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lana [24]
3 years ago
7

A large flat block of mass 7 m rests on a level, smooth tabletop (µk ≈ 0). On top of it is a much smaller block of mass m, with

a coefficient of static friction between the two blocks of µs. If the small block is pulled horizontally with constant force F, we want both blocks to slide together across the table without slipping on each other.
What is the largest acceleration that the system can be given in terms of µs and g?
1. F − µs g/ 8
2. F /8 m
3. µs g /7
4. µs g
5. 7 µs g /8
6. 8 µs g
Physics
1 answer:
mariarad [96]3 years ago
8 0

Answer:

3. µs g /7

Explanation:

The largest Force appear when the maximal friction Force is required.

Second Newton law for the small block:

F_friction=u_s*N=u_s*(mg)

F-F_friction=ma

F-u_s*(mg)=ma

Second Newton law for the Big Block:

F_friction=7ma

u_s*(mg)=7ma

a=u_s*g/7

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You are riding a bicycle. If you apply a forward force of 125 N, and you and
erica [24]

Answer:

1.52g

Explanation:

Given parameters:

Force  = 125N

Mass combined  = 82kg

Unknown:

Acceleration of the bicycle  = ?

Solution:

From Newton second law of motion suggests that:

   Force = mass x acceleration

  Acceleration = \frac{force}{mass}  = \frac{125}{82}   = 1.52g

8 0
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A joule is a unit of work is equal to ____?
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8 0
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Read 2 more answers
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are
Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

4 0
3 years ago
A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ
Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

8 0
3 years ago
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