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ZanzabumX [31]
3 years ago
10

A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.

4 and the block moves at constant velocity. the magnitude of is:
Physics
1 answer:
gulaghasi [49]3 years ago
4 0
The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force F_f acting against the motion. Since their resultant must be zero, we have:
F-F_f = 0
The frictional force is
F_f = \mu mg
where
\mu=0.4 is the coefficient of kinetic friction
mg=400 N is the weight of the block. 

Substituting these values, we find the magnitude of the force F:
F=F_f = \mu mg=(0.4 )(400 N)=160 N
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Talja [164]
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Is it possible to have sex then after a week you get your periods and get pregnant?
mrs_skeptik [129]

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2 years ago
In his​ motorboat, Bill Ruhberg travels upstream at top speed to his favorite fishing​ spot, a distance of 120120 ​mi, in 33 hr.
photoshop1234 [79]

Answer:

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

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x​ = the rate of the boat in still water

y​ = the rate of the current.

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Time taken upstream = 3 hr

Time taken downstream = 2.5 hr

Speed = Distance / Time

Speed upstream

\frac{120}{3}=x-y\\\Rightarrow 40=x-y

Speed downstream

\frac{120}{2.5}=x+y\\\Rightarrow 48=x+y

Adding both the equations

48+40=x-y+x+y\\\Rightarrow 88=2x\\\Rightarrow 44=x

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The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>

8 0
3 years ago
Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of
Fudgin [204]

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San  Francisco i.e 400 mi

considering ,

Alan's speed v_A=50mph

Beth's speed v_B=60mph

->For Alan:

The time required t_A= d/v_A= 400/50 => 8h

-> For beth:

The time required t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

3 0
2 years ago
PLEASE I REALLY NEED HELP!
Talja [164]

Answer:

v=\dfrac{1200}{t}\ m/s

Explanation:

Given that,

The car traveled a total of 1,200 meters during this test.

We need to find the average speed of the car. The average speed of the car is given by total distance covered divided by the time taken. So,

v=\dfrac{1200}{t}\ m/s

But putting the value of t we can find the average speed of the car.

3 0
3 years ago
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