A position-time graph of a spring is shown here. What is the period of the spring's motion.

Why did the producers and director decide to have the girls run outside to the water during the climactic moments of Act Three?

jolli1 [7]

Is this a book and most likely because the were cute

5 0

3 years ago

A remote-controlled car's wheel accelerates at 22.6 rad/s2 . If the wheel begins with an angular speed of 10.2 rad/s, what is th

Lapatulllka [165]

Answer:

70.9pi rad/s

Explanation:

angle turned through = 16

angular acceleration = 22.6 rad/s2

angular speed = 10.2 rad/s

time = ?

angular acceleration = angular speed/time

time = 10.2/22.6

time = 0.45 seconds

angular speed = angle turned through/time

angular speed = (16 × 2pi)/0.45

= 70.9pi rad/s

3 0

3 years ago

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately 33.23 m/s in a direction from the North of

approximately 9.18°.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at 5.3 m/s and North at 32.8 m/s

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ 33.23 m/s

Learn more here:

brainly.com/question/24430414

6 0

3 years ago

Help cant figure out which ones right

Ahat [919]

D.) It is an "Element".

[ Element cannot be separated by any means ]

Hope this helps!

4 0

4 years ago

The impulse experienced by a body is equivalent to the body’s change in

nika2105 [10]

<h2>Answer: </h2>

Momentum

<h2>Explanation: </h2>

The momentum of a particle is defined as the product of the particle mass and the particle velocity as follows:

$\overrightarrow{p}=m\overrightarrow{v}$

On the other hand, the impulse of a constant force is defined as:

$\overrightarrow{J}=\varSigma\overrightarrow{F}(t_{2}-t_{1})=\varSigma\overrightarrow{F}\Delta t$

We also know that the net force acting on a particle equals the rate of change of the particle’s momentum, so:

$\varSigma\overrightarrow{F}=m\overrightarrow{a}=m\frac{d}{dt}(\overrightarrow{v})=\frac{d}{dt}(m\overrightarrow{v})=\frac{d\overrightarrow{p}}{dt}$

If the force is constant, then $\frac{d\overrightarrow{p}}{dt}$ equals the total change in momentum over a period of time:

$\varSigma\overrightarrow{F}=\frac{\overrightarrow{p_{2}}-\overrightarrow{p_{1}}}{t_{2}-t_{1}} \\ \\ \varSigma\overrightarrow{F}(t_{2}-t_{1})=\overrightarrow{p_{2}}-\overrightarrow{p_{1}} \\ \\ \boxed{\overrightarrow{J}=\Delta \overrightarrow{p}}$

3 0

3 years ago