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Romashka [77]
3 years ago
13

What is the net force acting on a golf car travelling at a constiant speed of 5 mph?

Physics
2 answers:
pashok25 [27]3 years ago
5 0
0 Newtons, since it's at a constant speed. Hope this helps, and Brainliest answer would be appreciated!

Sholpan [36]3 years ago
5 0
It's D. A net force always produces acceleration. Since the speed is constant AND IN A STRAIGHT LINE, there's no net force on the car.
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An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. I
oksano4ka [1.4K]

Answer:

The deviation in path is 4.39 \times 10^{-3}

Explanation:

Given:

Velocity v = 1 \times 10^{6} \frac{m}{s}

Electric field E = 500 \frac{V}{m}

Distance x = 1 \times 10^{-2} m

Mass of electron m = 9.1 \times 10^{-31} kg

Charge of electron q = 1.6 \times 10^{-19} C

Time taken to travel distance,

    t = \frac{x}{v}

    t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }

    t = 10^{-8} sec

Acceleration is given by,

  F = qE

ma = qE

   a = \frac{qE}{m}

   a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }

   a = 8.77 \times 10^{13} \frac{m}{s^{2} }

For finding the distance, we use kinematics equations.

   y = vt + \frac{1}{2}  at^{2}

Where v = 0 because here initial velocity zero

   y = \frac{1}{2} at^{2}

   y = \frac{1}{2} \times  8.77 \times 10^{13 } \times (10^{-2} )^{2}

   y = 4.39 \times 10^{-3} m

Therefore, the deviation in path is 4.39 \times 10^{-3}

6 0
2 years ago
Capacitors in Combination: A 5.0-μF, a 14-μF, and a 21-μF capacitor are connected in series. How much capacitance would a single
tankabanditka [31]

The total capacitance is <em>C</em> such that

1/<em>C</em> = 1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)

Solve for <em>C</em> :

<em>C</em> = 1 / (1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)) ≈ 3.1 µF

7 0
3 years ago
Imagine you are charged to manage a project that aims to install wireless access points (aps) throughout the university campus.
Vitek1552 [10]

If you are charged to manage a project that aims to install wireless access points (aps) throughout the university campus, the steps to a foolproof plan for this project would be -

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The most crucial step in any WiFi installation is probably knowing what your network needs are.

2. Select the appropriate hardware for your wireless network

Finding the ideal access point is much simpler if your needs are clear, but the wide range of options might be difficult.

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It’s crucial to keep in mind that other factors besides your Internet connection and network hardware might affect how well your network performs.

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4 0
1 year ago
A hunter stands on a frozen pond (frictionless) and fires a 4.20g bullet at 965m/s horizontally.The mass of hunter + gun is 72.5
Afina-wow [57]

Answer:

The the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.

Explanation:

Given;

mass of bullet, m₁ = 4.2 g = 0.0042 kg

mass of hunter + gun = 72.5 kg

velocity of the bullet, u = 965 m/s

Momentum of the bullet when it was fired;

P = mv

P = 0.0042 x 965

P = 4.053 kg.m/s

Determine the recoil velocity of the hunter.

Total momentum = sum of the individual momenta

Total momentum = momentum of the bullet + momentum of the hunter

Apply the principle of conservation of momentum, sum of the momentum is equal to zero.

P_{hunter} + P_{bullet} = 0\\\\P_{hunter}  = -P_{bullet}\\\\72.5v = -4.053\\\\v = \frac{-4.053}{72.5} \\\\v = - 0.056 \ m/s\\\\Thus, the \ recoil \ velocity \ of \ the \ hunter \ is \ 0.056 \ m/s, \ in \ opposite \ direction \ of \ the \ bullet.

Therefore, the the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.

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A certain machine changes a large input force into a smaller output force. how will the machine affect the distance over which t
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A machine that lowers the input force is a step down machine. If the force is lowered, the distance at which the the force is applied will be lowered as well. Assuming that the work done by the force does not stay constant and is affected by force, the distance will be reduced.
6 0
3 years ago
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