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4 years ago

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Displacement vectors of 4 km north, 2km south, 5 km north, and 5km south combine to a total displacement of A 2 km south B 16 km

south C.16 km north D 2km north?

1 answer:

Phantasy [73]4 years ago

7 0

The north vectors add up as so the south vectors. Then subtract the two. For north its 4 + 5 = 9. South is 2 + 5 = 7. Then 9-7 = 2km North (D)

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A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

$0.374 L =x - 0.6 L$

x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

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3 years ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e. the length and radius have twice the

saul85 [17]

Answer:

The wire now has less (the half resistance) than before.

Explanation:

The resistance in a wire is calculated as:

$R=\alpha \frac{l}{s}$

Were:

R is resistance

$\alpha$ is the resistance coefficient

l is the length of the material

s is the area of the transversal wire, in the case of wire will be circular area ( $s=\pi r^{2}$ ).

So if the lenght and radius are doubled, the equation goes as follows:

$R=\alpha \frac{l}{\pi r^{2} } =\alpha \frac{2l}{\pi {(2r)}^{2} } =\alpha \frac{2l}{\pi 4 {r}^{2} }=\frac{1}{2} \alpha \frac{l}{\pi r^{2} }$

So finally because the circular area is a square function, the resulting equation is half of the one before.

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3 years ago

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3 years ago

If 35 J of work was performed in 70 seconds, how much power was used to do this task?

Irina18 [472]

$P = w \times delta \: T$
P = Power
W = Work
Delta T = Change of Time

So:
P = 35 Joules
Delta T = 70 secs
W = ?

$P = 35 \div 70 \\ P = 0.5 \: watts$

* Joule/Sec = Watt

Answer: The power used to do this task was 0.5 Watts.

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3 years ago

A baby stroller is a rest on top of a hill which is 10 m high. The stroller and baby have a mass of 20 kg. What is the potential

Anettt [7]

The potential energy is defined as the energy is contained in the body due to the height over the surface of the earth and it is calculated from the equation

$PE=mgh$

<em>where:</em>

PE: the potential energy in Joules.
m: the mass of the body in kg.
g: the acceleration due to the gravity in m/ $s^2$
h: the height of the body over the earth in meters.

<em>in our problem:</em>

$PE=20*9.8*10=1960 J$

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3 years ago