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3 years ago

A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?

Physics

1 answer:

rosijanka [135]3 years ago

5 0

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

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Explanation:

We use the Theorem of conservation of mechanical energy for finding the velocity when it strikes the ground:

Ei = Ef

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Ui = Kf

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v = sqrt(2gh)

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3 years ago

Which of the following describes one of Galileo Galilei's contributions to the development of the concept of gravity?

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Answer:

The correct option is;

c. He conducted experiments with balls and inclined planes

Explanation:

The inclined plane experiment conducted by Galileo Galilei was one of his most important experiments.

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3 years ago

Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a dista

luda_lava [24]

To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

$V = \frac{kq}{r}$

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

$V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}$

Replacing the values we have that

$V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}$

$V_1 = 215.12V$

Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

$V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}$

$V_2 = 328.2V$

Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say

$\frac{1}{2} mv^2 = q(V_2-V_1)$

Here

m = mass

v = Velocity

q = Charge

V = Voltage

Rearranging to find the velocity

$v = \sqrt{ \frac{2q(V_2-V_1)}{m}}$

Replacing,

$v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}$

$v = 6.3*10^6m/s$

Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is $6.3*10^6m/s$

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3 years ago

An AC adapter for a telephone-answering unit uses a transformer to reduce the line voltage of 120 V (rms) to a voltage of 9.0 V.

Illusion [34]

Answer:

1. The number of turns on the secondary output is 18

2. The root mean square power delivered to the transformer is 48 Watts.

Explanation:

A transformer is an electronic device that can be used for increasing or decreasing the value of a given voltage. It consists of primary coils and secondary coil of a definte number of turns. When voltage flows in the primary coil, it induces voltage in the secondary coil. The two types are: step-up and step down transformers.

1. For a given transformer,

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