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4 years ago

7

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second

oasis. what is the direction from the first oasis to the second oasis?

1 answer:

Vaselesa [24]4 years ago

8 0

The second oasis is slightly north-west of the first oasis.

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Please help!!!!!!!!!!

wlad13 [49]

Answer:

C. Acceleration due to gravity

Explanation:

which is 9.8 m/s2 on Earth. The formula for calculating weight is F = m × 9.8 m/s2.

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3 years ago

The force required to drag several objects across a floor was measured. Calculate the µK for these objects.

olganol [36]

Answer:

Explanation:

Case-A

Block weight $W=150\ lb$

Friction Force $f=30\ lb$

and friction force is

$f=\mu _kN$

where N=normal reaction (Usually equals to weight)

$\mu _k=\frac{30}{150}$

$\mu _K=\frac{1}{5}=0.2$

Case-B

Block weight $W=20\ N$

Friction Force $f=5\ N$

$\mu _k=\frac{f}{N}$

$\mu _k=\frac{5}{20}=\frac{1}{4}$

$\mu _k=0.25$

Case-C

Block weight $W=1500\ lb$

Friction Force $f=60\ lb$

$\mu _k=\frac{f}{N}$

$\mu _k=\frac{60}{1500}$

$\mu _k=0.04$

4 0

4 years ago

QUESTION 3 ( MARKS]

horrorfan [7]

Answer:

392 N

Explanation:

Draw a free body diagram of the rod. There are four forces acting on the rod:

At the wall, you have horizontal and vertical reaction forces, Rx and Ry.

At the other end of the rod (point X), you have the weight of the sign pointing down, mg.

Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.

Sum of the moments at the wall:

∑τ = Iα

(T sin θ) L − (mg) L = 0

T sin θ − mg = 0

T = mg / sin θ

Given m = 20 kg and θ = 30.0°:

T = (20 kg) (9.8 m/s²) / (sin 30.0°)

T = 392 N

7 0

4 years ago

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

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3 years ago

I need help with this science question, 50 points!

Gennadij [26K]

Answer:

an open circuit (A battery, bulb, wire,switch)

Explanation:

the battery supplies the power the wire carries the current from the battery to the bulb but the circuit is not completed if the switch is not connected (Switch )

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3 years ago