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3 years ago

Tracy scuffs her socked feet across a carpet. When she touches a doorknob, she gets a small shock.

Physics

2 answers:

lubasha [3.4K]3 years ago

6 0

I believe the answer would be B.

Brut [27]3 years ago

6 0

Answer:

Correct answer is b(Through friction between her feet and the carpet)

Explanation:

Whenever we rub our body against carpet or any other object made of fiber materials that time frictional force act. Due to frictional force our body get some extra electron.

When Tracy scuff her feet across a carpet, That time due to friction between her feet and carpet she got some extra electrons. These Electrons can move easily through conductor. When Tracy touched the doorknob which is made up of some metal(conductor) and has positive charge that time extra electrons moved from Tracy to the knob. Due to quick movement of electrons Tracy gets a small shock.

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Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc

svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

$v=v_{i}+gt\\$

Also get the equation of displacement

$S=v_{i}t+(1/2)gt^{2}$

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

$v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\ And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m$

For t₂

$v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\ And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m$

For t₃

$v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\ And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m$

For t₄

$v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\ And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m$

For t₅

$v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\ And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m$

4 0

4 years ago

An ambulance is currently traveling at 15m/s, and is accelerating with a constant acceleration of 5 m/s^2. The ambulance is atte

Lorico [155]

initial velocity=15m/s=u
Acceleration=a=5m/s^2
Final velocity=v=30m/s
Distance be s

Using 3rd equation of kinematics

$\\ \rm\longmapsto v^2-u^2=2as$

$\\ \rm\longmapsto s=\dfrac{v^2-u^2}{2a}$

$\\ \rm\longmapsto s=\dfrac{30^2-15^2}{2(5)}$

$\\ \rm\longmapsto s=\dfrac{900-225}{10}$

$\\ \rm\longmapsto s=\dfrac{675}{10}$

$\\ \rm\longmapsto s=67.5m$

5 0

3 years ago

The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be

eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0

2 years ago

Determine if the data are qualitative, quantitative, or neither. Zinc is a silver-gray metal. Chlorine has a density of 3.2 g/L.

AysviL [449]

Qualitative data gives the information of quality which can not be measured in numbers. For example: Color of eyes, softness of skin.

Quantitative data is information of quantity that can be represented in numbers. For example length and mass of any object.

Zinc is a silver-gray metal is a qualitative data, here silver gray color is quality of zinc metal which can not be measured in numbers.

Chlorine has a density of 3.2 g/L is a quantitative data. The value of density can be compared with other elements by comparing the numbers.

Gallium is not found in nature is neither qualitative nor quantitative.

Nitrogen has a melting point of –210.00 °C is a quantitative data because this is expressed in numbers.

Aluminum is a solid is a qualitative data because it tells about the state of element which can not be measured in numbers.

5 0

3 years ago

Read 2 more answers

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

miv72 [106K]

Answer:

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision

C. True. The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False. When they say bounces it implies the same speed with the opposite direction

4 0

3 years ago