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Ivan
3 years ago
15

When a force of 15 newtons is applied to a stationary chair, it starts moving. What can you say about the frictional force betwe

en the chair and the floor of the room?
Physics
1 answer:
harkovskaia [24]3 years ago
3 0
First you need to make a difference between friction while object is stationary and the friction while object is moving. Force required to start moving some object is slightly greater than force required to maintain objects movement. That means that to move a chair you need some force F1 but you can than slightly reduce force and chair will still be moving.

Now to the problem in this question: It can be said that "stationary friction force" is equal to 15 Newtons. Its also good to know that friction force between chair and floor while you are increasing your push is also increasing and is equal to force of your push. Once it reaches 15N which is it "critical value" for that chair, chair starts moving and friction force drops a little bit and now it is called friction force of moving chair.
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Answer:

268.2 m/s (1dp)

Explanation:

600 miles = 965,606m (1mile = 1609.34m)

965606/60 = 16,0943.43 ..... (metres traveled every minute)

16,0943.43/60 = 268.2238 .....

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Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
3 years ago
a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car​
uysha [10]

Answer:

= 201.53 meters

Explanation:

A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?

Use the formula  d = \frac{at^{2} }{2} ,

where d is the distance, t is the time and "a" is the acceleration.
d=\frac{9*54*6*5^{2} }{2} = 201.53 m

3 0
2 years ago
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