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3 years ago

6

The average force applied to move an initially stationary golf cart over a period of 25 s is 500 N. What is the change in moment

um of the cart?

1 answer:

Yuki888 [10]3 years ago

6 0

Answer:

12500(kg*m/s)

Explanation:

F=ma=mv/t=p/t

p=F*t=500N*25 s=12500(kg*m/s)

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3 years ago

I NEED HELP ON THIS QUESTION!

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The second option is the correct one. m/s^2

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3 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

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The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

<em>Current flowing in the wire, I = 4.00 mA</em>
<em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
<em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
<em>Length of wire, L = 2.00 m</em>
<em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

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RideAnS [48]

Correct order, from lowest potential energy to highest potential energy:

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Explanation:

The gravitational potential energy of the car is given by:

$U=mgh$

where

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g is the gravitational acceleration

h is the height of the car relative to the ground

In the formula, we see that m and g are constant, so the potential energy of the car depends only on its height above the ground, h. The higher the car from the ground, the larger its potential energy. Therefore, the position with least potential energy will be E, since the height is the minimum. Then, C will have more potential energy, because the car is at higher position, and so on: the position with greatest potential energy is A, because the height of the car is maximum.

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3 years ago

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Fehling's solution is added to the counters of test tube b after about 3 minutes and it turns brick red.what food substance was

hjlf

The food substance being used which turned Fehling's solution to brick red is a reducing sugar.

<h3>What is Fehling's solution?</h3>

This is an indicator which is used to test for the presence of reducing sugar or aldehydes in a solution.

Brick red precipitate of copper(I) oxide is formed when it tests positive to compounds such as glucose etc.

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2 years ago