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2 years ago

How do you get 5 minutes to seconds

2 answers:

5 0

There are 60 seconds in a minute.

This means that 5 minutes would be 60 times 5.
60×5 = 300

There are 300 seconds in 5 minutes.

RUDIKE [14]2 years ago

5 0

There are 60 seconds in 1 minute, and we're trying to find how long 5 minutes are.

Therefore, multiply 60 x 5, which equals 300.

There are 300 seconds in 5 minutes.

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Explain why asleel meedle does not csink on water

Juli2301 [7.4K]

Answer:

upthrust

Explanation:

i think it does not sink in water because of the force pulling it upwards

5 0

2 years ago

The battleship and enemy ships 1 and 2 lie along a straight line. Neglect air friction. battleship 1 2 Consider the motion of th

DaniilM [7]

Answer:

$\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}$

Explanation:

The vertical component of the initial velocities are

$v_v = v_0sin\theta$

If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation

$v_vt - gt^2/2 = s = 0$

$t(v_v - gt/2) = 0$

$v_v - gt/2 = 0$

$t = 2v_v/g = 2v_0sin\theta/g$

So the ratio of the times of the flights is

$t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}$

8 0

2 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):