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Mazyrski [523]
3 years ago
10

Question about the motion of a book, thanks

Physics
2 answers:
Andrej [43]3 years ago
7 0

Answer:

there is no picture

Explanation:

igomit [66]3 years ago
6 0
D ifnfifndisnsisndodjfifgigigjvivjviv
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lateral

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Moving company uses a machine to raise a 900 Newton refrigerator to the second floor of a building machine consists of a single
Juliette [100K]
Mechanical advantage of a machine is the ratio of the output force over the input force or M=Fo/Fi. Since M=1, Fi=Fo, or the input force is equal to the output force. This means that to raise the refrigerator that weighs 900 N, we need the same input force of 900 N, or Fo=Fi=900 N. 
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3 years ago
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Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force
xeze [42]

Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
7 0
3 years ago
What average force is required to stop a 1400 kg car in 6.0 s if the car is traveling at 90 km/h ? Express your answer to two si
devlian [24]

Answer:

F=5833.3 N N

Explanation:

Newton's second law applied to the car

F= m*a Formula (1)

F: Force in Newtons (N)

m : mass in kg

a: acceleration ( m/s²)

kinematics car

vf= v₀ + a*t  Formula (2)

vf : final velocity (m/s)

v₀ : final velocity  (m/s)

a : acceleration  ( m/s²)

t : time t

Equivalences

1 km= 1000m

1 h = 3600 s

Data

m= 1000kg

v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s

vf= 0

t= 6 s

Problem Development

We calculate the acceleration replacing the data in the formula (2) :

0 =  25 + a*6

a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)

We calculate the force is required to stop the car replacing the data  in the formula (1)

-F = 1400 kg*(-4.16 m/s²)

F=5833.3 N

8 0
3 years ago
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