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allochka39001 [22]
3 years ago
14

A 0.250 kg fan cart accelerates at 24 cm/ s^2 for 4.5 seconds. what is the net force (fan thrust minus drag and friction) on the

cart? How fast will it be going at the end of the 4.5 seconds?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

0.06 N

1.08 m/s

Explanation:

m = mass of the fan cart = 0.250 kg

a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²

F = Net force on the cart

Net force on the cart is given as

F = ma

F = (0.250) (0.24)

F = 0.06 N

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart

t = time interval = 4.5 s

Using the equation

v = v₀ + a t

v = 0 + (0.24) (4.5)

v = 1.08 m/s

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a man of mass 50 kg on the top floor of skyscraper step into an elevator what is the max weight as the elevator moves downward​
shtirl [24]
514.5 N is the answer
6 0
2 years ago
How will unbalanced forces affect the speed and direction of an object
azamat

Answer:

Explanation:

Unbalanced forces will result in the presence of acceleration. The formula

F net = ma

says that if there is a net force present and the object in question has a mass, then an acceleration is present. Now acceleration is constant in this situation because nowhere does it say the acceleration is changing. If acceleration is constant then the velocity is increasing at a steady pace (think linear function!).

The direction of the object depends on the direction that the net force is in. If the net force is to the left, then that object will accelerate to the left.

Hope this helps :)

3 0
3 years ago
An average hole drift velocity of 103 cm/sec results when 2 V is applied across a 1 cm long semiconductor bar. What is the hole
Helga [31]

Answer:

ε = 2 V/cm

Explanation:

To calculate the mobility inside this bar, we just need to apply the expression that let us determine the mobility. This expression is the following:

ε = ΔV / L

Where:

ε: Hole mobility inside the bar

ΔV: voltage applied in the bar

L: Length of the bar

We already have the voltage and the length so replacing in the above expression we have:

ε = 2 V / 1 cm

<h2>ε = 2 V/cm</h2><h2></h2>

The data of the speed can be used for further calculations, but in this part its not necessary.

Hope this helps

8 0
3 years ago
Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom
Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

   final velocity of Tarzan = v_f

law of conservation of energy

  PE_i + KE_i = PE_f + KE_f

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i = \dfrac{1}{2}mv_f^2

             v_f = \sqrt{2gh_i}

                   = \sqrt{2gL(1- cos\theta)}

                   = \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}

                          = 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the  = 3 m/s

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2

          gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2

             v_f = \sqrt{v_1^2+2gh_i}

             v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}

                       v_f= 11.29 m/s

3 0
3 years ago
JUST PLZ HELP!!! Why does the lightbulb in the right electrical circuit turn on but not the one on the left?
makkiz [27]
Because,

In left image pin is not touch to the wire.

In right image pin is touch to the wire.

Hope it helps you.....

Plz...Plz...Plz...Plz…Plz…

Mark be Brainliest.....

Please.....

And..

Please thanks me.....

Plz.....Plz.....
8 0
3 years ago
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