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Kruka [31]
4 years ago
11

What happens when someone is walking across the carpeted room then touches a metal door knob

Physics
1 answer:
igomit [66]4 years ago
6 0
Static electricity travels to the door knob because of the friction caused by the feet on the carpet. the friction traveled through the person, to their hand, and to the door knob because it is the best conductor.
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A typical reaction time to get your foot on the brake in your car is 0.2 second. If you are traveling at a speed of 60 mph (88 f
katrin [286]

Explanation:

Not enough information. It really depends on the technical details of the car ( the data provided is offering just the human factor of the reaction, not the time for getting the impulse through when using the breaks

5 0
3 years ago
Read 2 more answers
T or F: In situations of inhalation poisoning, you should call for help before you attempt a rescue.
siniylev [52]

Answer:

True

Explanation:

When attempting to rescue an individual who is suffering from inhalation poisoning you could possibly do more harm than good. It is advised to call professionals and experts within that field to prevent further damage.

5 0
3 years ago
You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
Dimas [21]

Answer:

Explanation:

mass of baseball, m = 0.145 kg

initial velocity, u = 12 m/s upward

(a) final velocity, v = u / 2 = 6 m/s

Let the height is h.

Use third equation of motion

v² = u² - 2gh

6 x 6 = 12 x 12 - 2 x 9.8 x h

36 - 144 = - 19.6 x h

h = 5.51 m

(b) initial kinetic energy, K = 0.5 x m x u² = 0.5 x 0.145 x 12 x 12 = 10.44 J

Final kinetic energy, K' = K/2

0.5 x m x v² = 10.44 /2

0.5 x 0.145 x v² = 5.22

v = 8.5 m/s

v² = u² - 2gh

8.5 x 8.5 = 12 x 12 - 2 x 9.8 x h

72.25 - 144 = - 19.6 x h

h = 3.66 m

7 0
4 years ago
What forces are being used when walking a dog and how ?
natta225 [31]
There are tons of forces that balance out on your body while you walk. Subsequent physics classes will tell you about each and how they are represented. Here are a few in order of how people usually learn them.

Gravity: The earth exerts a gravitational force on each particle in your body that has mass. Overall, this can be represented as a single force that pulls directly toward the center of the earth from the point called your center of mass.

Normal Force: The contact between your feet/shoes and the ground exerts a force normal (straight out from) the ground. If you are on flat ground, this force is directly opposite the force of gravity, and in most cases will be equal to it such that you have no vertical net force.

Friction: Friction between your shoes/feet and the ground, pointing parallel to the ground and in the direction of your walking motion creates the force necessary for you to move. The microscopic peaks and valleys of the ground and your feet/shoes create small normal forces that can sum into a direction of motion.

Air Buoyancy: Since you are in a fluid, the mass of the fluid you displace creates an upward force away from the center of the earth. Since the density of air is miniscule, this force is generally neglected except in the most precise of circumstances.

Drag and Air resistance: While you walk, as you move through a fluid, that fluid exerts friction on your body in the form of drag. It is usually small unless you’re moving very fast relative to the fluid.

Air pressure, blood pressure, body tensions: Your body has a balance of blood pressure, muscle tensions, which oppose outside air pressures which equalize out to form the shape your body is in.

Internal forces: Many forces act within you such as air pressure, other muscle tensions, and internal stresses which balance out. Usually in physics these are lumped under internal forces.
6 0
3 years ago
A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude
OverLord2011 [107]

\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

We would calculate the magnitude by applying pythagorean theorem:

\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}

\longrightarrow \sf{Magnitude = 33.12}

\longrightarrow \sf{The \:  vector \: is \: (- 14, 30)}

The angle between two vectors is given by the formula:

\sf{\longrightarrow \small  \cos \emptyset =  \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }

In two dimensional, the x axis of vector form is:

\small\sf{\longrightarrow (b1, b2) = (1, 0) }

\sf{\longrightarrow \small \cos \:  \emptyset =  \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }

\small\longrightarrow \sf{ \dfrac{14}{33.12} }

\small\longrightarrow \sf{\emptyset  \:  = arcCos (\dfrac{ - 14}{33.12} )}

\small\longrightarrow \sf{\emptyset= 115^\circ}

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

\small\bm{The \: angle  \: between  \: the  \: vector \: }

\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}

6 0
2 years ago
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