Why are special techniques and extinguishing agents are required to fight combustible metals fires?

One cereal bar has a mass of 37 g. What is the mass of 6 cereal bars? Is that more than or less than 1 kg? Explain your answer.

GuDViN [60]

The mass is 222g. No, it is less than 1kg. There are 1000 grams in a kilogram so it would be 0.222kg. Hope this Helps :D

7 0

3 years ago

An ice skater glides for two meters across ice is work done or no work done

garik1379 [7]

Work is done. work=forcexdisplacement. the ice skater glides 2 meters (displacement), so yes.

5 0

3 years ago

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol

il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.
In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:
$v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

6 0

3 years ago

What should a sailboat operator do when approaching a PWC head-on

nekit [7.7K]

In accordance with the International Regulation for the prevention of collisions at sea:

1.- A sailing boat has a passing preference over a motorized boat, except when the motor boat is limited by its draft.

2.- The sailboat must maintain its course and speed.

3.- If it is evident that the PWC does not respond, the sailboat must sound the warning signal, and change its course to starboard.

4.- All actions must be taken as soon as possible.

5.- If a sailboat is using its engine, the situation changes, and in that case, both ships must alter to starboard.

8 0

3 years ago

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

Kipish [7]

Answer:

The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

$\beta = 10Log(\frac{I}{I_o} )\\\\$

where;

I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²

I is the sound intensity

Intensity of sound at rock concert ;

$120 = 10Log(\frac{I}{10^{-12}} )\\\\12 = Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12} * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2$

The intensity of sound of a whisper;

$20 = 10Log(\frac{I}{10^{-12}} )\\\\2 = Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2} * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\$

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

$\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}$

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

3 0

2 years ago