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2 years ago

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A stationary skateboarder I with a mass of 50 kg pushes a stationary skateboarder II with a mass of 75 kg. After the push the sk

ateboarder II moves with a velocity of 2 m/s to the right. What is the velocity of the skateboarder I?

1 answer:

Mice21 [21]2 years ago

8 0

The final velocity of the skateboarder I after the collision is 3 m/s to the left.

<h3>Velocity of skateboarder I</h3>

The velocity of skateboarder I is determined from the principle of conservation of linear momentum.

m1u1 + m2u2 = m1v1 + m2v2

50(0) + 75(0) = 50v1 + 75(2)

0 = 50v1 + 150

v1 = (-150)/50

v1 = - 3

v1 = 3 m/s to the left

Thus, the final velocity of the skateboarder I after the collision is 3 m/s to the left.

Learn more about linear momentum here: brainly.com/question/7538238

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Answer:

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Explanation:

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An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45° relative to edge of the sidewalk. If it has

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Explanation:

Given

Velocity of ant is 0.1 m/s in a direction $45^{\circ}$

if it has traveled 1 m perpendicular to the edge of the sidewalk

i.e. from diagram

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A tennis player hits a ball at an angle of 50 degrees above horizontal so that it has an acceleration of 12 m/s2. What is the ho

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So we want to know what is the magnitude of the horizontal component of acceleration ah if we know that the overall acceleration a=12 m/s^2 and the angle of overall acceleration and the horizontal acceleration is α=50°. We know that ah=a*cosα. So now it isn't hard to get the horizontal component: ah=12*cos50=12*0.64=7.71 m/s^2. So the correct answer is ah=7.71 m/s^2.

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A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s) should the nozzle point in order that the

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So, based on the angle values that have been found, the angle of elevation of the nozzle can be <u>16° or 74°</u>.

<h3>Introduction</h3>

Hi ! This question can be solved using the principle of parabolic motion. Remember ! When the object is moving parabolic, the object has two points, namely the highest point (where the resultant velocity is 0 m/s in a very short time) and the farthest point (has the resultant velocity equal to the initial velocity). At the farthest distance, the object will move with the following equation :

$\boxed{\sf{\bold{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}}}$

With the following condition :

$\sf{x_{max}}$ = the farthest distance of the parabolic movement (m)
$\sf{v_0}$ = initial speed (m/s)
$\sf{\theta}$ = elevation angle (°)
g = acceleration due to gravity (m/s²)

<h3>Problem Solving :</h3>

We know that :

$\sf{x_{max}}$ = the farthest distance of the parabolic movement = 2.5 m
$\sf{v_0}$ = initial speed = 6.8 m/s
g = acceleration due to gravity = 9.8 m/s²

<h3>What was asked :</h3>

$\sf{\theta}$ = elevation angle = ... °

Step by Step :

Find the equation value $\sf{\bold{theta}}$ (elevation angle)

$\sf{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}$

$\sf{x_{max} \cdot g = (v_0)^2 \cdot \sin(2 \theta)}$

$\sf{\frac{x_{max} \cdot g}{(v_0)^2} = \sin(2 \theta)}$

$\sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)}$

$\sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)}$

$\sf{\frac{24.5}{46.24} = \sin(2 \theta)}$

$\sf{\sin(2 \theta) \approx 0.53}$

$\sf{\cancel{\sin}(2 \theta) \approx \cancel{\sin}(32^o)}$

Find the angle value of the equation by using trigonometric equations. Provided that the parabolic motion has an angle of elevation 0° ≤ x ≤ 90°.

First Probability

$\sf{2 \theta = 32^o + k \cdot 360^o}$

$\sf{\theta = 16^o + k \cdot 180^o}$

→ $\sf{k = 0 \rightarrow 16^o + 0 = 16^o}$ (T)

→ $\sf{k = 1 \rightarrow 16^o + 180^o = 196^o}$ (F)

Second Probability

$\sf{2 \theta = (180^o - 32^o) + k \cdot 360^o}$

$\sf{2 \theta = 148^o + k \cdot 360^o}$

$\sf{\theta = 74^o + k \cdot 180^o}$

→ $\sf{k = 0 \rightarrow 74^o + 0 = 74^o}$ (T)

→ $\sf{k = 1 \rightarrow 74^o + 180^o = 254^o}$ (F)

$\boxed{\sf{\therefore \theta \{16^o , 74^o\} }}$

<h3>Conclusion</h3>

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trapecia [35]

The correct answer to the question is : A) Tossing it horizontally at a faster speed.

EXPLANATION:

Before going to answer this question, first we have to know the horizontal distance covered by any body when it is fired parallel to the ground.

Let us consider a body of mass m is fired with velocity u parallel to the ground. Let it was projected at a height h from the ground.

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Here, g is known as the acceleration due to gravity.

From above, we see that more is the initial velocity, the more is the horizontal distance covered by it i.e the more is the widest curved path.

Hence, the correct answer to the question is that tossing a ball horizontally at a faster speed will trace out the widest curved path.

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