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andreev551 [17]
3 years ago
13

In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward

slope launches them into the air, where they perform acrobatic maneuvers. The end of a launch ramp is directed 63° above the horizontal. With this launch angle, a skier attains a height of 10.9m above the end of the ramp. What is the skier’s launch speed?
Physics
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

u = 11.6 m/s

Explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

H=\dfrac{u^2\ sin^2\theta}{g}

10.9=\dfrac{u^2\ sin^2(63)}{9.8}

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.

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Gravity on the surface of the moon is only 1/6 as strong as gravity on the Earth. What is the weight of a 19 kg object on the Ea
Dahasolnce [82]

The weight of anything in any place is

         (mass of the thing) x (acceleration of gravity in that place).

-- On Earth, the acceleration of gravity is about  9.807 m/s²

Weight of 19 kg of mass is  (19 kg) x (9.807 m/s²) =  <em>186.3 newtons</em>


-- On the Moon, the acceleration of gravity is about 1.623 m/s²

Weight of the same 19 kg of mass is  (19 kg) x (1.623 m/s²) = <em>30.8  newtons</em>

7 0
3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t
vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be m kg.

Given:

Initial velocity of the car is, u=25\ m/s

As the car stops, final velocity of the car is, v=0\ m/s

Change in momentum is, \Delta p=39400

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg

Therefore, the mass of the car is 1576 kg.

4 0
3 years ago
K-2/5=11k literal equations
krek1111 [17]
-2/5 = 11k - k
-2/5 = 10k
-2/5/10 = k
-2/5 * 10 = k
-2/50 = k
k = -1/25.

-1/25 - 2/5 = 11k is true.
6 0
3 years ago
The speed of a wave is 2ms, and its wavelength 0.4 meters. What is the period of the wave?
AnnyKZ [126]

Answer:

5

Explanation:

6 0
2 years ago
Read 2 more answers
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