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Anastasy [175]
3 years ago
12

An 80.0 kg skier slides down a hill shaped as shown. Assume

Physics
1 answer:
umka21 [38]3 years ago
8 0

The height above the ground from where the skier start is 11.5 m.

<h3>Conservation of energy</h3>

The height above the ground from where the skier start is determined by applying the principle of conservation of energy as shown below;

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

h = \frac{v^2}{2g} \\\\h = \frac{15^2}{2 \times 9.8} \\\\h = 11.5 \ m

Thus, the height above the ground from where the skier start is 11.5 m.

Learn more about conservation of energy here: brainly.com/question/166559

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A geologist sees a fault along which blocks of rock in the footwall have moved higher relative to blocks of rock in the hanging
Luda [366]

Answer: Normal fault

Explanation:

The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.

Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.

6 0
3 years ago
A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of
Neko [114]

Answer: 185.5672566

Explanation:  The friction is not relevant

Normal reaction is the force perpendicular to the surface.

this force resists the downwards forces applied which are gravity and a component of the applied force.

7 0
3 years ago
Read 2 more answers
If you are given distance and a period of time what can you calculate
Digiron [165]

If you are given distance and a period of time, you can calculate the speed. The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of its speed and direction of motion (e.g. 60 km/h to the north).

6 0
4 years ago
Read 2 more answers
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
where \omega_i and \omega_f are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
\alpha =  \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
where \omega_i = 0 is the initial angular speed. So, the angle covered after a time t=3.07 s is
\theta=  \frac{1}{2}  \alpha t^2 =  \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad
6 0
3 years ago
A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete
ale4655 [162]

Answer:

The  last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

cos X=\frac{x^2-y^2-z^2}{-2yz}

or

cos X=\frac{y^2 + z^2-x^2}{2yz}

substituting the values in the equation we get,

cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}

or

cos X=0.8951

or

X = 26.47°

similarly,

cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}

or

cos Y=0.649

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0
3 years ago
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