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wolverine [178]

3 years ago

13

Why do you experience a shock or even see a spark when you reach for a doorknob.

2 answers:

Verizon [17]3 years ago

4 0

Answer: static electricity

Explanation:

You see, you rubbed your body either against a coat or the carpet. It send negative electrons throughout your body and once you hit metal with your hand it creates an electric shock that’s called static electricity.

tresset_1 [31]3 years ago

3 0

When you touch a doorknob, all the charge wants to leave you and go to the doorknob. You see a spark and get a shock as the electrons leave you. Lightning is the result of static electricity.

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When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

dolphi86 [110]

Answer:

a) 1.725*10^5 N

b) 3.83*10^3 N

c) i) 173.24 kN

c) ii) 4.57 kN

Explanation:

See the attachment for calculations

6 0

3 years ago

A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin

anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

$\Delta L = L_0 \alpha \Delta T$

Where,

$L_0 =$ Initial Length

$\alpha =$ Thermal expansion coefficient

$\Delta T =$ Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

$0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C$

$103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C$

The coefficient of thermal expansion given is

$\alpha = 1.1*10^{-5}/\°C$

The initial length would be,

$L_0 = 565m$

Replacing we have to,

$\Delta L = L_0 \alpha \Delta T$

$\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))$

$\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))$

$\Delta L = 0.355m$

This means that the building will be 35.5cm taller

3 0

3 years ago

A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c

Anna35 [415]

The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

$h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2$

The direction of the acceleration of the ball is downwards

Learn more here: brainly.com/question/15407740

4 0

2 years ago

A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits th

Inessa05 [86]

initially coin is at rest and then it drop for total time t = 1.5 s

so here the speed of the coin at which it will hit the floor is to be find

$v_f = v_i + at$

here we know that

$v_i = 0$

a = 9.8 m/s^2

t = 1.5 s

now from above equation

$v_f = 0 + 1.5 (9.8)$

$v_f = 14.7 m/s$

so it will hit the floor with speed 14.7 m/s

3 0

3 years ago

The part of Earth's rocky outer layer that makes up the land masses is the

tangare [24]

D

Giddy UP!!!!!!!!!!!!!!!!!!!!!

4 0

3 years ago