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2 years ago

Sa fait combien 2×5÷6

Physics

2 answers:

Triss [41]2 years ago

8 0

Answer:

1.66666666667

Explanation:

divide the 5 and 6 first

jekas [21]2 years ago

7 0

Answer:

1.66666666667

Explanation:

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What is a disadvantage of using biomass as an energy resource?

Rama09 [41]

Answer:

hope this helps i think the answer is C

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3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

A body of mass 120 g is taken vertically upwards to reach the height of 5m,claculate work done

Anuta_ua [19.1K]

FOLLOW ME FOR CLEARING YOUR NEXT DOUBT

3 0

3 years ago

a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op

Kryger [21]

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒ $v_{rolling}=v_{translational}+v_{rotational}$

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒ $\omega=\frac{v_{translational}}{r}$

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒ $v_{rotational}=\omega \times r$

On substituting the estimated values, we get

⇒ $=\frac{v_{translational}}{r} \times r$

⇒ $=v_{translational}$

So that the velocity will be:

⇒ $v_{rolling}=v+v$

⇒ $=2v$

4 0

3 years ago

A football player with a mass of 85 kg wears a uniform and helmet that have a mass of 4.5 kg. The football player moves at 2.1 m

DochEvi [55]

The answer would be 187.95 kg.m/s.

To get the momentum, all you have to do is multiply the mass of the moving object by the velocity.

p = mv

Where:
P = momentum
m = mass
v = velocity

Not the question is asking what is the total momentum of the football player and uniform. So we need to first get the combined mass of the football player and the uniform.

Mass of football player = 85.0 kg
Mass of the uniform = <u> 4.5 kg</u>
TOTAL MASS 89.5 kg

So now we have the mass. So let us get the momentum of the combined masses.

p = mv
= (89.5kg)(2.1m/s)
= 187.95 kg.m/s

5 0

3 years ago

Read 2 more answers

Sa fait combien 2×5÷6​

Sa fait combien 2×5÷6