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2 years ago

Three examples that prove newton's law of motion

1 answer:

7 0

If you have a skateboard and you skate into a tree on accident the same amount of force you put onto that tree when you was on the skateboard will come back at you when you bounce back

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The temperature of a body of water influences _____

Maslowich

The temperature of the air above it

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2 years ago

Read 2 more answers

Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,

andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

D = - 0.5i - 0.25j

Therefore,

Dx = -0.5

Dy = -0.25

8 0

3 years ago

How does energy travel in a mechanical wave?

3241004551 [841]

Answer:

In a medium

Explanation:

8 0

2 years ago

Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr

castortr0y [4]

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

$R = \frac{2.525}{1000}\times 50$

$R = 0.126 ohm$

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

$V = iR$

$V = 11 \times 0.126$

$V = 1.4 Volts$

so most appropriate answer in given options is

A. 1.8 Volts

8 0

2 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

3 0

9 months ago

Three examples that prove newton's law of motion​

Three examples that prove newton's law of motion