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masya89 [10]
3 years ago
6

A box is 30 cm wide, 40 cm long and 25 cm high. Calculate the volume of the box in cubic centimeter.

Physics
1 answer:
diamong [38]3 years ago
6 0

Answer:

The correct answer would be A: 3,000.

Explanation:

To find the volume of a box, you have to do l * w * h. You multiply 30, 40, and 25 which gives you an answer of 3,000 cubic centimeters.

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Why are we mentley unstable to time travle
leva [86]

Answer:

Mental time travel has been studied by psychologists, cognitive neuroscientists, philosophers and in a variety of other academic disciplines. Major areas of interest include the nature of the relationship between memory and foresight, the evolution of the ability (including whether it is uniquely human or shared with other animals), its development in young children, its underlying brain mechanisms, as well as its potential links to consciousness, the self, and free will.

Explanation:

8 0
3 years ago
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a
Alex777 [14]

Answer: d. 5 m/s^2

Explanation:

Acceleration is the change in velocity in a given time.

a = (30-20)/2 = 5

7 0
3 years ago
Read 2 more answers
A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th
ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that  opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

     -Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

 a is the acceleration of the car

    a = \frac{frictional force}{mass} = \frac{10500}{1500} = -7m/s²

Now using;

   V² = U² + 2as

   V is the final velocity

   U is the initial velocity

   a is the acceleration

   s is the distance moved

  0² = 25² + 2 x 7 x s

  0 = 625 - 14s

  -625 = -14s

      s = 44.64m

   

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0
3 years ago
A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

7 0
3 years ago
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