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bija089 [108]
3 years ago
10

The manager of a department store is pushing horizontally with a force of magnitude 200N on a box of shirts. The box is sliding

across the horizontal floor with a forward acceleration. Nothing else touches the box. What must be true about the magnitude of the force of kineticfriction acting on the box (choose one)?
Physics
1 answer:
Andreas93 [3]3 years ago
8 0
Found the missing options on internet:
<span>a) it is less than 200 N; b) it is greater than 200 N; c) it is equal to 200 N; d) none of these statements are necessarily true
</span>
Solution:
The correct answer is a) it is less than 200 N.

In fact, the box is sliding with forward acceleration different from zero. This means that the net force acting on the box is different from zero, because Newton's second law states that:
\sum F = ma
the resultant of the forces acting on an object is equal to its mass times its acceleration.

In this case, we have 2 forces acting on the box:
- the force of 200 N of the man pushing the box
- the frictional force, acting against the direction of motion
The net acceleration is forward, this means that the net force acting on the box must be directed forward, therefore the magnitude of the frictional force must be smaller than the force of 200 N that pushes the box.

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Two long, straight wires are parallel and 10 cm apart.One carries a current of 2.0 A, the other a current of 5.0A. (a) If the tw
lubasha [3.4K]

Answer:

Explanation:

Given

Distance between two wires d=10 cm

Current value i_1=2 A

i_2=5 A

(a)If current flows in opposite direction

When current Flows in opposite direction the two wires will repel each other

Force due to current carrying wire F=\frac{\mu _0i_1i_2}{2\pi r}

F_{12}=F_{21}=\frac{\mu _0\times 2\times 5}{2\times \pi \times 0.1}

F_{12}=\frac{4\pi \times 10^{-7}\times 2\times 5}{2\pi \times 0.1}

F_{12}=2\times 10^{-5} N (Repulsive Force)

(b)If current is flowing in the same direction

When direction of current is same then force will be attractive in nature

F_{12}=F_{21}=\frac{\mu _0i_1i_2}{2\pi r}

F_{12}=\frac{4\pi \times 10^{-7}\times 2\times 5}{2\pi \times 0.1}

F_{12}=2\times 10^{-5} N (attractive Force)

5 0
4 years ago
1)the car's engine power is 44000W. Explain this number in a physical sense
Ratling [72]

Answer:

1) It expresses the rate (top speed) at which it can move with time.

2) P = 20 W

3) h = 18 km

Explanation:

1) Power is the rate of transfer of energy.

⇒ Power = \frac{Energy(or workdone)}{Time}

i.e P = \frac{E}{t}

Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.

2) m = 400g = 0.4 kg

    t = 20 s

h = 100m

g = 10 m/s^{2}

P = \frac{mgh}{t}

  = \frac{0.4*10*100}{20}

  = \frac{400}{20}

P = 20 W

3) u = 600 m/s

   g = 10 m/s^{2}

From the third equation of free fall,

V^{2} = U^{2} - 2gh

V is the final velocity, U is the initial velocity, h is the height.

0 = (600)^{2} - 2 x 10 x h

0 = 360000 - 20h

20h = 360000

h = \frac{360000}{20}

  = 18000

h = 18 km

The maximum height of the bullet would be 18 km.

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