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3 years ago

Cooking popcorn using a microwave is a form of what?

2 answers:

7 0

Cooking popcorn using a microwave is a form of "radiation".
Heat from a fire warming your hands in a form of, "convection"
Heat from the sun hitting a solar panel is a form of, "radiation"
The handle of a pot becoming too hot to grab as it cooks on the stove is a form of, "convection"

AVprozaik [17]3 years ago

3 0

It will be radiation since the microwave uses a lot of heat waves to heat up the popcorn

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two boxes sit on a frictionless surface and are in contact with one another. the first box has a mass of 7 kg and the second box

egoroff_w [7]

The acceleration of the boxes depends on the mass and weight.

we have a mass of 7 and 8 kilograms

if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2.

It will leave you with 100 N.

finally take the sq rt of 100 to get 10

7 0

4 years ago

2 * 1.5 * (.850/2)^2A small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The syste

grin007 [14]

Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

rotational velocity = 5060 rev/min

now calculating the rotational inertia of the system.

$I = m L^2$

where L is the length of road, we will take whole length of rod because mass is at the end of it.

$I = 1.5 \times 0.85^2$

I = 1.084 kg.m²

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

8 0

3 years ago

A baby carriage is rolling down a hill. Take into consideration, the carriage has 90J of kinetic energy and the mass of the carr

Sphinxa [80]

Answer:

YES. 6√10

Explanation:

apply E=1/2mv^2 and find the velocity of carriage which is 6√10 ms-1

as our velocity is higher than that of cart. we can catch the carriage

7 0

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,