A spring does 80 J of work launching a 1.85 kg rock into the air. Ignoring air resistance, how high will the rock go?

Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative

Nastasia [14]

Answer:

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Explanation:

It is given that,

$\vec{u}=2i-j-3k$

$\vec{v}=-3i+j-2k$

Taking the cross product of v and v such that,

$\vec{w}=u\times v$

$\vec{w}=(2i-j-3k)\times (-3i+j-2k)$

$\vec{w}=5i+13j-k$

$|w|=\sqrt{5^2+13^2(-1)^2}$

|w| = 13.92

Let $\hat{w}$ is the unit vector normal to the plane containing u and v. So,

$\hat{w}=\dfrac{\vec{w}}{|w|}$

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Hence, this is the required solution.

7 0

3 years ago

A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2

Tom [10]

the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

x = A cos (wt + Ф)

w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

w= $\sqrt{ \frac{200}{0.300} }$

w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

v= $\frac{dx}{dt}$

v= - A w sin (wt +Ф)

0 = -A w sin Ф

Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

4.00 = A cos 25.8t

Speed.

50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

Cos² 25.8t = $\frac{16}{A^2}$

sin² 25.8t = $\frac{3.756}{A^2 }$

1 = $\frac{1}{A^2} \ (16 + 3.756)$

A = $\sqrt{19.756}$

A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here: brainly.com/question/14311816

4 0

2 years ago

For Al, its atomic number is 13 and its mass number is 27. How many neutrons does it have?

ira [324]

Answer:

B.14 your welcome!

Explanation:

6 0

3 years ago

A baseball bat hits a baseball with a force of 100 newtons. What is the force and its direction exerted by the ball on the bat?

Brut [27]

Answer:

The force exerted by the ball on the bat has a magnitude of 100 N and its direction is exactly opposite to that of the force exerted by the bat on the ball.

Explanation:

Recall that Newton's third law tells us that : "For every action, there is an equal and opposite reaction."

Therefore if the bat acts on the ball with a force of 100 N, the ball acts on the bat with a similar magnitude of force (100 N) but direction opposite to the original force.

$F_{ab} = -\,F_{ba}$

7 0

3 years ago

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

$A = W + \dfrac{W}{2}$

$A =\dfrac{3W}{2}$

$A =\dfrac{3mg}{2}$

When the car is at the bottom, the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

$N = m g + \dfrac{mv^2}{r}$

$\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}$

$\dfrac{mg}{2} = \dfrac{mv^2}{r}$

$v = \sqrt{\dfrac{rg}{2}}$

$v = \sqrt{\dfrac{40\times 9.8}{2}}$

v = 14 m/s

8 0

3 years ago