Ask question

Ask question

All categories

Anastaziya [24]

3 years ago

7

A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is

the end of the ramp from the ground? m

1 answer:

Valentin [98]3 years ago

8 0

Answer:

<em>The end of the ramp is 38.416 m high</em>

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

If the maximum horizontal distance is known, we can solve the above equation for h:

$\displaystyle h=\frac {d^2g}{2v^2}$

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

$\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}$

$\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}$

h= 38.416 m

The end of the ramp is 38.416 m high

You might be interested in

The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization

ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0

4 years ago

Read 2 more answers

A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc

Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at $v_c = 23m/s$ speed is

$s_c = \Delta t v_c = 23\Delta t$

The distance traveled by the motorcycle after Δt (seconds) at $m_m = 23 m/s$ speed and acceleration of a = 8 m/s2 is

$s_m = \Delta t v_m + a\Delta t^2/2$

$s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2$

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

$s_m = s_c + 58$

$23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58$

$4 \Delta t^2 = 58$

$\Delta t^2 = 14.5$

$\Delta t = \sqrt{14.5} = 3.807s$

(b)

$s_m = 23 \Delta t + 4 \Delta t^2$

$s_m = 23*3.807 + 58 = 145.581 m$

5 0

3 years ago

A biologist formulates a hypothesis, performs experiments to test his hypothesis, makes careful observations, and keeps accurate

SVEN [57.7K]

<span>The next step of the scientific method is to analyze the results and draw conclusions. After that step, if the results agree with the hypothesis, then the results should be communicated. If the results do not support the original hypothesis, then the biologist must go back to the beginning and reformulate their hypothesis based on the results of the experiment.</span>

4 0

3 years ago

Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25

Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: $d_{1x} = 5.2 cos 25^{\circ}=4.7 mi$

North: $d_{1y} = 5.2 sin 25^{\circ}=2.2 mi$

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: $d_{2x}=0$

North: $d_{2y} = 5.0 mi$

So the components of the total displacement are

East: $d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi$

North: $d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi$

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

$d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi$

5 0

4 years ago

An external, frictionless pin only prevents rotation. A. True B. False

stepan [7]

Answer:

False

Explanation:

The roller displacement at an external, friction less rotation pin is found to be zero as it has a support which is no-yielding but still provides rotation to the object where as at an external pin a shear force is present and the moment is absent which means that at an external pin its not the rotation that is absent but its the moment which comes out to be zero.

7 0

3 years ago