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3 years ago

Are heat lamps are designed to reheat food when food falls under 135 degrees?

2 answers:

7 0

No. The heat lamps are not designed to reheat food when food falls under 135 degrees.

Explanation:

A heat lamp is an incandescent light bulb that is used for the principal purpose of creating heat. Heat lamps aren't counseled for holding doubtless venturous foods hot at a minimum of one hundred thirty-five degrees Fahrenheit.
In order to heat up a doubtless venturous food for decent holding the food ought to be cooked-over quickly to a minimum internal temperature of a hundred and sixty-five degrees.

Serjik [45]3 years ago

5 0

That statement is False.

most of them are designed to reheat food when it falls under 165 degrees

hope this helps

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Two or more velocities add by ____? Plz help

VladimirAG [237]

By vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
$R= \sqrt{(R_x)^2+(R_y)^2}$
where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by
$\tan \alpha = \frac{R_y}{R_x}$
where $\alpha$ is its direction with respect to the x-axis.

3 0

3 years ago

Read 2 more answers

Which is harder, air tight or water tight seals?

Vladimir [108]

Should be an air tight seal

4 0

3 years ago

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward a

tino4ka555 [31]

Answer:

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.

3 0

3 years ago

A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners

galina1969 [7]

Answer:

Explanation:

As the source is situated on x - axis , it must be situated in between the two listeners .

So the x coordinate of source is

(-7 + 3 )/2

= - 2 m

The equation of the wave- front will be that o a circle having centre at (-2,0)

and radius = distance between -2 and 3 , that is 5 m

equation of circle

=( x+2 )² + y² = 25

It cuts y axis when x = 0

Putting x = 0

4 + y² = 25

y² = 21

y = + √21 , or - √21

7 0

3 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$