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coldgirl [10]
4 years ago
11

A baggage handler at an airport applies a constant horizontal force with magnitude F1 to push a box, of mass m, across a rough h

orizontal surface with a very small constant acceleration a.
The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of 2a.
How does the magnitude of the force F2 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box?
Physics
1 answer:
Vikentia [17]4 years ago
6 0

Answer:

F_2 = 2F_1 - F where F is the friction force that is the same for both boxes

Explanation:

So the same boxes at the same mass m would create a same friction force, let this force be F. The net force in the first and 2nd cases would be

- 1st case: F1 - F

- 2nd case F2 - F

According to Newton 2nd law, this net would make acceleration

- 1st case a_1 = (F_1 - F)/m

- 2nd casea_2 = (F_2 - F)/m

Since a_2 = 2a_1

(F_2 - F)/m = 2(F_1 - F)/m

F_2 - F = 2F_1 - 2F

F_2 = 2F_1 - F

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Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you
dybincka [34]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.

So,

F = \frac{Kq1q2}{r^{2} }

Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.

Let's suppose, q1 = q

So,

if q1 = q

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Product of Charges = q x (Q-q)

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And the expression qQ - q^{2} is clearly a quadratic expression. And clearly its roots are 0 and Q.

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Answer:

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

Explanation:

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mass of the rocket = 600 grams

speed = 4.0 m/s

Step 2: Calculate weight

Fw = mg

with Fw = the weight (in Newton)

with m = the mass (in kg)

with g = the acceleration due to gravity (9.81 m/s²).

Fw = (0.600 kg)(9.81 m/s²)  = 5.886 N

Step 3: Calculate force available to provide acceleration

The rocket engine, when it is fired, exerts a 8.0 AND vertical thrust on the rocket.

5.886 N of that force will be used to counter the rocket's weight, leaving 2.114 N of force available to provide acceleration.  

Step 4: Calculate the rocket's upward acceleration:

Fnet = m*a

With Fnet = the net force (the force that remains after the rocket's weight is compensated)

with a = the rocket's acceleration (in m/s²)

2.114 N = (0.600 kg)*a

a = 3.52 m/s²  = the rocket's upward acceleration

Step 5: Calculate how long it will take to rise 20 meters into the air.

Δy = v0*t + 1/2 at²

with v0 = 0m/s

Δy = 1/2 at²

20 m = 1/2(3.52)t²

20 m = (1.76 m/s²)t²

11.36 = t²

t = 3.37 s

This means the rocket will take 3.37 seconds to reach the hoop.  It should be launched when the cart is 3.37 seconds away from being directly beneath the hoop.  

Step 6: Calculate the distance

v = Δx / t

4.0 m/s = Δx / 3.37 s

Δx = 13.48 m

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

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