4.67 m/s^2
∅ = tan-1 (a/g) = tan-1 (4.67/9.81) = 25.4 degrees.

4 0

3 years ago