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dem82 [27]
3 years ago
11

an automobile engineer found that the impact of a truck colliding at 16 km/h with a concrete pillar caused the bumper to indent

only 6.4 cm. The truck stopped. Determine the acceleration.
Physics
1 answer:
Zolol [24]3 years ago
8 0
<span>To find the acceleration we are given two facts to begin. The impact at 16 km/h and the dent of 6.4 cm, or 0.064 meters. In solving the problem uniform acceleration is assumed, which would mean the avg speed during the impact was 8 km/hr by taking 16/2. We know distance = rate*time (d=r*t) . So t = d / r, so 0.64/8 = 0.008hr for t. Now we can solve for acceleration by taking a = 16 / 0.008 = 2000 km/hr.</span>
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Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a
rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

\lambda = 656.46 \ nm

\Delta \lambda = 0.04

c=3\times 10^8

As we know,

⇒  \frac{\Delta \lambda}{\lambda} =\frac{v}{c}

On substituting the values, we get

⇒  \frac{656.46}{0.04} =\frac{v}{3\times 10^8}

⇒         v=\frac{656.46}{0.04} (3\times 10^8)

⇒            =16411.5\times 3\times 10^8

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8 0
2 years ago
Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p
DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = \frac{mv^{2}}{r}

Electrostatic force = \frac{kq^{2}}{r^{2}}

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}

v=\sqrt{\frac{kq^{2}}{mr}}

v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by  2.068 x 10^6 m / s.

6 0
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